Induction Proof of Taylor Series Formula

Question

I'm attempting to prove a formula for the taylor series of function from a differential equation. The equation is $$f(0)=1$$ $$f'(x) = 2xf(x)$$ I have found empirically that $$f(x) = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!}$$ I need to prove that this general formula works via induction.

Here is my attempt!

$$\mathrm{Show} \qquad1+x^2+\frac{x^4}{2}+\frac{x^6}{6}+...+\frac{x^{2k}}{k!}=\sum_{n=0}^k \frac{f^{(2n)}(0)x^{2n}}{(2n)!}$$

Prove true for $k=0$

$$1=\frac{f^{(0)}(0)x^{0}}{(0)!}$$ $$1 = 1 \ \checkmark$$

Assume true for $k=c$

$$1+x^2+\frac{x^4}{2}+...+\frac{x^{2c}}{c!}=\sum_{n=0}^c \frac{f^{(2n)}(0)x^{2n}}{(2n)!}$$

Prove true for $k=c+1$

$$\begin{align} 1+x^2+\frac{x^4}{2}+...+\frac{x^{2c}}{c!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1} \frac{f^{(2n)}(0)x^{2n}}{(2n)!} \\ \sum_{n=0}^c \frac{f^{(2n)}(0)x^{2n}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1} \frac{f^{(2n)}(0)x^{2n}}{(2n)!} \\ \frac{x^{2c+2}}{(c+1)!}&=\frac{f^{(2c+2)}(0)x^{2c+2}}{(2c+2)!} \end{align}$$

From there I don't know how to proceed. Maybe I shouldn't have broken apart the sum on the right? Any help would be appreciated.

Note: I know the differential is easily separable and solvable, but the project involves comparing the solutions

Answer 1

Note: This proof relies on a formula for the $n^{th}$ derivative of $f(x)$ that can be found HERE

The proof is rather extensive, and would roughly quadruple the length of this answer.

Proof of The Taylor Expansion of $f(x)$ $$\mbox{Prove true that } \ 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2k}}{k!}=\sum_{n=0}^{k}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}$$

By Induction:

$\circ \text{ Prove true for }k=0$ $$\frac{f^{(0)}(0)x^{0}}{0!} \\ 1=1 \ \checkmark$$

$\circ \text{ Assume true for }k=c$ $$ 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2c}}{c!}=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}$$

$\circ \text{ Prove true for }k=c+1$

\begin{equation} \begin{split} 1 + x^2 + \frac{x^4}{4} + \frac{x^6}{6} + \cdots + \frac{x^{2c}}{c!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \\ \sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!}&=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \mbox{By I.H} \\ =\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(c+1)!} \cdot \frac{(2c+2)!}{(2c+2)!} \\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot \frac{(2c+2)!}{(c+1)!} \\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot \frac{(2c+2)!}{(\frac{2c+2}{2})!} \\=\sum_{n=0}^{c}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!}+\frac{x^{2c+2}}{(2c+2)!} \cdot f^{(2c+2)}(0) \mbox{ From linked proof} \\=\sum_{n=0}^{c+1}\frac{f^{(2n)}(0)x^{(2n)}}{(2n)!} \checkmark \end{split} \end{equation}

$$f(x)=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!} \mathrm{Q.E.D}$$

Answer 2

Once you have your formula, the proof is straightforward.

If $f(x) = \sum_{k=0}^{\infty}\frac{x^{2k}}{k!} $, then

$\begin{array}\\ f'(x) &= \sum_{k=0}^{\infty}\frac{2kx^{2k-1}}{k!}\\ &= \sum_{k=1}^{\infty}\frac{2kx^{2k-1}}{k!} \qquad\text{since the }k=0\text{ term is zero}\\ &= \sum_{k=1}^{\infty}\frac{2x^{2k-1}}{(k-1)!} \qquad\text{cancelling }k\\ &= \sum_{k=0}^{\infty}\frac{2x^{2k+1}}{k!} \qquad\text{shifting }k\\ &= \sum_{k=0}^{\infty}\frac{2xx^{2k}}{k!}\\ &= 2x\sum_{k=0}^{\infty}\frac{x^{2k}}{k!}\\ &=2xf(x)\\ \end{array} $,