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RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5A

RS Aggarwal Class 8 Solutions Chapter 5 Playing with Numbers Ex 5A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A.

Other Exercises

Question 1.
Solution:
Let tens digit = x
Units digit = 3
∴Number = 3 + 10x
According to the condition,
7 (x + 3) = 3 + 10x
7x + 21 = 3 + 10x
21 – 3 = 10x – 7x
=> 3x = 18
x = $$\\ \frac { 18 }{ 3 }$$
∴Number = 3 + 10x
= 3 + 10 x 6 = 3 + 60 = 63

Question 2.
Solution:
Let ten’s digit = x
Then units digit = 2x
and number = 10x + 2x = 12x
According to the condition,
12x = x + 2x + 18
12x – x – 2x = 18
=> 9x = 18
x = $$\\ \frac { 18 }{ 9 }$$ = 2
∴Number = 12x = 2 x 12 = 24

Question 3.
Solution:
Let units digit = x
and tens digit = y
Number = x + 10y
Now x + 10y = 4 (x + y) + 3
=> x + 10y = 4x + 4y + 3
10y – 4y – 4x + x = 3
=> 6y – 3x = 3
2y – x = 1 ….(i)
∴Number by reversing the order of digits = y + 10x
=>x + 10y + 18 = y + 10x
=>10x – x + y – 10y = 18
=> 9x – 9y = 18
x – y = 2 ….(ii)
=> 2y – y = 3
y = 3
x = 2y – 1 = 2 x 3 – 1 = 6 – 1 = 5
∴Number = x + 10y = 5 + 3 x 10
= 5 + 30 = 35

Question 4.
Solution:
Sum of two digits of a number =15
Let units digit = x
Then tens digit = 15 – x
∴Number = 10 (15 – x) + x
= 150 – 10x + x = 150 – 9x
By interchanging the digits, the new number will be
= 10x + 15 – x = 9x + 15
According to the condition,
9x + 15 = 9 + 150 – 9x
9x + 9x = 159-15 = 144
18x = 144
=>x = $$\\ \frac { 144 }{ 18 }$$ = 8
∴Number = 150 – 9x = 150 – 9 x 8
= 150 – 72 = 78

Question 5.
Solution:
Let units place digit = x
and tens place digit = y
Then number = x + 10y
By interchanging the positions of the digits then
Units digits = y
and tens digit = x
∴Number = y + 10x
(x + 10y) – (y + 10x) = 63
=> x + 10y – y – 10x = 63
9y – 9x = 63
=> 9(y – x) = 63
y – x = $$\\ \frac { 63 }{ 9 }$$ = 7
∴Hence, difference of its digits = 7 Ans.

Question 6.
Solution:
Sum of three digits of a number = 16
Let units digit of a three-digit number = x
Then tens digit = 3x
and hundreds digit = 4x
∴Number = x + 10 x 3x + 100 x 4x
= x + 30x + 400x = 431x
But x + 3x + 4x = 16 => 8x = 16
∴x = $$\\ \frac { 16 }{ 8 }$$ = 2
∴Number = 431 x 862

Hope given RS Aggarwal Solutions Class 8 Chapter 5 Playing with Numbers Ex 5A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.


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# Is a simply connected elliptic space rationally homotopy equivalent to a loop space or a suspension?

Let $X$ be an elliptic simply connected space. Is it rationally homotopy equivalent to the suspension of some connected space $Y$? If not, is it rationally homotopy equivalent to a loop space?

