text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
5 초 512 MB 5 4 4 80.000%

## 문제

Dave is taking a multiple choice test on the Internet. Dave possibly gets many opportunities to submit his answers to the test, but he passes only if he gets all the questions correct. He must answer every question on the test to make a submission. The only information he receives after he submits is whether he has passed.

For each question, he estimates the probability that each of 4 responses is correct, independent of his responses to other questions. Given a fixed number of submissions he can make, Dave wants to choose his responses so that he maximizes the probability of passing the test.

What is the probability that Dave will pass the test if he chooses his responses optimally?

## 입력

The first line of input gives the number of cases, CC test cases follow.

Each test case starts with a line containing M and Q. Dave is allowed to make Msubmissions to solve the test. There are Q questions on the test. Q lines follow, each containing 4 probabilities of correctness. There will be at most 6 digits after the decimal point. The probabilities for each line are non-negative and sum to 1.

Limits

• 1 ≤ C ≤ 100
• 1 ≤ Q ≤ 30
• 1 ≤ M ≤ 10000

## 출력

For each test case, output one line containing "Case #XY" where X is the number of the test case (starting from 1), and Y is the probability of success.
Answers with a relative or absolute error of at most 10−6 will be considered correct.

## 예제 입력 1

3
10 2
0.25 0.25 0.25 0.25
0.25 0.25 0.25 0.25
64 3
0.3 0.4 0.0 0.3
1.0 0.0 0.0 0.0
0.2 0.2 0.2 0.4
3 2
0.5 0.17 0.17 0.16
0.5 0.25 0.25 0.0

## 예제 출력 1

Case #1: 0.625
Case #2: 1.0
Case #3: 0.5

## 채점 및 기타 정보

• 예제는 채점하지 않는다.


text:
Operators

Stream Processor operators.

An operator is a first-class building block in the stream processor API. Operators are strung together in a user's program, and provide the core functionality for the Stream Processor.

"operator" terminology

In the KX Stream Processor, the term "operator" can refer to either one of the basic operators like map, or filter, but can also refer to any component of a pipeline. In other words: all readers/writers/decoders/etc. are operators, but not all operators are readers/writers/decoders/etc.

kxi.sp.op aliases in kxi.sp

All operators in this module are available through aliases in kxi.sp for convenience. For example, you can access kxi.sp.op.map as kxi.sp.map.

 kxi.sp.op.OperatorParams (AutoNameEnum)

Specifies a parameter that will be provided to the function you provide to an operator.

kxi.sp.op.OperatorParams.data

Provide the message's data.

kxi.sp.op.OperatorParams.metadata

Provide the message's metadata, which may be required by other SP functions.

kxi.sp.op.OperatorParams.operator

Provide the operator's dictionary, which may be required by other SP functions.

 kxi.sp.op.Operator

Stream Processor operator interface.

An operator is a first-class building block in the stream processor API. Operators can be strung together to form a kxi.sp.Pipeline instance, which can then be run.

Pipeline and operator objects can be joined together using the | operator. Operators can also be joined to lists/tuples of pipelines or operators.

See Also: kxi.sp.Pipeline

Attributes:

Name Type Description
id str

The unique ID of this operator.

kxi.sp.op.Operator.as_pipeline property readonly

A new pipeline that only contains this operator.

kxi.sp.op.Operator.__new__ special staticmethod

Create and return a new object. See help(type) for accurate signature.

kxi.sp.op.accumulate

Aggregates a stream into an accumulator.

Aggregates a stream into an accumulator, which is updated and then emitted for each incoming batch. The value being emitted is passed to the output function, which can modify it before passing it to the next operator. If the accumulator is a dictionary, it may be necessary to enlist the result in the output function so the next operator receives a table.

By default, .qsp.accumulate emits the accumulator, but this value can be transformed with the output function. If the accumulator is a dictionary, an output function like {enlist x} could be used to emit tables.

Parameters:

Name Type Description Default
function Union[Callable, str]

An aggregator which takes the metadata, data, and the accumulator, and returns an updated accumulator.

required
initial Any

The initial state of the accumulator. The accumulation is performed in q, and as such this initial value must be convertible to q. This conversion can be performed manually with pykx.K.

None
output Union[Callable, str]

An optional function to transform output before emitting it. It gets passed the value of the accumulator.

None

Returns:

Type Description
Operator

An accumulate operator, which can be joined to other operators or pipelines.

kxi.sp.op.apply

Apply a function to incoming batches in the stream.

Unlike other operators, apply is asynchronous. To return data from it, you must use sp.push to push data through the pipeline when it is ready. Apply is commonly used with the state API to implement custom windowing, or with the task API.

When apply is used to buffer data, the onFinish option should be used to flush the buffer when the pipeline finishes. It will be called once for every key in the state including the default key, with the metadata dictionary passed to it containing only the key.

Parameters:

Name Type Description Default
function Union[Callable, str]

A function that will be called with three arguments: operator, metadata, and data. Operator and metadata can be used to get/set the state of the operator, among other uses.

required
on_finish Union[Callable, str]

A function run when the pipeline finishes or is torn down, to flush the buffer. It is passed the operator and metadata.

None

Returns:

Type Description
Operator

An apply operator, which can be joined to other operators or pipelines.

kxi.sp.op.filter

Filter elements out of a batch, or a batch from a stream.

Parameters:

Name Type Description Default
function Union[Callable, str]

A predicate function which will be applied to the data that passes through the operator. If the result is a boolean list, then only records that it flags in the batch progress through the stream. If the result is a singular boolean value, then either all records in the batch, or none of them progress through the stream.

required

Returns:

Type Description
Operator

A filter operator, which can be joined to other operators or pipelines.

kxi.sp.op.map

Apply a function to data passing through the operator.

The function provided will be called on every message that passes through the pipeline stream. The map operation can modify the data by performing aggregations, type changes, transformations, etc.

Parameters:

Name Type Description Default
function Union[Callable, str]

The function that will be applied to the data that passes through the operator. Can be provided either as a Python function, or as a string of q code that defines a q function.

required

Returns:

Type Description
Operator

A map operator, which can be joined to other operators or pipelines.

kxi.sp.op.merge

Merge two streams using a function to combine them.

Merge can be used to join and enrich streams with other streams, with static data, or to union two streams into a single stream. See also: union

Parameters:

Name Type Description Default
stream Pipeline

The separate pipeline that will be merged with.

required
function Union[Callable, str]

A function that takes two arguments, the data from each stream, and returns the data from both streams combined into one.

required

Returns:

Type Description
Operator

A merge operator, which can be joined to other operators or pipelines.

kxi.sp.op.reduce

Aggregates partial windows.

A window may include more records than can fit in memory. As such, it may be necessary to reduce the buffered records into a smaller, aggregated value at regular intervals. If the window operator uses the count_trigger option, a partial window will be emitted when the number of buffered records exceeds the count_trigger. Partial windows will also be emitted when a stream goes idle. These partial windows can be aggregated using this operator. When the window is complete, this operator will emit the result of reducing the partial windows.

The reduce operator runs a function on each incoming batch to update the accumulator for that window. Each window has a separate accumulator.

For partial windows, such as those emitted by count_trigger or idle streams, the accumulator will be updated but not emitted.

When a window is closed, such as when the high-water mark passes the end of the window, the accumulator will be updated for that final batch, and then it will be emitted.

By default, kxi.sp.reduce emits the accumulator, but this value can be transformed with the output function. If the accumulator is a dictionary, an output function like pykx.q.enlist could be used to emit tables.

Parameters:

Name Type Description Default
function Union[Callable, str]

An aggregator which takes the metadata, data, and the accumulator for a given window, and returns an updated accumulator.

required
initial Any

The initial state of the accumulator. The accumulation is performed in q, and as such this initial value must be convertible to q. This conversion can be performed manually with pykx.K.

None
output Union[Callable, str]

An optional function to transform output before emitting it. It gets passed the value of the accumulator.

None

Returns:

Type Description
Operator

A reduce operator, which can be joined to other operators or pipelines.

kxi.sp.op.split

Split the current stream.

The split operator allows a single stream to be split into arbitrarily many separate streams for running separate analytics or processing.

Split operators can be explicitly added, but are implicitly added if the same operator appears as a parent multiple times when resolving the streams given to sp.run

Returns:

Type Description
Operator

A split operator, which can be joined to other operators or pipelines.

kxi.sp.op.sql

Perform an SQL query on tabular data in the stream.

The query must:

If data in the stream is not table data, an error will be raised. Providing a schema will allow the query to be precompiled enabling faster processing on streaming data.

Queries run in a local worker process

SQL queries are not distributed across workers, and are currently run on each worker's substream.

Parameters:

Name Type Description Default
query Union[str, bytes, pykx.wrappers.CharVector]

An SQL query, which will be performed on tabular data in the stream.

required
schema Optional[pykx.wrappers.Table]

An empty table with the same schema as the data.

None

kxi.sp.op.udf

Retrieve a User-Defined-Function(UDF) from a defined 'KX_PACKAGE_PATH' package directory

Parameters:

Name Type Description Default
name Union[str, bytes, pykx.wrappers.CharVector]

Name of the UDF that is to be retrieved

required
package Union[str, bytes, pykx.wrappers.CharVector]

Name of the package from which to retrieve a UDF

required
version Union[str, bytes, pykx.wrappers.CharVector]

Version of the package from which to retrieve the UDF

None
params Optional[dict]

Additional optional input information to the UDF passed as the final argument to the function

pykx.Dictionary(pykx.q(''))

Returns:

Type Description
Operator

A function which is to be used as a UDF within a merge, filter, map operator

Examples:

Retrieve a UDF providing all necessary information

>>> from kxi import sp
>>> import pykx as kx
>>> import numpy as np
>>> import pandas as pd
# Set path for package retrieval
>>> kx.q.setenv('KX_PACKAGE_PATH', b'/tmp/packages')
# Retrieve the UDF function (in this example the UDF adds 10 to input)
>>> udf = sp.udf('map_udf', 'test', '1.0.0', {'param':10})
# Use the UDF within SP operator
| sp.map(udf)
| sp.write.to_console(timestamp='none'))
>>> data = pd.DataFrame({
'x': np.random.rand(5),
'x1': np.random.rand(5)
})
>>> kx.q('publish', data)
x        x1
-----------------
10.49318 10.39275
10.57852 10.51709
10.08389 10.51598
10.19599 10.40666
10.37564 10.17808
pykx.Identity(pykx.q('::'))

Retrieve a UDF using latest package and default parameters

>>> from kxi import sp
>>> import pykx as kx
>>> import numpy as np
>>> import pandas as pd
# Set path for package retrieval
>>> kx.q.setenv('KX_PACKAGE_PATH', b'/tmp/packages')
# Retrieve the UDF function
>>> udf = sp.udf('filter_udf', 'test')
# Use the UDF within SP operator (this UDF filters data for x<0.5)
| sp.filter(udf)
| sp.write.to_console(timestamp='none'))
>>> data = pd.DataFrame({
'x': np.random.rand(5),
'x1': np.random.rand(5)
})
>>> kx.q('publish', data)
x          x1
--------------------
0.07347808 0.7263142
0.3159526  0.9216436
0.3410485  0.1809536
pykx.Identity(pykx.q('::'))

kxi.sp.op.union

Unite two streams.

Like merge, but elements from both sides of the union are left as-is.

Parameters:

Name Type Description Default
stream Pipeline

The separate pipeline that will be unified with.

required

Returns:

Type Description
Operator

A union operator, which can be joined to other operators or pipelines.

format: plain


text:
Symbol
Problem
Aotivity $a1$ Factoring Polynomials the missing factor in each of the following. Find notebook. $1$ $x^{3}-8=\left(x-2\right)\left($ $→$ $2$ $2x^{3}+x^{2}-23x+20=\left(x+4\right)\left($ COPY $3$ $3x^{3}+2x^{2}-37x+12=\left(x-3\right)\left($ $△\right)$ $-$ $4$ $x^{3}-2x^{2}-x+2=\left(x-2\right)\left($ $5$ $2x^{3}-x^{2}-2x+1=\left(2x-1\right)\left($


text:
Is this possible to integrate?

1. Jan 20, 2006

jason17349

I haven't had any luck with mathcad and was wondering if this was possible to integrate...

$$\int_{0}^{2\pi} \frac {x+r*cos(\theta)}{(x^2+2r*x*cos(\theta)+r^2)^\frac {3}{2}} d\theta$$

Last edited: Jan 20, 2006
2. Jan 20, 2006

daveb

It's certainly integrable...now whether you can solve it by non-numerical methods may be another matter entirely.

3. Jan 22, 2006

jason17349

Does anybody have any suggestions on how to go about integrating this.... ? Or maybe explain why mathcad isn't able to integrate this..

4. Jan 22, 2006

StatusX

Do you have to integrate over theta first? If you could integrate over x first, you could use substitution. That's usually the easiest way to do integrals like these.

5. Jan 22, 2006

jason17349

Sorry, I should have said x and r are real non-negative constants..

