text:
One Step Closer to Ending Open Enrollment | James Howard One Step Closer to Ending Open Enrollment | James Howard

# One Step Closer to Ending Open Enrollment

D.C. has extended the special enrollment period for the District’s health exchange. If this progresses, we are getting closer and closer to ending open seasons for benefits signups. This can only be a good thing.


text:
Thread: Today's Favorite Mega Number View Single Post
2021-09-09, 02:09   #36
Dr Sardonicus

Feb 2017
Nowhere

3·1,657 Posts

Quote:
 Originally Posted by MattcAnderson Hi again all, My new favorite seven digit number is 9,999,999. 9,999,999 = 3*3*239*4649.
This brought to mind the following example illustrating the properties of the blocks of digits in the repeating decimals for k/p, k = 1 to p, p prime. Here, p = 239, a factor of 9999999. Since 239 divides 9999999, the repeating decimal for k/239, k = 1 to 238, has period 7. The blocks of digits may be further classified by the following 34 numbers:

[1004184, 2887029, 2050209, 1464435, 1046025, 1757322, 1255230, 1087866, 1506276, 3054393, 1380753, 1966527, 3765690, 2635983, 1882845, 1631799, 1422594, 1589958, 1129707, 4979079, 3556485, 3974895, 2426778, 4476987, 1213389, 2384937, 4560669, 1924686, 2803347, 2343096, 1673640, 1171548, 2008368, 4058577]

Each of these 34 numbers is the 7-digit block for the repeating decimal for a fraction k/239, for some k between 1 and 238. In each case, the value of k is obtained by dividing the 7-digit number by 41841. For example, dividing 1004184 by 41841 gives 24, and 24/239 = .10041841004184...

By cyclically permuting the digits in each block, the seven-digit blocks of the repeating decimals for the remaining fractions k/239, k = 1 to 238, are obtained.

For example, cyclically permuting the digits of the first block gives 0041841, and .00418410041841... = 1/239; permuting again, 10/239 = .04184100418410..., 100/239 = .41841004184100..., 44/239 = .18410041841004... and so on.

Each of the numbers is the smallest cyclic permutation of the block of digits whose leading digit is non-zero (i.e, a number between 1000000 and 9999999).

Last fiddled with by Dr Sardonicus on 2021-09-09 at 02:16 Reason: elaborate on permuting blocks of digits

format: plain


text:
## Formula

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

Emily Ng_4C
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am

### Formula

Why is there a negative in E=-hR/n^2?

JT Wechsler 2B
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

### Re: Formula

When an electron is still around the nucleus, the system is more stable and at a lower energy than if it were to not be there at all. Because of this, there is a negative sign in the equation.

Kelsey Warren 1I
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

### Re: Formula

The equation is (usually?) used for an electron losing energy and consequently emitting a photon. Because the electron loses energy, its energy is "negative" while the energy for the photon is positive. Dr. Lavelle also said in lecture that the negative sign means that the bound electron has lower energy than the free electron, in the case of an electron popping off the atom. I could use some more clarification on that as well.

Hailey Boehm 2H
Posts: 71
Joined: Fri Sep 28, 2018 12:24 am

### Re: Formula

The negative sign is there because a bound electron has lower energy than a free electron. When e- is completely removed, E=0. This equation makes 0 a reference point, and indicates that when the e- gets closer to nucleus/lower in orbitals, the e-'s energy is becoming more negative (or in other words it is losing energy).

Return to “Bohr Frequency Condition, H-Atom , Atomic Spectroscopy”

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text:
# Excluded Point Space is First-Countable

## Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Then $T$ is first-countable.

## Proof

Let $x \in S: x \ne p$.

By definition, the set $\left\{{x}\right\}$ is open in $T$.

Let $U \in \tau_{\bar p}: x \in U$.

Then $\left\{{x}\right\} \subseteq U$ and so $\left\{{\left\{{x}\right\}}\right\}$ is a local basis at $x$ which is trivially countable.

Now if $x = p$ there is only one $U \in \tau_{\bar p}: p \in U$, and that is $S$.

So $\left\{{S}\right\}$ is a local basis at $x$ which is trivially countable.

Hence the result, by definition of first-countable.

$\blacksquare$


text:
0th

Percentile

##### Copyrights of Files Used to Build R

R is released under the ‘GNU Public License’: see license for details. The license describes your right to use R. Copyright is concerned with ownership of intellectual rights, and some of the software used has conditions that the copyright must be explicitly stated: see the ‘Details’ section. We are grateful to these people and other contributors (see contributors) for the ability to use their work.

Keywords
misc
##### Details

The file R_HOME/COPYRIGHTS lists the copyrights in full detail.


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# Two racecar drivers, Abernathy and Berdoff, are driving around a

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Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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23 Jun 2015, 01:30
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Difficulty:

15% (low)

Question Stats:

74% (01:07) correct 26% (01:26) wrong based on 227 sessions

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Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

Kudos for a correct solution.

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Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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Updated on: 23 Jun 2015, 21:07
1
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.

This statement is clearly insufficient as we don't know the speeds of the drivers.
Insufficient

(2) Abernathy is driving 25 meters per minute faster than Berdoff.

let Berdoff speed be x meters/minute, Abernathy speed will be x+25 meters/minute

Abernathy and Berdoff are 200 meters apart, so the time taken for Abernathy to catch up to Berdoff = $$\frac{200}{(25+x-x)}$$= 200/25= 8 minutes
8*60=480 secnds

sufficient

Ans:B

Originally posted by ManojReddy on 23 Jun 2015, 04:41.
Last edited by ManojReddy on 23 Jun 2015, 21:07, edited 1 time in total.
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Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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23 Jun 2015, 09:42
2
Bunuel wrote:
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

Kudos for a correct solution.

Given: Relative Distance between A and B = 200 Meters

Time to catch up = Relative Distance / Relative Speed

So, we require the difference of the speeds of A and B to answer the question. [Relative Speed = Difference of Speeds]

Statement 1: The circumference of the racetrack is 1,400 meters.

DOesn't provide any information about the Relative Speed

Hence, NOT SUFFICIENT

Statement 2: Abernathy is driving 25 meters per minute faster than Berdoff.

i.e. Relative Speed = 25 Meters per Minute
i.e. Time = 200 / 25 = 8 Minutes = 8*60 = 480 Second

Hence, SUFFICIENT

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Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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23 Jun 2015, 09:45
ManojReddy wrote:
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.

This statement is clearly insufficient as we don't know the speeds of the drivers.
Insufficient

(2) Abernathy is driving 25 meters per minute faster than Berdoff.

let Berdoff speed be x meters/minute, Abernathy speed will be x+25 meters/minute

Abernathy and Berdoff are 200 meters apart, so the time taken for Abernathy to catch up to Berdoff = $$\frac{200}{(25+x-25)}$$= 200/x
Since we don't know the value of x, this statement alone is also insufficient.
Insufficient

even combining statements 1 and 2, we can't get the individual speeds of Abernathy and Berdoff.

Ans:E

Hi ManojReddy

Please check the highlighted part which as mistake

Relative Speed = Speed of A - Speed of B

Speed of B = x
Speed of A = x+25

Relative Speed = (x+25)-x = 25 meters per Minute
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Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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23 Jun 2015, 11:37
Bunuel wrote:
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

Kudos for a correct solution.

Statement 1 is irrelevant. The total racetrack is not needed in order to determine how long it will take for Abernathy to catch up to Berdoff. All we need is the difference in Speed between A and B.
Insufficient.

Statement 2 is sufficient for the above mentioned reason.

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Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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23 Jun 2015, 21:09
GMATinsight wrote:
ManojReddy wrote:
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.

This statement is clearly insufficient as we don't know the speeds of the drivers.
Insufficient

(2) Abernathy is driving 25 meters per minute faster than Berdoff.

let Berdoff speed be x meters/minute, Abernathy speed will be x+25 meters/minute

Abernathy and Berdoff are 200 meters apart, so the time taken for Abernathy to catch up to Berdoff = $$\frac{200}{(25+x-25)}$$= 200/x
Since we don't know the value of x, this statement alone is also insufficient.
Insufficient

even combining statements 1 and 2, we can't get the individual speeds of Abernathy and Berdoff.

Ans:E

Hi ManojReddy

Please check the highlighted part which as mistake

Relative Speed = Speed of A - Speed of B

Speed of B = x
Speed of A = x+25

Relative Speed = (x+25)-x = 25 meters per Minute

It's a miss from my side..
Thanks for correcting.

Regards,
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Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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25 Jun 2015, 15:10
1
statement 1: insufficient.

statement 2: speed = distance/time

25 =200/t

==> t =8 min = 480 seconds

Sufficient

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Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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28 Jun 2015, 09:14
1
Circular race track!
Distance difference/catch-up = 200meters!
St 1 - total circum/distance 1400 meters
NS
St2 Abernathy is driving 25 meters per minute faster than Berdoff.
in 1 min 25 meters more than berdoff
in 4 min 100 meters more than berdoff
in 8 min 200 meters more than berdoff
i.e 8x60=480 sec
Suff
Answr B
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Joined: 02 Sep 2009
Posts: 52227
Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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29 Jun 2015, 06:22
Bunuel wrote:
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

This question is much easier than most people assume on first glance! It doesn’t matter that the track is circular— if one driver is 200 meters behind another driver and both drive at constant rates, all you need to know in order to determine the time it takes for the rear driver to catch up is the rate at which the rear driver is “gaining” on the front driver.

While of course having the two drivers’ individual rates would be nice, you don’t actually even need that much information— all you need is the rate at which Abernathy is catching up. (When drivers are traveling in the same direction, you subtract their rates; you only need this relative rate.)

Thus, the rephrase is, “What is the difference in the two drivers’ rates?”

(1) INSUFFICIENT: Knowing the length of the track is insufficient and actually pretty useless, since your math problem really ends at the point that Abernathy catches up to Berdoff. Who cares how much track is in front of them when that happens?

(2) SUFFICIENT: This is exactly what you need— if Abernathy is “gaining” at 25 meters per minute, then it will take him 8 minutes to catch up (time = distance/rate, so time = 200/25, which is equal to 8).

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Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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29 Jun 2015, 08:38
Bunuel wrote:
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

Kudos for a correct solution.

Time to catch up is Gap / Difference in Speed
T = 200/diff in speed

State1: Gives the circumference of the track - Not Sufficient

State2: Is a direct answer. Hence Sufficient.
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Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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30 Sep 2015, 21:35
1
I saw this was a bit different then the question I had..

Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If
Abernathy is 200 meters behind Berdoff and both drivers drive at their respective
constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?
(1) The circumference of the racetrack is 1,400 meters.
(2) Berdoff is driving 25 meters per minute faster than Abernathy.

So

1) Alone NS
2) Alone NS

But C) interesting enough changes the question a bit, since the track is circular it changes the angle and now B is going to lap A and catch up.

