Homework Help: Not really a homework question, but

1. May 24, 2010

Jamin2112

1. The problem statement, all variables and given/known data

I need some more random, cool math tricks to bust out when I want to sound smart. So teach me how to do the following in the fanciest way possible.

Prove πe>eπ (knowing that π > e)

∫e-x2dx on interval (-∞,∞)

Prove 0!=1.

2. Relevant equations

?

3. The attempt at a solution

I remember being shown each of these a while ago, but I've now forgotten how to do them.

2. May 24, 2010

Cyosis

This one will make you sound smart alright.

3. May 24, 2010

Jamin2112

Let me try this.

π > e
---> ln(π) > ln(e)
----> ln(π) > 1
----> eln(π) > e1
----> π > e, as required.

Now do I sound smart? Did I make any mistakes?

4. May 24, 2010

Cyosis

I'm afraid this didn't make you sound smart. While there is nothing wrong with what you did, there is no connection with the original problem and it doesn't prove anything.

5. May 24, 2010

Jamin2112

I used to know a fancy way to prove [sin(x)]'=cos(x). I learned it from a Learning Company video. You draw a right triangle with one angle x, and then look at the change in the opposite side as x is increased. Can't remember how to do that either.

6. May 24, 2010

Jamin2112

If you guys are not going to answer my question, at least show me something fancy.

7. May 24, 2010

Cyosis

What is your point with this post Jamin2112, it doesn't really belong in this section of the forum.

Secondly my post #2 was sarcastic. I'll let you figure out why for a bit.

8. May 24, 2010

Staff: Mentor

Here you are starting by assuming that π > e, from which you conclude that π > e. Nothing earth-shattering there.

The integral you showed is usually done as a polar iterated integral, which is much easier than the corresponding Cartesian iterated integral.

For the third, 0! is defined to be 1. Definitions are never proved.

9. May 24, 2010

Cyosis

I suppose we could use the gamma function to show 0!=1, without defining it.

10. May 24, 2010

Tedjn

Jamin2112, if you just want something with which to impress your friends, you might as well memorize http://www.1729.com/blog/CubeRoots.html [Broken].

Last edited by a moderator: May 4, 2017
11. May 24, 2010

Jamin2112

Ha-ha! I totally forgot what I was trying to do halfway through the probel!

12. May 24, 2010

Jamin2112

Nah. I want something cooler. You know, something like approximating pi by dropping a pencil on the table and measuring the angle or something. Imagine how cool I'll be when I bring these things up at parties.

I'm waiting ........

Last edited by a moderator: May 4, 2017
13. May 24, 2010

joshsiret

Why don't you memorise pi to 100 digits...

14. May 24, 2010

Jamin2112

That takes no mathematical reasoning skills.

15. May 24, 2010

joshsiret

Neither does remebering an arbitrary proof.

16. May 25, 2010

Cyosis

What takes even less mathematical reasoning skills is proving $\pi^e>e^\pi$ since it's false, as hinted multiple times.

17. May 25, 2010

hellofolks

Hello,

Using double integrals and formulas for change of variables, you can show that

$$\int_{-\infty}^{+\infty}e^{-u^2/2} du=\sqrt{2\pi}.$$

This is a classic, useful result and you should know it by heart.
Now, setting $$x=u/\sqrt{2}$$, you get $$x^2=u^2/2$$ and $$du=\sqrt{2}\,dx$$, from which

$$\int_{-\infty}^{+\infty}e^{-x^2} \sqrt{2}\,dx=\sqrt{2\pi}.$$

Thus,

$$\int_{-\infty}^{+\infty}e^{-x^2} dx=\frac{\sqrt{2\pi}}{\sqrt{2}}=\sqrt{\pi}.$$

I was rather sleepy when I wrote this, so please correct me if you find any mistakes. And Jamin2112, why are you so interested in these things? Do you study math, statistics, physics, engineering or computing in college? Are you in high school? Do you just want to impress your friends, because, if they don't study the kind of stuff I mentioned in college, far from amusing them, you will scare them!
Qualitative information usually works better in a talk than deriving results. Instead of writing out the equations I showed you, you could tell them that this integral is of central importance in statistics and that it is mathematically interesting, too, since it can't be evaluated in a closed form in compact intervals, but can be shown to have the value $$\sqrt{\pi}$$ when evaluated on the whole real line.
I hope that helps.

18. May 25, 2010

Mentallic

Oh, I'm imagining something alright.

19. May 25, 2010

vaibhav1803

What parties..?..(elaborate plz.)
Q.what is the probability that on choosing any two numbers the two are relatively prime(no common factor)
Q.In an automotive factory, the vehicles are numbered serially, you a common person on an average spot the vehicle with serial number 60, give me the most reasonable guess for the number of vehicles produced, with rigorous proof.
..if you can explain me the answer to this, i'll show u stuff in math you wouldn't have seen in a lifetime+ show you the fancy ways to solve your given problems, ;)..i mean it.

this is an open question mail solutions to me, pointless for those who know but all the more fun :P

Last edited: May 25, 2010
20. May 25, 2010

Count Iblis

Parties and math (however interesting) don't go together very well. You may have more success with a physics demonstration, which you can base on math. E.g. with a deck of cards there are some interesing demonstrations possible, http://arxiv.org/abs/0707.0093" [Broken]

Last edited by a moderator: May 4, 2017
21. May 25, 2010

Jamin2112

Studying Math & Statistics in college.

22. May 25, 2010

Jamin2112

Hmmm.....

I'll assume you mean a unique pair of integers.

If two integers a and b are relatively prime, there exists another integer c (≠ 1, a, b) such that c divides a and b.

If I restricted a and b to {2,3,4},

The possibilities for (a,b) would be

(2,2), (2,3), (2,4), (3,3), (3,4), (4,4).

The only unique pairs that are relatively prime are (2,3) and (3,4).

So, restricting a and b to 3 integers gave me 2 relatively prime combinations. It must follow that as I open up the set to n integers, I have (n - 1) relatively prime combinations.

The answer is: ∞ - 1

23. May 25, 2010

Count Iblis

Well, you can't draw a sweeping conclusion like that....

Think of this line of reasoning. If p is a prime number, then one in p numbers will be divisble by p, so the probability that two randomly chosen numbers are both divisible by p is 1/p^2. The probability that they don't have p as a common divisor is thus 1-1/p^2....

24. May 25, 2010

LCKurtz

So wait until you are at a party with 40 or more guests and make a bet with someone that there are two people in the room sharing the same month/day birthdate. Give them 2 to 1 odds. Make yourself the life of the party for fun and profit.

25. May 25, 2010

Jamin2112

I already know the birthday problem .....

C'mon, you guys need to show me something cool.