6. Jan 22, 2006

StatusX

Even so, you can pretend it's a variable. Then, integrate with respect to x to get some function F, so that your integral becomes:

$$\int \left( \frac{\partial}{\partial x} F \right) d\theta = \frac{\partial}{\partial x} \left( \int F d\theta \right)$$

I don't know if that helps, but it's a cool trick.

7. Jan 22, 2006

benorin

8. Jan 23, 2006

benorin

Maple gives an answer involving Elliptic integrals of the first and second kind, but in terms of those integrals, the answer is not that bad (though long enough for me not to post it.)


text:
# Re: [tlaplus] A predictable choice

On 18.11.2015 16:44, fl wrote:
> I suppose it is a question that has already been answered and I
> apologize but if
> somebody could enlighten me anyway.
>
> If I have a:
>
> CONSTANT GETC(_)
>
> and in the model if I have
>
> GETC(x) <- CHOOSE y \in { A, NEWLINE, NULL } : TRUE
>
> and if I launch TLC
>
> then I always get NULL
>
> why ?

Hi Frederic,

you might want to read section 13.5 "The choose Operator" of the hyperbook:

"In mathematics, there is no such thing as a nondeterministic operator
or a nondeterministic function. If some expression equals 42 today, then
it equaled 42 yesterday and it will still equal 42 next year. [...] The
semantics of TLA + do not specify which value in 1..	10; but it is the
same one every time.".

Cheers
Markus


text:
## Tokyo Journal of Mathematics

### Braids and Nielsen-Thurston Types of Automorphisms of Punctured Surfaces

#### Abstract

Let $F$ be a compact, orientable surface with negative Euler characteristic, and let $x_1, \cdots, x_n$ be $n$ fixed but arbitrarily chosen points on $\mathrm{int}F$, each of which has a (small) diskal neighborhood $D_i \subset F$. Denote by $\mathcal{S}_n(F)$ a subgroup of $\mathrm{Diff}(F)$ consisting of "sliding" maps $f$ each of which satisfies

(1) $f(\{x_1, \dots , x_n\}) = \{x_1, \dots , x_n\}, f(D_1\cup \cdots \cup D_n) = D_1 \cup \cdots \cup D_n$ and

(2) $f$ is isotopic to the identity map on $F$

Then by restricting such automorphisms to $\hat{F} = F - \mathrm{int}(D_1 \cup \cdots \cup D_n)$, we have automorphisms $\hat{f} : \hat{F} \to \hat{F}$, which form a subgroup $\mathcal{S}_n(\hat{F})$ of $\mathrm{Diff}(\hat{F})$. We give a Nielsen-Thurston classification of elements of $\mathcal{S}_n(\hat{F})$ using braids in $F \times I$ which characterize the elements of $\mathcal{S}_n(\hat{F})$.

#### Article information

Source
Tokyo J. Math., Volume 28, Number 2 (2005), 527-538.

Dates
First available in Project Euclid: 5 June 2009

https://projecteuclid.org/euclid.tjm/1244208205

Digital Object Identifier
doi:10.3836/tjm/1244208205

Mathematical Reviews number (MathSciNet)
MR2191064

Zentralblatt MATH identifier
1100.37025

#### Citation

ICHIHARA, Kazuhiro; MOTEGI, Kimihiko. Braids and Nielsen-Thurston Types of Automorphisms of Punctured Surfaces. Tokyo J. Math. 28 (2005), no. 2, 527--538. doi:10.3836/tjm/1244208205. https://projecteuclid.org/euclid.tjm/1244208205

#### References

• J. S. Birman, Braids, links, and mapping class groups, Annals of Mathematics Studies, No. 82 Princeton University Press, Princeton, N.J. (1974).
• A. J. Casson and S. A. Bleiler, Automorphisms of surfaces after Nielsen and Thurston, London Mathematical Society Student Texts, 9 Cambridge University Press, Cambridge, 1988.
• C. J. Earle and J. Eells, A fibre bundle description of Teichmüller theory, J. Diff. Geom. 3 (1969), 19–43.
• C. J. Earle and A. Schatz, Teichmüller theory for surfaces with boundary, J. Diff. Geom. 4 (1970), 169–185.
• E. Fadell and L. Neuwirth, Configuration spaces, Math. Scand. 10 (1962), 111–118.
• A. Fathi, F. Laudenbach and V. Poenaru, Travaux de Thurston sur les surfaces, Asterisque 66-67 (1991/1979).
• J. Hass and P. Scott, Intersections of curves on surfaces, Israel J. Math. 51 (1985), 90–120.
• Y. Imayoshi, M. Ito and H. Yamamoto, On the Nielsen-Thurston-Bers type of some self-maps of Riemann surfaces with two specified points, Osaka J. Math. 40 (2003), 659–685.
• Y. Imayoshi, M. Ito and H. Yamamoto, A reducibility problem for monodromy of some surface bundles, preprint.
• I. Kra, On the Nielsen-Thurston-Bers type of some self-maps of Riemann surfaces, Acta Math. 146 (1981), 231–270.
• W. P. Thurston, On the geometry and dynamics of diffeomorphisms of surfaces, Bull. Amer. Math. Soc. 19 (1988), no. 2, 417–431.


text:
# A Simple Salt Solution Mixing Problem - yet stuck

1. Jun 9, 2013

### Rokas_P

Hi, here is a very simple solution mixing problem that I can't solve which I am really ashamed of.

Problem. A vessel whose capacity is 5 liters contains 2 liters of 15% salt solution. How many liters of 20% salt solution have to be mixed to the 15% solution to produce a solution with as high salt concentration as possible?

My solution. I constructed this function for salt concentration in the new solution:

$$f(x)=\frac{0.3+0.2x}{2+x}$$

(The amount of salt in the first solution is 2*0.15 = 0.3)

I obtained the derivative of the above function:

$$f'(x)=\frac{0.2(2+x)-(0.3+0.2x)}{\left(2+x\right)^2}=\frac{0.4+0.2x-0.3-0.2x}{\left(2+x\right)^2}=\frac{0.1}{\left(2+x\right)^2}$$

I'm stuck at this point. If I try to find points of extremum, there are none. It means that I cannot increase salt concentration in the original solution by adding the said 20% salt solution. But I know the answer is 3 liters, so I must be doing something wrong.

--

As a sidenote, this problem is taken from a national maths exam. As such, you are not expected to think of mass, volume and density of water and salt. Also, I didn't make use of the fact that the vessel can contain up to 5 liters of liquid but just don't see where it fits in. Finally, I am not sure if this problem is even supposed to be solved using optimization techniques (derivatives and minima/maxima) but that was the first idea I got ...

2. Jun 9, 2013

### SteamKing

Staff Emeritus
Well, 2 + x must be less than or equal to 5 liters, since that is the maximum capacity of the vessel.

3. Jun 9, 2013

### Staff: Mentor

Think about it - the more 20% solution you add, the closer to 20% is the concentration of the salt in the mixture. So you have to add as much as possible. How much can you add?

4. Jun 9, 2013

### etudiant

Alternatively, the answer is to toss out the 15% solution. Just use the full 20% salt solution.

5. Jun 10, 2013

### Rokas_P

Thanks everyone for the explanations.

The problem here was that I expected the function to have a local maximum and when the derivative showed me it had none, I thought I must have done something wrong but now I realize that there is no reason for the function to have a maximum.

We can consider this problem solved.

6. Jun 10, 2013


text:
# K-purity and orthogonality

## Michel Hebert

Adamek and Sousa recently solved the problem of characterizing the subcategories K of a locally $\lambda$-presentable category C which are $\lambda$-orthogonal in C, using their concept of K$\lambda$-pure morphism. We strengthen the latter definition, in order to obtain a characterization of the classes defined by orthogonality with respect to $\lambda$-presentable morphisms (where $f : A \rightarrow B is called$\lambda$-presentable if it is a$\lambda$-presentable object of the comma category A/C). Those classes are natural examples of reflective subcategories defined by proper classes of morphisms. Adamek and Sousa's result follows from ours. We also prove that$\lambda$-presentable morphisms are precisely the pushouts of morphisms between$\lambda\$-presentable objects of C.

Keywords: pure morphism, othogonality, injectivity, locally presentable categories, accessible categories

2000 MSC: 18A20, 18C35, 03C60, 18G05

Theory and Applications of Categories, Vol. 12, 2004, No. 12, pp 355-371.

http://www.tac.mta.ca/tac/volumes/12/12/12-12.dvi
http://www.tac.mta.ca/tac/volumes/12/12/12-12.ps
http://www.tac.mta.ca/tac/volumes/12/12/12-12.pdf
ftp://ftp.tac.mta.ca/pub/tac/html/volumes/12/12/12-12.dvi
ftp://ftp.tac.mta.ca/pub/tac/html/volumes/12/12/12-12.ps

TAC Home


text:
# How do I attach a file to an e-mail message?

## In Zimbra Webmail:

1. Click on New Message on the left-hand side of the screen
2. Type the e-mail address you want to send to on the To: line
3. Click on Attach on the left-hand side.
4. This will bring up a window where you can browse your folders and select the file(s) you want to attach. When you done selecting the file, click on the Open button. (Alternatively, you can drag and drop file(s) to the Attach section)
5.  You will see the attached file in the Attachments section of the e-mail.

# How do I decode a base 64 encoded 'attachment' that is embedded in a mail message?

If you receive an e-mail with an attachment embedded within:

.. 1) save the e-mail as a plain text file

2) edit the e-mail and remove any headers

Delete everything before and including the lines that look like:

Content-Disposition: attachment; filename='13234.pdf' ^M
^M

Delete everything including and after the lines that look like:

------------- End Forwarded Message -------------^M
^M
^M

3) At a terminal line, type:

mimencode -u name_of_text_file -o new_file

# How do I open up an e-mail attachment with a .max extension?

If you receive an e-mail with an attachment containing the extension max, here is how you can open the file.

1) If you are on a Unix/Linux machine:
Click on the Windows icon on the bottom of your screen.

If you are on a Windows machine:
Go to Step 3. If you do not see Paperport Viewer, you may need to have
it installed for you.
2) Login, using the password you were assigned when you first arrived.

3) Click on Start->Programs->Paperport->Paperport Viewer. This will open the viewer.


text:
# Calculate probability using brute-force method

I have to calculate a probability using the following probabilities from a Bayesian network: $$P(+a)=...$$ $$P(+a|+b)=..., P(+a|¬b)=...$$ $$P(+b|+a)=..., P(+b|¬a)=...$$ $$P(+d|+b)=..., P(+d|¬b)=...$$ $$P(+f|+b,+c)=..., P(+f|¬b,+c)=..., P(+f|¬b,¬c)=...$$ $$P(+g|+c)=..., P(+g|¬c)=...$$

I have to calculate $$P(a|+d,+f,¬g)$$.

I have no idea about how to do it, but I guess it is something like that:

$$P(a|+d,+f,¬g)=\frac{P(a,+d,+f,¬g)}{P(+d,+f,¬g)}$$

And then, I have to calculate $$P(a,+d,+f,¬g)$$ and $$P(+d,+f,¬g)$$ using the previous probabilities. Is that correct?

But the $$P(a|$$ is confusing me because it hasn't a + or a ¬.

NOTE: you don't have to elaborate each formula, only point me to the right direction.

• Please, do not downvote. I can read. Your comment is enough to make me remove the image. Thanks. – VansFannel Apr 30 '19 at 15:17
• Downvote converted to upvote. Please note that downvote indicates my personal dislike for text images, since this is inconsiderate for those using screen readers and other assistive technologies, notably those with Disorders of Reading and Writing, visually impaired, etc. Besides, it's not searching on search engines like Approach0 – GNUSupporter 8964民主女神 地下教會 Apr 30 '19 at 15:28

## 1 Answer

The Bayesian network tells you that the joint distribution factorizes according to $$P(a,b,c,d,f,g) = P(a)P(b|a)P(c|a) P(d|b)P(f|b,c) P(g|c),$$ So you can compute any truth assignments using the conditional distributions you have presented.

For instance $$P(+a, \neg b, +c, \neg d, +f, +g) = P(+a)P(\neg b|+ a)P(+ c| + a) P(\neg d| \neg b)P(+ f|\neg b,+c) P(+g|+ c).$$ Note that you have all these probabilities: for those that you do not have explicit, consider the normalization constraint that tells you that $$P(\neg E) = 1 - P(E)$$.

You have correctly written that $$P(a| +d, +f, \neg g) = \frac{P(a,+d,+f,\neg g)}{P(+d,+f,\neg g)}.$$ Note that this is a short-hand that represents two probabilities: $$P(+a|+d,+f,\neg g)$$ and $$P(\neg a|+d,+f,\neg g)$$. By normalization, you only need to compute one of them, and the other one follows by $$P(+a|+d,+f,\neg g) = 1 - P(\neg a|+d,+f,\neg g)$$; so we only look at $$P(+a| +d, +f, \neg g)$$.

