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I'm trying to understand the remainder in Taylor's theorem from this source: https://proofwiki.org/wiki/Taylor%27s_Theorem/One_Variable/Integral_Version

enter image description here

I don't understand the very last parts of the proof (the parts where the remainder $R_n$ is solved). Can someone explain why does it follow from:

$$\frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}+\int_a^x \frac{f^{(n+2)}(t)}{(n+1)!}(x-t)^{n+1} \;dt$$

that $$R_N = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}?$$

I think the changing between letters $a$ and $\xi$ confuses me...?

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1 Answer 1

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We use this result: there's $c\in(a,b)$ such that

$$\int_a^b f(x)g(x)dx=f(c)\int_a^b g(x)dx$$ so in our case we have

$$\int_a^x\frac{f^{(n+1)}(t)}{n!}(x-t)^ndt=f^{(n+1)}(\xi)\int_a^x\frac{(x-t)^n}{n!}dt=f^{(n+1)}(\xi)\frac{(x-a)^{n+1}}{(n+1)!}=:R_n$$

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  • $\begingroup$ +1 Thank you for your help! =) Could you please also tell me what is name of the result you used? I can then google it and search for proof etc. =) $\endgroup$
    – jjepsuomi
    Jul 15, 2014 at 18:56
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    $\begingroup$ The name is Second mean-value theorem for the integral. $\endgroup$
    – user63181
    Jul 15, 2014 at 23:02
  • $\begingroup$ Nice work, my friend! $\endgroup$
    – amWhy
    Jul 16, 2014 at 11:04

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