Find $f^{(97)}(0)$ of Maclaurin series for $f(x) = x^2\cos{9x}$

Find $f^{(97)}(0)$ of Maclaurin series for $f(x) = x^2\cos{9x}$

I tried finding for the Maclaurin series for $f(x) = x^2\cos{9x}$ and I got $$\sum_{n=0}^\infty \frac{(-1)^n(81)^nx^{2n+2}}{(2n)!}$$
I'm not sure if that's correct. If that is true, or if wrong (please kindly correct it), I want to get the 97th derivative of that evaluated at 0. Thank you!

• Oh yeah, my mistake. Thank you for noticing that – Marcus Mar 19 '13 at 8:55
• So you can conclude $f^{(97)}(0)=0$. – user63181 Mar 19 '13 at 8:57
• We don't need to calculate at all. Our function is an even function, infinitely differentiable. Any odd-numbered derivative of an even function is an odd function, and therefore $0$ at $0$. – André Nicolas Mar 19 '13 at 9:12

In this case, take the first few derivatives and discern a pattern. In this case, for $n \ge 2$, even derivatives will be nonzero at $x=0$, while odd derivatives will be zero. Why?

Consider the first few derivatives of the function $f(x) = x^2 \cos{a x}$:

$$f'(x) = -a x^2 \sin{a x} + 2 x \cos{a x}$$

$$f''(x) = -a^2 x^2 \cos (a x)-4 a x \sin (a x)+2 \cos (a x)$$

$$f'''(x)=a^3 x^2 \sin (a x)-6 a^2 x \cos (a x)-6 a \sin (a x)$$

Note that the factors $x^2 \cos{a x}$, $x \sin{a x}$, and $\cos{a x}$ appear in even derivatives, and $x^2 \sin{a x}$, $x \cos{a x}$, and $\sin{a x}$ appear in odd derivatives. This will be the case no matter what.

To see this, take the second derivative of $A x^2 \cos{a x} + B x \sin{a x} + C \cos{a x}$: you will get another linear combination of these terms. Same for $A x^2 \sin{a x} + B x \cos{a x} + C \sin{a x}$. Therefore $f^{(97)}(0) = 0$ because $97$ is odd.

• Oh yes, I just realized that cosine is an "even function". T But how about if we're looking for the 96th derivative? I just want to figure out how to find for the f^(n) (0) for any given maclaurin series. – Marcus Mar 19 '13 at 9:08
• That's a different, tougher problem. In this case, I would form a recurrence relation between successive derivative values at $x=0$ and solve the recurrence. It will be a very large number for the $96$th derivative. – Ron Gordon Mar 19 '13 at 9:16
• Anyway, thank you for the help! – Marcus Mar 19 '13 at 9:20

With a few tricks, this is quite easy. The Taylor expansion at $0$ of $\cos x$ is $$\sum _{n=0}^\infty \frac{(-1)^n\cdot x^{2n}}{(2n)!}$$ and so (it easily follows) that the Taylor expansion at $0$ for $\cos (9x)$ is obtained from that of $\cos x$ by substituting $9x$ for $x$: $$\sum _{n=0}^\infty \frac{(-1)^n\cdot 81\cdot x^{2n}}{(2n)!}$$ and now it further (easily) follows that the Taylor expansion at $0$ of $x^2\cdot \cos (9x)$ is obtained from that of $\cos (9x)$ by multiplying term by term by $x^2$: $$\sum _{n=0}^\infty \frac{(-1)^n\cdot 81\cdot x^{2n+2}}{(2n)!}$$

and now derivatives are performed term by term (as for any power series within its radius of convergence). The radius of convergence for the result above is positive so we can use it to compute derivaties at $0$. As all monomials in the result above are even, any odd derivative will result in all monomials having odd degree. So, the free coefficient will be $0$, and so upon evaluation at $x=0$, the 97th derivative will be $0$. (This is no surprise of course as the original function $x^2\cdot \cos (9x)$ is even.)

$f^{(97)}(0)$ is a coefficient at $x^{97}$ multiplied by $(97)!$ , since $$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n =\sum_{n=0}^\infty {a_nx^n} \Rightarrow f^{(97)}(0)=n!a_n (=0)$$