let $\theta_{x}$ be the operator : $$\theta_{x}=x\frac{d}{dx}$$
What is the closed form for :
$$\theta_{x}^{n}\left[x\cos(x)\right]$$ $n$ being an positive integer.
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Try computing the case $n=1,2$ and understanding the resulting pattern – Avitus Dec 11 '13 at 20:43
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i did ... for $n=1,2,3; \text{and}; 4$ .. but i wasn't able to detect the pattern ! – Mohammad Al Jamal Dec 11 '13 at 20:44
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Let $\theta:=\theta_x$, $f(x):=x\cos(x)$ and $g(x)=x^2\sin x$. Then
$$\theta(f)=f-g, $$ $$\theta^2(f)=\theta(f)-\theta(g)=f-g-\theta(g), $$ $$\theta^3(f)=\theta(f)-\theta(g)-\theta^2(g)=f-g-\theta(g)-\theta^2(g); $$
in summary
$$\theta^n(f)=f-g-\theta(g)-\dots-\theta^{n-1}(g).$$
Avitus
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So the problem $\theta^n x\cos(x)$ is "reduced" to $\theta^n x^2\sin(x)$ ... – flonk Dec 11 '13 at 21:01
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This is the easiest form I found: I began to compute the operators $\theta^i(g)$ but I got tired quite soon...(addedunm: all depends on the specific problem of the OP and its level of detail) – Avitus Dec 11 '13 at 21:02
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You can have the closed form
$$ \sum_{k=0}^{n} {n\brace k} x^k \left(x\cos\left(x+\frac{k\pi}{2}\right)+k\sin\left(x+\frac{k\pi}{2}\right)\right),$$
where ${n\brace k}$ is the Stirling numbers of the second kind. For the $n$th derivative of $x\cos(x)$, see here.
Note: We used the identity
$$ (xD)^n = \sum_{k=0}^{n}{n\brace k} x^k D^k. $$
Mhenni Benghorbal
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I probably missunderstood, but for $n=1$ we should arrive at $xD(x\cos x)=x\cos x-x^2\sin x$, while using your formula I arrive at $-x\sin x+x\cos x$, as ${1\brace 0}=0$ and ${1\brace 1}=1$ – Avitus Dec 11 '13 at 21:13
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No no, I did not see $x$ in front of $\cos(x+...)$: so the power is $x^{k+1}$. Now it is clear to me: thank you for feedback.! – Avitus Dec 11 '13 at 21:15
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I did not know that decomposition involving the Stirling numbers of second kind: it is interesting indeed – Avitus Dec 11 '13 at 21:16
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@Avitus: No problem. You made me to check the formula which is good. Thanks for your comments. – Mhenni Benghorbal Dec 11 '13 at 21:23
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2@MhenniBenghorbal ... there is a typo in the formula above ... in the expression between brackets, n should be k. – Mohammad Al Jamal Dec 12 '13 at 18:30
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Something must be wrong as the result should be an odd function of $x$. Furthermore $\theta^2 x\cos = x\cos -3x^2\sin-x^3\cos$ but your formula gives $-x^2\cos-2x^2\sin-x\sin-x^3\cos$. – flonk Dec 13 '13 at 08:23
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