Test 1 Sample Part 2 Problems - #1

Hey guys, so I worked out the first problem but I was wondering if anybody knew how to answer the second part of the question. "How is this related to our solution of the heat problem?" I'll show my steps my below but I was wondering if somebody could help me relate the two.

Note - the original post I had up was completely wrong so I went through and reworked it with what Dylan added down below and this should be right now:

So the problem asks us to show that $$\lim_{n\to\infty} \int_{-\infty}^{\infty} \cos(y)K_n(x-y) dy = \cos(x)$$ where $$K_n(x) = \begin{cases}
\frac{n}{2} & |x|\leq \frac{1}{n} \\
0 & \text{otherwise}
\end{cases}$$

I don't generally explain much but of course, if I skip too many steps or you have questions about anything, feel free to ask and I'll try to explain to the best of my abilities.
$$\lim_{n\to\infty} \int_{x-\frac{1}{n}}^{x+\frac{1}{n}} \frac{n}{2}\cos(y)dy$$
$$\lim_{n\to\infty} \frac{n}{2} \int_{x-\frac{1}{n}}^{x+\frac{1}{n}} \cos(y)dy$$
$$\lim_{n\to\infty} \frac{n}{2} \sin(y) \Bigg|_{y = x-\frac{1}{n}}^{y =x+ \frac{1}{n}}$$
$$\lim_{n\to\infty} \frac{n}{2}\bigg[ \sin\bigg(x+\frac{1}{n}\bigg) - \sin\bigg(x-\frac{1}{n}\bigg)\bigg]$$
I will interject here to say that I used the following trig identities/properties $\sin(-u) = -\sin(u)$ and $\sin(u\pm v) = \sin(u)\cos(v) \pm \cos(u)\sin(v)$ and that's how I got to these next few steps:
$$\lim_{n\to\infty} \frac{n}{2}\bigg[ \sin\bigg(\frac{1}{n}\bigg)\cos(x) + \cos\bigg(\frac{1}{n}\bigg)\sin(x) + \sin\bigg(\frac{1}{n}\bigg)\cos(x) - \cos\bigg(\frac{1}{n}\bigg)\sin(x)\bigg]$$
$$\lim_{n\to\infty} \frac{n}{2}\bigg[ 2\sin\bigg(\frac{1}{n}\bigg)\cos(x)\bigg]$$
$$\lim_{n\to\infty} n\cdot\sin\bigg(\frac{1}{n}\bigg)\cos(x)$$
This seems like a reasonable place to plug in $n = \infty$ into the limit but if we do that, we see that we get something that resembles $\infty\cdot0$ which means we have to manipulate the equation a bit to use L'Hopital's rule.
$$\lim_{n\to\infty} n\cdot\sin\bigg(\frac{1}{n}\bigg)\cos(x) = \lim_{n\to\infty} \frac{\sin\Big(\frac{1}{n}\Big)\cos(x)}{\frac{1}{n}} = \lim_{n\to\infty} \frac{\sin(n^{-1})\cos(x)}{n^{-1}}$$
Now that we can apply L'Hopital's rule, the equation becomes:
$$\lim_{n\to\infty}\frac{-n^{-2}\cos(n^{-1})\cos(x)}{-n^{-2}} = \cos(x)$$
The $-n^{-2}$ in the numerator and denominator cancel out and $\cos(n^{-1}) = \cos\big(\frac{1}{n}\big) = 1$ since $n\to\infty, \frac{1}{n} = 0$ and $\cos(0) = 1$.

So we've shown that $$\lim_{n\to\infty} \int_{x-\frac{1}{n}}^{x+\frac{1}{n}} \frac{n}{2}\cos(y)dy = \cos(x)$$
I realize that the steps where I turned the fractions into negative exponents were unnecessary but they make taking the respective derivatives easier.

In any case, that's how I worked this problem out and I feel silly asking how this relates to the heat kernel but any input would be appreciated!

It looks good overall, but I think there is a small issue with your bounds. The kernel is defined so that it is equal to $n/2$ when the argument is in the interval $[-1/n,1/n]$. In the convolution operation, we are integrating $K_n(x-y)$, so for the kernel to equal $n/2$, it is $x-y$ which must be between $-1/n$ and $1/n$. Explicitly,

$$K_n(x-y)=\frac{n}{2} \; \; \mathrm{when} \; \; -\frac{1}{n}\leq x-y \leq\frac{1}{n}.$$

This can be turned into an inequality for $y$, which will give the bounds for the integral. Caitlin has worked out a similar problem with the same kernel here.

Regarding the relation to the heat kernel, I believe it is this: The heat kernel is a solution to the heat equation (we showed this in class), and if we convolute an initial condition function $\varphi(x)$ with it, getting

$$\int_{-\infty}^{\infty} G(x-y,t) \varphi(y) \; d y,$$

this is still a solution of the heat equation (I think we also showed this in class). The heat kernel has the property that

$$\lim_{t \rightarrow 0} \int_{-\infty}^{\infty} G(x-y,t) \varphi(y) \; d y = \varphi(x).$$

This was engineered to be true because we want it to solve the initial condition $u(x,0) = \varphi(x)$. The kernel $K_n$ is a much simpler function which has the same sort of property when we convolute and take a limit

$$\lim_{n \rightarrow \infty} \int_{-\infty}^{\infty} K_n(x-y) f(y) \; dy = f(x).$$

Does that check out with you guys?

Hmmm, I see what you're saying but I was under the impression that n/2 is just an arbitrary numerical value so the (x - y) bit wouldn't affect it. I don't know if that addresses what you're saying so hopefully you understand what I mean, I'm not sure how to better explain it.

Thank you for the feedback though, Dylan, it's much appreciated! Your explanation of the relation makes perfect sense to me.

Maybe this is a better way to say it: $K_n(x)$ is a function of one variable. When we write $K_n(x-y)$, this has now become a function of two variables, one of which is the variable of integration, $y$. Thus, the kernel's value now depends on both $x$ and $y$:

$$K_n(x-y) = \begin{cases} \frac{n}{2} &\mathrm{if} \; |x-y|\leq \frac{1}{n} \\ 0 & \mathrm{otherwise} \end{cases}$$

The bounds of the integral change to where the kernel is nonzero, but now where the kernel is nonzero is when

$$|x-y|\leq \frac{1}{n}.$$

Another way of writing this is $\frac{-1}{n} \leq x - y \leq \frac{1}{n}$. Subtracting $x$ from each part of the inequality, this implies

$$-x-\frac{1}{n} \leq -y \leq -x + \frac{1}{n}.$$

We can multiply everything by $-1$ to get $+y$, but this will flip the inequality around (I always have to think about multiplying $1<2<3$ by $-1$ to double check myself here):

$$x+\frac{1}{n} \geq y \geq x - \frac{1}{n}.$$

Rearranging this so that it's in order of smaller to bigger, this says

$$ x-\frac{1}{n} \leq y \leq x +\frac{1}{n}.$$

So the interval where the kernel is not equal to zero is from $y=x-1/n$ to $y=x+1/n$, and these should be the bounds of the integral.

I think I might have accidentally messed up my own clarity before, but hopefully this is a bit better.

No, you were beyond clear, I royally screwed up. Thanks for catching my mistake, I fixed it! : )

Trig identities are going to kill me