Now we need observe that the numerator and denominator probability involve a subset of the variables. This means that we must marginalize. We need to remove the variables that do not appear using the sum-rule of probability theory:

$$P(A) = \sum_B P(A, B).$$

In your case, we have that $$P(+a, +d, +f, \neg g) = \sum_{b,c} P(+a, b, c +d, +f, \neg g),$$ where the summation sign means that we sum over all possible outcomes of $$b$$, and $$c$$: $$P(+a, +d, +f, \neg g) = P(+a, +b, +c +d, +f, \neg g) + P(+a, +b, \neg c +d, +f, \neg g) + P(+a, \neg b, +c +d, +f, \neg g) + P(+a, \neg b, \neg c +d, +f, \neg g).$$ To compute each one of these probabilities, you use the definition of the joint at the beginning of the answer.

Similarly, you can compute $$P(\neg a, +d, +f, \neg g)$$. Then, you use the same rule to compute $$P(+d, +f, \neg g) = P(+a, +d, +f, \neg g) + P(\neg a, +d, +f, \neg g).$$

It's a lot of products and sums :) good luck!

• Thanks for your answer. I don't understand this sentence "We need to remove the variables that do not appear using the sum-rule of probability theory". I think you are saying that to get the probability of A, I have to sum all the probabilities where A appears. – VansFannel Apr 30 '19 at 15:13
• Quite the opposite, if you want the probability of $A$, but only know the joint $P(A,B)$, then you need to "remove" the $B$ from the joint, to find $P(A)$. You do this by summing the joint over all the possible outcomes of $B$: for instance $P(A) = P(A,+B) + P(A,\neg B)$. – Riccardo Sven Risuleo Apr 30 '19 at 15:15
• Maybe I misunderstood your point: a lot of terms in the summation may disappear; the only ones that matter are the ones where the variable you are interested in appears (together with all the ones they are conditioned upon). – Riccardo Sven Risuleo Apr 30 '19 at 15:19
• Thanks. In case of $P(+d, +f, \neg g)$, why don't you use $b$ and $c$? In other words, why do you use only $a$ to calculate it? Thanks. – VansFannel Apr 30 '19 at 15:26
• I think you have forgotten the $c$ on the first formula. – VansFannel Apr 30 '19 at 15:49


text:
## Zoltán Fülöp ; Dávid Kószó ; Heiko Vogler - Crisp-determinization of weighted tree automata over strong bimonoids

dmtcs:5943 - Discrete Mathematics & Theoretical Computer Science, June 15, 2021, vol. 23 no. 1
Crisp-determinization of weighted tree automata over strong bimonoids

Authors: Zoltán Fülöp ; Dávid Kószó ; Heiko Vogler

We consider weighted tree automata (wta) over strong bimonoids and their initial algebra semantics and their run semantics. There are wta for which these semantics are different; however, for bottom-up deterministic wta and for wta over semirings, the difference vanishes. A wta is crisp-deterministic if it is bottom-up deterministic and each transition is weighted by one of the unit elements of the strong bimonoid. We prove that the class of weighted tree languages recognized by crisp-deterministic wta is the same as the class of recognizable step mappings. Moreover, we investigate the following two crisp-determinization problems: for a given wta ${\cal A}$, (a) does there exist a crisp-deterministic wta which computes the initial algebra semantics of ${\cal A}$ and (b) does there exist a crisp-deterministic wta which computes the run semantics of ${\cal A}$? We show that the finiteness of the Nerode algebra ${\cal N}({\cal A})$ of ${\cal A}$ implies a positive answer for (a), and that the finite order property of ${\cal A}$ implies a positive answer for (b). We show a sufficient condition which guarantees the finiteness of ${\cal N}({\cal A})$ and a sufficient condition which guarantees the finite order property of ${\cal A}$. Also, we provide an algorithm for the construction of the crisp-deterministic wta according to (a) if ${\cal N}({\cal A})$ is finite, and similarly for (b) if ${\cal A}$ has finite order property. We prove that it is undecidable whether an arbitrary wta ${\cal A}$ is crisp-determinizable. We also prove that both, the finiteness of ${\cal N}({\cal A})$ and the finite order property of ${\cal A}$ are undecidable.

Volume: vol. 23 no. 1
Section: Automata, Logic and Semantics
Published on: June 15, 2021
Submitted on: December 6, 2019
Keywords: Computer Science - Formal Languages and Automata Theory


text:
Functions and algebra: Manipulate and simplify algebraic expressions

# Unit 1: Multiplying algebraic fractions

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Multiply algebraic fractions using factorisation.
• Divide algebraic fractions using factorisation.

## What you should know

Before you start this unit, make sure you can:

• Manipulate and simplify algebraic expressions. Refer to level 2 subject outcome 2.2 units 1 and 3 if you need help with this.
• Simplify algebraic expressions by means of factorisation. Refer to level 2 subject outcome 2.2 unit 2 if you need help with this.

Here is a short self-assessment to make sure you have the skills you need to proceed with this unit:

1. Simplify $\scriptsize 2(3a+b)(3a-b)-{{(3a-b)}^{2}}$.
2. Factorise $\scriptsize {{(x+y)}^{2}}-16{{y}^{2}}$.
3. Simplify $\scriptsize \displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\times \displaystyle \frac{{6a}}{{3a-2b}}$.

Solutions

1. .
\scriptsize \begin{align*}2(3a+b)(3a-b)-{{(3a-b)}^{2}}&=2(9{{a}^{2}}-{{b}^{2}})-(9{{a}^{2}}-6ab+{{b}^{2}})\\&=18{{a}^{2}}-2{{b}^{2}}-9{{a}^{2}}+6ab-{{b}^{2}}\\&=9{{a}^{2}}+6ab-3{{b}^{2}}\end{align*}
2. .
\scriptsize \begin{align*}{{(x+y)}^{2}}-16{{y}^{2}}&=\left( {\sqrt{{{{{(x+y)}}^{2}}}}-\sqrt{{16{{y}^{2}}}}} \right)\left( {\sqrt{{{{{(x+y)}}^{2}}}}+\sqrt{{16{{y}^{2}}}}} \right)\\&=(x+y-4y)(x+y+4y)\\&=(x-3y)(x+5y)\end{align*}
3. .
\scriptsize \begin{align*}\displaystyle \frac{{9{{a}^{2}}-4{{b}^{2}}}}{{18ab}}\times \displaystyle \frac{{6a}}{{3a-2b}}&=\displaystyle \frac{{(3a-2b)(3a+2b)}}{{18ab}}\times \displaystyle \frac{{6a}}{{(3a-2b)}}\\&=\displaystyle \frac{{(3a+2b)}}{{3\text{ }\cancel{1}\cancel{8}\cancel{a}b}}\times \displaystyle \frac{{\cancel{6}\cancel{a}}}{1}&& \text{Factors of }(3a-2b)\text{ cancel}\\&=\displaystyle \frac{{3a+2b}}{{3b}}\end{align*}

## Introduction

Algebraic expressions are the bread and butter of much of mathematics. To mix metaphors, they are the nuts and bolts of equations. In order to solve equations, you need to be able to simplify them and the algebraic expressions from which they are made.

Remember that an algebraic expression is made up of constants and variables joined together by addition, subtraction, multiplication and division.
In the expression $\scriptsize 3x+2$:

• the $\scriptsize 2$ is a constant because its value does not change
• the $\scriptsize x$ is a variable because its value can change
• the $\scriptsize 3$ in front of the variable $\scriptsize x$ is called the coefficient of $\scriptsize x$ and is being multiplied by the $\scriptsize x$
• the $\scriptsize 3x$ and $\scriptsize 2$ are called the terms of the algebraic expression. Terms are separated by addition or subtraction. In this algebraic expression we have two terms.

We usually have to simplify algebraic expressions to make them easier to work with or calculate with or use in some other way.

## Factorisation recap

Before we dive in, let’s spend some time revising what we know about factorisation.

In the same way that we can split any number into its factors (numbers that multiply together to produce the original number) we can also factorise algebraic expressions. For example, the factors of $\scriptsize 12$, in their pairs, are $\scriptsize 1\text{ and }12$; $\scriptsize 2\text{ and }6$; $\scriptsize 3\text{ and }4$. These are just the whole number factor pairs. $\scriptsize \displaystyle \frac{1}{2}\text{ and }24$ are also factors of $\scriptsize 12$ because $\scriptsize \displaystyle \frac{1}{2}\times 24=12$.

Similarly, we can say that the factors of $\scriptsize 4{{x}^{2}}$ are $\scriptsize 4\text{ and }x\text{ and }x$ because $\scriptsize 4\times x\times x=4{{x}^{2}}$.

### Factorising by taking out the highest common factor

Taking out the (usually) highest common factor is the first step of factorising algebraic expressions. It is the simplest to do but is also the first thing you should try when you need to split an algebraic fraction into its factors (i.e. factorise it). Here is an example:

$\scriptsize 4{{x}^{3}}+2{{x}^{2}}+6x$

We can see the common factors in each term more easily if we expand each term.

$\scriptsize \left( {2\times 2\times x\times x\times x} \right)+\left( {2\times x\times x} \right)+\left( {2\times 3\times x} \right)$

The highest common factor we can take out of each term is $\scriptsize 2\times x$.

$\scriptsize \left( {2\times 2\times x\times x\times x} \right)+\left( {2\times x\times x} \right)+\left( {2\times 3\times x} \right)$

If we take this highest common factor out of each term we get:

\scriptsize \begin{align*}&4{{x}^{3}}+2{{x}^{2}}+6x\\&=2x\left[ {\left( {2\times x\times x} \right)+\left( x \right)+\left( 3 \right)} \right]\\&=2x\left( {2{{x}^{2}}+x+3} \right)\end{align*}

### Factorising by grouping

Factorising by grouping is another way to create highest common factors in different parts of an algebraic expression. This is done by grouping these different parts together. Here is an example:

$\scriptsize 3ax+bx-3ay-by-9a-3b=\left( {3ax+bx} \right)-\left( {3ay+by} \right)-\left( {9a+3b} \right)$

You need to be careful of the signs inside the brackets. If you multiply the brackets out, you must always be left with the original expression.

Now we can take a common factor out of each bracket:

\scriptsize \begin{align*}3ax+bx-3ay-by-9a-3b&=\left( {3ax+bx} \right)-\left( {3ay+by} \right)-\left( {9a+3b} \right)\\ &=x(3a+b)-y(3a+b)-3(3a+b)\end{align*}

Can you see that we now have a common factor in each term of $\scriptsize (3a+b)$? Let’s take it out:

\scriptsize \begin{align*}3ax+bx-3ay-by-9a-3b&=\left( {3ax+bx} \right)-\left( {3ay+by} \right)-\left( {9a+3b} \right)\\ &=x(3a+b)-y(3a+b)-3(3a+b)\\ &=(3a+b)(x-y-3)\end{align*}

### Factorising the difference of two squares

This factorisation technique only applies to algebraic expressions where there are two terms subtracted from each other. The pattern is always the same.

$\scriptsize {{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$

Sometimes, as in the next example, you need to do some extra work before you can get to the difference of two squares.

\scriptsize \begin{align*}&2a({{a}^{2}}-9)-7({{a}^{2}}-9)&&\text{Take out the common factor of }({{a}^{2}}-9)\\ &=({{a}^{2}}-9)(2a-7)&&\text{Now factorise the difference of two squares}\\&=(a+3)(a-3)(2a-7)&& \text{Don't forget the original common factor of }(2a-7)\end{align*}

You don’t need to have perfect squares to factorise using this method. Just follow the basic pattern:

$\scriptsize x-y=\left( {\sqrt{x}+\sqrt{y}} \right)\left( {\sqrt{x}-\sqrt{y}} \right)$

A trinomial is an algebraic expression with three terms. A quadratic trinomial is an algebraic expression of the form $\scriptsize a{{x}^{2}}+bx+c$.

In level 2, we saw how to factorise quadratic trinomials using trial and error or a method called grouping. You can use whichever method you find easiest. Her are two examples:

\scriptsize \begin{align*}&2{{y}^{2}}+5y-3&&\text{This quadratic trinomial is in the standard form aleady}\\ &=(2y+1)(y-3)&& \text{Multiply the binomial together to make sure you get the original}\end{align*}

\scriptsize \begin{align*}&7x-6+5{{x}^{2}}&&\text{This quadratic trinomial is not yet in standard form}\\ &=5{{x}^{2}}+7x-6\\ &=(5x-3)(x+2)&& \text{Multiply the binomial together to make sure you get the original expression}\end{align*}

## Multiply algebraic fractions using factorisation

Multiplying algebraic fractions together works the same way as multiplying any other fractions together. We multiply the numerators to find the numerator of the product, and then we multiply the denominators to find the denominator of the product.

However, if we factorise the numerators and denominators of the fractions, we may find factors that we can ‘cancel’, making the multiplication of numerators and denominators simpler.

Remember, we never actually cancel anything. If we have the same factor in the numerator and the denominator, we recognise that the value of this is equal to $\scriptsize 1$ and therefore it has no influence on the value of the final product.

