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I am taking a Calculus II course in college, and because we learned about power series today, I've been trying to work out a proof for Taylor's Remainder Theorem as a practice exercise. However, I've been stuck on part (b) for a while now, and I'm getting quite frustrated.

My solution for part (a):

$$f(x)-f(a) = \int_a^x f'(t) dx$$ $$f(x) = f(a) + \int_a^x f'(t) dx$$

My attempt for part (b) using integration by parts:

$$f(x) = f(a) + f'(x)x-f'(a)a-\int_a^x f''(t)t dt$$

I'm not sure where to go from there, or if my integration is even correct in that step. Can someone please at least give me a hint?

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    $\begingroup$ Induction should work. $\endgroup$ – hamam_Abdallah Oct 28 '16 at 22:06
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when you do the integration by parts you should have t-x as the anti derivative of 1 with respect to t not just t, because x is a constant with respect to t.

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There is a small trick, namely to make the choice of a power of $x-t$ for the antiderivative in the integration by parts: $$ f(x)-f(a)=\int_a^x 1 \cdot f'(t)dt = - \left[(x-t) f'(t) \right]_a^x + \int_a^x (x-t)f''(t) dt = \ ... \ {\rm etc}$$ You then end up with a Taylor expansion only involving derivatives at $a$.

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