The distance between them is now 1200
and the distance is speed in 25

Making B and A together in 2880 seconds?

SO C

IS this right logic?
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Re: Two racecar drivers, Abernathy and Berdoff, are driving around a  [#permalink]

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06 Jan 2018, 02:43
Bunuel wrote:
Two racecar drivers, Abernathy and Berdoff, are driving around a circular track. If Abernathy is 200 meters behind Berdoff and both drivers drive at their respective constant rates, how long, in seconds, will it take for Abernathy to catch up to Berdoff?

(1) The circumference of the racetrack is 1,400 meters.
(2) Abernathy is driving 25 meters per minute faster than Berdoff.

Kudos for a correct solution.

I understand that this is an MGMAT question and there was an error in the second statement which is corrected in the question mentioned in this thread.
But assuming that there was no error and the question is changed to "how long, in seconds, will it take for Berdoff to catch up to Abernathy?", what would be the ideal way to solve the question?
Re: Two racecar drivers, Abernathy and Berdoff, are driving around a &nbs [#permalink] 06 Jan 2018, 02:43
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text:
$h$—Principle and fluid dynamics

-
Camillo De Lellis, Universität Zürich
Fine Hall 110

In the early nineties Scheffer produced a complicated example of a nontrivial weak solution to the incompressible Euler equations, having compact support in space and time. Subsequent papers by Shnirelman produced other examples of quite irregular solutions by different, yet complicated, methods.In a recent joint work with László Székelyhidi we have used a suitable "$h$—principle'' to produce solutions with the same behavior in a relatively simple way. Our approach answers to further questions left open by the works of Scheffer and Shnirelman and might be relevant in understanding a long-standing conjecture of Onsager. The same kind of analysis has relevant applications also to the theory of hyperbolic systems of conservation laws and shares some surprising similarities with aspects of the theory of fully developed turbulence.
***Please note that there will be an additional talk by the speaker at IAS


text:
#### Archived

This topic is now archived and is closed to further replies.

# Trees and depth

This topic is 5839 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Given an arbitrary tree (not necessarily binary) that has nodes with random weights (e.g. distances from each other), how would I find the root or center node? This would ideally minimize the height of the tree.

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If it''s a tree, then you just find the node with no parent - the weights are irrelevant. Perhaps you were thinking of a graph?

[ MSVC Fixes | STL | SDL | Game AI | Sockets | C++ Faq Lite | Boost ]

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You''re probably right.

Ok, if I picked a random node on this graph, how would I know if this was
the root node of the graph? Let''s just say I forgot about weights for a moment,
and just looked at the depths from this node. I could probably do it by
looping through all nodes and assign each one root, and traverse the entire
graph for each one, so in the end, I would know which one is the best pick,
but that would be too time consuming if I have a lot of nodes.

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Well, technically a graph has no root - you could start anywhere on the graph. So I think that when you say "root" you mean "a chosen node which minimises the distances to all other nodes"?

I have no idea, not being big on algorithms, but maybe a "minimum spanning tree" is what you''re after.

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quote:

you mean "a chosen node which minimises the distances to all other nodes"?

That's what I mean. But I don't think a MST will give me
the center node. I think all it does is create a graph,
which really doesn't say where the center node is.

Basically, I want to keep the graph structure connected how it
already is. Something like that would try to change it.

Edited by - jtech on February 22, 2002 11:33:04 PM

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Sorry, I know little of graph/tree theory too.

Correction: Yes, I do have a tree, not a graph.

I think the main difference between a tree and a graph is that a
graph can have loops, but a tree cannot. Take any branch of a tree,
and you will eventually come to a leaf. But, follow a path on a graph,
and you could end up going in circles. A MST is a way to convert
a graph to a tree, but I already have a tree, so this step is not needed.

Better if I give an example of what I want to do:

Given this tree structure:

A
\\\
BCD

How would I find out that A is the best root? The relationship
between parent and child can be reversed. So, looking at this, you can
say A has 3 parents B,C,D. Or you can say A has no parent, and 3 children,
B,C,D.

If I pick the best root of my tree, I can optimize my searching routines.

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Red-black trees and Splay Trees are designed to minimize the height of a tree, though I''ve only used them for binary trees (Splay could probably work on n-ary.. red-black might be a little weird if not impossible to generalize)

They''re for rooted trees though... nodes in a rooted tree only have one parent, so what you describe with A having no children and three parents would be more like a graph... and most algorithms for graphs have pretty poor running times like O(n^2)

Take a look at http://mathworld.wolfram.com/Tree.html and look up other information on Dijkstra''s Algorithm and Dijkstra Trees

...Personally, I''d find a better structure or a new way to optimize your search algorithms

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The only way you can know if the current node you''re at is the root is:
#1, Each node has a Parent pointer that points to it''s parent, and, if the Parent pointer is NULL, you''re at root.
#2, You have a variable declared Root, and the Node you''re at == Root.

Most implementatoins of trees have a root pointer, and each ndoe only has children pointers. However, it begs the question, how did you get to the node you''re at if you don''t know the root?

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I think you''re more confused than I am.

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All I want is a balanced tree (the tree is not binary), and Red-Black,
Splay trees are for binary trees only.

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OK, so you have some random tree, and you want to find the best possible root for that tree?

If it''s a Binary tree, look into B+ Trees, and if its a n-ary tree, look up Dijkstra, Kruskal, and Prim. all 3 of these guys have done alot of research on minimum distances and such.

hope that helps

Waramp.

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quote:

its a n-ary tree, look up Dijkstra, Kruskal, and Prim. all 3 of these guys have done alot of research on minimum distances and such.

These algo''s don''t find the root...they only find a tree and
give you an arbitrary node as root. They don''t give you the
best root.

With Dijkstra, you have to pick a node to start, it doesn''t
give it to you at the end.

Don''t worry about it. I''m looking into something else.

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Some of you are looking at the literal description of a tree in memory as being the only description of that tree. Here''s a dumb example:

Think of a 3d model of the hierarchy of joints in a human. The root might be the pelvis, and everything else is a descendant of the pelvis. So the literal description to the program has the pelvis as being the root.

But any tree can be viewed from any node as being the root. What jtech wants is to be able to find which node optimizes tree depth to the minimum.

Back to the human model. We could easily view the left big toe as the parent, and everything connected to that as being a descendant. This in fact would be very handy if we were doing a calculation with the big toe rooted to the floor.

Anyway, any tree can be viewed with any node as the root.

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This may work:

Start at the existing parent and traverse the tree. When you get to the leaves, return back, saving maximum descendant distance at each node.

Now, traverse the tree again from the root. Take the maximum descendant distance of the root with you to each node, adding in the maximum distance to the maximum descendant distance saved at that node and the distance traversed to that node. That number should give you the maximum depth for that node if it was the root. Touch all of the nodes. Choose the one with the least total.

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Ignore my post. I thought you were asking an almost mindless question.

Turns out, it''s a very good question. Good luck.

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Given this tree, with all distances equal from node to node, say 1 unit:

  B |E - A - C | D

A is obviously the best choice, but let''s make D root.

If I record all distances from D:
D to A = 1
D to B = 2
D to C = 2
D to E = 2

D has one branch with max dist of 2.

Now traverse again, and add the max dist to all nodes:

D to A = 1 + 2 = 3
D to B = 2 + 2 = 4
D to C = 2 + 2 = 4
D to E = 2 + 2 = 4

3 the the least one, so A is the best root. Does this sound
correct?

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quote:
Original post by jtech

  B K | | E - A - C G J | | | D - - - F - H

Umm, look:

D is the root. Traverse the tree and compute the max chain of descendants at each link.

D -> A = 2
A -> B = 1
A -> E = 1
A -> C = 1
D -> F = 4
F -> G = 1
F -> H = 3
H -> J = 2
J -> K = 1

Now:

Start at D. Traverse the tree again. Start by going to A. D's max depth is D -> F = 4. Carry this to A, which is 1 away, so add 4 to 1, and we get 5. Since 5 is greater than any of A's max depths, which all equal 1, A's true depth is 5.

Keep traversing the tree as above. When we traverse from D to F, don't take the D -> F depth value. Instead, take the greatest of D's other depth values. In this case, it would be D -> A, which equals 2. Add 2 to the distance we traversed from D to F, which is 1. 1 + 2 = 3, which is no greater than F's maximum recorded depth of 3, so we conclude F's true depth is 3.

Keep doing this, and keep track of the node with the smallest value.

___________________________________

Edited by - bishop_pass on February 26, 2002 8:42:03 PM

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By the way, as far as I can tell, my solution above works. In the example above, F is the optimal root.

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quote:
Original post by WarAmp
If it''s a Binary tree, look into B+ Trees, and if its a n-ary tree, look up Dijkstra, Kruskal, and Prim. all 3 of these guys have done alot of research on minimum distances and such.

Actually, B-Trees are not binary trees, they are m-ary trees where each node can have at most m child nodes, and at least ciel(m/2) child nodes. Except in the case of the root. B-Trees can be really good at minimizing the depth of a tree, and you can gaurantee that all data in the tree will be at the same level.

J.W.

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ahhh. When I was taught B+Trees, I was told they were for binary, but now that I think of it, I suppose they could easily be extended for m-branches. Thanks for clearing that up.

Waramp.

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Well, gosh darn it. I think I provided a solution above. Is anybody going to comment on it, criticize it, or at least thank me?

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I''m still trying to figure it out

I got the code written to find the max num of descendants for
each node, but the second part is little murky.

Here''s what I got:

  typedef struct _NODE { INT num_nodes; struct _NODE *nodes;} NODE;INTGet_Num_Descendants( NODE *parent, INT parent_depth ) { if(NULL == parent) { return parent_depth; } INT max=0; for(INT i=0; i < parent->num_nodes; i++) { if(parent->nodes[i]) { INT d = Get_Num_Descendants( parent->nodes[i], (parent_depth+1) ); max = MAX( max, d ); } } max = MAX( max, parent_depth ); return max;}


text:
## Squares

DanishDynamite
Posts: 2608
Joined: Mon Jun 07, 2004 4:58 pm
Location: Copenhagen

### Squares

Prove that for any whole number N > 5, a square can be divided into N squares.

(I suspect this is the type of problem where you either see the answer fairly quickly or it takes you hours to see the solution. I was clearly in the last category.)

gnome
Posts: 22960
Joined: Tue Jun 29, 2004 12:40 am
Location: New Port Richey, FL

### Re: Squares

DanishDynamite wrote:Prove that for any whole number N > 5, a square can be divided into N squares.

(I suspect this is the type of problem where you either see the answer fairly quickly or it takes you hours to see the solution. I was clearly in the last category.)
Ok... for even N's, take a square from the lower right corner and adjust its size so that the remaining space can be filled with an odd number of squares. The larger the corner square, the more small squares can fill the rest.