• $\mathbb{CP}^2$ is elliptic, it's definitely not a rational suspension, and I think it's not a rational loop space. – Qiaochu Yuan Oct 31 '17 at 0:11
• Cup products on a suspension vanish. Given that elliptic simply connected spaces satisfy Poincaré duality over $\mathbb Q$, that severely limits your options for being rationally equivalent to a suspension. – Arun Debray Oct 31 '17 at 0:12
• that's right but it can exist – tarik Oct 31 '17 at 0:19
• my question about loop space is equivalent to when the rational cohomology algebra of X is a free graded commutative algebra. – tarik Oct 31 '17 at 0:31

As Qiaochu points out, $\mathbb{C}\mathbb{P}^2$ is a simply-connected elliptic space (its only non-zero rational homotopy groups being $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$ and $\pi_5(\mathbb{C}\mathbb{P}^2) \otimes \mathbb{Q} = \mathbb{Q}$) that is not homotopy equivalent to a suspension, since it has non-trivial cup product. It is also not rationally homotopy equivalent to a loop space. Indeed, a loop space has the rational homotopy type of a product of Eilenberg-MacLane spaces $K(\mathbb{Q}, k)$. Since $\pi_2(\mathbb{C}\mathbb{P}^2)\otimes \mathbb{Q} = \mathbb{Q}$, there would be a factor of $K(\mathbb{Q}, 2)$ in the rational homotopy type of $\mathbb{C}\mathbb{P}^2$, but $K(\mathbb{Q}, 2)$ has infinite cohomology.

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# Math Help - Import images from several location

1. ## Import images from several location

Does anyone know how I can tell latex to include an image from a location other than the working directory? I have several images all in different locations.

Thanks in anticipation

2. Originally Posted by kungal
Does anyone know how I can tell latex to include an image from a location other than the working directory? I have several images all in different locations.

Thanks in anticipation
Include the absolute path in the file specification

CB

3. ## Including graphics

Hi kungal,

Originally Posted by kungal
Does anyone know how I can tell latex to include an image from a location other than the working directory?
you could use the environment variables TEXINPUTS or the command \graphicspath, see the UK TeX FAQ: Importing graphics from “somewhere else”.

Stefan


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# Postgresql: Scripting psql execution with password

How can I call psql so that it doesn't prompt for a password?

This is what I have:

psql -Umyuser < myscript.sql

However, I couldn't find the argument that passes the password, and so psql always prompts for it.

P.S.: I'm using PostgreSQL 9.0

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I ended up going for the PGPASSWORD environment variable. This fitted my usecase perfectly. Simple and self-contained in the script. –  Axel Fontaine Jan 1 '12 at 5:03

There are several ways to authenticate to postgresql. You may wish to investigate alternatives to password authentication at http://www.postgresql.org/docs/current/static/client-authentication.html.

There is no option to provide the password as a command line argument because that information is often available to all users, and therefore insecure. However, in Linux/Unix environments you can provide an environment variable for a single command like this:

-
I think PGPASSWORD is deprecated but still works, btw. Just FYI –  Scott Marlowe Jun 29 '11 at 17:16
Yep, it's deprecated (and so noted in one of the links). Since it's come up, it's probably also worth noting that the deprecation is hotly contested because it's extremely useful for many people yet can be used in some circumstances without serious security concerns. It seems to me that it's no worse than storing .pgpass on an NFS filesystem, for example. I use PGPASSWORD routinely. –  Reece Jun 29 '11 at 18:35

You can add this command line at the begining of your script:

set PGPASSWORD=[your password]

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in my case set command did not work but export PGPASSWORD=[password] did work –  Can Kavaklıoğlu Jun 3 '13 at 9:09
SET should work on Windows. –  Hosam Aly Jan 23 '14 at 19:05

-
PGPASSWORD=[your password] psql -Umyuser < myscript.sql

-

Given the security concerns about using the PGPASSWORD environment variable, I think the best overall solution is as follows:

1. Write your own temporary pgpass file with the password you want to use.
2. Use the PGPASSFILE environment variable to tell psql to use that file.
3. Remove the temporary pgpass file

There are a couple points of note here. Step 1 is there to avoid mucking with the user's ~/.pgpass file that might exist. You also must make sure that the file has permissions 0600 or less.

