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If we have the polynomial $$P(x-a)=f(a)+f'(a)(x-a),$$ I can't get how does differentiating the polynomial $k$ times we obtain $$P_k(x-a)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\ldots+\frac{f^k(a)}{k!}(x-a)^k$$

kindly help...

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  • $\begingroup$ The second is a Taylor polynomial, which suggests that something else should be differentiated. Could you check it? $\endgroup$ – Przemysław Scherwentke Nov 8 '14 at 7:03
  • $\begingroup$ @patang I think it's the $k^{th}$ degree taylor polynomial of $f$ at $a$.. $\endgroup$ – spectraa Nov 8 '14 at 7:06
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    $\begingroup$ It is not differentiating the polynomial $k$ times. It is just the Taylor expansion truncated at the $k^{th}$ term. $\endgroup$ – Claude Leibovici Nov 8 '14 at 7:11
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Here is another way to approach the problem. Consider, instead, the polynomial

$P(x-a)=f(a)+f^{\prime}(a)(x-a)+\int^{x}_{a}(x-t)f^{\prime\prime}(t) dt$.

Then on the latter term: $\int^{x}_{a}(x-t)f^{\prime\prime}(t) dt$ (using Taylor's theorem) apply integration by parts several times ($k$-times).

Namely, let $u=f^{\prime\prime}(t)$, then $du=f^{\prime\prime\prime}(t)$, and $dv=(x-t)dt$ so that $v=-\frac{1}{2}(x-t)^{2}$, and use by parts:

$uv-\int vdu.$

Thusly, continue with this procedure to obtain: $P(x-a)=f(a)+f^{\prime}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-t)^{2}+\cdots+\frac{f^{k}(a)}{k!}(x-t)^{k}.$

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  • $\begingroup$ thanks for the answer ...but why did you change the original polynomial to $P(x-a)=f(a)+f^{\prime}(a)(x-a)+\int^{x}_{a}(x-t)f^{\prime\prime}(t) dt$ ... $\endgroup$ – patang Nov 8 '14 at 9:03
  • $\begingroup$ Thank you. I did it so that you can see that you are not actually differentiating the polynomial. One only needs Taylor's theorem which is just integration by parts multiple times. $\endgroup$ – Rene Cabrera Nov 8 '14 at 9:05

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