For odd N's... cut out a large square in the center of the bottom edge. Adjust size to need as with the even solution.

Hardly rigorous, but I think it makes it clear enough.

DanishDynamite
Posts: 2608
Joined: Mon Jun 07, 2004 4:58 pm
Location: Copenhagen

### Re: Squares

gnome wrote:
DanishDynamite wrote:Prove that for any whole number N > 5, a square can be divided into N squares.

(I suspect this is the type of problem where you either see the answer fairly quickly or it takes you hours to see the solution. I was clearly in the last category.)
Ok... for even N's, take a square from the lower right corner and adjust its size so that the remaining space can be filled with an odd number of squares. The larger the corner square, the more small squares can fill the rest.

For odd N's... cut out a large square in the center of the bottom edge. Adjust size to need as with the even solution.

Hardly rigorous, but I think it makes it clear enough.
"Hardly rigorous" seems right.

I can understand your solution for even N's as it is the one I found, but your solution for odd N's leaves me puzzled.

Could you post a diagram?

gnome
Posts: 22960
Joined: Tue Jun 29, 2004 12:40 am
Location: New Port Richey, FL

### Re: Squares

DanishDynamite wrote:
gnome wrote:Ok... for even N's, take a square from the lower right corner and adjust its size so that the remaining space can be filled with an odd number of squares. The larger the corner square, the more small squares can fill the rest.

For odd N's... cut out a large square in the center of the bottom edge. Adjust size to need as with the even solution.

Hardly rigorous, but I think it makes it clear enough.
"Hardly rigorous" seems right.

I can understand your solution for even N's as it is the one I found, but your solution for odd N's leaves me puzzled.

Could you post a diagram?
...

Actually I'm incorrect, you still get an even number. Let me work on it some more.

ceptimus
Posts: 1166
Joined: Wed Jun 02, 2004 11:04 pm
Location: UK
Trying to hide this in a spoiler ('select' it, or Ctrl-A to view)

Given any existing arrangement, any existing square can be subdivided into four equal ones, yielding an extra three squares in total. So if we can solve for any three consecutive numbers, then all numbers greater than that are proven soluble by induction.

+-+-+-+
| | | |
+-+-+-+
| | | = 6 (and therefore 9, 12, 15...)
| +-+
| | |
+---+-+
+-+-+---+
| | | |
+-+-+ |
| | | |
+-+-+---+ = 7 (and therefore 10, 13, 16...)
| | |
| | |
| | |
+---+---+
+-+-+-+-+
| | | | |
+-+-+-+-+
| | |
| +-+ = 8 (and therefore 11, 14, 17...)
| | |
| +-+
| | |
+-----+-+

DanishDynamite
Posts: 2608
Joined: Mon Jun 07, 2004 4:58 pm
Location: Copenhagen
The cepter is a must, for ceptimus!

A different inductive proof than mine, but just as valid. Well done!

(My proof was based on dividing the case into even and odd numbers of squares. The bit which took me forever to see was how to make a division into 7 squares.)

gnome
Posts: 22960
Joined: Tue Jun 29, 2004 12:40 am
Location: New Port Richey, FL
Beautiful. I was approaching that, but I don't think I could have made the argument so elegantly.


text:
시간 제한메모리 제한제출정답맞힌 사람정답 비율
5 초 (추가 시간 없음) 512 MB111100.000%

## 문제

In the Datastring Research Corporation, the string $s$ is called good, if it has even number of distinct subsequences.

The string $t$ is called a subsequence of the string $s$, if it can be derived from $s$ by crossing out a certain set of symbols. The string $s$ and the empty string are also considered to be subsequences of $s$. In a string of the length $l$, there is a total of $2^l$ subsequences, however, some of them may coincide. For instance, a three-letter string $abb$ has only 6 distinct subsequences: empty, $a$, $b$, $ab$, $bb$, $abb$.

The string $s$ was lying on the String Researcher's desk, and he was trying to figure out whether it was a good one. He did realize that counting the number of distinct subsequences is a tedious task, so, instead of tackling the task right away, he went on a coffee break --- with donuts.

Upon his return, the String Researcher discovered that someone had made an $N-1$ cuts in the string $s$, and it fell apart into $N$ non-empty substrings $s_1, s_2, \dots, s_N$ scattered all over the table. And he absolutely does not remember the initial string $s$. However, he is curious about how many permutations $p = (p_1, \dots, p_N)$ exist, such that the recovered string $s_{p_1}s_{p_2}\dots s_{p_N}$ is good. Note that there is a total of $N!$ permutations (the factorial of $N$), and all these permutations are considered different, even if some of the substrings $s_i$ match.

## 입력

The first line of the input file contains a single integer $N$ --- the number of substrings ($2 \leq N \leq 20$). The $i$th of the following $N$ lines contains the substring $s_i$.

All of the $s_i$ strings  are non-empty and consist exclusively of lower-case Latin letters. It is guaranteed that the sum of lengths of all the lines $s_i$ does not exceed $10^5$.

## 출력

In the only line of the output file, print a single integer number: the number of permutations such that the concatenation of the strings $s_i$ in the order of permutation is a good one.

## 예제 입력 1

3
a
a
baa

## 예제 출력 1

4

## 힌트

The string $a + a + baa$ has 14 subsequences, the string $a + baa + a$ has 13 subsequences, and the string $baa + a + a$ has 10 subsequences. Each of these lines can be obtained with the aid of only two permutations, since there are two occurrences of the substring $a$.


text:
# Thread: product and quotient of cos and sin

1. ## product and quotient of cos and sin

I have the quantities 2(cos pi/3 + i sin pi/3) and 4(cos pi/4 + i sin pi/4) and am looking to see if I have the correct product and quotient respectively.
I come up with:
8(cos 7pi/12 + i sin 7pi/12) and 1/2(cos 7pi/12 + i sin 7pi/12)
correct or not quite? Thanks for the help in advance!

2. How did you get these answers?

Did you use these methods Complex numbers : De Moivre's theorem : Multiplication using polar form ?

3. ## sin and cos

yeah something like that...


text:
# Cauchy's Integral Formula

## Theorem

Let $D = \left\{{z \in \C : \left|{z}\right| \le r}\right\}$ be the closed disk of radius $r$ in $\C$.

Let $f: U \to \C$ be holomorphic on some open set containing $D$.

Then for each $a$ in the interior of $D$:

$\displaystyle f \left({a}\right) = \frac 1 {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z-a}\right)} \ \mathrm d z$

where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.

### General Result

$\displaystyle f^{\left({k}\right)} \left({a}\right) = \frac{k!} {2 i \pi} \int_{\partial D} \frac {f \left({z}\right)} {\left({z-a}\right)^{k+1}} \ \mathrm dz$

for $k = 1, 2, 3, \ldots$

## Source of Name

This entry was named for Augustin Louis Cauchy.


text:
## 414/514 The theorems of Riemann and Sierpiński on rearrangement of series

November 16, 2014

I.

Perhaps the first significant observation in the theory of infinite series is that there are convergent series whose terms can be rearranged to form a new series that converges to a different value.

A well known example is provided by the alternating harmonic series,

$\displaystyle 1-\frac12 +\frac13-\frac14+\frac15-\frac16+\frac17-\dots$

and its rearrangement

$\displaystyle 1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\dots$

According to

Henry Parker Manning. Irrational numbers and their representation by sequences and series. John Wiley & Sons, 1906,

Laurent evaluated the latter by inserting parentheses (see pages 97, 98):

$\displaystyle \left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\dots$ $\displaystyle=\frac12\left(1-\frac12+\frac13-\frac14+\dots\right)$

A similar argument is possible with the rearrangement

$\displaystyle 1+\frac13-\frac12+\frac15+\frac17-\frac14+\dots$,

which can be rewritten as

$\displaystyle 1+0+\frac13-\frac12+\frac15+0+\frac17+\dots$ $\displaystyle =\left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\dots\right)$ $\displaystyle +\left(0+\frac12+0-\frac14+0+\frac16+0-\dots\right)$ $\displaystyle =\frac32\left(1-\frac12+\frac13-\frac14+\dots\right).$

The first person to realize that rearranging the terms of a series may change its sum was Dirichlet in 1827, while working on the convergence of Fourier series. (The date is mentioned by Riemann in his Habilitationsschrift, see also page 94 of Ivor Grattan-Guinness. The Development of the Foundations of Mathematical Analysis from Euler to Riemann. MIT, 1970.)

Ten years later, he published

G. Lejeune Dirichlet. Beweis des Satzes, dass jede unbegrenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftlichen Factor sind, unendlich viele Primzahlen enthält. Abhandlungen der Königlich Preussischen Akademie der Wissenschaften von 1837, 45-81,

where he shows that this behavior is exclusive of conditionally convergent series:

Theorem (Dirichlet). If a series converges absolutely, all its rearrangements converge to the same value.

Proof. Let $u_0,u_1,\dots$ be the original sequence and $u_{\pi(0)},u_{\pi(1)},\dots$ a rearrangement. Denote by $U_0,U_1,\dots$ and $V_0,V_1,\dots$ their partial sums, respectively. Fix $\epsilon>0$. We have that for any $n$, if $m$ is large enough, then for all $i\le n$ there is some $j\le m$ with $\pi(j)=i$. Also, there is a $k$ such that for all $j\le m$ there is a $i\le k$ with $\pi(j)=i$, so

$|U_m-V_m|\le\sum_{i=n+1}^m|u_i|+\sum_{i=n+1}^k|u_j|.$

Choosing $n$ large enough, and using that $\sum_i|u_i|$ converges,  we can ensure that the two displayed series add up to less than $\epsilon$. This gives the result. $\Box$

## 414/514 Homework 3 – Continuity and series

November 11, 2014

This set is due in three weeks, on Friday, December 5th, at the beginning of lecture.

1. Given a finite set $A\subseteq\mathbb T=\{z\in\mathbb C\mid |z|=1\}$, find (with proof) a power series $f(x)=\sum_{n=0}^\infty a_n x^n$ with radius of convergence $1$ and such that if $z\in\mathbb T$, then $f(z)$ converges iff $z\notin A$.

Find (with proof) a countably infinite set  $X\subseteq \mathbb T$ and a power series $f(x)=\sum_{n=0}^\infty a_n x^n$ with radius of convergence $1$ and such that if $z\in\mathbb T$, then $f(z)$ converges iff $z\notin X$.