Some have suggested leveraging bash to shortcut this as follows:

PGPASSFILE=<(echo myserver:5432:mydb:jdoe:password) psql -h myserver -U jdoe -p 5432 mydb

This uses the <() syntax to avoid needing to write the data to an actual file. But it doesn't work because psql checks what file is being used and will throw an error like this:

WARNING: password file "/dev/fd/63" is not a plain file

-

If you intend on having multiple hosts/database connections, the ~/.pgpass file is the way to go.

Steps:

2. Do not add string quotes around your field values. You can also use * as a wildcard for your port/database fields.
3. Create an alias in your bash profile that runs your psql command for you. For example:alias postygresy='psql --host hostname database_name -U username' The values should match those that you inputted to the ~/.pgpass file.
5. Type your alias from the command line.

Note that if you have an export PGPASSWORD='' variable set, it will take precedence over the file. It is also wise to change the permissions of your file so that the contents are obscured from other users. This can be achieved with chmod 600 ~/.pgpass

-

If you're having problems on windows like me (I'm using Windows 7 64-bit) and set PGPASSWORD=[Password] did not work.

Then, as Kavaklioglu said in one of the comments,

export PGPASSWORD=[password]

Certainly does work on windows :)

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export PGPASSWORD=[password] does not work for me using the command line (cmd.exe) at all. Are you sure you weren't using cygwin or something similar? –  Devin Snyder Nov 18 '14 at 23:57
Only works with cl, you added it to the file right? Now just typed it into command? –  Jamie Hutber Nov 19 '14 at 9:06


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# How do you write an equation of a line that passes through points (0,5), (-3,5)?

##### 2 Answers
Feb 9, 2017

$y = 5$

#### Explanation:

The equation of a line can be written as

$y - {y}_{0} = m \cdot \left(x - {x}_{0}\right)$

where $\left({x}_{0} , {y}_{0}\right)$ is any point that lies on the line.

The gradient, $m$, of the line can be found using any two non-identical points that lie on the line.

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

$= \frac{5 - 5}{- 3 - 0}$

$= 0$

A gradient of zero indicates that the line is horizontal.

The equation of the line can thus be simplified to

$y = {y}_{0}$

In this case, both points have a $y$ coordinate of 5. The equation of this line is therefore

$y = 5$.

Feb 9, 2017

$\textcolor{g r e e n}{y = 5}$

#### Explanation:

For this particular example, we could note that the value of the $y$ coordinate is a constant: $5$, so the equation is:
$\textcolor{w h i t e}{\text{XXX}} y = 5$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For the general case:
$\textcolor{w h i t e}{\text{XXX}}$Given two points color(red)(""(x_1,y_1) and color(blue)(""(x_2,y_2))
$\textcolor{w h i t e}{\text{XXX}}$A two-point equation can be written as:
$\textcolor{w h i t e}{\text{XXXXXX}} \frac{y - \textcolor{red}{{y}_{1}}}{x - \textcolor{red}{{x}_{1}}} = \frac{\textcolor{red}{{y}_{1}} - \textcolor{b l u e}{{y}_{2}}}{\textcolor{red}{{x}_{1}} - \textcolor{b l u e}{{x}_{2}}}$

Substituting
$\textcolor{w h i t e}{\text{XXX")color(red)(} \left(0 , 5\right)}$ for color(red)(""(x_1,y_1)) and
$\textcolor{w h i t e}{\text{XXX")color(blue)(} \left(- 3 , 5\right)}$ for color(blue)(""(x_2,y_2))
we have:
$\textcolor{w h i t e}{\text{XXX")(y-color(red)5)/(x-color(red)0)=(color(red)5-color(blue)(5))/(color(red)(0)-color(blue)(} \left(- 3\right)}$
Simplifying:
$\textcolor{w h i t e}{\text{XXX}} \frac{y - 5}{x} = 0$

$\textcolor{w h i t e}{\text{XXX}} y = 5$


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Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
 Already signed up? New to KöMaL?

# Problem S. 68. (January 2012)

S. 68. One method to compress data is Huffman coding. In problem I. 276, you had to make a presentation about the coding process. Now you have to create a program to do the Huffman coding.