### Example 1.1

Multiply the algebraic fractions and simplify:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+4x-5}}{{3x+18}}\times \displaystyle \frac{{x+6}}{{x+5}}$

Solution

The first thing we need to do is factorise as many of the numerators and denominators as we can:

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+4x-5}}{{3x+18}}\times \displaystyle \frac{{x+6}}{{x+5}}\\&=\displaystyle \frac{{(x-1)(x+5)}}{{3(x+6)}}\times \displaystyle \frac{{x+6}}{{x+5}}\end{align*}

Now you can either multiply the numerators and denominators and then ‘cancel’, or ‘cancel’ and then multiply the numerators and denominators.

\scriptsize \begin{align*}&\displaystyle \frac{{(x-1)(x+5)}}{{3(x+6)}}\times \displaystyle \frac{{x+6}}{{x+5}}\\&=\displaystyle \frac{{(x-1)\bcancel{{(x+5)}}\bcancel{{(x+6)}}}}{{3\bcancel{{(x+6)}}\bcancel{{(x+5)}}}}\\&=\displaystyle \frac{{(x-1)}}{3}\end{align*}
OR
\scriptsize \begin{align*}&\displaystyle \frac{{(x-1)\bcancel{{(x+5)}}}}{{3\bcancel{{(x+6)}}}}\times \displaystyle \frac{{\bcancel{{x+6}}}}{{\bcancel{{x+5}}}}\\&=\displaystyle \frac{{(x-1)}}{3}\end{align*}

### Example 1.2

Simplify the algebraic expression:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+11x+30}}{{{{x}^{2}}+5x+6}}\cdot \displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+8x+16}}$

Solution

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+11x+30}}{{{{x}^{2}}+5x+6}}\cdot \displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+8x+16}}\\&=\displaystyle \frac{{(x+5)(x+6)}}{{(x+2)\bcancel{{(x+3)}}}}\cdot \displaystyle \frac{{\bcancel{{(x+4)}}\bcancel{{(x+3)}}}}{{\bcancel{{(x+4)}}(x+4)}}\\&=\displaystyle \frac{{(x+5)(x+6)}}{{(x+2)(x+4)}}\end{align*}

Note: While we could multiply the numerator and denominator out, we tend to leave them in factorised form. This is usually considered the simplest form of an expression.

### Take note!

If you have an expression such as $\scriptsize \displaystyle \frac{{x+3}}{{x+3}}$, you can ‘cancel’ the whole factor of $\scriptsize (x+3)$ in the numerator with the whole factor of $\scriptsize (x+3)$ in the denominator.
$\scriptsize \displaystyle \frac{{x+3}}{{x+3}}=\displaystyle \frac{{\bcancel{{(x+3)}}}}{{\bcancel{{(x+3)}}}}=1$

If you have an expression such as $\scriptsize \displaystyle \frac{{x+3}}{{x-3}}$, you cannot ‘cancel’ the $\scriptsize x$’s or the $\scriptsize 3$‘s because these are separate terms separated by $\scriptsize +$ and $\scriptsize -$ signs.

$\scriptsize \displaystyle \frac{{x+3}}{{x-3}}\ne \displaystyle \frac{{\bcancel{x}+\bcancel{3}}}{{\bcancel{x}-\bcancel{3}}}$

Convince yourself that this is true by substituting different values for $\scriptsize x$.

### Exercise 1.1

Simplify the following algebraic expressions:

1. $\scriptsize \displaystyle \frac{{n-12}}{{{{n}^{2}}-144}}$
2. $\scriptsize \displaystyle \frac{{{{x}^{2}}-x-6}}{{{{x}^{2}}-9}}\cdot \displaystyle \frac{{2{{x}^{2}}+7x-15}}{{2{{x}^{2}}+x-6}}$
3. $\scriptsize \displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{6}$
4. $\scriptsize \displaystyle \frac{{{{t}^{2}}-1}}{{{{t}^{2}}+4t+3}}\cdot \displaystyle \frac{{{{t}^{2}}+2t-15}}{{{{t}^{2}}-4t+3}}$
5. $\scriptsize \displaystyle \frac{{36{{x}^{2}}-25}}{{5{{x}^{2}}+70x+120}}\cdot \displaystyle \frac{{3{{x}^{2}}+36x}}{{2{{x}^{2}}-18}}\cdot \displaystyle \frac{{{{x}^{2}}-2x-3}}{{6{{x}^{2}}+11x+5}}$

The full solutions are at the end of the unit.

## Divide algebraic fractions using factorisation

Dividing algebraic fractions works the same way as the division of other fractions. To divide one algebraic fraction by another one, multiply the first expression by the reciprocal of the second. In other words:

$\scriptsize \displaystyle \frac{1}{x}\div \displaystyle \frac{{{{x}^{2}}}}{3}=\displaystyle \frac{1}{x}\times \displaystyle \frac{3}{{{{x}^{2}}}}$

Once the division expression has been rewritten as a multiplication expression, we can multiply the fractions as normal.

### Example 1.3

Divide the following algebraic expression and simplify:

$\scriptsize \displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\div \displaystyle \frac{{{{x}^{2}}-4}}{{{{x}^{2}}+2x+1}}$

Solution

The first step is to rewrite the division expression as a multiplication expression.

\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\div \displaystyle \frac{{{{x}^{2}}-4}}{{{{x}^{2}}+2x+1}}\\&=\displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\times \displaystyle \frac{{{{x}^{2}}+2x+1}}{{{{x}^{2}}-4}}\end{align*}

Now we can proceed as normal.

\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x-6}}{{{{x}^{2}}-1}}\times \displaystyle \frac{{{{x}^{2}}+2x+1}}{{{{x}^{2}}-4}}\\&=\displaystyle \frac{{(2x-3)\bcancel{{(x+2)}}}}{{\bcancel{{(x+1)}}(x-1)}}\times \displaystyle \frac{{\bcancel{{(x+1)}}(x+1)}}{{\bcancel{{(x+2)}}(x-2)}}\\&=\displaystyle \frac{{(2x-3)(x+1)}}{{(x-1)(x-2)}}\end{align*}

### Example 1.4

Divide the following algebraic expression and simplify:

$\scriptsize \displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\div \displaystyle \frac{{3{{x}^{2}}-10x+8}}{{{{x}^{2}}+5x-14}}$

Solution

Rewrite the division expression as a multiplication expression by multiplying by the reciprocal of the second fraction.

\scriptsize \begin{align*}&\displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\div \displaystyle \frac{{3{{x}^{2}}-10x+8}}{{{{x}^{2}}+5x-14}}\\&=\displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\times \displaystyle \frac{{{{x}^{2}}+5x-14}}{{3{{x}^{2}}-10x+8}}\end{align*}

Now proceed as normal to factorise the numerators and denominators.

\scriptsize \begin{align*}&\displaystyle \frac{{9{{x}^{2}}-16}}{{3{{x}^{2}}+25x+28}}\times \displaystyle \frac{{{{x}^{2}}+5x-14}}{{3{{x}^{2}}-10x+8}}\\&\text{=}\displaystyle \frac{{\bcancel{{(3x+4)}}\bcancel{{(3x-4)}}}}{{\bcancel{{(3x+4)}}\bcancel{{(x+7)}}}}\times \displaystyle \frac{{\bcancel{{(x-2)}}\bcancel{{(x+7)}}}}{{\bcancel{{(3x-4)}}\bcancel{{(x-2)}}}}\\&=1\end{align*}

### Note

The reciprocal of a number is its multiplicative inverse. If $\scriptsize x$ is any non-zero number, then its reciprocal is $\scriptsize \displaystyle \frac{1}{x}$ such that $\scriptsize x\times \displaystyle \frac{1}{x}=1$.

### Exercise 1.2

Divide the following algebraic expressions and simplify.

1. $\scriptsize \displaystyle \frac{{2{{x}^{2}}+x-3}}{{7{{x}^{2}}-7}}\div \displaystyle \frac{{2{{x}^{2}}-x-6}}{{3x+3}}$
2. $\scriptsize \displaystyle \frac{{{{q}^{2}}-9}}{{{{q}^{2}}+6q+9}}\div \displaystyle \frac{{{{q}^{2}}-2q-3}}{{{{q}^{2}}+2q-3}}$
3. $\scriptsize \displaystyle \frac{{144{{b}^{2}}-25}}{{72{{b}^{2}}-6b-10}}\div \displaystyle \frac{{24{{b}^{2}}+22b+5}}{{36{{b}^{2}}-9}}$

The full solutions are at the end of the unit.

### Simplify complex fractions

Let’s now look at how to simplify complex fractions.

### Example 1.5

Simplify the algebraic expression:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+x-6}}{{{{x}^{2}}-2x-3}}\cdot \displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-x-2}}\div \displaystyle \frac{{10{{x}^{2}}+27x+18}}{{{{x}^{2}}+2x+1}}$

Solution

In this case, we can either multiply the first two fractions and then deal with the division of the third fraction or we can rewrite the division as a multiplication straight away. Rewriting it as a multiplication straight away will allow us to ‘cancel’ in one step.

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+x-6}}{{{{x}^{2}}-2x-3}}\cdot \displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-x-2}}\div \displaystyle \frac{{10{{x}^{2}}+27x+18}}{{{{x}^{2}}+2x+1}}\\&=\displaystyle \frac{{{{x}^{2}}+x-6}}{{{{x}^{2}}-2x-3}}\cdot \displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-x-2}}\times \displaystyle \frac{{{{x}^{2}}+2x+1}}{{10{{x}^{2}}+27x+18}}\\&=\displaystyle \frac{{\bcancel{{(x-2)}}(x+3)}}{{\bcancel{{(x+1)}}\bcancel{{(x-3)}}}}\times \displaystyle \frac{{\bcancel{{(2x+3)}}\bcancel{{(x-3)}}}}{{\bcancel{{(x+1)}}\bcancel{{(x-2)}}}}\times \displaystyle \frac{{\bcancel{{(x+1)}}\bcancel{{(x+1)}}}}{{(5x+6)\bcancel{{(2x+3)}}}}\\&=\displaystyle \frac{{x+3}}{{5x+6}}\end{align*}

### Example 1.6

Simplify the algebraic expression:

$\scriptsize \displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+x-6}}\div \displaystyle \frac{{{{x}^{2}}-16}}{{{{x}^{2}}+4x+3}}\div \displaystyle \frac{{3x+9}}{{3{{x}^{2}}-12}}$

Solution

When you have an expression with two division signs, you can rewrite both of them by multiplying by the reciprocal of each fraction.

\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+x-6}}\div \displaystyle \frac{{{{x}^{2}}-16}}{{{{x}^{2}}+4x+3}}\div \displaystyle \frac{{3x+9}}{{3{{x}^{2}}-12}}\\&=\displaystyle \frac{{{{x}^{2}}+7x+12}}{{{{x}^{2}}+x-6}}\times \displaystyle \frac{{{{x}^{2}}+4x+3}}{{{{x}^{2}}-16}}\times \displaystyle \frac{{3{{x}^{2}}-12}}{{3x+9}}\\&=\displaystyle \frac{{(x+3)(x+4)}}{{(x-2)(x+3)}}\times \displaystyle \frac{{(x+1)(x+3)}}{{(x+4)(x-4)}}\times \displaystyle \frac{{3({{x}^{2}}-4)}}{{3(x+3)}}\\&=\displaystyle \frac{{\bcancel{{(x+3)}}\bcancel{{(x+4)}}}}{{\bcancel{{(x-2)}}\bcancel{{(x+3)}}}}\times \displaystyle \frac{{(x+1)\bcancel{{(x+3)}}}}{{\bcancel{{(x+4)}}(x-4)}}\times \displaystyle \frac{{\bcancel{3}(x+2)\bcancel{{(x-2)}}}}{{\bcancel{3}\bcancel{{(x+3)}}}}\\&=\displaystyle \frac{{(x+1)(x+2)}}{{(x-4)}}\end{align*}

### Exercise 1.3

Simplify the following:

1. $\scriptsize \displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\div \displaystyle \frac{{5{{x}^{3}}-20x}}{6}$
2. $\scriptsize \left( {\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}} \right)\left( {\displaystyle \frac{{2{{x}^{3}}+8{{x}^{2}}+6x}}{{16{{x}^{3}}+24{{x}^{2}}}}} \right)\div \displaystyle \frac{{{{x}^{2}}-x-2}}{{{{x}^{2}}+3x-10}}$
3. $\scriptsize \displaystyle \frac{{\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)}}{{y-4}}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to multiply algebraic fractions by first factorising the numerators and denominators in order to ‘cancel’ like factors.
• How to divide algebraic fractions by multiplying by the reciprocal of the fraction.

