The integration of e^(-x^2) (2 Viewers)

[Drsoccerball]

It's basically the same as the question from last year's BOS. Converting it into a (polar) volume integral by exchanging variables and then taking the limits. Trapping the squared value between two bounds which become identical. Taking the square root of both sides.

If you don't state rank next year ill chop my balls off.

 

[braintic]

If you don't state rank next year ill chop my balls off.

Is that an "if and only if" statement?

 

[Drsoccerball]

Is that an "if and only if" statement?

ahahahah no. Its more like in inequalities

 

[Paradoxica]

Is that an "if and only if" statement?

He didn't specify so I'll assume he has other reasons to chop his balls off.

You don't need to sub in -inf to inf.
Integrating from -3 to 3 gives 99.998% of the total value.
The number of terms you add has a much bigger effect on the estimate than the size of the limits.

How would you know it's the square root of pi though? I mean, you can take the approximation to as many terms as you like, but proving identicality is another matter entirely. Evidence is not proof. (pls don't kill me science kids)

 

[braintic]

He didn't specify so I'll assume he has other reasons to chop his balls off.



How would you know it's the square root of pi though? I mean, you can take the approximation to as many terms as you like, but proving identicality is another matter entirely. Evidence is not proof. (pls don't kill me science kids)

Check out this link:
https://www.youtube.com/watch?v=nqNzKeVCYBU

But ... you will need to understand double integrals and polar coordinates.

 

[Drsoccerball]

Check out this link:
https://www.youtube.com/watch?v=nqNzKeVCYBU

But ... you will need to understand double integrals and polar coordinates.

#Braintics youtube account.
The only thing i didnt get was how the r coefficient came?

 

[Carrotsticks]

and the improper integral from -infinity to positive infinity is (pi)^1/2.
is there a way to use drsoccerball's genius to find an approximation for pi?? subbing infinities into that sexy polynomial dont look so good at this moment...
anybody got ideas??

Of course you can.

Find any improper integral within some finite domain that evaluates to some relatively simple expression involving pi, where it can be 'made the subject' later.

Now integrate the Taylor expansion for the first however many terms you like (depending on the degree of accuracy you desire) over the same domain.

The two values are 'approximately' (using the term 'approximately' very loosely here) equal to each other, and work from there.

 

[Carrotsticks]

#Braintics youtube account.
The only thing i didnt get was how the r coefficient came?

It is pretty much the same R as what you are familiar with when you use Rcis(theta). We are converting from a cartesian coordinate system to a polar coordinate system.

 

[braintic]

It is pretty much the same R as what you are familiar with when you use Rcis(theta). We are converting from a cartesian coordinate system to a polar coordinate system.

I think it's simpler to think of it as coming from l=rθ

The area of a thin annular sector is approximately equal to the arc length times the change in radius.

 

[Carrotsticks]

It's basically the same as the question from last year's BOS. Converting it into a (polar) volume integral by exchanging variables and then taking the limits. Trapping the squared value between two bounds which become identical. Taking the square root of both sides.

Pretty much this. I had to modify it slightly though to acquire neater bounds. The conversion to a polar system naturally results in a upper/lower circular bound, which was rather messy to work with (algebraically, it was quite terrible). So instead of bounding a circular domain by two concentric circular domains, I changed them to square domains, which resulted in much cleaner expressions.

How would you know it's the square root of pi though? I mean, you can take the approximation to as many terms as you like, but proving identicality is another matter entirely. Evidence is not proof. (pls don't kill me science kids)

Can you elaborate further on what you meant by this?

 

[braintic]

.

Just wondering why the little light next to your name never comes on to indicate you are logged in?

 

[Carrotsticks]

Just wondering why the little light next to your name never comes on to indicate you are logged in?

Perhaps because I have set my browsing to private, so I do not appear in the list of users who are browsing the thread.

 

[braintic]

Perhaps because I have set my browsing to private, so I do not appear in the list of users who are browsing the thread.

Ahhh .... so no-one can tell you've been visiting the Neighbours and Home and Away forums?

 

[RealiseNothing]

So all of these integrals such as can be found by using their Taylor series?

This can be found using complex analysis in a VERY cool way.

 

[Carrotsticks]

Ahhh .... so no-one can tell you've been visiting the Neighbours and Home and Away forums?

I guess someone had to discover my little secret some day.

 

[Physicklad]

Check out this link:
https://www.youtube.com/watch?v=nqNzKeVCYBU

But ... you will need to understand double integrals and polar coordinates.

every step of that proof titillates me

 

[Paradoxica]

Pretty much this. I had to modify it slightly though to acquire neater bounds. The conversion to a polar system naturally results in a upper/lower circular bound, which was rather messy to work with (algebraically, it was quite terrible). So instead of bounding a circular domain by two concentric circular domains, I changed them to square domains, which resulted in much cleaner expressions.
Can you elaborate further on what you meant by this?

We can't compute an infinite number of terms. You can't stop at some arbitrary degree of precision and then declare it converges to simply because it appears to be converging to it. I (and probably a whole bunch of other people) would demand proof.

This can be found using complex analysis in a VERY cool way.

You mean the improper integral, right?

 

[RealiseNothing]

You mean the improper integral, right?

Yep that, could also replace the with 0.

 

[RealiseNothing]

We can't compute an infinite number of terms. You can't stop at some arbitrary degree of precision and then declare it converges to simply because it appears to be converging to it. I (and probably a whole bunch of other people) would demand proof.

Carrotsticks knows that, he pretty much topped real/complex analysis at usyd.

He's saying that drsoccerball asked for an approximation, not an exact value necessarily. So what braintic was saying about only using a few terms is fine.

 

[Drsoccerball]

What did he study anyway? Has he finished? Does he have a job?