Find $f^{(2016)}(0)$ if $f(x)=\sin(x^2)$.

From the Maclaurin series, $$\sin(x^2)=\sum_{n=0}^\infty\frac{(-1)^nx^{4n+2}}{(2n+1)!}$$

Comparing coefficient, $$\frac{f^{(j)}(0)}{j!}=\frac{(-1)^n}{(2n+1)!}$$

Does it mean that if $j$ is not a multiple of $4n+2$, then the coefficient of $j$ term is $0$, consequently the $j$th derivative at that point is $0$?

So to find $f^{(2016)}(0)$ , $2016=4n+2$ implies that $n=503.5$, not an integer, so $f^{(2016)}(0)=0$?

  • $\begingroup$ You mean $f^{(2016)}(x)$ and not $f^{2016}(x)$ right ? $\endgroup$ – Theodoros Mpalis Mar 29 '16 at 14:03
  • $\begingroup$ yes, the 2016th derivative, sorry for not stating clearly $\endgroup$ – Dave Clifford Mar 29 '16 at 14:03
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    $\begingroup$ @DaveClifford You are right. $\endgroup$ – Jean-Claude Arbaut Mar 29 '16 at 14:07
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    $\begingroup$ Yes, your answer is zero as you guess. $\endgroup$ – GEdgar Mar 29 '16 at 14:10
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    $\begingroup$ The coefficient of $x^{2016}$ in that series is zero. No need to mention "factorials" and whether they are "undefined". $\endgroup$ – GEdgar Mar 29 '16 at 14:21

Let $(a_j)_{j=0}^\infty$ be the coefficients, then $$ f^{(j)}(0) = j! a_j $$ $a_j$ is nonzero if $j = 4n + 2$ for some $n$, but that is not the case for $j = 2016$ since $2016 \equiv 0 \mod 4$, so $a_{2016} = 0$


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