# Unit 1: Assessment

#### Suggested time to complete: 25 minutes

Simplify the following:

1. $\scriptsize \displaystyle \frac{{{{a}^{2}}+9a+18}}{{{{a}^{2}}+3a-18}}$
2. $\scriptsize \displaystyle \frac{{{{c}^{2}}+2c-24}}{{{{c}^{2}}+12c+36}}\cdot \displaystyle \frac{{{{c}^{2}}-10c+24}}{{{{c}^{2}}-8c+16}}$
3. $\scriptsize \displaystyle \frac{{16{{a}^{2}}-24a+9}}{{4{{a}^{2}}+17a-15}}\div \displaystyle \frac{{16{{a}^{2}}-9}}{{4{{a}^{2}}+11a+6}}$
4. $\scriptsize \displaystyle \frac{{2{{x}^{2}}+x}}{{2x-1}}\times \displaystyle \frac{{4{{x}^{2}}-1}}{{{{x}^{2}}}}\div \displaystyle \frac{{2x+1}}{x}$
5. $\scriptsize \displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\div \displaystyle \frac{{x-7}}{{4x+5}}\cdot \displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\div \displaystyle \frac{{3x-4}}{{x-1}}$

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{n-12}}{{{{n}^{2}}-144}}\\&=\displaystyle \frac{{\bcancel{{n-12}}}}{{(n+12)\bcancel{{(n-12)}}}}\\&=\displaystyle \frac{1}{{n+12}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}-x-6}}{{{{x}^{2}}-9}}\cdot \displaystyle \frac{{2{{x}^{2}}+7x-15}}{{2{{x}^{2}}+x-6}}\\&=\displaystyle \frac{{\bcancel{{(x+2)}}\bcancel{{(x-3)}}}}{{(x+3)\bcancel{{(x-3)}}}}\cdot \displaystyle \frac{{\bcancel{{(2x-3)}}(x+5)}}{{\bcancel{{(2x-3)}}\bcancel{{(x+2)}}}}\\&=\displaystyle \frac{{(x+5)}}{{(x+3)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{2}{{{{{(x-y)}}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}-{{y}^{2}}}}{6}\\&=\displaystyle \frac{{\bcancel{2}}}{{\bcancel{{(x-y)}}(x-y)}}\times \displaystyle \frac{{(x+y)\bcancel{{(x-y)}}}}{{\bcancel{6}3}}\\&=\displaystyle \frac{{(x+y)}}{{3(x-y)}}\end{align*}
4. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{t}^{2}}-1}}{{{{t}^{2}}+4t+3}}\cdot \displaystyle \frac{{{{t}^{2}}+2t-15}}{{{{t}^{2}}-4t+3}}\\&=\displaystyle \frac{{\bcancel{{(t+1)}}\bcancel{{(t-1)}}}}{{\bcancel{{(t+1)}}(t+3)}}\cdot \displaystyle \frac{{\bcancel{{(t-3)}}(t+5)}}{{\bcancel{{(t-1)}}\bcancel{{(t-3)}}}}\\&=\displaystyle \frac{{(t+5)}}{{(t+3)}}\end{align*}
5. .
\scriptsize \begin{align*}&\displaystyle \frac{{36{{x}^{2}}-25}}{{5{{x}^{2}}+70x+120}}\cdot \displaystyle \frac{{3{{x}^{2}}+36x}}{{2{{x}^{2}}-18}}\cdot \displaystyle \frac{{{{x}^{2}}-2x-3}}{{6{{x}^{2}}+11x+5}}\\&=\displaystyle \frac{{\bcancel{{(6x+5)}}(6x-5)}}{{5\bcancel{{(x+12)}}(x+2)}}\cdot \displaystyle \frac{{3x\bcancel{{(x+12)}}}}{{2(x+3)\bcancel{{(x-3)}}}}\cdot \displaystyle \frac{{\bcancel{{(x+1)}}\bcancel{{(x-3)}}}}{{\bcancel{{(6x+5)}}\bcancel{{(x+1)}}}}\\&=\displaystyle \frac{{3x(6x-5)}}{{10(x+2)(x+3)}}\end{align*}

Back to Exercise 1.1

### Exercise 1.2

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x-3}}{{7{{x}^{2}}-7}}\div \displaystyle \frac{{2{{x}^{2}}-x-6}}{{3x+3}}\\&=\displaystyle \frac{{2{{x}^{2}}+x-3}}{{7{{x}^{2}}-7}}\times \displaystyle \frac{{3x+3}}{{2{{x}^{2}}-x-6}}\\&=\displaystyle \frac{{\bcancel{{(2x+3)}}\bcancel{{(x-1)}}}}{{7\bcancel{{(x+1)}}\bcancel{{(x-1)}}}}\times \displaystyle \frac{{3\bcancel{{(x+1)}}}}{{\bcancel{{(2x+3)}}(x-2)}}\\&=\displaystyle \frac{3}{{7(x-2)}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{q}^{2}}-9}}{{{{q}^{2}}+6q+9}}\div \displaystyle \frac{{{{q}^{2}}-2q-3}}{{{{q}^{2}}+2q-3}}\\&=\displaystyle \frac{{{{q}^{2}}-9}}{{{{q}^{2}}+6q+9}}\times \displaystyle \frac{{{{q}^{2}}+2q-3}}{{{{q}^{2}}-2q-3}}\\&=\displaystyle \frac{{\bcancel{{(q+3)}}\bcancel{{(q-3)}}}}{{\bcancel{{(q+3)}}\bcancel{{(q+3)}}}}\times \displaystyle \frac{{(q-1)\bcancel{{(q+3)}}}}{{(q+1)\bcancel{{(q-3)}}}}\\&=\displaystyle \frac{{(q-1)}}{{(q+1)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{144{{b}^{2}}-25}}{{72{{b}^{2}}-6b-10}}\div \displaystyle \frac{{24{{b}^{2}}+22b+5}}{{36{{b}^{2}}-9}}\\&=\displaystyle \frac{{144{{b}^{2}}-25}}{{72{{b}^{2}}-6b-10}}\times \displaystyle \frac{{36{{b}^{2}}-9}}{{24{{b}^{2}}+22b+5}}\\&=\displaystyle \frac{{\bcancel{{(12b+5)}}\bcancel{{(12b-5)}}}}{{2\bcancel{{(12b-5)}}(3b+1)}}\times \displaystyle \frac{{9\bcancel{{(2b+1)}}(2b-1)}}{{\bcancel{{(12b+5)}}\bcancel{{(2b+1)}}}}\\&=\displaystyle \frac{{9(2b-1)}}{{2(3b+1)}}\end{align*}

Back to Exercise 1.2

### Exercise 1.3

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\div \displaystyle \frac{{5{{x}^{3}}-20x}}{6}\\&=\displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\times \displaystyle \frac{6}{{5{{x}^{3}}-20x}}\\&=\displaystyle \frac{{{{x}^{2}}+6x-7}}{{x+1}}\cdot \displaystyle \frac{{{{x}^{2}}-x-2}}{{2x-14}}\times \displaystyle \frac{6}{{5x({{x}^{2}}-4)}}\\&=\displaystyle \frac{{\bcancel{{(x+1)}}\bcancel{{(x-7)}}}}{{\bcancel{{(x+1)}}}}\cdot \displaystyle \frac{{(x+1)\bcancel{{(x-2)}}}}{{2\bcancel{{(x-7)}}}}\times \displaystyle \frac{6}{{5x(x+2)\bcancel{{(x-2)}}}}\\&=\displaystyle \frac{{6(x+1)}}{{10x(x+2)}}\\&=\displaystyle \frac{{3(x+1)}}{{5x(x+2)}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}\times \displaystyle \frac{{2{{x}^{3}}+8{{x}^{2}}+6x}}{{16{{x}^{3}}+24{{x}^{2}}}}\div \displaystyle \frac{{{{x}^{2}}-x-2}}{{{{x}^{2}}+3x-10}}\\&=\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}\times \displaystyle \frac{{2{{x}^{3}}+8{{x}^{2}}+6x}}{{16{{x}^{3}}+24{{x}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}+3x-10}}{{{{x}^{2}}-x-2}}\\&=\displaystyle \frac{{2{{x}^{2}}-3x-9}}{{{{x}^{2}}-9}}\times \displaystyle \frac{{2x({{x}^{2}}+4x+3)}}{{16{{x}^{3}}+24{{x}^{2}}}}\times \displaystyle \frac{{{{x}^{2}}+3x-10}}{{{{x}^{2}}-x-2}}\\&=\displaystyle \frac{{\bcancel{{(2x+3)}}\bcancel{{(x-3)}}}}{{\bcancel{{(x+3)}}\bcancel{{(x-3)}}}}\times \displaystyle \frac{{2x\bcancel{{(x+3)}}\bcancel{{(x+1)}}}}{{8{{x}^{2}}\bcancel{{(2x+3)}}}}\times \displaystyle \frac{{\bcancel{{(x-2)}}(x+5)}}{{\bcancel{{(x+1)}}\bcancel{{(x-2)}}}}\\&=\displaystyle \frac{{2x(x+5)}}{{8{{x}^{2}}}}\\&=\displaystyle \frac{{(x+5)}}{{4x}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)}}{{y-4}}\\&=\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)\div \displaystyle \frac{{y-4}}{1}\\&=\left( {\displaystyle \frac{{3{{y}^{2}}-10y+3}}{{3{{y}^{2}}+5y-2}}\cdot \displaystyle \frac{{2{{y}^{2}}-3y-20}}{{2{{y}^{2}}-y-15}}} \right)\times \displaystyle \frac{1}{{y-4}}\\&=\displaystyle \frac{{\bcancel{{(3y-1)}}\bcancel{{(y-3)}}}}{{\bcancel{{(3y-1)}}(y+2)}}\cdot \displaystyle \frac{{\bcancel{{(2y+5)}}\bcancel{{(y-4)}}}}{{\bcancel{{(2y+5)}}\bcancel{{(y-3)}}}}\times \displaystyle \frac{1}{{\bcancel{{(y-4)}}}}\\&=\displaystyle \frac{1}{{(y+2)}}\end{align*}

Back to Exercise 1.3

### Unit 1: Assessment

1. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{a}^{2}}+9a+18}}{{{{a}^{2}}+3a-18}}\\&=\displaystyle \frac{{(a+3)\bcancel{{(a+6)}}}}{{(a-3)\bcancel{{(a+6)}}}}\\&=\displaystyle \frac{{a+3}}{{a-3}}\end{align*}
2. .
\scriptsize \begin{align*}&\displaystyle \frac{{{{c}^{2}}+2c-24}}{{{{c}^{2}}+12c+36}}\cdot \displaystyle \frac{{{{c}^{2}}-10c+24}}{{{{c}^{2}}-8c+16}}\\&=\displaystyle \frac{{\bcancel{{(c-4)}}\bcancel{{(c+6)}}}}{{(c+6)\bcancel{{(c+6)}}}}\cdot \displaystyle \frac{{(c-6)\bcancel{{(c-4)}}}}{{\bcancel{{(c-4)}}\bcancel{{(c-4)}}}}\\&=\displaystyle \frac{{(c-6)}}{{(c+6)}}\end{align*}
3. .
\scriptsize \begin{align*}&\displaystyle \frac{{16{{a}^{2}}-24a+9}}{{4{{a}^{2}}+17a-15}}\div \displaystyle \frac{{16{{a}^{2}}-9}}{{4{{a}^{2}}+11a+6}}\\&=\displaystyle \frac{{16{{a}^{2}}-24a+9}}{{4{{a}^{2}}+17a-15}}\times \displaystyle \frac{{4{{a}^{2}}+11a+6}}{{16{{a}^{2}}-9}}\\&=\displaystyle \frac{{\bcancel{{(4a-3)}}\bcancel{{(4a-3)}}}}{{\bcancel{{(4a-3)}}(a+5)}}\times \displaystyle \frac{{\bcancel{{(4a+3)}}(a+2)}}{{\bcancel{{(4a+3)}}\bcancel{{(4a-3)}}}}\\&=\displaystyle \frac{{(a+2)}}{{(a+5)}}\end{align*}
4. .
\scriptsize \begin{align*}&\displaystyle \frac{{2{{x}^{2}}+x}}{{2x-1}}\times \displaystyle \frac{{4{{x}^{2}}-1}}{{{{x}^{2}}}}\div \displaystyle \frac{{2x+1}}{x}\\&=\displaystyle \frac{{2{{x}^{2}}+x}}{{2x-1}}\times \displaystyle \frac{{4{{x}^{2}}-1}}{{{{x}^{2}}}}\times \displaystyle \frac{2}{{2x+1}}\\&=\displaystyle \frac{{x\bcancel{{(2x+1)}}}}{{\bcancel{{(2x-1)}}}}\times \displaystyle \frac{{(2x+1)\bcancel{{(2x-1)}}}}{{{{x}^{2}}}}\times \displaystyle \frac{2}{{\bcancel{{(2x+1)}}}}\\&=\displaystyle \frac{{2x}}{{{{x}^{2}}}}\\&=\displaystyle \frac{2}{x}\end{align*}
5. .
\scriptsize \begin{align*}&\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\div \displaystyle \frac{{x-7}}{{4x+5}}\cdot \displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\div \displaystyle \frac{{3x-4}}{{x-1}}\\&=\left( {\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\div \displaystyle \frac{{x-7}}{{4x+5}}} \right)\cdot \left( {\displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\div \displaystyle \frac{{3x-4}}{{x-1}}} \right)&&\text{Placing the fractions being divided in brackets makes things clearer}\\&=\left( {\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\times \displaystyle \frac{{4x+5}}{{x-7}}} \right)\cdot \left( {\displaystyle \frac{{18{{x}^{3}}-32x}}{{4{{x}^{2}}+x-5}}\times \displaystyle \frac{{x-1}}{{3x-4}}} \right)\\&=\displaystyle \frac{{3{{x}^{2}}-21x}}{{18{{x}^{2}}+24x}}\times \displaystyle \frac{{4x+5}}{{x-7}}\cdot \displaystyle \frac{{2x(9{{x}^{2}}-16)}}{{4{{x}^{2}}+x-5}}\times \displaystyle \frac{{x-1}}{{3x-4}}\\&=\displaystyle \frac{{3x\bcancel{{(x-7)}}}}{{6x\bcancel{{(3x+4)}}}}\times \displaystyle \frac{{\bcancel{{(4x+5)}}}}{{\bcancel{{(x-7)}}}}\cdot \displaystyle \frac{{2x\bcancel{{(3x+4)}}(3x-4)}}{{\bcancel{{(4x+5)}}(x-1)}}\times \displaystyle \frac{{(x-1)}}{{(3x-4)}}\\&=\displaystyle \frac{{6{{x}^{2}}}}{{6x}}\\&=x\end{align*}

