Quiz-B P2

[Victor Ivrii]

Simplify (write as the sum of $\delta (x)$, $\delta '(x)$, $\delta''' (x)$, ... with the numerical coefficients)
\begin{equation}
\sin(x)\delta'''(x).
\tag{1}
\end{equation}

[Jingxuan Zhang]

Observe for nice $\varphi$
$$(\sin(x) \varphi(x))'''=-\cos(x) \varphi(x)-3\sin(x)\varphi'(x)+3\cos(x)\varphi''(x)+\sin(x)\varphi'''(x).$$
Therefore
$$\langle\sin(x)\delta'''(x),\varphi(x)\rangle=\langle\delta'''(x),\sin(x) \varphi(x)\rangle=-\langle\delta(x),(\sin(x) \varphi(x))'''\rangle=\varphi(0)-3\varphi''(0).$$
So $$\sin(x)\delta'''(x)=\delta(x)-3\delta''(x).$$

[Victor Ivrii]

Typical error: calculated $\int \sin(x)\delta''' (x)\,dx$ thus applying $\sin(x)\delta''' (x)$ to the single test function $1$ instead of an arbitrary $\varphi$.

Another solution (some thought about it): $\displaystyle{\sin(x)=x-\frac{x^{3}}{3!}+\ldots}$ and therefore $\displaystyle{\sin(x)\delta''=x\delta'''-\frac{x^{3}}{3!}\delta'''}$ since $x^{k}\delta^{(n)}(x)=0$ for $k>n$. But one need to know that
$$
x^k\delta^{(n)}(x)= k! (-1)^k \delta^{(n-k)}(x)\qquad\text{for }\ n\ge k.
\tag{*}$$


Bonus: Prove (*)

[Andrew Hardy]

We have the definition that $$  (\partial f)(\phi) = -(f)(\partial\phi) $$

so $$ ( x^k \delta^{(1)}(x) ) = -k  x^{(k-1) }\delta^{(1-1)} (x) $$
assume this holds true for  n-1
 $$ x^{(k+1)} δ^{(n-1)}(x)=(k-1)! (−1)^{(k-1)}δ^{(n-1−k)}(x) $$
via induction  and by the definition $ x^{k+1} δ^{(n-1)}(x) = - x^k δ^{(n)}(x) $exactly, how? and what you make induction with respect to? V.I.
 $$ x^k δ^{(n)}(x)=  k!(−1)^kδ^{(n−k)}(x) $$
 I have to add the condition $$ n\geq k $$ because if n was less than, I'd have a negative derivative which is undefined.

[Victor Ivrii]

Actually I imposed $n\ge k$ but the negative derivative are defined as primitives: $\delta^{(-1)}(x)=\theta (x)$ and
$\delta^{(-k)}= \frac{x^{k-1}}{(k-1)!} \theta(x)$ for $k\ge 1$. We can even deal with $k\in \mathbb{C}$...

[Andrew Hardy]

Dr. Ivrii,

I believe I updated the induction to be more explicit

[Tristan Fraser]

Here's my initial attempt for the induction:

We start with n =  1 case:

$< \partial f, \phi (x)> = <-f, \partial\phi (x)> $ for the general function $\phi(x)$. We can integrate by parts to get from LHS to get to RHS. This also holds for $\phi(x) = \delta(x)$. Then, for the n = 2 case, we can redefine a new function:

$<\partial^2 f, \phi(x)> = <-F, \partial^2\phi(x)>$ for the general function $\phi(x)$. We could also just redefine the functions in such a manner that we get the expression:

$<\partial F, \phi(x)> = <-f,\partial \phi(x)>$, it also holds for $\delta(x)$

Then generalizing to order  n = k, we can show it holds. Where F is the primitive for f. We can always reduce it to just a first order derivation instead of kth order.

[Victor Ivrii]

Again, how you do it without "we can show"?

BTW, not $<,>$ but $\langle,\rangle$

[Zhongnan Wu]

I got an idea to prove the formula * shown in the reply part.

[Victor Ivrii]

In (*) there is no $f$ (or, rather, very specific $f$)

[Zhongnan Wu]

So at the last part, I let f(x)=x^n and take n times derivatives to get (*).
The first half is aimed at finding a general conclusion.

[Victor Ivrii]

, everything is much simpler : There is a very standard formula (basically binomial for derivatives)
$$
(uv)^{(n)}= \sum_{j=0}^n \frac{n!}{k!((n-j)!} u^{(j)}v^{(n-j)}.
$$
Therefore
$$
\langle x^k\delta^{(n)} ,\varphi\rangle = \langle \delta^{(n)} ,x^k\varphi\rangle=
(-1)^n(x^k \varphi)^{(n)} (0) = (-1)^n\Bigl[\sum_{j} (x^k)^{(j)} \varphi ^{(n-j)}\Bigr](0).
\tag{**}$$
However,  $j>k\implies (x^k)^{(j)}=0$, $j\le k\implies  (x^k)^{(j)}=\frac{k!}{(k-j)!}x^{k-j}$ and therefore in (**) only term with $j=k$ is not $0$. So we get
$$
(-1)^n k! \varphi^{(n-k)}(0) = (-1)^{k}k!\langle \delta^{(n-k)},\varphi\rangle.
$$

QED