Is it always possible to express Taylor series with summation notation?

Question

Is it always possible to express the Taylor series with summation notation?

For example, for $f(x) = x^2 \cos(x^2)$, the series will be as:

$$f(x) = x^2 - \frac{360}{6!}x^6 + \frac{151200}{10!}x^{10}+\dots$$

and I can't find the correct summation notation.

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EDIT:

Thanks for the two great answers. One thing remains that I pick the correct answer, is it always true, or a conjecture?

Answer 1

It is always possible to express a Taylor series in summation notation, but maybe not in a form you'd like. This is because the definition of Taylor series is itself in summation notation. That is, we can always express the Taylor series of $f$ at $a$ (provided $f$ is infinitely differentiable at $a$) by: $$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x - a)^n.$$ Writing it more explicitly than this can be tricky (and I'm guessing sometimes impossible to do so with elementary functions of $n$).

In your case, we can find a nice closed form for the coefficients. Start with the series for $\cos$: $$\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}.$$ Then, substitute $x^2$ into the function: \begin{align*} \cos(x^2) &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}(x^2)^{2n} \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{4n}. \end{align*} Finally, multiply both sides by $x^2$: \begin{align*} x^2\cos(x^2) &= x^2\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{4n} \\ &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{4n + 2}. \end{align*} The Taylor series centred at $0$ is the only power series representation of a function centred at $0$, so the above series must be the Taylor series of $x^2 \cos(x^2)$ centred at $x = 0$.

If you want to find $f^{(m)}(0)$ for arbitrary $m$, note that the $x^m$ term is $0$ unless $m = 4n + 2$ for some $n$. So, if $m$ is not $2$ modulo $4$, then $f^{(m)}(0) = 0$. If $m = 4n + 2$ for some $n$, then $$\frac{f^{(m)}(0)}{m!} = \frac{(-1)^n}{(2n)!} \implies f^{(m)}(0) = (-1)^n \frac{(4n+2)!}{(2n)!}.$$

Answer 2

I realise it's a conjecture but I would argue in theory all functions can be approximated with Taylor series written as summation notation, since the purpose of a Taylor series is to approximate a function through an infinite sum of monomial terms (so a stupid-long polynomial). I would urge you to recheck your Taylor series for $cos(x^2)$, it appears you may have used a Taylor series for an inverse trigonometric function, I don't believe your coefficients are Bernoulli numbers but they sure are reminiscient of them.

Let's start off easy. We know the Taylor series for $cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$ all the way to infinity (tip: if you ever forget a common Taylor series approximation you can always find them here). Looking at our pattern we can deduce a Taylor series for $cos(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$. If we want the Taylor series for $cos(x^2)$ the process isn't much different: all we do is replace $x$ with $x^2$, obtaining $cos(x^2)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(2n)!}$. If I wanted to find the Taylor series for $x^2cos(x^2)$, all I have to do is multiply in a term of $x^2$ into the approximation of $cos(x^2)$. Now, $x^2cos(x^2)=x^2-\frac{x^6}{2!}+\frac{x^8}{4!}-...$, where each successive term has a power of $x^{4n+2}$ (i.e. when n=1 our power is $x^{4(1)+2}=x^6$). Therefore, the function $f(x)=x^2cos(x^2)$ has a Taylor series approximation expressed as $f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n+2}}{(2n)!}$.

Hope this helps!