Back to Unit 1: Assessment


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Resistors - Maple Help

Resistors

Background

 Resistance What is resistance in a circuit?   By definition, an electrical circuit is a closed path where electrons flow. Resistance is the hindrance of the flow of electrons in a circuit. Electrons do not flow through the circuit in a direct path. Rather, electrons traverse the circuit in a zigzag fashion dictated by the collisions with the atoms of the medium of travel. These collisions hinder the movement of electrons and hence add resistance to the circuit.   What are the factors that affect resistance?   Length of path - The longer the path, the more places where collisions can occur and the more resistance occurs in the circuit. Cross sectional area of path - A wider path means that more current can flow through the path at the same time. Because electrons flow at higher rates through paths with greater cross sectional areas, less resistance occurs in the circuit. Material of path - Various materials have different conductive abilities. Consequently, bad conductors will increase resistance in a circuit, compared to good conductors.   What is a resistor?   A resistor is a passive electronic component (it does not have its own source of power) that reduces the electron flow through a circuit, thereby adding resistance to the circuit. It does this by using the aforementioned factors and is also engineered to maintain a specific resistance value over a range of conditions. Energy that the resistors remove from the circuit is lost in the form of heat. An ideal resistor behaves as dictated by Ohm's law: $I=\frac{\mathrm{ΔV}}{R}$ , where $\mathrm{ΔV}$ is the potential difference or voltage across the resistor in volts (V), $R$ is the resistance value of the resistor in ohms (Ω) and $I$ is the current that flows through the resistor in amps (A).   Resistors can be combined in series and parallel to make more complicated circuits. On a physical resistor component, the colors of the bands on the resistor describe the resistance value (Ω) of the resistor. Typically, there are four bands and the color values are read from left to right. The last or fourth band is usually gold or silver. The first two bands represent the value of the resistor and the third band represents the multiplier. The fourth band represents the tolerance value of the resistor and is a percentage indicating the range over that the actual resistance value may deviate from the intended one. For example, a 5% tolerance means the resistor is within ±5% of the intended value.

In the activity below, use the left hand set of Combo Boxes to find out the resistance of a resistor based on its color code, or use the right hand set of Combo Boxes to find out what the specific resistance of a resistor looks like.

 First Band: BlackBrownRedOrangeYellowGreenBlueVioletGrayWhite   Second Band: BlackBrownRedOrangeYellowGreenBlueVioletGrayWhite   Third Band: SilverGoldBlackBrownRedOrangeYellowGreenBlueViolet   Fourth Band: SilverGoldBrownRedGreenBlueVioletGray First Digit: 0123456789   Second Digit: 0123456789   Multiplier: 1101001000100001000001000000100000000.10.01   Tolerance: 105210.50.250.10.05%

To verify, you can use the resistance values from the resistor color code chart below.

 Color 1st Digit (Band 1) 2nd Digit (Band 2) Multiplier (Band 3) Tolerance % (Band 4) Black 0 0 1 - Brown 1 1 10 ±1 Red 2 2 100 ±2 Orange 3 3 1k - Yellow 4 4 10k - Green 5 5 100k ±0.5 Blue 6 6 1M ±0.25 Violet 7 7 10M ±0.1 Gray 8 8 - ±0.05 White 9 9 - - Gold - - 0.1 ±5 Silver - - 0.01 ±10

 More MathApps


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# Derivative Mathematical Formulas

## Derivative Mathematical Formulas

LIMIT EVALUATION AT +- ∞

• $\lim\limits_{x \to \infty } e^{x} = \infty$
• $\lim\limits_{x \to -\infty } e^{x} = 0$
• $\lim\limits_{x \to \infty} lnx = \infty$
• $\lim\limits_{x \to \infty} \frac{c}{x^{n}} = 0 \left ( n > 0 \right )$
• $\lim\limits_{x \to \infty} \frac{x}{\sqrt[x]{x!}} = e$
• $\lim\limits_{x \to \infty} \left ( 1 + \frac{k}{x} \right )^{x} = e^{k}, e = 2.71$
• $\lim\limits_{x \to \infty} \left ( 1 - \frac{1}{x} \right )^{x} = \frac{1}{e}$
• $\lim\limits_{x \to \infty} x \left ( \frac{\sqrt{2\Pi x}}{x!} \right )^{\frac{1}{x}} = e$
• $\lim\limits_{x \to \infty} \frac{x!}{x^{x}e^{-x}\sqrt{x}} = \sqrt{2\Pi }$
• $\lim\limits_{x \to \infty} log_{a}\left ( 1 + \frac{1}{x} \right )^{x} = log_{a} e$

### PROPERTIES OF LIMITS

• $\lim\limits_{x \to a}[Cf\left ( x \right )] = C \lim_{x \to a}[f\left ( x \right )]$
• $\lim\limits_{x \to a}[f\left ( x \right )]^{n} = [\lim_{x \to a} f\left ( x \right )]^{n}$

LIMIT EVALUATION AT ZERO

• $\lim\limits_{x \to 0} a^{x} = 1$
• $\lim\limits_{x \to 0} \frac{log_{e}\left ( 1+x \right )}{x} = 1$
• $\lim\limits_{x \to 0} \frac{x}{log_{a}\left ( 1+x \right )} = \frac{1}{log_{a}e}$
• $\lim\limits_{x \to 0} \frac{a^{x}-1}{x} = ln a, a > 0$
• $\lim\limits_{x \to 0} \frac{sin x}{x} = 1$
• $\lim\limits_{x \to 0} \frac{tan x}{x} = 1$
• $\lim\limits_{x \to 0} \frac{1 - cos x}{x} = 0$
• $\lim\limits_{x \to 0} \frac{1 - cos x}{x^{2}} = \frac{1}{2}$
• $\lim\limits_{x \to 0} \frac{arcsin x}{x} = 1$
• $\lim\limits_{x \to 0} \frac{arctan x}{x} = 1$
• $\lim\limits_{x \to 1} \frac{ \left ( arccos x \right )^{2}}{1-x} = 2$

## Derivative Definition

• $\frac{d}{dx}\left ( f\left ( x \right ) \right ) = f'\left ( x \right ) = \lim_{h \to 0}\frac{f\left ( x+h \right )-f\left ( x \right )}{h}$

## Basic Properties

• $\left ( cf\left ( x \right ) \right )' = c\left ( f'\left ( x \right ) \right )$
• $\left ( f\left ( x \right )\pm g\left ( x \right ) \right )' = f'\left ( x \right )\pm g'\left ( x \right )$
• $\frac{d}{dx}\left ( c \right )=0$

## Mean Value Theorem

• $f'\left ( c \right ) = \frac{f\left ( b \right )-f\left ( a \right )}{b-a}$

## Quotient Rule

• $\frac{d}{dx}\left ( \frac{f\left ( x \right )}{g\left ( x \right )} \right ) = \frac{f'\left ( x \right )g\left ( x \right )-f\left ( x \right )g'\left ( x \right )}{\left [ g\left ( x \right ) \right ]^{2}}$

## Power Rule

• $\frac{d}{dx}\left ( x^{n} \right ) = nx^{n-1}$

Chain Rule

• $\frac{d}{dx}\left ( f\left ( g\left ( x \right ) \right ) \right ) = f'\left ( g\left ( x \right ) \right )g'\left ( x \right )$

## Limit Evaluation Method

• $\lim\limits_{x \to -3}\frac{x^{2}-x-12}{x^{2}+3x} = \lim\limits_{x \to -3}\frac{\left ( x+3 \right )\left ( x-4 \right )}{x\left ( x+3 \right )} = \lim\limits_{x \to -3}\frac{\left ( x-4 \right )}{x} = \frac{7}{3}$

## L’Hopital’s Rule

• $\lim\limits_{x \to a}\frac{f\left ( x \right )}{g\left ( x \right )} = \frac{0}{0} or \frac{\pm \infty }{\pm \infty } then \lim\limits_{x \to a}\frac{f\left ( x \right )}{g\left ( x \right )} = \lim\limits_{x \to a}\frac{f'\left ( x \right )}{g'\left ( x \right )}$

## Common Derivatives

• $\frac{d}{dx}\left ( x \right ) = 1$
• $\frac{d}{dx}\left ( sin x \right ) = cos x$
• $\frac{d}{dx}\left ( cos x \right ) = -sin x$
• $\frac{d}{dx}\left ( tan x \right ) = sec^{2} x$
• $\frac{d}{dx}\left ( sec x \right ) = sec x tan x$
• $\frac{d}{dx}\left ( csc x \right ) = -csc x cot x$
• $\frac{d}{dx}\left ( cot x \right ) = -csc^{2} x$
• $\frac{d}{dx}\left ( sin^{-1} x \right ) = \frac{1}{\sqrt{1-x^{2}}}$
• $\frac{d}{dx}\left ( cos^{-1} x \right ) = -\frac{1}{\sqrt{1-x^{2}}}$
• $\frac{d}{dx}\left ( tan^{-1} x \right ) = -\frac{1}{1+x^{2}}$
• $\frac{d}{dx}\left ( a^{x} \right ) = a^{x} ln\left ( a \right )$
• $\frac{d}{dx}\left ( e^{x} \right ) = e^{x}$
• $\frac{d}{dx}\left ( ln\left ( x \right ) \right ) = \frac{1}{x}, x>0$
• $\frac{d}{dx}\left ( ln\left | x \right | \right ) = \frac{1}{x}$
• $\frac{d}{dx}\left ( log_{a}\left ( x \right ) \right ) = \frac{1}{x ln\left ( a \right )}$

## Chain Rule and Other Examples

• $\frac{d}{dx}\left ( \left [ f\left ( x \right ) \right ]^{n} \right ) = n\left [ f\left ( x \right ) \right ]^{n-1}f'\left ( x \right )$
• $\frac{d}{dx}\left ( e^{f\left ( x \right )} \right ) = f'\left ( x \right )e^{f\left ( x \right )}$
$\frac{d}{dx}\left ( ln\left [ f\left ( x \right ) \right ] \right ) = \frac{f'\left ( x \right )}{f\left ( x \right )}$
• $\frac{d}{dx}\left ( sin\left [ f\left ( x \right ) \right ] \right ) = f'\left ( x \right )cos\left [ f\left ( x \right ) \right ]$
• $\frac{d}{dx}\left ( cos\left [ f\left ( x \right ) \right ] \right ) = -f'\left ( x \right )sin\left [ f\left ( x \right ) \right ]$
• $\frac{d}{dx}\left ( tan\left [ f\left ( x \right ) \right ] \right ) = f'\left ( x \right )sec^{2}\left [ f\left ( x \right ) \right ]$
• $\frac{d}{dx}\left ( sec\left [ f\left ( x \right ) \right ] \right ) = f'\left ( x \right )sec\left [ f\left ( x \right ) \right ]tan\left [ f\left ( x \right ) \right ]$
• $\frac{d}{dx}\left ( tan^{-1}\left [ f\left ( x \right ) \right ] \right ) = \frac{f'\left ( x \right )}{1+\left [ f\left ( x \right ) \right ]^{2}}$
• $\frac{d}{dx}\left ( f\left ( x \right )^{g\left ( x \right )} \right ) = f\left ( x \right )^{g\left ( x \right )}\left ( \frac{g\left ( x \right )f'\left ( x \right )}{f\left ( x \right )}+ln \left ( f\left ( x \right ) \right )g'\left ( x \right ) \right )$

## Properties of Limit

• $\lim\limits_{x \to a}\left [ f\left ( x \right )\pm g\left ( x \right ) \right ] = \lim\limits_{x \to a} f\left ( x \right )\pm \lim\limits_{x \to a} g\left ( x \right )$
• $\lim\limits_{x \to a}\left [ f\left ( x \right ) g\left ( x \right ) \right ] = \lim\limits_{x \to a} f\left ( x \right ) \lim\limits_{x \to a} g\left ( x \right )$
• $\lim\limits_{x \to a}\left [ \frac{f\left ( x \right )}{g\left ( x \right )} \right ] = \frac{\lim\limits_{x \to a}f\left ( x \right )}{\lim\limits_{x \to a}g\left ( x \right )} if \lim_{x \to a}g\left ( x \right )\neq 0$

We provide few shortcut tricks on this topic. Please visit this page to get updates on more Math Shortcut Tricks. You can also like our facebook page to get updates.

