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


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


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.

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Graduate student

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