# A derivation of the the Taylor expansion formula

I have tried, for long, to prove Taylor’s Theorem on my own. The only way that this proof is different from the hundreds of proofs online is that I have written $\int_0^a f'(x)dx$ as $\int_0^a f'(a-x)dx$. This solves a lot of the problems I was facing in developing the Taylor expansion.

Let $f\in C^{k+1}[0,a]$. Then we have

$f(a)-f(0)=\int_0^a{f'(x)dx}=\int_0^a{f'(a-x)dx}=xf'(a-x)|_0^a+\int_0^axf''(a-x)dx$.

We can easily calculate that $xf'(a-x)|_0^a=af'(0)$. Also,

$\int_0^axf''(a-x)dx=\frac{x^2}{2}f''(a-x)|_0^a+\int_0^a \frac{x^2}{2}f'''(a-x)dx$.

Clearly, $\frac{x^2}{2}f''(a-x)|_0^a=\frac{a^2}{2}f''(0)$.

Continuing in this fashion, we get that

$f(a)-f(0)=af'(0)+\dots+\frac{a^n}{n!}f^{(n)}(0)+\dots+\int_0^a\frac{x^k}{k!}f^{(k+1)}(x)dx$

Assuming that $f$ is smooth, if as $k\to\infty$, $\int_0^a \frac{x^k}{k!}f^{(k+1)}(x)dx\to 0$ then we can recover the standard Taylor expansion formula.