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$ f(x) = 6x^7\sin^2(x^{1000}) e^{x^2} $

Find $ f^{(2013)}(0) $

A math forum friend suggest me to use big O symbol, however have no idea what that is, so how does that helping?

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    $\begingroup$ Where did you find this problem? 2013 is an odd year to choose... :) $\endgroup$
    – apnorton
    Commented Dec 30, 2012 at 20:58
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    $\begingroup$ $\sin^2(x^{1000}) = x^{2000}+O(x^{4000})$ reduces the complexity of the computation considerably... $\endgroup$
    – WimC
    Commented Dec 30, 2012 at 21:04

2 Answers 2

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Note that,

$$ 6\,x^{7} \sin\left(x^{1000}\right)\sin\left(x^{1000}\right)e^{x^2} $$

$$ = 6\,x^{7} \left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left( x^{1000}-\frac{x^{3000}}{3!}+\dots \right)\left(1+\frac{x^2}{1!}+\frac{x^4}{4!}+\dots\right) $$

$$ = 6x^7x^{2000}\left( 1-\frac{x^{2000}}{3!} +\dots\right)^2\left(1+\frac{x^2}{1!}+\frac{x^4}{2!}+\dots\right) $$

$$ = 6x^{2007}\left(1+\frac{x^2}{1!}+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots\right)\left( 1-\frac{x^{2000}}{3!} +\dots\right)^2 $$

Now, it is clear that the coefficient of $x^{2013}$ is $1$, which implies that

$$ \frac{f^{(2013)}(0)}{(2013)!} = 1 \implies f^{(2013)}(0)=(2013)!. $$

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    $\begingroup$ Nice job there (+1) $\endgroup$ Commented Dec 31, 2012 at 20:56
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    $\begingroup$ @Chris'ssister: Thank you. $\endgroup$ Commented Dec 31, 2012 at 23:08
  • $\begingroup$ Best answer as usual.Mhenni $\endgroup$ Commented Sep 30, 2017 at 16:29
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Hint:

Consider the Taylor expansion of $f$

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    $\begingroup$ Short and sweet...+1 :-) $\endgroup$
    – amWhy
    Commented Dec 30, 2012 at 21:14

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