 Author Topic: problem 3  (Read 13055 times)

Thomas Nutz

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• Karma: 1 problem 3
« on: October 07, 2012, 10:15:42 PM »
Could anyone give me a hint on how to integrate problem 3a)? In order to get rid of the absolute value I have to split up the integral and then the -inf to inf formulas don't work any more...

Thanks a lot!

Calvin Arnott Re: problem 3
« Reply #1 on: October 08, 2012, 12:18:45 AM »
Try completing the square and using the error function. It's a really messy integral.

Victor Ivrii Re: problem 3
« Reply #2 on: October 08, 2012, 01:12:46 AM »
Try completing the square and using the error function. It's a really messy integral.
$\newcommand{\erf}{\operatorname{erf}}$
Yes, you need to consider  $\erf(z)$ as an elementary function (and there is no compelling arguments why trigonometric functions are considered as such but not many others. In fact there are plenty of important special functions coming often from PDE, more precisely, from separation of variables--not $\erf$ but many others).

Peishan Wang

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• Karma: 6 Re: problem 3
« Reply #3 on: October 08, 2012, 07:43:01 AM »
Should part (c) and part (d) be: solve the IBVP for x>0 or for all x? Thanks!

Victor Ivrii Re: problem 3
« Reply #4 on: October 08, 2012, 08:12:31 AM »
Should part (c) and part (d) be: solve the IBVP for x>0 or for all x? Thanks!

For $x>0$ only, I adjusted problem to make it explicit
« Last Edit: October 08, 2012, 08:20:21 AM by Victor Ivrii »

Shu Wang

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• Karma: 0 Re: problem 3
« Reply #5 on: October 08, 2012, 07:07:43 PM »
Try completing the square and using the error function. It's a really messy integral.
$\newcommand{\erf}{\operatorname{erf}}$
Yes, you need to consider  $\erf(z)$ as an elementary function (and there is no compelling arguments why trigonometric functions are considered as such but not many others. In fact there are plenty of important special functions coming often from PDE, more precisely, from separation of variables--not $\erf$ but many others).

When you integrate erf(z) it always gives you zero because it's an odd function. When multiplied to any integrated function (and as alpha -> 0), the resulting functions are always 0. Does that make any sense?

Calvin Arnott Re: problem 3
« Reply #6 on: October 08, 2012, 07:46:14 PM »
Try completing the square and using the error function. It's a really messy integral.
$\newcommand{\erf}{\operatorname{erf}}$
Yes, you need to consider  $\erf(z)$ as an elementary function (and there is no compelling arguments why trigonometric functions are considered as such but not many others. In fact there are plenty of important special functions coming often from PDE, more precisely, from separation of variables--not $\erf$ but many others).

When you integrate erf(z) it always gives you zero because it's an odd function. When multiplied to any integrated function (and as alpha -> 0), the resulting functions are always 0. Does that make any sense?

The error function itself isn't integrated here, the function G(x,y,t)*g(y) is integrated.

It's true that an odd function will integrate to 0 on a domain symmetric about 0, and anything multiplied by this evaluated integral will be 0, but because we're multiplying G(x,y,t) inside of the integral sign that fact isn't too relevant here.

Consider the analogue to the function f(x) = x. f is an odd function, so the integral on say (-a,a) is 0. But when we multiply inside of the integral sign by another odd function g(x) = x^3, we have g*f(x) = x^4- certainly not 0 on the interval (-a,a) when a != 0.
« Last Edit: October 08, 2012, 07:47:52 PM by Calvin Arnott »

Bowei Xiao

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• Karma: 2 Re: problem 3
« Reply #7 on: October 08, 2012, 09:26:55 PM »
just wondering is that sounds true if i try to change the integral to some kind of PDF of a normal distribution?

Victor Ivrii Re: problem 3: OT--Integrating erf(z)
« Reply #8 on: October 08, 2012, 09:42:55 PM »
$\newcommand{\erf}{\operatorname{erf}}$
Integrating by parts
\begin{equation*}
\int _0^z \erf(z)\,dz= z \erf(z)-\int_0^z z\erf '(z)\,dz=z\cdot \erf(z)-\sqrt{\frac{2}{\pi}}\int_0^z  z \cdot e^{-\frac{z^2}{2}}\,dz=
z \cdot \erf(z)-\sqrt{\frac{2}{\pi}}\Bigl[e^{-\frac{z^2}{2}}-1\Bigr]
\end{equation*}

Peishan Wang

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• Karma: 6 Re: problem 3
« Reply #9 on: October 10, 2012, 10:52:54 PM »
Q3 part(a)

(Sorry this part is kind of long so I have to attach 4 pictures. Please let me know if there's anything wrong with the answers.)
« Last Edit: October 15, 2012, 05:56:54 AM by Peishan Wang »

Peishan Wang

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« Reply #10 on: October 10, 2012, 10:57:53 PM »
Q3 part(b) which used the same method as part(a).
« Last Edit: October 15, 2012, 05:54:30 AM by Peishan Wang »

Peishan Wang

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• Karma: 6 Re: problem 3
« Reply #11 on: October 10, 2012, 11:00:01 PM »
Q3 part(c)

Peishan Wang

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« Reply #12 on: October 10, 2012, 11:03:51 PM »
Q3 part(d)

We get the same answer as in part (a) because g(x) is itself an even function, so if we take even reflection we still get g(x).
« Last Edit: October 15, 2012, 05:57:15 AM by Peishan Wang »

Zarak Mahmud Re: problem 3
« Reply #13 on: October 10, 2012, 11:04:48 PM »
Hahah, yeah I think I would have had to spend about 3 hours typing that all up.

Peishan Wang

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• Karma: 6 Re: problem 3
« Reply #14 on: October 11, 2012, 02:47:33 PM »
Yeah my answers are really long but I don't know other ways to solve the problem.... 