Is there a more concise way?

Question

I am currently trying to prove Taylor's Theorem, this is the proof I am given:

I am finding this very long and hard to follow, can anyone give me a more concise, clear proof?

Thanks

Answer 1

We have by the fundemental theorem of calculus: $$f(x)-f(a)=\int_a^xf'(t)dt$$ and we integrate by parts we find $$f(x)-f(a)=-(x-t)f'(t)\Bigg|_a^x+\int_a^x(x-t)f''(t)dt=(x-a)f'(a)+\int_a^x(x-t)f''(t)dt$$ Now if we integrate again we find $$\int_a^x(x-t)f''(t)dt=-\frac{(x-t)^2}{2!}\Bigg|_a^x+\int_a^x\frac{(x-t)^2}{2!}f^{(3)}(t)dt=\frac{(x-a)^2}{2!}+\int_a^x\frac{(x-t)^2}{2!}f^{(3)}(t)dt$$ and the induction is clear.

Notice that at the $n$ steps we find the remainder on the form $$R_n=\int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)dt$$ and by the first mean value theorem for integration we find $\xi\in(a,x)$ such that $$R_n=\frac{(x-\xi)^n}{n!}f^{(n+1)}(\xi)(x-a)$$ and this is the Cauchy form of the remainder or we write the remainder on the form $$R_n=\int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)dt=f^{(n+1)}(\xi)\int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)dt=f^{(n+1)}(\xi)\frac{(x-a)^{n+1}}{(n+1)!}$$ and this is the Lagrange form of the remainder.