The first part is easy, the second one may be tricky. If after a few honest efforts you find yourself stuck, take a look at the literature. For instance, at either Stefan Mazurkiewicz, Sur les séries de puissances, Fundamenta Mathematicae, 3, (1922), 52–58, or Fritz Herzog, George Piranian, Sets of convergence of Taylor series. I. Duke Math. J., 16, (1949), 529–534. Both papers prove more general results, by explicit constructions. Present a detailed adaptation of their argument that solves the question. Do not simply reproduce the proof in the papers for the more general cases. Other relevant references on this topic are Fritz Herzog, George Piranian, Sets of convergence of Taylor series. II. Duke Math. J., 20, (1953), 41–54, and Thomas W. Körner, The behavior of power series on their circle of convergence. In Banach spaces, harmonic analysis, and probability theory (Storrs, Conn., 1980/1981), 56–94, Lecture Notes in Math., 995, Springer, Berlin-New York, 1983.

2. Present a detailed proof of Weierstrass theorem that for any compact interval $[a,b]$, any continuous function $f:[a,b]\to\mathbb R$ is the uniform limit of a sequence of polynomials. The proof should be different from the one presented in lecture (based on Bernstein polynomials). Again, although many books have a proof of this result, make sure your write up is your own.

## 414/514 Examples of Baire class two functions

November 3, 2014

Previously, we listed some examples of Baire class one functions. Here we do the same for functions in the next class of Baire. Recall that if $I$ is an interval, the function $f:I\to\mathbb R$ is (in) Baire class two ($\mathcal B_2$) iff it is the pointwise limit of a sequence of Baire one functions.

This post comes from an answer I posted on Math.Stackexchange about a year ago.

Here are three examples:

1. Let $C$ be the Cantor set. For each interval $(a,b)$ contiguous to $C$, define $f$ on ${}[a,b]$ by

$f(x)=\frac{2(x-a)}{b-a}-1$,

so $f$ maps the interval to $[-1,1]$. Otherwise, let $f(x)=0$.

2. Write each $x\in(0,1)$ in binary: $x=0.a_1a_2a_3\dots$, not terminating in a string of $1$s, and define

$\displaystyle f(x)=\limsup_{n\to\infty} \frac{a_1+\dots+a_n}n$.

3. Conway’s base 13 function.

The first two examples come from Bruckner’s book Differentiation of real functions. All three are examples of functions that are not derivatives but have the intermediate value property.

The first one is discontinuous precisely at the points of $C$, and it is “almost” Baire class $1$, in that one can turn it into a Baire class $1$ function by only modifying its values (carefully) at the endpoints of intervals contiguous to $C$. But if one does this, then the function no longer has the intermediate value property.

The second function has the property that the image of any subinterval of $(0,1)$, no matter how small, is all of $(0,1)$. The third function is in the same spirit, but it behaves even more dramatically: The image of every open interval is all of $\mathbb R$.

To verify that the functions are indeed in Baire class at most $2$:

1. For example 1, use that the limit of $x^n$ on ${}[0,1]$ is $0$ for $x<1$, and $1$ at $x=1$, to get for each open interval $(a,b)$ contiguous to $C$ a Baire class $1$ function $f_{[a,b]}$ that is zero everywhere except on ${}[a,b]$, where it coincides with $f$. Now use that the sum of finitely many Baire class $1$ functions is Baire class $1$.
2. For example 2, there are several ways to proceed. Here is one, which I do not think is optimal, but (I believe) is correct: Recall that a limsup is the infimum (over $m$) of a supremum (over all $n>m$), so it is enough to see that each $f_m(x)= \sup_{n>m}g_n$ is Baire class $1$, where

$\displaystyle g_n(x)=\frac{a_1+\dots+a_n}n$.

The point is that each $g_n$ has finitely many discontinuities, all of which are jump discontinuities. Any such function is Baire class $1.$ This would appear to mean that $f_m$ is Baire class $2$, but we are saved by noting that $f_m$ is the uniform limit of the $g_n$, $n>m$. (The point is that each Baire class is closed under uniform limits.)

3. The argument for example 3 is similar. (Note that this function is unbounded.)

To see that the functions are not Baire class $1$: The functions in examples 2 and 3 are discontinuous everywhere, but the set of points of continuity of a Baire class $1$ function is dense. For example 1, use Baire’s extension of this result giving us that, in fact, if $f$ is Baire class $1$, then for any perfect set $P$, the set of points of continuity of $f\upharpoonright P$ is comeager relative to $P$. In example 1 this fails (by design) when $P=C$. (All we need is that, for any closed set $D$, the restriction of a Baire one function to $D$ has at least one continuity point on $D$. Baire also showed that this characterizes Baire one functions.)

Example 2 is also discussed in the van Rooij-Schikhof book (see their Exercise 9.M).

To close, let me include some examples that do not have the intermediate value property. Note first that if $A\subseteq\mathbb R$ and $\chi_A$ is its characteristic (or indicator) function, then $\chi_A$ is continuous iff $A=\emptyset$ or $\mathbb R$. More interestingly, $\chi_A$ is Baire class $1$ iff $A$ is both an $F_\sigma$ and a $G_\delta$ set.

Recall that a set is $F_\sigma$ iff it is the countable union of closed sets, and it is $G_\delta$ iff it is the countable intersection of open sets. The notation $F_\sigma$ is pronounced F-sigma. Here, the F is for fermé, “closed” in French, and the $\sigma$ is for somme, French for “sum”, “union”. Similarly, the notation $G_\delta$ stands for G-delta. Here, the G is for Gebiet, German for “area”, “region”— neighborhood—, and the $\delta$ is for Durchschnitt, German for “intersection”.

Note that, in particular, open sets are both: They are clearly $G_\delta$, and any open interval (and therefore, any countable union of open intervals) is a countable union of closed intervals. It follows that closed sets are also both. In particular, the characteristic function of the Cantor set is Baire class $1$. More generally, a function $f$ is Baire class $1$ iff the preimage $f^{-1}(U)$ of any open set is $F_\sigma$.

For the more general case where $A$ is $F_\sigma$ or $G_\delta$, then $\chi_A$ is Baire class $2$. For any $A$ which is either, but not both, $\chi_A$ is an example of a properly Baire class $2$ function. For instance, this is the case with $A=\mathbb Q$. In fact, $\chi_A$ is Baire class $2$ iff $A$ is both an $F_{\sigma\delta}$ and a $G_{\delta\sigma}$ set ($G_{\delta\sigma}$ sets are countable unions of $G_\delta$ sets, that is, countable unions of countable intersections of open sets, and $F_{\sigma\delta}$ sets are countable intersections of $F_\sigma$ sets, that is, countable intersections of countable unions of closed sets).

More generally, $f$ is Baire class $2$ iff for any open $U$, the set $f^{-1}(U)$ is $G_{\delta\sigma}$. For details, and a significant generalization due to Lebesgue, that characterizes each Baire class and relates it to the hierarchy of Borel sets, see section 24 in Kechris’s book Classical descriptive set theory.

## 414/514 References on continuous nowhere differentiable functions

October 19, 2014

Just as the last two times I have taught 414/514, I am assigning a final project on the topic of continuous nowhere differentiable functions (see here and here for the previous times).

The project requires that you choose an example of a continuous nowhere differentiable function, and to write a report describing the function, indicating who first introduced it, and presenting complete proofs of its continuity and nowhere differentiability. Additional information relevant for context is highly encouraged.

I am including links to two encyclopedic references on the subject. Feel free to follow the arguments there closely if needed, or to consult other sources, but make sure that what you turn in is your own version of the details of the argument, and that full details (rather than a sketch) are provided.

1. Johan Thim’s Master thesis (Continuous nowhere differentiable functions), written under the supervision of Lech Maligranda.
2. A.N. Singh’s short book on The theory and construction of non-differentiable functions. (See here for a short review.)

As I mentioned before,

Please take this project very seriously (in particular, do not copy details from books or papers, I want to see your own version of the details as you work through the arguments). Feel free to ask for feedback as you work on it; in fact, asking for feedback is a good idea. Do not wait until the last minute.

The project should be typeset and is due Wednesday, December 17 (though I strongly encourage you to turn it in earlier).

Please contact me by email as soon as you have chosen the topic you are going to cover, and I’ll list it here, to avoid repetitions.

• Stephanie Potter: Wen’s function.
• Jeremy Siegert: Orlicz functions.
• Stuart Nygard: Besicovitch’s function.
• Monica Agana: Koch’s snowflake.

## 414/514 Homework 2 – Monotone and Baire one functions

October 10, 2014

This set is due in three weeks, on Monday, November 3, at the beginning of lecture.

1. Let $f:[a,b]\to\mathbb R$ be increasing. We know that $f(x-)$ and $f(x+)$ exist for all $x\in[a,b]$, and that $f$ has at most countably many points of discontinuity, say $t_1,t_2,\dots$ For each $i$ let $I_i,J_i$ be the intervals $(f(t_i-),f(t_i))$ and $(f(t_i),f(t_i+))$. Some of these intervals may be empty, but for each $i$ at least one of them is not. (Here we follow the convention that $f(a-)=f(a)$ and $f(b+)=f(b)$.) Let $\mathrm{lh}(I)$ denote the length of the interval $I$, and say that an interval $(\alpha,\beta)$ precedes a point $t$ iff $\beta\le t$.

Verify that $\sum_i(\mathrm{lh}(I_i)+\mathrm{lh}(J_i))<+\infty$ and, more generally, for any $x$,

$s(x):=\sum\{\mathrm{lh}(I_i)\mid I_i$ precedes $f(x)\}$ $+\sum\{\mathrm{lh}(J_i)\mid J_i$ precedes $f(x)\}<+\infty$.

Define a function $f_0:[a,b]\to\mathbb R$ by setting $f_0(x)=f(x)-s(x)$. Show that $f_0$ is increasing and continuous.

Now, for each $n>0$, define $f_n:[a,b]\to\mathbb R$ so that $f_n\upharpoonright[a,t_n)=f_{n-1}\upharpoonright[a,t_n)$, $f_n(t_n)=f_{n-1}(t_n)+\mathrm{lh}(I_n)$, and $f_n(x)=f_{n-1}(x)+\mathrm{lh}(I_n)+\mathrm{lh}(J_n)$ for all $x\in(t_n,b]$. Show that each $f_n$ is increasing, and its only discontinuity points are $t_1,\dots,t_n$.

Prove that $f_n\to f$ uniformly.

Use this to provide a (new) proof that increasing functions are in Baire class one.

2. Solve exercise 3.Q in the van Rooij-Schikhof book: If $f:[a,b]\to\mathbb R$ is such that for all $x$, we have that $f(x-)$ and $f(x+)$ exist, then $f$ is the uniform limit of a sequence of step functions. The approach suggested in the book is the following:

Show that it suffices to argue that for every $\epsilon>0$ there is a step function $s$ such that $|f(x)-s(x)|<\epsilon$ for all $x$.

To do this, consider the set $A=\{x\in[a,b]\mid$ there is a step function $s$ on $[a,x]$ such that $|f(t)-s(t)|<\epsilon$ for all $t\in[a,x]\}$.

