please solve a 2013 th derivative question?

$ 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?

Solution 1:

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

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

Solution 2:

Hint:

Consider the Taylor expansion of $f$