### Author Topic: Lecture 19  (Read 5556 times)

#### Aida Razi

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##### Lecture 19
« on: November 13, 2012, 03:12:17 PM »
In heat equation: We know IFT of e^ (-(Î¾^2)/2) is (âˆš2pi)e (-(x^2)/2). Then when we try to find IFT of e^ (-(ktÎ¾^2)), we scale Î¾ to âˆš(2kt)Î¾ and therefore x scale to âˆš(2kt)x. But in the lecture note, it is mentioned that x scale to (2kt)^(-1/2)x. I believe it should be just square root of 2kt and not (-square root) of 2kt.

#### Victor Ivrii

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##### Re: Lecture 19
« Reply #1 on: November 13, 2012, 03:29:59 PM »
No, if you scale $x\mapsto \sigma x$ you scale $\xi \mapsto \sigma^{-1}\xi$ to keep $x\xi$ the same..

#### Miranda Jarvis

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##### Re: Lecture 19
« Reply #2 on: December 19, 2012, 11:27:42 AM »
Just to clarify, I'm assuming that in the lead up to equation 25 it was meant to be $-|\xi|^{-1}e^{-|\xi|y}$ not $x$ and $i$  . . .
« Last Edit: December 19, 2012, 01:15:50 PM by Miranda Jarvis »

#### Miranda Jarvis

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##### Re: Lecture 19
« Reply #3 on: December 19, 2012, 11:53:40 AM »
I'm sorry to nit pick but as someone who is having a hard enough time following much of this material, I was wondering in the lines shown in the screen capture attached (the lead up to equation 39 is there an extra $i$ in the denominator of the first equation? I think this is supposed to be equal to the second term in the right side of equation 38 . . .

#### Victor Ivrii

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##### Re: Lecture 19
« Reply #4 on: December 19, 2012, 12:12:30 PM »
I'm sorry to nit pick but as someone who is having a hard enough time following much of this material, I was wondering in the lines shown in the screen capture attached (the lead up to equation 39 is there an extra $i$ in the denominator of the first equation? I think this is supposed to be equal to the second term in the right side of equation 38 . . .

You are correct. Fixed

#### Miranda Jarvis

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##### Re: Lecture 19
« Reply #5 on: December 19, 2012, 01:25:42 PM »
again, not to be a pain, but is it in fact supposed to be $\xi$ not $xi$ in the lead up to equation 25?

#### Victor Ivrii

again, not to be a pain, but is it in fact supposed to be $\xi$ not $xi$ in the lead up to equation 25?