last day (17 days later) » 

22:33
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A: Compute $\lim_{t\searrow 0}\frac{F(x,t)-f(x)}{t}$ where $F(x,t)$ is a complicated integral.

Oliver DiazThis is a suggestion. The function \begin{aligned} u(x;t)=\frac{1}{\sqrt{2\pi t}}\int_\mathbb{R}e^{-\frac{|y-x|^2}{2t}}f(y)\,dy=W_t*f(x) \end{aligned} where $W_t(y)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{y^2}{2t}}$ is solution to the heat equation problem with initial condition $u(x,0)=f(x)$ \begin{...

so in your notation, $$\lim_{t\searrow 0}\dfrac{u(x,t)-u(x)}{t}$$ is the infinitesimal generator of a brownian motion? what should I do to involve the $u(xe^{-t},\sqrt{1-e^{-2t}})$?
It should be $(1-e^{-2t})$. In any event, $F(x,t)$ of the problem is $u(xe^{-t},1-e^{-2t})$, where $u$ is solution to the Cauchy problem above.
I got this, but then how do I compute my limit? I understand $F(x,t)=u(xe^{-t},1-e^{-2t})$, so that $$\lim_{t\searrow 0}\dfrac{F(x,t)-f(x)}{t}=\lim_{t\searrow 0}\dfrac{u(xe^{-t},1-e^{-2t})-f(x)}{t}.....$$ I still don't know how to compute it...
Use the chain rule. The limit you are looking for is $\partial_tF$ at $t=0$. Notice that $u(x,0)=f(x)$.
$F(x,0)=u(x,0)=f(x)$
I got it, so $\partial_{t}F=\partial_{t} u(xe^{-t},1-e^{-2t})=u_{t}(xe^{-t}, 1-e^{-2t})\cdot 2e^{-2t}$, so at $t=0$, we have the limit is $2u_{t}(x,0)|_{t=0}$? but what is this thing? I am really sorry for these many confusions...
22:34
I just made further edits to my answer. I hope I did not make silly mistakes in applying the chain rule.
hello
sorry for replying late
I went to out to throw away trash..
let me take a look
okay
I got it
you are correct I believe
thank you so much

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