Show that $A$ is non-empty. Show that if $a\le y\le x$ and $x\in A$, then also $y\in A$. This shows that $A$ is an interval ${}[a,\alpha)$ or ${}[a,\alpha]$, with $\alpha\le b$. Show that in fact the second possibility occurs, that is, $\alpha\in A$. For this, the assumption that $f(\alpha-)$ exists is useful. Finally, show that $\alpha=b$. For this, use now the assumption that $f(\alpha+)$ exists.

3. (This problem is optional.) Find a counterexample to the following statement: If $f:[a,b]\to\mathbb R$ is the pointwise limit of a sequence of functions $f_1,f_2,\dots$, then there is a dense subset $X\subseteq [a,b]$ where the convergence is in fact uniform. What if $f$ and the functions $f_n$ are continuous?  Can you find a (reasonable) weakening of the statement that is true?

4. (This is example 1.1 in Andrew Bruckner’s Differentiation of real functions, CRM monograph series, AMS, 1994. MR1274044 (94m:26001).) We want to define a function $f:[0,1]\to\mathbb R$. Let $C$ be the Cantor set in ${}[0,1]$. Whenever $(a,b)$ is one of the components of the complement of $C$, we define $f(x)=(2(x-a)/(b-a))-1$ for $x\in[a,b]$. For $x$ not covered by this case, we define $f(x)=0$. Verify that $f$ is a Darboux continuous function, and that it is discontinuous at every point of $C$.

Verify that $f$ is not of Baire class one, but that there is a Baire class one function that coincides with $f$ except at (some of) the endpoints of intervals $[a,b]$ as above.

Verify that $f$ is in Baire class two.

## 414/514 Simple examples of Baire class one functions

October 6, 2014

Recall that a real-valued function $f$ defined on an interval $I$ is (in) Baire class one ($\mathcal B_1$) iff it is the pointwise limit of continuous functions.

Examples are continuous functions, of course, but functions in $\mathcal B_1$ do not need to be continuous. An easy example is the function $f:[0,1]\to\mathbb R$ given by $f(x)=0$ if $x\ne 1$ and $f(x)=1$ if $x=1$. This is the pointwise limit of the functions $f_n(x)=x^n$. By the way, an easy modification of this example shows that any function that is zero except at finitely many points is in $\mathcal B_1$.

Step functions are another source of examples. Suppose that $a=x_0 and that $s:[a,b]\to\mathbb R$ is constant on each $(x_i,x_{i+1})$. Then $s$ is the pointwise limit of the functions $s_k$, defined as follows: Fix a decreasing sequence $\epsilon_k$ converging to $0$, with $\epsilon_k\le 1/k$ and $2\epsilon_k for all $i$. Now define $\hat s_k$ as the restriction of $f$ to

$\displaystyle \{x_0,x_1,\dots,x_n\}\cup\bigcup_{i=0}^{n-1}[x_i+\epsilon_k,x_{i+1}-\epsilon_k]$,

and let $s_k:[a,b]\to\mathbb R$ extend $\hat s_k$ by joining consecutive endpoints of the components of its domain with straight segments.

An important source of additional examples is the class of derivatives. Suppose $f:\mathbb R\to\mathbb R$ and $f(x)=g'(x)$ for all $x$. This is the pointwise limit of the functions $f_n(x)$ given by

$f_n(x)=\displaystyle\frac{g\left(x+\frac1n\right)-g(x)}{\frac1n}.$

This simple construction does not quite work if $f$ is defined on a bounded interval (as $x+1/n$ may fall outside the interval for some values of $x$). We can modify this easily by using straight segments as in the case of step functions: Say $f:[a,b]\to\mathbb R$. For $n$ large enough so $1/n, define $f_n(x)$ as above for $x\in[a,b-1/n]$, and now set $f_n(b)=f(b)$ and extend $f_n$ linearly in the interval ${}[b-1/n,b]$.

Additional examples can be obtained by observing, first, that $\mathcal B_1$ is a real vector space, and second, that it is closed under uniform limits (the latter is not quite obvious). This gives us, for instance, that all monotone functions are in $\mathcal B_1$, since monotone functions are the uniform limit of step functions on bounded intervals: Given an increasing $f:[a,b]\to\mathbb R$, let $f_n(x)=\lfloor nf(x)\rfloor/n$. It follows that all functions of bounded variation are in $\mathcal B_1$, since any such function is the difference of two increasing functions.

Another interesting source of examples is characteristic functions. Given $X\subseteq\mathbb R$, the function $\chi_X$ is in $\mathcal B_1$ iff $X$ is both an $\mathbf F_\sigma$ and a $\mathbf G_\delta$ set.

On the other hand, $\chi_{\mathbb Q}$ is not in $\mathcal B_1$, since it is discontinuous everywhere while Baire class one functions are continuous on a comeager set.

## 414/514 Homework 1 – The reals

September 11, 2014

This set is due in two weeks, on Friday September 26, at the beginning of lecture.

1. Recall that in the construction of the reals via Dedekind cuts, a real is simply the left set of a pair $(A\mid B)$ in a cut of $\mathbb Q$, that is, a “real” is a set $A\subset \mathbb Q$ that is non-empty, bounded above, closed to the left (meaning, if $x\in A$, $y\in\mathbb Q$, and $y, then $y\in A$), and such that $A$ has no maximum. We also have a copy $\mathbb Q^*$ of $\mathbb Q$ inside $\mathbb R$, given by the identification $q\mapsto q^*:=\{t\in\mathbb Q\mid t. We left a few details to be verified when we presented this construction.

Let $r$ be a real (in the sense just described). Define carefully the real $-r$ (meaning, describe $-r$ as a specific set of rationals, and verify that it is a real in the sense under discussion), and verify that $-r+r=0$, and that $-r$ is the only real with this property.

Define carefully the product $rs$ of reals $r$ and $s$, and verify that the distributive property holds.

Check that $\mathbb R$ is Dedekind-complete, that is, any cut of $\mathbb R$ is realized. (S0, ignoring the formal difference between $\mathbb Q$ and $\mathbb Q^*$, this version of $\mathbb R$ is the Dedekind-completion of $\mathbb Q$, and this gives us that it is also the Dedekind completion of itself. )

2. More generally, define the Dedekind completion of a dense order, and verify its existence and uniqueness (up to isomorphism). In particular, the field $\mathbb R(x)$ of  rational functions admits a completion, call it $\hat{\mathbb R}(x)$. Can we extend the addition operation on $\mathbb R(x)$ so it is defined in all of $\hat{\mathbb R}(x)$ and makes it into an abelian group? Can we extend the order so $\hat{\mathbb R}(x)$ is in fact an ordered group? What, if any, is the problem trying to extend multiplication?

3. Recall that in the construction of the reals via Cauchy sequences, a real is an equivalence class of Cauchy sequences of rationals, under the equivalence relation that states that two Cauchy sequences $q_0,q_1,\dots$ and $r_0,r_1,\dots$ are equivalent iff $q_0,r_0,q_1,r_1,\dots$ is a Cauchy sequence.

Verify that this is indeed an equivalence relation, and that, given equivalent sequences $\vec q$ and $\vec r$, the resulting interleaving sequence is equivalent to both. Verify that the (pointwise) definitions of addition and multiplication make sense, and that the resulting set equipped with these operations is indeed a field. Define carefully the ordering relation, and prove that it gives us a field ordering. Finally, verify that the resulting ordered field is indeed Dedekind complete.

4. Recall the construction of the reals described in Street’s paper An efficient construction of real numbers. His short note makes many claims that are not proved there. Provide as many of the missing details as possible.

5. Given a linear order $(X,<)$, in the order topology the open sets are (by definition) those subsets of $X$ that are union of (bounded or unbounded) open intervals in $X$. Show that a linear order $(X,<)$ is order isomorphic to $\mathbb R$ iff the following three properties are verified:

• $X$ has no first or last elements.
• $X$ is connected, that is, we cannot write $X=A\cup B$ where $A$ and $B$ are open, nonempty, and disjoint.
• $X$ is separable, that is, there is a countable subset of $X$ that is dense in $X$.


text:
# Monte Carlo Simulations of Radiative Transfer: Project Overview

The goal of this project was to simulate radiative transfer in the circumbinary disk GG Tau by using statistical methods (i.e. Monte Carlo methods). To provide some context for the project, I will describe the astrophysical system that I considered, and I will describe the objectives of the project

Of the many stars in the night sky, a significant number of them, when you are able to resolve them, are in fact binary stars. These are star systems in which two stars orbit a common center of mass known as a barycenter. For this reason, we regard GG Tau to be a close binary system; a system in which two stars are separated by a significant distance comparable to the diameters of each star.

The system of GG Tau has four known components with a hypothesized fifth component. There is a disk of material encircling the entire system which we call a circumbinary disk. Additionally, there is also an inner disk of material that surrounds the primary and secondary components referred to as a circumstellar disk.

The goal of this project was to use Monte Carlo methods to simulate radiative transfer processes present in GG Tau. This is accomplished through the use of importance sampling two essential quantities over cumulative distribution function of the form

$\displaystyle \xi=\int_{0}^{L} \mathcal{P}(x)dx\equiv \psi(x_{0}), (1)$

using FORTRAN code. I will discuss the mathematical background of probability and Monte Carlo methods (stochastic processes) in more detail in a later post. The general process of radiative transfer is relatively simple. Starting with an emitter,  radiation travels from its emission point, travels a certain distance, and at this point the radiation could either be scattered into an angle different from the incidence angle, or it can be absorbed by the material. In the context of this project, the emitter (or emitters) are the primary and secondary components of GG Tau. The radiation emitted is electromagnetic radiation, or light. This light will travel a distance $L$ after which the light is either scattered or absorbed. The code uses Eq.(1) (cumulative distribution function) to sample optical depths and scattering angles to calculate the distance traveled $L$. I will reserve the exact algorithm for a future post as well. In the next post, I will discuss the basics of radiative transfer theory as presented in the text “Radiative Transfer” by S. Chandrasekhar, 1960.

The following are the sources  used in this project

Wood, K., Whitney, B., Bjorkman, J., and Wolff M., 2013. Introduction to Monte Carlo Radiative Transfer.

Whitney, B. A., 2011. Monte Carlo Radiative Transfer. Bull. Astr. Soc. India.

Kitamura, Y., Kawabe, R., Omodaka, T., Ishiguro, M., and Miyama, S., 1994. Rotating Protoplanetary Gas Disk in GG Tau. ASP Conference Series. Vol. 59.

Piètu, V., Gueth, F., Hily-Blant, P., Schuster, K.F., and Pety, J., 2011. High Resolution Imaging of the GG Tauri system at 267 GHz. Astronomy & Astrophysics. 528. A81.