The Huffman coding replaces each byte of the original file with a sequence of bits of variable length: less frequent bytes will be replaced by longer, whereas more frequent bytes by shorter bit sequences. To create the Huffman code, a binary tree should be created first according to the frequencies of the original bytes, then this tree should be traversed appropriately. The tree is built up as follows.

1. First, each byte is considered as an isolated node of the graph. The node has no branches. The value of the node is the frequency of the actual character.

2. Then the two nodes with the least frequencies are merged: they will be the left and right branches of a new node whose value is the sum of the frequencies of the original two nodes. Then the original two nodes are deleted.

3. The previous step is repeated until only one node remains.

Now the Huffman code of a character is obtained by walking from the root to the actual character: each left step is a 0, while a right step is a 1 bit.

Your program s68 should create a possible Huffman code. The first command line argument is the name of the input file to be encoded. The second command line argument is the name of the output text file: each line should contain the sequence of bits (and their length) corresponding to each original byte. Lines of the output should be sorted according to the length of the bit sequences, then according to alphabetical order of the original bytes, as in the example. Non-printable ASCII characters and the space character should be displayed as their corresponding ASCII codes. Your program should also display the length in bits of the original file and the encoded file on the standard output.

In the table, Bemeneti állomány'' is the input file, Kimeneti állomány'' is the output file, Standard kimenet'' is the standard output, Eredeti hossz'' is the original length, while Kódolt hossz'' is the length of the encoded text.

The source code (s68.pas, s68.cpp, ...) without the .exe or any other auxiliary files generated by the compiler but with a short documentation (s68.txt, s68.pdf, ...) -- also describing which developer environment to use for compiling, further a brief description of your solution -- should be submitted in a compressed file s68.zip.

(10 pont)

Deadline expired on February 10, 2012.

Sorry, the solution is available only in Hungarian. Google translation

Mintamegoldásként Adrián Patrik debreceni 12. osztályos tanuló kritika és dokumentációját (s68ap-kritika-dokumentacio.pdf) valamint programját (s68ap.c), és Szabó Attila pécsi 11. osztály diák dokumentációját (s68sza.pdf) valamint programját(s68sza.c) adjuk közre.

### Statistics:

 15 students sent a solution. 10 points: Adrián Patrik, Hoffmann Áron, Jákli Aida Karolina, Kucsma Levente István, Szabó 928 Attila. 9 points: Havasi 0 Márton, Marussy Kristóf, Strenner Péter. 8 points: 1 student. 7 points: 4 students. 6 points: 2 students.

Problems in Information Technology of KöMaL, January 2012


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# Birthday list

I'm making this note for people that want to know others birthdays and say HAPPY BIRTHDAY to them. If you want you can write your birthday in the comments below and then i will write them here. And you can reshare it so more people can see it. And if you want you can write if your first or second. Waiting!

• Mohmmad Farhan - 20. 04. 2008.
• Ram Mohith - 17. 03. 2003.
• Annie Li - 23. 12.

Note by B D
1 month, 1 week ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

- 1 month ago

Mine is on 20th of August.

- 3 weeks, 6 days ago

I am on 23rd of dec

- 1 month ago

Thanks!

- 1 month ago

Np

- 1 month ago

Near my mom's 19th of dec

- 1 month ago

Haha. Coincidence or not?

- 1 month ago

I am not so sure

- 1 month ago

Haha, i treat coincidences more like clues

- 1 month ago

OH

- 1 month ago

$\large {\color{green}Ram} \quad {\color{blue}Mohith} \quad - \quad {\color{red}17/03/2003}$

- 1 month, 1 week ago

Thanks Ram!

- 1 month, 1 week ago

My birthday is on 20th April 2008 (You can see that this not my account but rather my brother's). The Day Hitler was born was on my birthday

- 1 month, 1 week ago

- 1 month, 1 week ago

you Misspelled my brother's name

- 1 month, 1 week ago

Your right. I will correct it.

