Compute Maclaurin expansion of $f(x)=e^{5x^2}$ and find a type for $f^{(n)}(0)$

Question

I'm practising to improve my skills at solving Taylor-Maclaurin Expansion problems and, currently, I'm trying to solve the following:

1) Based on the Maclaurin expansion of $e^x$, write $f(x)=e^{5x^2}$ as a power series around $x=0$.

2) For every $n \in \mathbb{N}$, find a type for $f^{(n)}(0)$.

My Attempt:

1. Using the known expansion of $e^x$, $f(x)$ can be written as: $$ e^x=\sum_{k=0}^{\infty}{\frac{x^k}{k!}}\Rightarrow f(x)=\sum_{k=0}^{\infty}{\frac{(5x^2)^k}{k!}} =\sum_{k=0}^{\infty}{\frac{5^k x^{2k}}{k!}} =1+5x^2+\frac{25x^4}{2}+\frac{125x^6}{6}+O(x^7) $$

2. I have calculated the first 6 derivatives of $f(x)$:

   • $f^{(1)}(x)=e^{5x^2}(0+10x)$
   • $f^{(2)}(x)=e^{5x^2}(10+0x+100x^2)$
   • $f^{(3)}(x)=e^{5x^2}(0+300x+0x^2+10^3 x^3)$
   • $f^{(4)}(x)=e^{5x^2}(300+0x+6\cdot10^3x^2+0x^3+10^4 x^4)$
   • $f^{(5)}(x)=e^{5x^2}(0+1.5\cdot10^4 x+0x^2+10^5 x^3+0x^4+10^5 x^5)$
   • $f^{(6)}(x)=e^{5x^2}(1.5\cdot10^4+0x+4.5\cdot10^5x^2+0x^3+1.5\cdot10^6 x^4+0x^5+10^6x^6)$

   
   Then, I calculated the value of the derivatives for $x=0$:

   • $f^{(1)}(0)=0$
   • $f^{(2)}(0)=10$
   • $f^{(3)}(0)=0$
   • $f^{(4)}(0)=300$
   • $f^{(5)}(0)=0$
   • $f^{(6)}(0)=15000$

   
   By observing the above results for, I have come up with: $f^{(n)}(0)=f^{(n-2)}(0)\cdot(n-1)$.

Question:

The current type I've found for $f^{(n)}(0)$ is based on instinct and observation. Is there a better type or, if not, is there a more rigorous way to find this type?

Answer 1

Look back at the definition of the MacClaurin series $$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = \sum_{k=0}^\infty \frac{5^k}{k!}x^{2k} $$

Since $f(x)$ is even, $f^{(n)}(0) = 0$ if $n$ is odd. Otherwise for $n=2k$ $$ f^{(n)}(0) = \frac{5^kn!}{k!} $$

Answer 2

Since you have McLaurain expansion, you directly get that $$\frac{f^{(n)}(0)}{n!}=\begin{cases}\frac{5^{n/2}}{(n/2)!}&n\in 2\mathbb N\\ 0&n\in2\mathbb N+1\end{cases}$$