Skrutskie, M.F., Snell, R.L., Strom, K.M., Strom, S.E., and Edwards, S., 1993. Detection of Circumstellar Gas Associated with GG Tauri. The Astrophysical Journal. 409:422-428.

Choudhuri, A.R., 2010. Astrophysics for Physicists. 2.

Carroll B.W., Ostlie, D.A., 2007. An Introduction to Modern Astrophysics. 9,10.

Chandrasekhar, S., 1960. Radiative Transfer. 1.

Schroeder, D., 2000. Introduction to Thermal Physics. 8.

Ross, S., 2010. Introduction to Probability Model. 2,4,11.

I obtained the FORTRAN code from K. Wood via the following link. (For the sake of completeness the URL is: www-star.st-and.ac.uk/~kw25/research/montecarlo/montecarlo.html)


text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Mathematical research inspired in fundamental part by mathoverflow

Mathoverflow has led in several instances to new mathematical work that arose directly from mathematical ideas, questions or answers posted here. Articles containing such work, inspired in fundamental part by mathoverflow material, might be characterized as having been born on mathoverflow.

Let us collect together here references to such work. In each case, please include a link to the relevant mathoverflow post or posts, a link to the mathematical work, for example an article at the math arxiv or the relevant journal, and a very brief summary abstract. (More detailed abstracts would presumably be available for those following the links to the article.)

In order that this question will not become burdened with excessive posts, let's agree to the following principles:

1. Please makes posts here only about essentially completed articles, rather than works-in-progress.
2. Please make posts here only about articles whose main existence was inspired by mathoverflow, such as a case where the topic of a question or the essence of an answer became the main substance of an article.
3. Please do not make posts here concerning articles simply because they cite mathoverflow, or simply because mathoverflow was involved at a critical step of the article, since such citations will eventually become so numerous as to be unremarkable. Rather, make a post here only in connection with an article that could be characterized as essentially about the topic or ideas expressed on mathoverflow and the authors were directly inspired by that mathoverflow material.

Please note the related meta-thread for more general discussion about references to mathoverflow, and Gower's question on breakthroughs that seeks examples of situations where mathoverflow helped a researcher make a critical advance.

-
I don't see why we cannot keep this on meta. – Thierry Zell Feb 15 2012 at 21:21
MO has a great impact on the mathematical community as a whole - highly disputable, and I speak as someone who likes the site. – Yemon Choi Feb 15 2012 at 21:31
It seems fine to me to have this list here on the main site, perhaps with the (modified) understanding that only completed papers, with links to journals or to the math arxiv, would be appropriate for posting. I would prefer it to the meta site discussion, in part because it will be nice to see it periodically pop up to the top, when someone completes a new MO-inspired paper. But with the modified understanding I proposed, this won't be so often as to become annoying. So I voted to re-open. – Joel David Hamkins Feb 16 2012 at 0:56
In particular, the meta-site forum for this information is rather discursive, and doesn't organize the information as well as MO question/answer format would. – Joel David Hamkins Feb 16 2012 at 0:59
Rather than having this comment thread get out of control, I recommend we move the discussion to meta. I've started a thread: meta.mathoverflow.net/discussion/1312/… Please upvote this comment so it appears "above the fold" – David White Feb 16 2012 at 20:10

The paper "The mate-in-n problem of infinite chess is decidable" by D. Brumleve, J. D. Hamkins, and me was inspired by a question asked on Mathoverflow. The paper is available on the arxiv. Please see J. D. Hamkins' blog for the abstract or if you would like to post a comment, here's a short version of the abstract:

"Infinite chess is chess played on an infinite edgeless chessboard. The familiar chess pieces move about according to their usual chess rules, and each player strives to place the opposing king into checkmate. The mate-in-n problem of infinite chess is the problem of determining whether a designated player can force a win from a given finite position in at most n moves. The main theorem of this article, confirming a conjecture of the second author and C. D. A. Evans, establishes that the mate-in-n problem of infinite chess is computably decidable, uniformly in the position and in n. The proof proceeds by showing that the mate-in-n problem is expressible in what we call the first-order structure of chess, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable."

I'm looking forward to hearing about other papers inspired by Mathoverflow.

-

Ben Green's paper on (not) computing the Möbius function arose from this question on MathOverflow.

Abstract. Any function $F : \{1,\dots,N\} \rightarrow \{-1,1\}$ such that $F(x)$ can be computed from the binary digits of $x$ using a bounded depth circuit is orthogonal to the Möbius function $\mu$ in the sense that $\frac{1}{N} \sum_{x \leq N} \mu(x)F(x) = o_{N \rightarrow \infty}(1)$. The proof combines a result of Linial, Mansour and Nisan with techniques of Kátai and Harman-Kátai, used in their work on finding primes with specified digits.

-
 Thanks, Joel: I updated my post accordingly. – GH Mar 4 2012 at 20:58


text:
# 13. Loss Functions

Loss functions are responsible for calculating some measure of wrongness. We usually use these to inform the neural network and us, just how good or bad our predictions $$\hat{y}$$ are compared to the ground truth $$y$$.

Loss Functions Status

Category

Name

Docs

Forward

Backward

loss

Categorical Cross Entropy (CCE)

Categorical Cross Entropy (CCE)

loss

Mean Absolute Error (MAE)

Mean Absolute Error (MAE)

loss

Mean Squared Error (MSE)

Mean Squared Error (MSE)

## 13.4. Loss Abstraction

Neural Network Loss Functions.

class fhez.nn.loss.loss.Loss

Abstract loss class to unify loss function format.

Calculate gradient of loss with respect to $$\hat{y}$$.

property cache

Get caching dictionary of auxilary data.

disable_cache()

Disable caching.

enable_cache()

Enable caching.

abstract forward(signal: numpy.ndarray, y: numpy.ndarray, y_hat: numpy.ndarray)

Calculate loss(es) given one or more truths.

property inputs

Get cached input stack.

Neural networks backpropogation requires cached inputs to calculate the gradient with respect to x and the weights. This is a utility method that initialises a stack and allows you to easily append or pop off of it so that the computation can occur in FILO.

property is_cache_enabled

Get status of whether or not caching is enabled.

update()

Loss funcs have no params so do nothing.


text:
# Gravitation and Space Time warps

1. Aug 17, 2011

### Lazernugget

Okay, so here's my questions:

1. In the theory of the expanding universe, the universe, well, expands. But space it's self would have to expand faster than light could travel or at the same speed for that to work. Umm, Hello? The space would travel faster than light. Would it not? If so, explain that please.

2. In general relativity, Gravity is explained by warps in space time. What exactly causes those warps? The warps depend on mass, so does the Higgs Boson have affect on the size and effect of those warps in space? Or is it the mysterious force carrier Graviton that exists within all matter with mass, meaning if we could somehow contain free Gravitons we could warp space by first neutralizing the gravitons and then dispersing them to fold and bend space?

-Lazer

2. Aug 17, 2011

### WannabeNewton

General Relativity does not deal with the Higgs Boson nor with gravitons (even though it implies their existence). To put it simply, the source of curvature in GR is the Energy - Momentum tensor and this affects the geometry/curvature. You have to hold on until a QFT framework incorporating gravity is fully developed.

Last edited by a moderator: May 5, 2017
3. Aug 17, 2011

### jfy4

Don't you think rather that GR assumes their existence, as opposed to implies their existence? While the development of a QFT of gravity would hopefully have a graviton (or an analogue) I don't think it's classical limit, GR, implies that one exists, I think it assumes it. As far as the Higgs, I don't see that falling out of GR, or a quantization of GR either, although, I couldn't know since it hasn't been done in full...

What do you think?

4. Aug 17, 2011

### WannabeNewton

Maybe I worded it wrong. I meant more along the lines of implied it from the symmetry of $h_{\mu \nu }$ and the group velocity being equal to unity from $\square \bar{h_{\mu \nu }} = 0$. As far as the higgs goes I agree with you that it doesn't have much going for it right now.

Last edited: Aug 17, 2011
5. Aug 17, 2011

### bcrowell

Staff Emeritus
Huh? GR was developed before quantum mechanics, so it can't have assumed the existence of a graviton.

6. Aug 17, 2011

### Lazernugget

Thanks for the replies, they're all great. So, I have Q #1 explained, but I still think I could use a bit more clarification on my Q #2

7. Aug 18, 2011

### jfy4

Looks like I worded it wrong! How I meant the above was in the sense that, given a correct quantum theory of gravity (and the graviton which we think will be part of it), it implies GR. On the other hand, given a classical limit, it does not imply the quantum theory.

Are you okay with that?

8. Aug 18, 2011

### jfy4

The stress-energy-momentum tensor, $T_{\alpha\beta}$, is the source for the Einstein tensor, $G_{\alpha\beta}=R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R$, which gives a description of the curvature of space-time. For Einstein's field equations, the curvature described is due to matter (anything that isn't the gravitational field).

9. Aug 19, 2011

### Naty1

That's the way it usually works. As h approaches zero, QM approaches classical.

10. Aug 19, 2011

### bcrowell

Staff Emeritus
Sure, makes sense to me.

11. Aug 19, 2011

### K^2

In GR, speed of light is a local limit, because it only makes sense in flat space-time. If your space-time sufficiently curved, two remote objects can be moving relative to each other much faster than speed of light. Consider Alcubierre Warp Drive for an extreme example.


text:
Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.