- 1 month, 1 week ago


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The set of points should be given by the second choice, i.e., the set of $\operatorname{Spec} \mathbb{C}$-points, over $\operatorname{Spec} \mathbb{C}$. However, there is an additional step you need to do before defining $\pi_1$ (besides choosing a basepoint), which is applying an analytification functor to endow the set with a suitable topology. This functor takes locally finite type schemes over $\operatorname{Spec} \mathbb{C}$ to complex analytic spaces. There is a brief exposition of analytification in SGA 1, Exp. 12, and in Serre's GAGA.
Regarding your comment about making a topological space from a variety defined over $\mathbb{Z}$, a $\mathbb{C}$-point of the base change to $\mathbb{C}$ over $\mathbb{C}$ is the same as a $\mathbb{C}$-point of the original scheme. This is the universal property of fiber product.
The set of points should be given by the second choice, i.e., the set of $\operatorname{Spec} \mathbb{C}$-points, over $\operatorname{Spec} \mathbb{C}$. However, there is an additional step you need to do before defining $\pi_1$ (besides choosing a basepoint), which is applying an analytification functor to endow the set with a suitable topology. This functor takes locally finite type schemes over $\operatorname{Spec} \mathbb{C}$ to complex analytic spaces. There is a brief exposition of analytification in SGA 1, Exp. 12, and in Serre's GAGA.


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## Singularities of the Green function of the nonstationary Schrödinger equation.(English. Russian original)Zbl 0920.35008

Funct. Anal. Appl. 32, No. 2, 132-134 (1998); translation from Funkts. Anal. Prilozh. 32, No. 2, 80-83 (1998).
The Green function $$\psi= \psi(x,y,t)$$ of the nonstationary Schrödinger equation is determined by the relations $ih\psi_t= -\textstyle{{1\over 2}} h^2\Delta\psi+ v(x)\psi,\quad \psi(x,y,0)= \delta(x- y),\;t\geq 0,\;y\in\mathbb{R}^d.$ In what follows, we assume that $$v$$ is a smooth real function. For a chosen $$t>0$$, the function $$\psi$$ can have singularities with respect to $$x$$. These singularities depend on the rate of growth of $$v(x)$$ as $$x\to\infty$$. If $$v$$ satisfies the condition $| v(x)|\leq c| x|^{2\gamma},\quad \gamma<1,$ for sufficiently large $$x$$, then $$\psi$$ is smooth. This assertion, as well as several subsequent assertions, is given here in a simplified form.
The main objective of the present paper is to describe the formula mechanism responsible for the form of the singularities of $$\psi$$ for $$\gamma>1$$ in the case $$d=1$$. Under certain restrictions, we shall be able to say something about the character of the singularities of $$\psi$$ for $$d>1$$.

### MSC:

 35A20 Analyticity in context of PDEs 81Q05 Closed and approximate solutions to the Schrödinger, Dirac, Klein-Gordon and other equations of quantum mechanics
Full Text:

### References:

 [1] S. Zeldich, Commun. Math. Phys.,90, 1–26 (1983). · Zbl 0554.35031 [2] W. Craig, T. Kapeller, and W. Strauss, Microlocal Dispersive Smoothing for the Schrödinger Equation, Preprint, 1996. [3] K. Yajima, Smoothness and Nonsmoothness of the Fundamental Solution for Initial Value Problem for Time-Dependent Schrödinger Equations, Preprint, 1995. [4] L. Kapitanski and I. Rodnianski, IMRN, No. 2, 41–54, 1996. · Zbl 0859.35146 [5] L. Kapitanski, I. Rodnianski, and K. Yajima, On the Fundamental Solution of a Perturbed Harmonic Oscillator, Preprint, 1996. · Zbl 0892.35035 [6] V. P. Maslov and M. V. Fedoryuk, Semiclassical Approximation for Equations of Quantum Mechanics [in Russian], Nauka, Moscow, 1976. · Zbl 0449.58002 [7] V. S. Buslaev, Funkts. Anal. Prilozhen.,3, No. 3, 17–31 (1969). [8] H. L. Montgomery, Ten lectures on the interface between analytic number theory and harmonic analysis, CBMS, Regional Conference Series in Math., No. 84, Amer. Math. Soc. (1994).
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Are there any distributions with only polynomially many non-zero Fourier coefficients and a small support?