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Article

# Globular Cluster Abundances from High-Resolution, Integrated-Light Spectroscopy. IV. The Large Magellanic Cloud: $\alpha$, Fe-peak, Light, and Heavy Elements

(Impact Factor: 5.99). 11/2011; 746(1). DOI: 10.1088/0004-637X/746/1/29
Source: arXiv

ABSTRACT

We present detailed chemical abundances in 8 clusters in the Large Magellanic
Cloud (LMC). We measure abundances of 22 elements for clusters spanning a range
in age of 0.05 to 12 Gyr, providing a comprehensive picture of the chemical
enrichment and star formation history of the LMC. The abundances were obtained
from individual absorption lines using a new method for analysis of high
resolution ($R\sim$25,000) integrated light spectra of star clusters. This
method was developed and presented in Papers I, II, and III of this series. In
this paper, we develop an additional integrated light $\chi^2$-minimization
spectral synthesis technique to facilitate measurement of weak ($\sim$15 m\AA)
spectral lines and abundances in low signal-to-noise ratio data (S/N$\sim$30).
Additionally, we supplement the integrated light abundance measurements with
detailed abundances that we measure for individual stars in the youngest
clusters (Age$<$2 Gyr) in our sample. In both the integrated light and stellar
abundances we find evolution of [$\alpha$/Fe] with [Fe/H] and age. Fe-peak
abundance ratios are similar to those in the Milky Way, with the exception of
[Cu/Fe] and [Mn/Fe], which are sub-solar at high metallicities. The heavy
elements Ba, La, Nd, Sm, and Eu are significantly enhanced in the youngest
clusters. Also, the heavy to light s-process ratio is elevated relative to the
Milky Way ([Ba/Y]$>+0.5$) and increases with decreasing age, indicating a
strong contribution of low-metallicity AGB star ejecta to the interstellar
medium throughout the later history of the LMC. We also find a correlation of
integrated light Na and Al abundances with cluster mass, in the sense that more
massive, older clusters are enriched in the light elements Na and Al with
respect to Fe, which implies that these clusters harbor star-to-star abundance
variations as is common in the Milky Way.

### Full-text

Available from: Janet Colucci, Aug 08, 2014
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• Source
• "Gray points are from [12], and measured using Lick system line indexes. synthesis code [7] for the best possible accuracy in our final measurements for Ca, Si, Eu, Ba and Y. Abundances for Ba and Eu include corrections for hyperfine splitting, as described in [7]. The IL abundance analysis is described in detail in [3] [4] [5]. "
##### Article: The Detailed Chemical Abundance Patterns of M31 Globular Clusters
[Hide abstract]
ABSTRACT: We present detailed chemical abundances for $>$20 elements in $\sim$30 globular clusters in M31. These results have been obtained using high resolution ($\lambda/\Delta\lambda\sim$24,000) spectra of their integrated light and analyzed using our original method. The globular clusters have galactocentric radii between 2.5 kpc and 117 kpc, and therefore provide abundance patterns for different phases of galaxy formation recorded in the inner and outer halo of M31. We find that the clusters in our survey have a range in metallicity of $-2.2<$[Fe/H]$<-0.11$. The inner halo clusters cover this full range, while the outer halo globular clusters at R$>$20 kpc have a small range in abundance of [Fe/H]$=-1.6 \pm 0.10$. We also measure abundances of alpha, r- and s-process elements. These results constitute the first abundance pattern constraints for old populations in M31 that are comparable to those known for the Milky Way halo.
• Source
##### Article: Comparison of Convective Overshooting Models and Their Impact on Abundances from Integrated Light Spectroscopy of Young ($<$ 3 Gyr) Star Clusters
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ABSTRACT: As part of an ongoing program to measure detailed chemical abundances in nearby galaxies, we use a sample of young to intermediate age clusters in the Large Magellanic Cloud with ages of 10 Myr to 2 Gyr to evaluate the effect of isochrone parameters, specifically core convective overshooting, on Fe abundance results from high resolution, integrated light spectroscopy. In this work we also obtain fiducial Fe abundances from high resolution spectroscopy of the cluster individual member stars. We compare the Fe abundance results for the individual stars to the results from isochrones and integrated light spectroscopy to determine whether isochrones with convective overshooting should be used in our integrated light analysis of young to intermediate age (10 Myr -3 Gyr) star clusters. We find that when using the isochrones from the Teramo group, we obtain more accurate results for young and intermediate age clusters over the entire age range when using isochrones without convective overshooting. While convective overshooting is not the only uncertain aspect of stellar evolution, it is one of the most readily parametrized ingredients in stellar evolution models, and thus important to evaluate for the specific models used in our integrated light analysis. This work demonstrates that our method for integrated light spectroscopy of star clusters can provide unique tests for future constraints on stellar evolution models of young and intermediate age clusters.
The Astrophysical Journal 02/2012; 749(2). DOI:10.1088/0004-637X/749/2/124 · 5.99 Impact Factor
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##### Article: GNOMOS: The Gemini NIR-Optical Multi Object Spectrograph
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ABSTRACT: This paper is a response to a call for white papers solicited by Gemini Observatory and its Science and Technology Advisory Committee, to help define the science case and requirements for a new Gemini instrument, envisaged to consist of a single-object spectrograph at medium resolution simultaneously covering optical and near-infrared wavelengths. In this white paper we discuss the science case for an alternative new instrument, consisting instead of a multi-object, medium-resolution, high-throughput spectrograph, covering simultaneously the optical and near-infrared slices of the electromagnetic spectrum. We argue that combination of wide wavelength coverage at medium resolution with moderate multiplexing power is an innovative path that will enable the pursuit of fundamental science questions in a variety of astrophysical topics, without compromise of the science goals achievable by single-object spectroscopy on a wide baseline. We present a brief qualitative discussion of the main features of a notional hardware design that could conceivably make such an instrument viable.


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# Descriptivist theory of names

(Redirected from Descriptive theory of names)

Descriptivist theory of names is a view of the nature of the meaning and reference of proper names generally attributed to Gottlob Frege and Bertrand Russell. The theory consists essentially in the idea that the meanings (semantic contents) of names are identical to the descriptions associated with them by speakers, while their referents are determined to be the objects that satisfy these descriptions.

In the 1970s, this theory came under strong attack from causal theorists such as Saul Kripke, Hilary Putnam and others. However, it has seen something of a revival in recent years, especially under the form of what are called two-dimensional semantic theories. This latter trend is exemplified by the theories of David Chalmers, among others.

## The descriptive theory and its merits

A simple descriptivist theory of names can be thought of as follows: for every proper name p, there is some collection of descriptions D associated with p that constitute the meaning of p. For example, the descriptivist may hold that the proper name Saul Kripke is synonymous with the collection of descriptions such as

1. the man who wrote Naming and Necessity
2. a person who was born on November 13, 1940 in Bay Shore, New York
3. the son of a leader of Beth El Synagogue in Omaha, Nebraska
etc ...

The descriptivist takes the meaning of the name Saul Kripke to be that collection of descriptions and takes the referent of the name to be the thing that satisfies all or most of those descriptions.

A simple descriptivist theory may further hold that the meaning of a sentence S that contains p is given by the collection of sentences produced by replacing each instance of p in S with one of the descriptions in D. So, the sentence such as "Saul Kripke stands next to a table" has the same meaning as the following collection of sentences:

1. The man who wrote Naming and Necessity stands next to a table.
2. A person who was born on November 13, 1940 in Bay Shore, New York stands next to a table.
3. The son of a leader of Beth El Synagogue in Omaha, Nebraska stands next to a table.
etc ...

A type of simple descriptivism was originally formulated by Frege in reaction to problems that confronted the predominant theory of names of the 19th century due to John Stuart Mill. Mill's theory is often referred to as the "Fido"-Fido theory because it suggests that the meaning of a proper name is simply its bearer in the external world (its direct referent, as we would say now). There are several significant problems with this proposal, however. First, it does not explain how and why names without bearers can still be meaningful even though they have no reference. Take the following two sentences:

• (A) There is no Santa Claus.
• (B) Santa Claus does not exist.

According to Mill's theory, these sentences must be meaningless. This is the case because sentences obtain their meanings compositionally and one of the main constituents of these sentences--the predicate in the first and the subject in the second-- is meaningless. But such sentences obviously seem perfectly meaningful to most human beings and they are used constantly in everyday language to express true statements about reality. More evidence for the meaningfulness of sentences such as the ones presented above consists in the fact that they are synonymous. Furthermore, the conjunction of (A) and (B):

• (C) There is no Santa Claus and Santa Claus does not exist.

is redundant. While the following sentence:

• (D) Santa Claus does not exist but there is a Santa Claus.

Frege set about to resolve this problem, among others, with his famous distinction between sense and reference. In the case of proper names, the sense (or Sinn) of a term consists in the (usually) definite description that speakers associate with it. Thus, the sense of the proper name Santa Claus may be something like “The benevolent, bearded elf that brings gifts to children at Christmas time.” This sense of a term is objective (it is an abstract object) for Frege and is definitely not to be confused with its subjective representation in the mind of each individual speaker. However, a proper name can have more than one sense associated with it. The name Santa Claus could be associated with “The benevolent, bearded elf…” as well as with the description “The fat, old gentleman with the red cape…” In Frege, the relationship between sense and representation is one of determination: the references of names are determined by their senses as modes of presentation. If referents are objects in the external world, then senses are simply different ways of grasping the same object through different means. An object need not necessarily have a referent either in the external world or in the realm of abstract objects but it will always have a sense in the objective realm of thought for Frege. Hence, problems concerning the meaningfulness of sentences like (A) and (B) as well as the types of problems associated with sentences (C) and (D) above do not arise on this view.

Russell’s approach is somewhat different. First of all, Russell makes an important distinction between what he calls “ordinary” proper names and “logically” proper names. Logically proper names are indexicals such as this and that, which directly refer (in a Millian sense) to sense-data or other objects of immediate acquaintance. For Russell, ordinary proper names are abbreviated definite descriptions. Here definite description refers again to the type of formulation “The…” which was used above to describe Santa Claus as “the benevolent, bearded….’’ According to Russell, the name “Aristotle” is just a sort of shorthand for a definite description such as “The last great philosopher of ancient Greece” or “The teacher of Alexander the great” or some conjunction of two or more such descriptions. Now, according to Russell’s theory of definite descriptions, such descriptions must, in turn, be reduced, to a certain very specific logical form of existential generalization as follows:

• "The king of France is bald".

becomes

• $\exists x (K(x) \land \forall y (K(y) \rightarrow x=y) \land B(x))$

This says that there is exactly one object ‘’x’’ such that ‘’x’’ is King of France and ‘’x’’ is bald. Notice that this formulation is entirely general: it says that there is some x out in the world that satisfies the description, but does not specify which one thing ‘’x’’ refers to. Indeed, for Russell, definite descriptions (and hence names) have no reference at all and their meanings (senses in the Fregean sense) are just the truth conditions of the logical forms illustrated above. This is made clearer by Russell’s example involving ‘’Bismarck’’:

• (G) ”The Chancellor of Germany...”

In this case, Russell suggests that only Bismarck himself can be in a relation of acquaintance such that the man himself enters into the proposition expressed by the sentence. For any other than Bismarck, the only relation that is possible with such a proposition is through its descriptions. Bismarck could never have existed and the sentence (G) would still be meaningful because of its general nature described by the logical form underlying the sentence.

Notwithstanding these differences however, descriptivism and the descriptive theory of proper names came to be associated with both the views of Frege and Russell and both address the general problems (names without bearers, Frege’s puzzles concerning identity and substitution in contexts of intentional attitude attributions) in a similar manner.

Another problem for Millianism is Frege’s famous puzzles concerning the identity of co-referring terms. For example:

• (V) ”Hesperus is Phosphorus.”

In this case, both terms (“Hesperus” and “Phosphorus”) refer to the same entity: Venus. The Millian theory would predict that this sentence is trivial, since meaning is just reference and “Venus is Venus” is not very informative. Suppose, however, that someone did not know that Hesperus and Phosphorus both referred to Venus. Then it is at least arguable that the sentence (V) is an attempt to inform someone of just this fact.

Other problems for Millianism are those of negative existentials (e.g., “Batman does not exist”) and statements such as “Fred believes that Cicero, but not Tully, was Roman.”

## Kripke’s objections and the causal theory

In his book Naming and Necessity, Saul Kripke struck several powerful blows against the descriptivist theory. At the end of Lecture I (in Naming and Necessity, p. 64-70) Kripke sets out what he believes to be the tenets of the descriptivist theory. Kripke formally states a number of theses as the core of the descriptivist theory, with these theses explaining the theory in terms of reference (rather than the sense or meaning). As he explains before stating the theory, "There are more theses if you take it in the stronger version as a theory of meaning" (p. 64).