(1) $$\frac{\pi}{180}$$ cos x°
(2) $$\frac{1}{90}$$ cos x°
(3) $$\frac{\pi}{90}$$ cos x°
(4) $$\frac{2}{\pi}$$ cos x°
(2) $$\frac{1}{90}$$ cos x°

Explaination:

Question 2.
If y = f(x2 + 2) and f'(3) = 5 , then $$\frac{\mathrm{dy}}{\mathrm{d} x}$$ at x = 1 is
(1) 5
(2) 25
(3) 15
(4) 10
(4) 10

Explaination:
y = f(x2 + 2)
$$\frac{\mathrm{dy}}{\mathrm{d} x}$$ = f’ (x2 + 2) × 2x
$$\frac{\mathrm{dy}}{\mathrm{d} x} / x=1$$ = f’ (12 + 2) × 2 × 1
= f’(3) × 2
= 5 × 2 = 10

Question 3.
If y = $$\frac{1}{4}$$u4, u = $$\frac{2}{3}$$ x3 + 5, then $$\frac{\mathrm{dy}}{\mathrm{d} x}$$ is
(1) $$\frac{1}{27}$$ x2 (2x3 + 15)3
(2) $$\frac{2}{27}=$$ x (2x3 + 5)3
(3) $$\frac{2}{27}=$$ x2 (2x3 + 15)3
(4) – $$\frac{2}{27}=$$ x (2x3 + 5)3
(3) $$\frac{2}{27}=$$ x2 (2x3 + 15)3

Explaination:

Question 4.
If f(x) = x2 – 3x, then the points at which f (x) = f’ (x) are
(1) both positive integers
(2) both negative integers
(3) both irrational
(4) one rational and another irrational
(3) both irrational

Explaination:
f(x) = x2 – 3x
f’(x) = 2x – 3
f(x) = f'(x)
⇒ x2 – 3x = 2x – 3
x2 – 3x – 2x + 3 = 0
x2 – 5x + 3 = 0

Question 5.
If y = $$\frac{1}{a-z}$$, then $$\frac{\mathrm{d} \mathrm{z}}{\mathrm{d} \mathrm{y}}$$ is
(1) (a – z)2
(2) – (z – a)2
(3) (z + a)2
(4) – (z + a)2
(1) (a – z)2

Explaination:

Question 6.
If y = cos (sin x2), then $$\frac{\mathrm{dy}}{\mathrm{d} x}$$ at x = $$\sqrt{\frac{\pi}{2}}$$ is
(1) – 2
(2) 2
(3) – 2 $$\sqrt{\frac{\pi}{2}}$$
(4) 0
(4) 0

Explaination:

Question 7.
If y = mx + c and f(0) = f’(0) = 1, then f(2) is
(1) 1
(2) 2
(3) 3
(4) – 3
(3) 3

Explaination:
y = mx+c
$$\frac{d y}{d x}$$ = m
y = x + c (i.e.) f(x) = x + c
y(a tx = 0) = f(0) 0 + c = 1 ⇒ c = 1
y = x + 1 ⇒ f(x) = x + 1
f(2) = 2 + 1 = 3

Question 8.
If f(x) = x tan-1 x then f'(x) is
(1) $$1+\frac{\pi}{4}$$
(2) $$\frac{1}{2}+\frac{\pi}{4}$$
(3) $$\frac{1}{2}-\frac{\pi}{4}$$
(4) 2
(2) $$\frac{1}{2}+\frac{\pi}{4}$$

Explaination:

Question 9.
$$\frac{d}{d x}$$ (ex + 5 log x) is
(1) ex . x4 (x + 5)
(2) ex . x (x + 5)
(3) ex + $$\frac{5}{x}$$
(4) ex – $$\frac{5}{x}$$
(1) ex . x4 (x + 5)

Explaination:

Question 10.
If the derivative of (ax – 5)e at x = 0 is – 13 , then the value of a ¡s
(1) 8
(2) – 2
(3) 5
(4) 2
(4) 2

Explaination:
y = (ax – 5)e3x
$$\frac{d y}{d x}$$ = y’ = (ax – 5) (3e3x) + e3x (a)
= e3x[3ax – 15 + a]
Given $$\frac{d y}{d x}$$ = -13 at x = 0
⇒ [-15 + a] = -13
⇒ a = -13 + 15
a = 2

Question 11.
x = $$\frac{1-t^{2}}{1+t^{2}}$$, y = $$\frac{2 t}{1+t^{2}}$$ then $$\frac{d y}{d x}$$ is
(1) – $$\frac{\mathbf{y}}{x}$$
(2) $$\frac{\mathbf{y}}{x}$$
(3) – $$\frac{x}{y}$$
(4) $$\frac{x}{y}$$
(3) – $$\frac{x}{y}$$

Explaination:

Question 12.
If x = a sin θ and y = b cos θ, then  is
(1) $$\frac{\mathbf{a}}{\mathbf{b}^{2}}$$ sec2 θ
(2) $$-\frac{\mathbf{b}}{\mathbf{a}}$$ sec2 θ
(3) $$-\frac{b}{a^{2}}$$ sec3 θ
(4) $$-\frac{b^{2}}{a^{2}}$$ sec3 θ
(3) $$-\frac{b}{a^{2}}$$ sec3 θ

Explaination:

Question 13.
The differential coefficient of log10 x with respect to log10 x is
(1) 1
(2) – (log10 x)2
(3) (log10 x)2
(4) $$\frac{x^{2}}{100}$$
(2) – (log10 x)2

Explaination:

Question 14.
If f(x) = x + 2 , then f’ (f(x)) at x= 4 is
(1) 8
(2) 1
(3) 4
(4) 5
(2) 1

Explaination:
f(x) x + 2
f’ (f(x)) = $$\frac{\mathrm{d}}{\mathrm{d} x}$$ (f(x))
= $$\frac{\mathrm{d}}{\mathrm{d} x}$$ (x + 2) = 1

Question 15.
If y = $$\frac{(1-x)^{2}}{x^{2}}$$, then $$\frac{\mathrm{dy}}{\mathrm{d} x}$$ is
(1) $$\frac{2}{x^{2}}+\frac{2}{x^{3}}$$
(2) $$-\frac{2}{x^{2}}+\frac{2}{x^{3}}$$
(3) $$-\frac{2}{x^{2}}-\frac{2}{x^{3}}$$
(4) $$-\frac{2}{x^{3}}+\frac{2}{x^{2}}$$
(4) y = $$\frac{(1-x)^{2}}{x^{2}}$$

Explaination:

Question 16.
If pv = 81, then $$\frac{\mathbf{d} \mathbf{p}}{\mathbf{d v}}$$ at v = 9 is
(1) 1
(2) – 1
(3) 2
(4) – 3
(2) – 1

Explaination:

Question 17.

(1) 0
(2) 2
(3) 3
(4) 4
(3) 3

Explaination:

Question 18.
It is given that f’ (a) exists, then

(1) f(a) – af'(a)
(2) f'(a)
(3) – f'(a)
(4) f(a) + af'(a)
(1) f(a) – af'(a)

Explaination:

Question 19.
If f(x) = then f’ (2) is
(1) 0
(2) 1
(3) 2
(4) does not exist
(3) 2

Explaination:

Question 20.
If g(x) = (x2 + 2x + 3) f(x) and f(0) – 5 and then g'(0) is
(1) 20
(2) 14
(3) 18
(4) 12

Question 21.
If f(x) = then at x = 3, f'(x) is
(1) 1
(2) – 1
(3) 0
(4) does not exist
(2) – 1

Explaination:

From equations (1) and (2) we get
f’ (3) ≠ f’ (3+)
∴ limit of f(x) does not exist at x = 3
f’ (x) does not exist at x = 3

Question 22.
The derivative of f(x)= x|x| at x = – 3 is
(1) 6
(2) – 6
(3) does not exist
(4) 0
(1) 6

Explaination:
f(x) = x|x|
f(x) = x(-x) ⇒ f(x) = – x2
f ‘(x) = -(2x)
f ‘(-3) = -(2) (-3) = 6

Question 23.
If f(x) = then which one of the following is true?
(1) f(x) is not differentiable at x = a
(2) f(x) is discontinuous at x = a
(3) f(x) is continuous for all x in R
(4) f(x) is differentiable for all x ≥ a
(1) f(x) is not differentiable at x = a

Explaination:

f’ (a+) = 3 ………. (2)
From equations (1) and (2) we get
f'(a) ≠ f'(a+)
∴ f’ (x) does not exist at x = a
∴ f(x) is not differentiable at x = a

Question 24.
If f(x) = is differentiable at x = 1, then
(1) a = $$\frac{1}{2}$$, b = $$\frac{-3}{2}$$
(2) a = $$\frac{-1}{2}$$, b = $$\frac{3}{2}$$
(3) a = $$-\frac{1}{2}$$, b = $$-\frac{3}{2}$$
(4) a = $$\frac{1}{2}$$, b = $$\frac{3}{2}$$
(3) a = $$-\frac{1}{2}$$, b = $$-\frac{3}{2}$$

Explaination:

Question 25.
The number of points in R in which the function f(x) = |x – 1| + |x – 3| + sin x is not differentiable, is
(1) 3
(2) 2
(3) 1
(4) 4


text:
# Body Type Calculator + Online Solver With Free Steps

The Body Type Calculator is a helpful tool for quickly determining the important characteristics of a human body. The calculator takes some basic details of the body height, weight, etc.

This calculator displays a piece of detailed information about different kinds of body masses, body fluids, and essential organ properties.

## What Is a Body Type Calculator?

A Body Type Calculator is an online calculator that is specifically designed to find useful information about a person’s body based on age, gender, mass, and height.

Health is the most significant element of our life. To maintain health one needs to be updated about the state of the body. There are certain parameters that evaluate the status of health.

For a single parameter, you need to perform a number of calculations. Also, you should have knowledge of biological concepts for each of them. But the Body Type Calculator is an all-in-one calculator that gives you comprehensive body information.

It is an extremely useful tool for doctors, nutritionists, fitness trainers, and every person that cares about his health. It is entirely free and may be accessed endlessly using your browser.

## How To Use the Body Type Calculator?

You can use the Body Type Calculator by entering the required four parameters in their respective fields. The calculator needs the age, gender, mass, and height of the person.

For each input, there is a labeled box. The detailed instructions on the use of the calculator are given below.

### Step 1

Enter the age of the person in the first box and select the gender of the person in the second box.

### Step 2

Then insert the weight of the person. It only accepts values in pounds. It is a unit of weight and its symbol is ‘lbs.

### Step 3

Now enter the height of the person. As height is measured in feet and inches collectively, therefore it has boxes for both of them.

### Step 4

Click the Submit button to start the computations.

### Result

The first section gives the BMI, different body masses, ideal weight, and body surface area. The calculator gives extra information about each of them in brackets by comparing them with standards.

The next section provides the number of various body fluids present in the body. The final section gives the properties of three essential organs that include the lungs, heart, and brain.

## How Does the Body Type Calculator Work?

The body type calculator works by determining the body characteristics, body fluids, and typical organ properties. This calculator is based on four basic measurements of the body that are age, sex, weight, and height.

These four measurements should be provided as input and as a result, the calculator displays the required information. An understanding of the body’s properties is essential to using this calculator to avoid any health risks.

### What Is the Body Mass Index?

The body mass index (BMI) is a measure of body fat that uses a person’s height and weight. It uses an equation that divides a person’s weight and height to calculate the estimated body fat.

BMI is used to find out whether a person is a healthy weight or unhealthy. A greater value of BMI indicates too much body fat whereas a low value indicates little body fat.

The high value of BMI leads to hazardous diseases such as heart disease, diabetes, high blood pressure, gallstones, sleep apnea, and depression. A person with a high BMI should take measures to lower his BMI to avoid these health risks.

The BMI with a low value also results in bad health conditions that include bone loss, decreased immune function, anemia, malnutrition, and infertility.

The accuracy of BMI is helpful in determining the weight problems in children and adults but it also has limits. It exaggerates the measure of body fat in athletes or people having muscular bodies. It also underrates the body fat in older people.

