### Author Topic: Quiz-B P2  (Read 4265 times)

#### Victor Ivrii ##### Quiz-B P2
« on: April 02, 2018, 08:26:18 PM »
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

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• Karma: 20 ##### Re: Quiz-B P2
« Reply #1 on: April 02, 2018, 09:46:14 PM »
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 ##### Re: Quiz-B P2
« Reply #2 on: April 04, 2018, 07:04:13 AM »
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 (*)
« Last Edit: April 04, 2018, 11:17:19 AM by Victor Ivrii »

#### Andrew Hardy

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« Reply #3 on: April 04, 2018, 11:25:54 AM »
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.
« Last Edit: April 04, 2018, 12:07:29 PM by Andrew Hardy »

#### Victor Ivrii ##### Re: Quiz-B P2
« Reply #4 on: April 04, 2018, 11:33:04 AM »
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

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« Reply #5 on: April 04, 2018, 11:41:15 AM »
Dr. Ivrii,

I believe I updated the induction to be more explicit
« Last Edit: April 04, 2018, 12:07:54 PM by Andrew Hardy »

#### Tristan Fraser

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• Karma: 11 ##### Re: Quiz-B P2
« Reply #6 on: April 04, 2018, 12:08:01 PM »
Here's my initial attempt for the induction:

$< \partial f, \phi (x)> = <-f, \partial\phi (x)>$ 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 ##### Re: Quiz-B P2
« Reply #7 on: April 04, 2018, 12:10:29 PM »
Again, how you do it without "we can show"?

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

#### Zhongnan Wu

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« Reply #8 on: April 08, 2018, 01:51:31 PM »
I got an idea to prove the formula * shown in the reply part.

#### Victor Ivrii ##### Re: Quiz-B P2
« Reply #9 on: April 08, 2018, 02:22:58 PM »
In (*) there is no $f$ (or, rather, very specific $f$)

#### Zhongnan Wu

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« Reply #10 on: April 08, 2018, 07:04:47 PM »
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 ##### Re: Quiz-B P2
« Reply #11 on: April 08, 2018, 07:44:48 PM »
, 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