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#1 Re: This is Cool » Physics engines + Graphical engines + fun » 2007-07-06 14:24:13

Very nice, I'm looking forward to seeing more of your work in the future.

#2 Re: Euler Avenue » GRE subject test » 2007-06-21 14:31:13

Sounds like a good plan. It's nice to have extra chances if you feel that you didn't prepare yourself well enough the first time around. I don't think that anything I'm taking this coming semester will help me too much on the exam, so I'm thinking about taking it in August too, and possibly a retake if I feel that I could have done better. My only problem is that I need to figure out if I'll be at home or at school when the test is administered. I'll probably be at school for TA training though. I think I have about 2 weeks still before I need to be certain of that so I can register by the deadline.

#4 Re: Euler Avenue » GRE subject test » 2007-06-21 09:40:43

When are you taking it? I need to take mine very soon as well.

#5 Re: Help Me ! » Integration » 2007-05-22 06:34:43

Is this a homework assignment for a class? If so, what class? Are there any limits of integration for this integral? This integral is known as a Fresnel integral. If the integration is done from 0 to ∞, it's not very hard to calculate the integral if you know what the gamma function is. There isn't really much to do with this if it is an indefinite integral though.

#6 Re: Euler Avenue » Number Giants » 2007-05-21 11:37:49

I don't know if I said this here, but if all the matter in the universe were converted into ink, it wouldn't be enough to fully write out Graham's number. It's pretty big.

#7 Re: Help Me ! » Advice » 2007-05-12 21:20:42

I use a program called WinShell to type my math papers. It's a free program that you type text and LaTeX into and it will make it into a PDF or PS file. If you want to try it out you'll need Miktex too:

http://www.miktex.org/
http://www.winshell.org/

You'll need to know a little more about LaTeX to make a document, but there is a demo file included with WinShell that you can use to get started.

As for understanding topology, it may be best for you to take an actual course on it instead of self-study. Have you only been learning it from Rudin's text? Perhaps you should look into chapter 3 of Apostol's "Mathematical Analysis" or the Munkres text "Topology". If you want a free topology resource, consider this online text:

http://uob-community.ballarat.edu.au/~s … opbook.pdf

It's probably the most basic introduction to topology you'll find. If even that is hard to follow, you might just need a teacher. It is sometimes easier to see what's going on when someone else is writing it on a board in front of you.

#8 Re: This is Cool » 9,8,0,1 » 2007-05-12 20:39:20

Man, that's too much work. I'm not too big on calculations like that. I hate when people are like "you're a mathematician? What is 47236 times 2923.424?" as if all math is just arithmetic. You don't need special mathematical insight to do such things. Dividing 13-digit numbers in your head isn't the same as being able to prove a theorem or understand a mathematical concept. Anyway I just wanted to say that. Have a nice day.

#10 Re: Help Me ! » numbers of sequences we can do ! » 2007-05-05 09:35:14

HallsofIvy wrote:

If mean sequences of those six numbers, in increasing order, then the answer is 1: {1, 2, 3, 4, 5, 6} itself!  Any other sequence would change the order so the numbers would not be increasing.  That seems rather obvious.  Perhaps I am misunderstanding.

I would imagine sequences such as (1, 1, 1, 1, 1, 1), (1, 1, 2, 2, 3, 3), etc. would work as well. The requirement is x[sub]1[/sub] ≤ x[sub]2[/sub] ≤ x[sub]3[/sub] ≤ x[sub]4[/sub] ≤ x[sub]5[/sub] ≤ x[sub]6[/sub], not x[sub]1[/sub] < x[sub]2[/sub] < x[sub]3[/sub] < x[sub]4[/sub] < x[sub]5[/sub] < x[sub]6[/sub]. So try looking at it again. smile

#11 Re: This is Cool » easy proof of the chain rule? » 2007-05-01 17:56:16

Silly me, if a function f is differentiable at c, then f is continuous at c. Hohoho, good teamwork *high fives*

#12 Re: This is Cool » easy proof of the chain rule? » 2007-05-01 12:07:44

luca-deltodesco wrote:

the proof relys on g being continuous does it not?

This is the problem with your proof, Mikau. It's somewhat subtle, so I'll point out where:

Mikau wrote:

We know that Δg = g(u + Δu) - g(u) therefore as Δu approache zero, Δg approaches zero

This is the same as saying that if |Δu| = |(u + Δu) - u| < δ, then for any ε > 0, |Δg| = |g(u + Δu) - g(u)| < ε, or in other words that g is continuous. The chain rule does not require continuity, only differentiability.

#13 Re: This is Cool » easy proof of the chain rule? » 2007-05-01 07:06:33

I suppose it is the same logic, but of course they have to write the proof SOMEWHERE in the book, since the chain rule is such a common thing, and it's not just a direct corollary. Which reminds me that I hate when a book states a less popular theorem and says "see this author for a proof".