There are different BMI ranges that are used to classify weight types:

1. Less than 18.5 = Underweight
2. 18.5 to 24.9 = Optimum range
3. 25 to 29.9 = Overweight
4. 30 to 34.9 = Class 1 obesity
5. 35 to 39.9= Class 2 obesity
6. Greater than 40= Class 3 obesity

The body fat mass is the weight of the fat in the body. It is essential to keep the body warm or secure the body organs. However, the fat mass should be in the optimum range since the surplus of this leads to health problems for instance type 2 diabetes or obesity.

The scarcity of fat mass is also not good for health because it causes loss of bone mass and osteoporosis. This calculator calculates the body fat mass that helps to keep the check of it.

If the percentage is too high then a person should do more workouts and take a healthy diet whereas if the fat mass is too low it means that there is a requirement for taking a balanced diet that includes fats and protein.

### What is the Body Surface Area?

The body surface area (BSA) is the measure of the total surface area of a body used to determine the quantity of fluid and the drug dosage.

BSA has many important applications in medical sciences. It applies to the diagnosis of chronic disorders that result in loose pigments in patches of skin. It also predicts the mortality rate in patients that have chronic heart failure.

The optimum BSA is 1.7$m^2$ however it also depends on the age and gender of a person. The body type calculator calculates the BSA following the Mosteller formula by taking the square root of the product of weight and height divided by 3600.

The body characteristics are fully explained above, now we will discuss the body fluids since this calculator also provides the information regarding body fluids and the most important one is body water.

### What Is Body Water?

The body water is the quantity of water content contained in the human body. The human body carries 60 percent of the water.

Water is an essential element of the person’s body because it assists to carry oxygen and nutrients, regulating body temperature, lubricating joints, and protecting organs and tissues.

A full-grown man contains 60 percent of the water whereas an adult woman has 55 percent of the water in her body. The ideal percentage of the water body depends on several factors such as aging, hormones, and body composition.

A better measurement of ideal body water is determined by the ratio of extracellular water to the total body water. The total body water is divided into two types that are extracellular water and intracellular water.

This calculator determines the total body water along with both types.

#### Extracellular Water

The water that is present outside the body cells is known as extracellular water. This water is located in the blood, interstitial fluid, transcellular fluid, and blood plasma.

Extracellular water is essential since it manages the motion of electrolytes, permits oxygen delivery to the cells, and removes the waste products from the metabolic process.

#### Intracellular Water

Intracellular water prevails inside the body cells and consists of 70 percent of cytosol. This kind of water is important because it transports the molecules to the different organelles in the cells. It comprises 2/3 of total body water.

The other body fluids that are determined by the body type calculator are whole blood, blood plasma, and red blood cells.

## Solved Examples

### Example 1

Frank is a national athlete. He is 23 years old with a height of 5.8 feet and his weight is around 67 kg. He wants to know his BMI and body characteristics to maintain his health for the upcoming national games.

### Solution

The calculator gives the following information about Frank’s body.

#### Body Characteristics

The characteristic of the body is given in Table 1.

Table 1

#### Body Fluids

The details of body fluid are shown in Table 2.

Table 2

#### Organ Properties

Table 3 shows the properties of the important organs of the body.

Table 3

### Example 2

A nutritionist wants to know the complete information about the body and the organs of a female patient to prepare a healthy diet plan for her. The details of the patient are given below.

Age = 17 years, Height = 5.3 feet, Weight = 48 kg

### Solution

According to the calculator, the patient’s details are given below.

Table 4

Table 5

Table 6


text:
Special Relativity

After establishing that light travels with a finite speed and discovering Fraunhofer diffraction, $19^\text{th}$-century scientists had every reason to believe that light was a wave. Like sound traveling through air, or shock waves through Earth's crust, light appeared to be a disturbance in a medium it was travelling through. They called this medium luminiferous aether, which was thought to permeate all of space.

Proponents of aether theory had been inspired by Rømer's work centuries earlier. They reasoned that Earth moved relative to the aether as Earth orbited the Sun. Consequently, they hoped to deduce aether's properties by observing changes in the light from distant stars throughout the year.

In this quiz, we will first look at how observers moving relative to a medium detect waves. Then we will extend our insights to light moving through the mysterious luminiferous aether.

The Mystery of the Luminiferous Aether

Imagine a submarine traveling through a long tank of water. You already know from experience that if we want the submarine to move faster, we can push on it with greater force, independent of the water it moves through.

However, propagating waves in water are different. They are small changes in water pressure that move along at the speed of sound. They are not material objects like the submarine.

Consider the sound waves generated by a sonar transmitter in the submarine's nose. How can we make the sound waves travel faster with respect to the ground?

The Mystery of the Luminiferous Aether

Nineteenth-century scientists called the medium that light propagates through luminiferous aether.

The problem with this wave model of light was that no one could say for sure what the aether was. An obvious question they asked to get to the bottom of this mystery is, “At what speed is the luminiferous aether traveling?" or rather, "What speed are we traveling at relative to it?”

In the 1880s, experimentalists Albert Michelson and Edward Morley set out to answer this question. Their idea was simple: if Earth is moving with respect to the aether, then we should be able to detect an asymmetry in how long it takes to send a light signal one way versus another.

Let's illustrate with a thought experiment... on bikes.

The Mystery of the Luminiferous Aether

Even when he's on a lazy bike ride, Michelson's mind is never liberated from plans of his next experiment, and he's about to collide with a tree. Morley calls out a warning from a distance of $L_0$ behind Michelson, but it might be too late.

Sound moves at a known speed of $c_\text{sound}$ relative to the air. If they were at rest in the air's reference frame, the sound would take time $\displaystyle \Delta T_0 =\frac{L_0}{c_\textrm{sound}}$ to travel distance $L_0$.

However, they are both moving relative to the air with speed $V_0.$ How long does it take Michelson to hear Morley?

The Mystery of the Luminiferous Aether

A few minutes later, Michelson and Morley are riding with speed $V_0$ parallel to each other, a distance of $L_0$ from each other.

Compared to $\Delta T_0,$ the propagation time when they are at rest relative to air, how long does it take Michelson to hear Morley?

Recall $\displaystyle \Delta T_0 = \frac{L_0}{c_\textrm{sound}}.$

The Mystery of the Luminiferous Aether

Michelson and Morley's idea to prove (or disprove) the existence of aether was a straightforward extension of their recreational biking misadventures. If light moves at a speed $c$ through the aether, and Earth moves through the aether as it orbits the Sun, then light takes different amounts of time to propagate the same distance along perpendicular directions.

So, they built a device called an interferometer with two identical, long arms at $90^\circ$ to one another and a coherent light source (the equivalent of a modern-day laser). They used a beam-splitter (a special kind of optical device made from metal-coated prisms) to divide the beam and send it toward mirrors at each end of each arm.

After they are reflected, the beams move back along the arms and are combined again before they enter a detector. If the propagation time along each arm is different, Michelson and Morley would detect interference (a mismatch between the waves' peaks and troughs) between the two beams—as long as the interferometer was moving relative to the aether.

The Mystery of the Luminiferous Aether

Let's help Michelson and Morley predict the time difference between the arms, assuming the experiment moves with speed $v$ relative to the aether.

Imagine, for the sake of argument, we've managed to align one arm of the interferometer with $v$ as in the image below. The time $T_\text{parallel}$ for the beam to propagate from the beam-splitter to the mirror and back again depends on the setup's speed relative to the aether.

Assuming the length of the interferometer's arm (from the beam-splitter to the mirror) is $L_0,$ which expression gives $T_\text{parallel}?$

Hint: Each leg of the trip will be the form of a fraction with different denominators. Find a common denominator and add the fractions.

The Mystery of the Luminiferous Aether

We can compare this prediction to the time, $T_\textrm{perp},$ to travel back and forth along the other arm.

When the interferometer is moving with speed $v$ relative to the aether, what is the propagation time along the arm perpendicular to the velocity?

The Mystery of the Luminiferous Aether

Since $T_\textrm{parallel} \propto L_0$ and also $T_\textrm{perp} \propto L_0$, the predicted time difference between the arms is proportional to $L_0,$ too: $\Delta T = \big\lvert T_\textrm{parallel} - T_\textrm{perp} \big\rvert \propto L_0.$ Thus, the bigger your interferometer, the easier it is to detect the effect of moving through aether.

The interferometer Michelson and Morley used in their original experiment had $L_0=\SI{11}{\meter}.$ The split beam takes a different time to propagate along each arm. What is the time difference $\Delta T$ between the arms?

Assume that $v \sim \SI[per-mode=symbol]{30}{\kilo\meter\per\second},$ the orbital speed of Earth. At the time of this experiment, no one had any idea of how fast Earth was moving relative to the aether, but people agreed this was a reasonable lower limit. Also, use $c=\SI[per-mode=symbol]{3e5}{\kilo\meter\per\second}.$

The Mystery of the Luminiferous Aether

Although this seems like a small time difference, it's long enough for the peaks and troughs of a light wave to become misaligned before they come back together in an interferometer the size of Michelson and Morley's.

Interference can be detected when the waves are misaligned by as little as a tenth of a wavelength. The wavelength of light is very short $\big($about $\lambda_\textrm{light} \approx \SI{700}{\nano\meter}=\SI{7e-7}{\meter}\big)$ and light travels fast, so the beams accrue a detectable wavelength difference $\Delta \lambda \gt 0.1 \lambda_\textrm{light}$ quickly.

The wavelength misalignment (also known as "decoherence") using the $\Delta T$ you calculated in the last problem is $\Delta \lambda = c\Delta T \approx \SI{1.2e-7}{\meter}.$ This is definitely above the detection threshold of Michelson and Morley's device.

The Mystery of the Luminiferous Aether

If, indeed, Earth is moving through a mysterious luminiferous aether, then Michelson and Morley's interferometer was almost certain to detect its effect.

In this quiz, we derived an expression for the propagation time difference between the arms, $\Delta T,$ and we checked that our interferometer was the right size to detect such a minute effect.

However, regardless of how Michelson and Morley oriented their device relative to Earth's motion, they measured no time difference. The beams always recombined in such a way that no interference pattern was produced.

Which of the following could produce this result?

The Mystery of the Luminiferous Aether

Michelson and Morley knew what they expected to see, and their experiment had the precision to see it. Their experiments convinced the scientific community near the turn of the $20^\text{th}$ century that both beams took exactly the same time to propagate along perpendicular arms of the interferometer.

For 18 years, the community of physicists speculated that Earth shared a "special" rest frame with the undetectable aether. None of the speculations could be validated by experiment.

Then, in 1905, a young German patent clerk named Einstein posited that, indeed, the length of an interferometer's arms might actually depend on how fast Earth moves around the Sun, simultaneously killing the numerous theories of luminiferous aether and changing the way we view the universe completely.

The Mystery of the Luminiferous Aether

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