A derivation of the the Taylor expansion formula

Posted byayushkhaitan3437Posted inUncategorized

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.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush

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