#14 Re: Guestbook » hi » 2007-05-01 07:01:21

Also note that my words towards Jane were sarcastic. Maybe a moderator should move this thread to the Jokes section.

#15 Re: This is Cool » easy proof of the chain rule? » 2007-04-30 12:25:16

One reason they are shorter is because they use another theorem or two previously stated in the same chapter. They're definitely not imprecise or informal. Could you give me an example of a lengthy chain rule proof?

#16 Re: This is Cool » easy proof of the chain rule? » 2007-04-30 10:48:59

When g is not continuous you have to prove it otherwise. smile

I don't know why everyone thinks the proofs of the chain rule are lengthy or overly complicated. I can think of two or three proofs what are only about a paragraph. There is also a proof using the hyperreal number system that is only a line or two.

#17 Re: Guestbook » hi » 2007-04-30 10:35:15

Thanks for pointing that out Jane, you have been reported

#19 Re: Guestbook » hi » 2007-04-29 06:48:36

Devantè wrote:

That is never really my intention.

Ok, how about this then. If you post DIRECTLY after the post you are commenting on, you don't need to quote them. If it is the first post in the thread, it is obvious who you are talking to, and if it isn't, you can say "^", "To the above poster" or "@someone" to get their attention. If you absolutely need to quote something someone said, then just quote that part (like what I just did in this post). Surely all the smilies and such in the post aren't important things you need to quote. I'm sure this shouldn't be a problem for you since it isn't your intention and you're so against small posts intended to raise post count.

Is it going to be possible sometime in the future to make quotes not count towards your post count?

Also, rida, I think the intention was irony. And if justlooking's intention wasn't so, mine was. I thought it was funny for a one time thing sad

#20 Re: Guestbook » hi » 2007-04-28 20:40:44

justlookingforthemoment wrote:
Devantè wrote:
muskaan wrote:

uproflol cool:)hello..........

nice website

i think it is the best website i had ever seen.up
thanking uuuuuu.........................

bye bye ...............................wave
sleep

Thanks, muskaan!

Hey, Devant\\\e, stop quoting unnecessary text to get past the 100 character limit! tongue

lol

#21 Re: Help Me ! » Hard problems » 2007-04-28 07:46:02

1. To make this simple, let us see if we can do the problem with only two reciprocals, so 1/2006 = 1/a + 1/b, with -2006 < a, b < 2006. Multiplying both sides by 2006ab gives ab = 2006a + 2006b. We manipulate the expression so that we can factor:

Since -2006 < a, b < 2006, we have that 0 < 2006 - a < 2 × 2006 and 0 < 2006 - b < 2 × 2006. So we want two factors (2006 - a and 2006 - b) of 2006² that are between 0 and 2 × 2006. We know that 2006 = 2 × 17 × 59. So our factors are 2² × 17² (= 1156) and 59² (= 3481). Then 2006 - a = 1156 and 2006 - b = 3481, so a = 850 and b = -1475. Thus

2. You're perfect so far smile. The next step is to realize that 5 times the digits of our number (I'll call it N) must be divisible by 25, since 5 is its last digit. The multiples of 25 end in 00, 25, 50, or 75. Since N can't have 0 or an even number as a digit, it must end in 75. Once again, since N is 5 times the product of its digits, it is divisible by 5 × 7 × 5 = 175. Starting at 1 × 175, every 4th multiple of 175 ends in 75. So 1 × 175 = 175, 5 × 175 = 875, 9 × 175 = 1575, ..., 57 × 175 = 9975 (we stop at 57 since N must be less than 10000). If you write out all these products you see that the only numbers there with only odd digits are 175, 1575, 5775, 7175 and 9975. By direct calculation the only one of these which is equal to 5 times the product of its digits is 175.

You will get a calculator for this test, right? It might be a little tedious to do this all by hand.

#22 Re: Help Me ! » Pi Identity » 2007-04-26 21:22:08

The problem is very easy if you know how Euler solved the Basel problem (finding the exact value of 1/1² + 1/2² + 1/3² + ...). He cleverly manipulated the Taylor expansion of sin x. To solve your problem, you do virtually the same thing but with the Taylor expansion for cos x.

We start with the Maclaurin series for cos x:

This function will have zeros at (2n+ 1)π, so factor it according to that fact:

This simplifies to

Now if you were to multiply this all out, the coefficient of x² would be

But in our original expression for cos x, the coefficient for x² was 1/2! = 1/2, so the sum must be equal to 1/2:

Now, simplify:


So

Ninja edit: My solution is more creative than Jane's. swear

Real edit: But Jane's uses the fact that 1/1² + 1/2² + 1/3² + ... = π²/6, so that might be better to use if this is for an assignment.

#23 Re: Help Me ! » Logarithms ;] » 2007-04-26 20:59:28

I used the logarithm property

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