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**Final Exam / Re: problem 5**

« **on:**December 20, 2012, 02:50:21 PM »

Hopeful solution for problem 5

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sorry the original image is vertical, I don't know why after upload it rotate to the left

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http://www.math.toronto.edu/courses/apm346h1/20129/HA9.html#problem-9.1

WTH: Can you rotate before posting?!!!! - V.I.

Hopeful solution for problem 1:

WTH: Can you rotate before posting?!!!! - V.I.

Hopeful solution for problem 1:

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Hopeful solution for Question 2:

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part(d)

Jinchao--are we supposed to rotate a monitor? Sure one can download and rotate an image but methinks it would be simpler for you to rotate it before posting. Also, why all these weird colours--it is a home work in math, not in visual arts V.I.

Jinchao--are we supposed to rotate a monitor? Sure one can download and rotate an image but methinks it would be simpler for you to rotate it before posting. Also, why all these weird colours--it is a home work in math, not in visual arts V.I.

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Solutions for part (a) and part(b)

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Dear professor, I am wondering that will you provide basic Fourier transform table in the exam?

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in part a, it seems that if we want the expansion to be real domain, the only way we can do is first manipulate in complex domain by the exponential expansion, then use Euler formula to convert into real domain? Because there is no integral formula for $\int sin(x)exp(x)$

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We are only asked to draw the graph but not the functional form, don't we?

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From part (a) we have

$u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{4kt} } f(y) dy $

Rewrite the formula as:

$u(x,t) = \frac{1}{\sqrt{2\pi } \sqrt{2kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{2 \sqrt{2kt}^2 } } f(y) dy $

We see this is a normal density of random variable x centered at y with standard error $ \sqrt{2kt}$ . When $t \to 0$, the standard error approach to 0 as well. The random variable x approaches to a deterministic form. So $lim_{t \to 0} u(x,t) = f(y) $.

I think the argument can be written more accurate if we take the functional form of f(x) in the whole real line. But then we are not directly prove the result based on the formula of u(x,t) we get from part (a).

$u(x,t) = \frac{1}{\sqrt{4\pi kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{4kt} } f(y) dy $

Rewrite the formula as:

$u(x,t) = \frac{1}{\sqrt{2\pi } \sqrt{2kt}} \int_{-10}^{10} e^{ \frac{-(x-y)^2}{2 \sqrt{2kt}^2 } } f(y) dy $

We see this is a normal density of random variable x centered at y with standard error $ \sqrt{2kt}$ . When $t \to 0$, the standard error approach to 0 as well. The random variable x approaches to a deterministic form. So $lim_{t \to 0} u(x,t) = f(y) $.

I think the argument can be written more accurate if we take the functional form of f(x) in the whole real line. But then we are not directly prove the result based on the formula of u(x,t) we get from part (a).

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Subqueston (d):

$ \frac{dt}{1} = \frac{dx}{x^2} $

$ t = -x^{-1}+c $

so the general solution is $ u(t,x)=f(t+x^{-1})$.

$u(0,x)=f(x^{-1})=g(x)$

$f(y)=g(y^{-1})$

Since $y=x^{-1}$, so when $x>0$ we have $y>0$ as well.

$u(t,x)=f(t+x^{-1})$

We need $t+x^{-1}>0$

Since $x>0$,

therefore $tx+1>0$

so the domain be defined is $\{(t,x) | tx>-1 \}.

$ \frac{dt}{1} = \frac{dx}{x^2} $

$ t = -x^{-1}+c $

so the general solution is $ u(t,x)=f(t+x^{-1})$.

$u(0,x)=f(x^{-1})=g(x)$

$f(y)=g(y^{-1})$

Since $y=x^{-1}$, so when $x>0$ we have $y>0$ as well.

$u(t,x)=f(t+x^{-1})$

We need $t+x^{-1}>0$

Since $x>0$,

therefore $tx+1>0$

so the domain be defined is $\{(t,x) | tx>-1 \}.

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