For a distribution X over $\{0,1\}^n$, we can define the Fourier coefficient of the distribution as $\hat{Y}(s)= \textbf{E}_{y\in Y}({\chi_s(y)})$.
The question I have is, do there exist distributions that can be efficiently computed such that at most polynomially many Fourier coefficients are non-zero?

• Akshay, would you clarify what requirements you have on the pseudorandom distribution? Do you merely require that the distribution be indistinguishable from a uniformly distributed random variable, by any polynomial-time algorithm? Do you have any additional requirements, e.g., on the amount of stretch or the size of the seed? – D.W. Oct 23 '13 at 19:33
• @D.W. My bad, I actually require the distribution to have a small size. I don't really want it to derandomize BPP but rather constant width branching programs. I have edited the question. Thanks! – Akshay Oct 24 '13 at 16:33
• Hmm. I'm still not sure if I've got the requirements right. So you don't need the distribution to be pseudorandom? Do you need the distribution to be efficiently computable, or to have small support? The body of the question says the former, the title says the latter. Those two conditions are not equivalent. For instance, I suspect there are efficiently computable distributions that have polynomially many non-zero Fourier coefficients; but there are not distributions that have polynomially many non-zero Fourier coefficients and that have small support. – D.W. Oct 24 '13 at 17:57
• I guess "efficiently computable" is not clear itself. – Sasho Nikolov Oct 24 '13 at 23:15
• Usually "pseudorandom distribution" means (i) it approximates the uniform distribution in some sense and (ii) it is statistically far from uniform. Here (i) is measured in terms of the sparsity of the Fourier transform (uniform has support 1). And I'm interpreting (ii) to mean small seed length. – Thomas supports Monica Oct 25 '13 at 5:16

If you want $Y$ to have entropy less than $0.99 n$ bits, the answer is no, by the uncertainty principle: Either $Y$ has high entropy or its Fourier transform has large support.

Theorem. Let $H(Y)$ be the Shannon entropy of $Y$ and let $F \subset \{0,1\}^n$ be the support of $\hat{Y}$. Then $H(Y) \geq n - \log |F|$.

Proof. Consider the collision probability of $Y$ (the probability that two independent samples of $Y$ are the same). By Parseval's identity, $$CP(Y) = \sum_y \text{Pr}[Y=y]^2 = 2^{-n} \sum_s \hat{Y}(s)^2 \leq 2^{-n} |F|,$$ as $|\hat{Y}(s)| \leq 1$. On the other hand, $CP(Y) = 2^{-H_2(Y)}$, where $H_2(Y)$ is the Renyi entropy of $Y$. Noting that $H_2(Y) \leq H(Y)$ gives the result. Q.E.D.

If $|F| = \mathrm{poly}(n)$, then $H(Y) = n-O(\log n)$. So, if $Y$ is the output of a pseudorandom generator, the seed length is at least $n-O(\log n)$.

• I don't follow the reasoning here, especially the last sentence. Why does the fact that $Y$ has large support mean that it isn't pseudorandom? Actually, that sounds backwards. Indeed, I'd expect that any pseudorandom distribution had darn well better have large support, if it is to have any hope of being pseudorandom (if $|S|$ is small, then it's easy to distinguish $Y$ from random, which means the distribution isn't pseudorandom). – D.W. Oct 23 '13 at 6:22
• @D.W. I think we often want a much bigger stretch, e.g. we would need to stretch a $\log(n)$-bit seed to an $n$-bit output in order to derandomize BPP (but such generators would have support only of size $n$). So maybe the exact answer is "maybe, if you consider stretching $n-\log(n)$ bits to $n$ bits sufficient, no otherwise". – usul Oct 23 '13 at 13:24
• @D.W. $\log |S|$ is a lower bound on the seed length needed to sample $Y$. One wants a pseudorandom distribution to have small seed length and thus small support. The point is that we want $Y$ to `look like' something with large support, whilst actually having sparse support. – Thomas supports Monica Oct 23 '13 at 14:17
• If the seed length is $\lg n$ bits, then the generator is not pseudorandom: there is a trivial polynomial-time attack to distinguish its output from true random (simply enumerate all seeds to see if the value of $Y$ is attained by any of them, i.e., enumerate its support). – D.W. Oct 23 '13 at 17:25
• @D.W. agreed that this was a bad example, mainly because of the efficiently sample-able requirement. I should instead have just said that we would like a PRG to stretch for instance $o(n)$-bit seeds to $n$-bit outputs. – usul Oct 23 '13 at 18:03