As he states it, the descriptivist theory is "weaker," i.e., the claims it makes do not assert as much as a stronger theory would, this actually makes it harder to refute. The descriptivist theory of meaning would include these theses and definitions however, thus refuting these would suffice for refuting the descriptivist theory of meaning as well. Kripke formulates them as follows:

1. To every name or designating expression 'X', there corresponds a cluster of properties, namely the family of those properties φ such that [speaker] A believes 'φX'
2. One of the properties, or some conjointly, are believed by A to pick out some individual uniquely.
3. If most, or a weighted most, of the φ's are satisfied by one unique object y, then y is the referent of 'X'.
4. If the vote yields no unique object, 'X' does not refer.
5. The statement, 'If X exists, then X has most of the φ's [corresponding to X]' is known a priori by the speaker.
6. The statement, 'If X exists, then X has most of the φ's [corresponding to X]' expresses a necessary truth (in the idiolect of the speaker).[1]

(1) States the properties or concepts related to any given proper name, where a name 'X' has a set of properties associated with it. The set of properties are those that a speaker, on inquiry of "Who is Barack Obama?" would respond "The President of the U.S., former Senator of Illinois, husband of Michelle Obama, etc." (1) does not stipulate that the set of properties φ is the meaning of X. (2) stipulates the epistemic position of the speaker. Note (2) says "believed by A to pick out".

(3) Takes the properties in (1) and (2) and turns them into a mechanism of reference. Basically, if a unique object satisfies the properties associated with 'X' such that A believes that 'X has such-and-such properties', it picks out or refers to that object. (4) states what happens when no object satisfies the properties (Kripke talks in terms of taking a "vote" as to the unique referent).

(5) Follows from (1)-(3). If there is a set of properties that speaker A believes to be associated with X, then these properties must be already known by the speaker. In this sense they are a priori. To know what a bachelor is, an individual must know what an unmarried male is; likewise an individual must know who is 'The President of the U.S., former Senator of Illinois, husband of Michelle Obama, etc.' to know who Obama is. (6) However is not a direct product of the theses. Kripke notes "(6) need not be a thesis of the theory if someone doesn't think that the cluster is part of the meaning of the name" (p. 65). However when the descriptivist theory is taken as a theory of reference and meaning, (6) would be a thesis.

Taken as a theory of reference, the following would be true:

• If someone fits the description 'the author who wrote, among other things, 1984 and Animal Farm' uniquely, then this someone is the George Orwell. (Thesis 3)
• 'George Orwell wrote, among other things, 1984 and Animal Farm' is known a priori by the speaker. (Thesis 5)

The idea in the second sentence is that one can't refer to something without knowing what he or she is referring to. Taken as a theory of reference and meaning, the following would be true:

After breaking down the descriptivist theory, he begins to point out what's wrong with it. First, he offered up what has come to be known as “the modal argument” (or argument from rigidity) against descriptivism. Consider the name ‘’Aristotle’’ and the descriptions “the greatest student of Plato’’, “the founder of logic” and “the teacher of Alexander.” Aristotle obviously satisfies all of the descriptions (and many of the others we commonly associate with him), but it is not a necessary truth that if Aristotle existed then Aristotle was any one, or all, of these descriptions, contrary to thesis (6). Aristotle might well have existed without doing any single one of the things he is known for. He might have existed and not have become known to posterity at all or he might have died in infancy.

Suppose that Aristotle is associated by Mary with the description “the last great philosopher of antiquity” and (the actual) Aristotle died in infancy. Then Mary’s description would seem to refer to Plato. But this is deeply counterintuitive. Hence, names are ‘’rigid designators’’, according to Kripke. That is, they refer to the same individual in every possible world in which that individual exists.

This is the counterintuitive result of thesis (6). For descriptivists Aristotle means “the greatest student of Plato’’, “the founder of logic” and “the teacher of Alexander.” So the sentence “the greatest student of Plato, etc., was the greatest student of Plato,” is equivalent to “Aristotle was the greatest student of Plato, etc.” Of course a sentence like “x=x” is necessary, but this just isn't the case with proper names and their descriptions. Aristotle could have done something else, thus he is not necessarily identical to his description.

The second argument employed by Kripke has come to be called the ‘’epistemic argument” or “the argument from unwanted necessity.” This is simply the observation that if the meaning of ‘’Angela Merkel’’ is “the Chancellor of Germany”, then “Angela is the Chancellor of Germany” should seem to the average person to be a priori, analytic, and trivial, as if falling out of the meaning of “Angela Merkel” just as “unmarried male” falls out of the meaning of “bachelor.” If thesis (5) is to hold, the properties of Angela Merkel should be known a priori by the speaker. But this is not true. We had to go out into the world to see who the Chancellor of Germany is.

Kripke’s third argument against descriptive theories consisted in pointing out that people may associate inadequate or inaccurate descriptions with proper names. Kripke uses Kurt Gödel as an example. The only thing most people know about Gödel is that he proved the incompleteness of arithmetic. Suppose he hadn't proved it, and really he stole it from his friend Schmidt. Thesis (3) says that if most of the properties associated with 'Gödel' are satisfied by one unique object, in this case Schmidt, then Schmidt is the referent of 'Gödel.' This means that every time someone (in the world where Gödel stole the incompleteness theorem from Schmidt) says 'Gödel' he or she is actually referring to Schmidt. This is far too counter-intuitive for the descriptivist theory to hold.

Such arguments seem to have convinced the majority of philosophers of language to abandon descriptive theories of proper names.

## Revival of descriptivism and two-dimensionalism

In recent years, there has been something of a revival in descriptivist theories, including descriptivist theories of proper names. Metalinguistic description theories have been developed and adopted by such contemporary theorists as Kent Bach and Jerrold Katz. According to Katz, “metalinguistic description theories explicate the sense of proper nouns - but not common nouns - in terms of a relation between the noun and the objects that bear its name.” Differently from the traditional theory, such theories do not posit a need for sense to determine reference and the metalinguistic description mentions the name it is the sense of (hence it is "metalinguistic") while placing no conditions on being the bearer of a name. Katz's theory, to take this example, is based on the fundamental idea that sense should not have to be defined in terms of, nor determine, referential or extensional properties but that it should be defined in terms of, and determined by, all and only the intensional properties of names.

He illustrates the way a metalinguistic description theory can be successful against Kripkean counterexamples by citing, as one example, the case of ‘’Jonah’’. Kripke’s Jonah case is very powerful because in this case the only information that we have about the Biblical character Jonah is just what the Bible tells us. Unless we are fundamentalist literalists, it is not controversial that all of this is false. Since, under traditional descriptivism, these descriptions are what define the name Jonah; these descriptivists must say that Jonah did not exist. But this does not follow. But under Katz's version of descriptivism, the sense of Jonah contains no information derived from the Biblical accounts but contains only the term "Jonah" itself in the phrase "the thing that is a bearer of 'Jonah'." Hence, it is not vulnerable to these kinds of counterexamples.

The most common and challenging criticism to metalinguistic description theories was put forth by Kripke himself: they seem to be an ad hoc explanation of a single linguistic phenomenon. Why should there be a metalinguistic theory for proper nouns (like names) but not for common nouns, count nouns, verbs, predicates, indexicals and other parts of speech.

Another recent approach is two-dimensional semantics. The motivations for this approach are rather different from those that inspired other forms of descriptivism, however. Two–dimensional approaches are usually motivated by a sense of dissatisfaction with the causal theorist explanation of how it is that a single proposition can be both necessary and a posteriori or contingent and a priori.


text:
# Powers of 2 whose Digits are Powers of 2

## Open Question

The only known powers of $2$ whose digits are also all powers of $2$ are:

$1, 2, 4, 8, 128$

Are there any more?

## Progress

Demonstrated up to at least $2^{10 \, 000 \, 000}$.

This can be achieved by looking at the lowest $20$ digits only, by calculating those of successive powers of $2$ after applying the modulo operation.


text:
# 7th Homework

This homework can be prepared individually or in teams of up to two students. It is up to you to find a partner in case you prefer teamwork. Download hw_7.tar.gz and extract it. Then add your homework solutions to the directory. Rename the directory according to the rules in the syllabus before submitting it as compressed archive. Don't forget to add the correct subject to the email when submitting.

Properly document the files you hand in to receive full credit. Each module (file), class and function needs a docstring. The purpose of each function needs to be summarized as a single sentence and then, if required, your intent and the behavior of the function can be described in more detail.

All of the tasks below need to

• run with the provided input data without crashing
• produce correct results
• have a comprehensive amount of unit tests
• have proper docstrings
• be implemented without using non-English function and variable names, docstrings etc.

in order to receive (full) points. The grader will only award a part of the total points. The rest will be awarded if all your functions etc. have a comprehensive amount of proper unit tests.

1. Austrian Social Security Numbers (1.5 points)
Your company requires a tool to validate Austrian social security numbers (SSNs). As you wanted to save the company some money while minimizing your own effort, you tried to find an existing solution. Unfortunately, you couldn't find anything that really met your companies needs. However, from one of your friends in Sweden you obtained a part of a small GUI program (validate_ssn_gui.py) to validate Swedish SSNs. Using this Python module allows you to reuse the GUI and only implement the logic to validate an Austrian social insurance number. Such a number consists of 10 digits of which the 4th denotes a checksum allowing simple validation. The format of such a number is "xxxCddmmyy" where "xxx" are consecutively assigned digits, "C" is the checksum digit, "dd" donates the day, "mm" the month and "yy" the year of birth. The following formula for calculating the checksum uses the this notation: "x1" is the first of the three "x" digits, "m2" is the second month digit, and so on.
C =(x1*3 + x2*7 + x3*9 + d1*5 + d2*8 + m1*4 + m2*2 + y1*1 + y2*6) modulo 11

The required validation module (validate_ssn.py) should contain at least one function is_valid(digit_list) that returns True if the given list of digits represents a valid SSN. Otherwise, the function must return False. Note that your code should not make any assumptions about the correctness of the given input.
Name the module: validate_ssn.py
2. Flow Shop Scheduling (2.5 points)
Your boss used a little tool to calculate the scheduling for your company's machines. As the IT department failed to realize a proper strategy for backing up important data, parts of the program got lost. Apparently the UI is still working, but when she wants to solve a particular problem the program crashes with a weird error she doesn't understand. Please run johnson_gui.py to understand what's going on and make the program work again by implementing the missing file.
Your boss tried to collect all useful information for you. She found some old tests for the missing file (the grader for hw7). She also said she recalled that when she initially got the program, the techies had to install some Python libraries to make the GUI work. Furthermore, she found a part of the documentation of the lost file:
"""Return a flow shop schedule using Johnson's rule.

Args:
jobs: A dictionary of jobs. Job IDs are keys, a list of machine times
are the values.
E.g. {'3': [5, 7]} means that job '3' needs 5 time units on the
first machine and 7 time units on the second.

Returns:
A list of job IDs.
"""

Name the module: johnson.py
3. Tests (5 points)
Write a comprehensive suite of unittest for the previous tasks using Python's unittest framework. Write enough tests to get comfortable that your code works as expected.


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Limit problem.

• December 27th 2008, 08:54 PM
arvindshetty86
Limit problem.
Can any1 help me to find the solution for this problem

lim a sin square x b log cosx /4
x->0

equals to half i.e 1/2
Find a and b.
• December 27th 2008, 08:59 PM
Chris L T521
Quote:

Originally Posted by arvindshetty86
Can any1 help me to find the solution for this problem

2
lim a sin^ x b log cosx 1
x->0 ------------------------- = ---
4 2

Find a and b.

$\lim_{x\to0}\frac{a\sin^2\left(x\right)\cdot b\log\left(\cos\left(x\right)\right)}{4}=\frac{1}{ 2}$

The way you posted the question is unclear and hard to interpret. Is this what you mean?
• December 27th 2008, 08:59 PM
kalagota
Quote:

Originally Posted by arvindshetty86
Can any1 help me to find the solution for this problem

2
lim a sin^ x b log cosx 1
x->0 ------------------------- = ---
4 2

Find a and b.

hi can you fix your post? try using parenthesis and stuffs.. :) tnx
• December 27th 2008, 09:12 PM
arvindshetty86
yes Chris that's the question.Thanks for the help.
• December 27th 2008, 10:50 PM
Isomorphism
L'Hospitals
Quote:

Originally Posted by For Arvind by Chris
$\lim_{x\to0}\frac{a\sin^2\left(x\right)\cdot b\log\left(\cos\left(x\right)\right)}{4}=\frac{1}{ 2}$

Hello Arvind,

Can you recheck the question? Something is not right here...

Why would they keep constants like that (a and b) when they can group it and call it one constant ( like c = ab) ?

And secondly, a simple L'Hospital will show that:

$\lim_{x\to0}\sin^2 x\cdot \log\left(\cos x\right) = 0$

So I dont think a constant ab exists (Thinking)

All this leads me to believe there is a typo (Nod)