Not really a homework question, but

[Jamin2112]

Homework Statement

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 π^e>e^π (knowing that π > e)

∫e^-x2dx on interval (-∞,∞)

Prove 0!=1.

Homework Equations

?

The Attempt at a Solution

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

 

[Cyosis]

Prove π^e>e^π (knowing that π > e)

This one will make you sound smart alright.

 

[Jamin2112]

This one will make you sound smart alright.

Let me try this.

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

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

 

[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.

 

[Jamin2112]

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.

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.

 

[Jamin2112]

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

 

[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.

 

[Mark44]

Let me try this.

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

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

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.

 

[Cyosis]

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

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

 

[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 .

 

[Jamin2112]

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

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

 

[Jamin2112]

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

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 ...

 

[joshsiret]

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

 

[Jamin2112]

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

That takes no mathematical reasoning skills.

 

[joshsiret]

Neither does remebering an arbitrary proof.

 

[Cyosis]

That takes no mathematical reasoning skills.

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

 

[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.

 

[Mentallic]

Imagine how cool I'll be when I bring these things up at parties.

Oh, I'm imagining something alright.

 

[vaibhav1803]

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 ...

What parties..?..(elaborate please.)
answer any 1 simple question,
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.
by your curiosity i expect answers..:D

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

 

[Count Iblis]

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 ...

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"

 

[Jamin2112]

Jamin2112, why are you so interested in these things? Do you study math, statistics, physics, engineering or computing in college?

Studying Math & Statistics in college.

 

[Jamin2112]

What parties..?..(elaborate please.)
answer any 1 simple question,
Q.what is the probability that on choosing any two numbers the two are relatively prime(no common factor)

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

 

[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...

 

[LCKurtz]

Studying Math & Statistics in college.

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.

 

[Jamin2112]

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.

I already know the birthday problem ...

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

 

[Jamin2112]

http://arxiv.org/abs/0707.0093"

Can you rewrite that stuff for me? I'm confused about all the notation on that site.

 

[Count Iblis]

Can you rewrite that stuff for me? I'm confused about all the notation on that site.

But even without reading these articles in detail, it isn't that hard to take a deck of cards, place it on the edge of a table and then create an overhang such that the top card is no longer above the table.