Finding the $n$th derivative of a function when $x=0$ using Maclaurin series.

Question

CONTEXT: Uni question made up by lecturer

So I have found the Maclaurin series of $f(x)=x^2e^{x^2}$ to be $\sum_{n=0}^\infty \frac{x^{2n+2}}{n!}$ which can be rewritten as $\sum_{n=0}^\infty x^{n+2}\frac{x^n}{n!}$, which means $f^{(n)}(0)=x^{n+2}$.

From this, I've found $f^{(6)}(0)=x^{6+2}=x^{8}$, which I know to be incorrect as a calculator will tell you that $f^{(6)}(0)=360$.

Can someone please direct me as to where I went wrong? Thanks!

Answer 1

It is not correct to say that $\sum_{n=0}^\infty x^{n+2}\frac{x^n}{n!}$ means that $f^{(n)}(0)=x^{n+2}$. Note that $f^{(n)}(0)$ is supposed to be a number. On the other hand,$$\sum_{n=0}^\infty\frac{x^{2n+2}}{n!}=\sum_{n=1}^\infty\frac{x^{2n}}{(n-1)!}.$$So, $f^{(n)}(0)=0$ if $n$ is odd or if $n=0$. On the other hand, if $n$ is even and greater than $0$, then$$\frac{f^{(n)}(0)}{n!}=\frac1{\left(\frac n2-1\right)!}.$$In other words,$$f^{(n)}(0)=\frac{n!}{\left(\frac n2-1\right)!}.$$In particular,$$f^{(6)}(0)=\frac{6!}{2!}=\frac{720}2=360.$$

Answer 2

The equation $$ f^{(n)}(0) = x^{n+2} $$ does not make sense. The left side is a number. The right side is a function of $x$.

I suggest you start without the $\Sigma$ notation and write the power series with an ellipsis $$ \begin{align} f(x) & = x^2\left( 1 + x^2 + \frac{1}{2!}(x^2)^2 + \frac{1}{3!}(x^2)^3 + \cdots \right) \\ & = x^2 + x^4 + \frac{1}{2!}x^6 + \frac{1}{3!}x^8+ \cdots \end{align} $$ Then you can see that the $n$th derivative at $x=0$ is $0$ when $n$ is odd or $0$. When $n=2k > 0$ is even it's $n!/(k-1)!$.

Answer 3

$\sum_{n=0}^{\infty}x^{n+2}\frac{x^n}{n!} = \sum_{n=0}^{\infty}f^{(n)}(0)\frac{x^n}{n!}$ does not imply that $x^{n+2} = f^{(n)}(0)$.

The $x^{2n+2}$ term in your sum is actually a way to filter out the fact that for all odd n: $f^{(n)}(0) = 0$.

This becomes clear if you look at the actual expansion of the series in the form of $\sum_{n=0}^{\infty}f^{(n)}(0)\frac{x^n}{n!}$

$x^2e^{x^2} = \frac{0}{0!}x^0 + \frac{0}{1!}x^1 + \frac{2}{2!}x^2 + \frac{0}{3!}x^3 + \frac{24}{4!}x^4 + \frac{0}{5!}x^5 + \frac{360}{6!}x^6 + \frac{0}{7!}x^7 + \frac{6720}{8!}x^8 + \frac{0}{9!}x^9 + \frac{151200}{10!}x^{10} + \dots$ $x^2e^{x^2} =\frac{2}{2!}x^2 + \frac{24}{4!}x^4 + \frac{360}{6!}x^6 + \frac{6720}{8!}x^8 + \frac{151200}{10!}x^{10} + \dots$

This sum happens to be $\sum_{n=0}^{\infty}\frac{x^{2n+2}}{n!}$ but with a different 'step size'.

Answer 4

Here's the way I think about it: note that a Maclaurin series is essentially:

$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}\cdot x^n= f(0)+\frac{f^{(1)}(0)}{1!}\cdot x+\frac{f^{(2)}(0)}{2!}\cdot x^2+\ldots$$

The Maclaurin Series that you computed is:

$$x^2e^x = \sum_{n=0}^\infty \frac{x^{2n+2}}{n!} = x^2+\frac{x^4}{1!}+\frac{x^6}{2!}+\ldots $$

We observe that the coefficient of the $x^6$ term corresponds to $\frac{f^{(6)}(0)}{6!}$. Thus we have that $$\frac{f^{(6)}(0)}{6!} = \frac{1}{2!}$$ $$f^{(6)}(0) = \frac{6!}{2!} = \frac{720}{2} = 360$$ So $f^{(6)}(0)=360$.