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# Homework Help: Separable differential equation

1. Jun 10, 2004

### Luminous Blob

I'm doing some revision on differential equations, and am getting stuck on what should be a simple problem.

The question is:

Solve: dP/dt = 0.2P(1000 - P)

Find particular solutions when (i) P(0) = 1000 and (ii) P(0) = 2000

The answers are supposed to be:

i) P(t) = 1000
ii) P(t) = 1000[1 - (1/2)exp(-200t)]^(-1)

I keep getting the wrong answers. The general solution I came up with (after doing the partial fraction decomposition and integrating) is:

t + C = 1/200[ ln|P| - ln|1000 - P| ]
exp(200t + C) = exp(ln|P| - ln|1000 - P|)
Aexp(200t) = P/(1000 - P) {where A = exp(C)}
P = 1000Aexp(200t) - (P)Aexp(200t)
P[ 1 + Aexp(200t) ] = 1000Aexp(200t)

P = 1000Aexp(200t)/[ 1 + Aexp(200t)]

But I don't get the right answers when I solve for A with the initial condition.

I'd appreciate it if anyone can tell me where I'm going wrong here.

2. Jun 10, 2004

### arildno

i) YOU HAVE PERFORMED DIVISION BY ZERO HERE!!
(Right when you started to solve the diff. eq.)

$$\frac{1000Ae^{200t}}{1+Ae^{200t}}=\frac{1000}{1+\frac{1}{A}e^{-200t}}, A\neq0$$


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# More Fourier Series

1. Nov 10, 2005

### stunner5000pt

Find the Fourier series for f(x)=1 on the interval 0 <=x <= pi in temrs of phi = sin nx. By integrating thi series find a convergent series for hte function g(x) =x oin this interval assuming that the set {sin nx} is complete

i can find the Fourier series for f(x) =1. But i would like to know why n is supposed to be odd becuase the answer i get is
$$\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1 - \cos(n \pi)}{n} \sin(nx)$$
the text ssumes n is odd. Is it becuase if n was even then the summand would be zero?
the answer in the text is $$\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\sin(2k-1) x}{2k-1}$$

How would i integrate the series?
f i integrate the answer for the text i get $$\frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{\cos(2k-1) x}{(2k-1)^2}$$
but hte answer in the book i
$$\frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1-\cos(2k-1) x}{(2k-1)^2}$$

Thank you

2. Nov 11, 2005

### quasar987

Precisely. Try plugging different values of n in cos(n pi). You will see that cos(n pi) = (-1)^n. Hence, for n even, cos(n pi)=1 and the summand vanishes. Now your sum is over all odd positive integers. But you can restore the sum to be over all positive integer by noting that the change of variable n-->2k-1 (with k running over the naturals) spans exactly every odd positive integers. This is what the book did.

If you've seen how to prove the uniform convergence of Fourier series, do it. If not, assume it does converge uniformly such that the integral of the sum is the serie of the integrals.

I think it comes from the constants of integration and then evaluating the serie at x=0, but I don't understand the mechanism exactly. It is true that setting one constant 0 and the other 1 / (2k-1)^2 makes the serie work when evaluating at x = 0, but how can we be sure that these are the right constant?