Let $f \in C^k[a,b]$.Show that for $x,x_0 \in [a,b]$, $$f(x)=\sum\limits_{j=0}^\mathbb{k-1}{{1\over j!}f^{(j)}(x_0)(x-x_0)^j}+{1\over k!}{\int_{x_0}^x f^{(k)}(t)(x-t)^k \,dt}$$ and after this use this result for proving the following forms of Taylor's theorem:

$i)$ if we assume the strongest condition of derivative $f \in C^{k+1}[a,b]$ show that $$f(x)=\sum\limits_{j=0}^\mathbb{k}{{1\over j!}f^{(j)}(x_0)(x-x_0)^j}+O(x-x_0)^{k+1}$$

$ii)$ if we do not have condition of higher derivative show that $$f(x)=\sum\limits_{j=0}^\mathbb{k}{{1\over j!}f^{(j)}(x_0)(x-x_0)^j}+o(x-x_0)^{k+1}$$

I am trying to find a complete answer(proof) for this parts of Taylors theorem but i did not manage to find anything since on internet.Any proof for this would be apreciated.


Set $$ \varphi(x)=\sum_{j=0}^{k-1}\frac{f^{(j)}(x)}{j!}(z-x)^j $$ then $$ φ'(x)=\frac{f^{(k)}(x)}{(k-1)!}(z-x)^{k-1} $$ and thus $$ φ(z)-φ(x)=\int_x^z\frac{f^{(k)}(t)}{(k-1)!}(z-t)^{k-1}\,dt $$ Now change the variables $x$ to $x_0$ and then $z$ to $x$ to obtain $$ f(x)=\sum_{j=0}^{k-1}\frac{f^{(j)}(x_0)}{j!}(x-x_0)^j+\frac{1}{(k-1)!}\int_{x_0}^x f^{(k)}(t)(x-t)^{k-1}\,dt $$ So it seems the powers and factorials in the error term in your first formula are somewhat off.

For the conclusions, take the (second) most simple case as illustration. $$ f(x)=f(x_0)+f'(\tilde x)(x-x_0),\qquad \tilde x\in [x_0,x], $$ gives directly $f(x)=f(x_0)+O((x-x_0)^1)$ and with a virtual zero \begin{align} f(x)&=f(x_0)+f'(x_0)(x-x_0)+\Bigl(f'(\tilde x)-f(x_0)\Bigr)(x-x_0)\\ &=f(x_0)+f'(x_0)(x-x_0)+o((x-x_0)^1) \end{align} for $f\in C^1(\Bbb R)$.

  • $\begingroup$ this is for the first part what about the two others? $\endgroup$ – Paris Lamp Sep 28 '15 at 16:38
  • $\begingroup$ Apply the mean value theorem of integration. Note that in formula a), the summation range should be 0..k and in formula b) 0..k+1. $\endgroup$ – LutzL Sep 28 '15 at 16:55
  • $\begingroup$ @LutzL your first answer is for showing $f(x)=\sum_{j=0}^{k-1}\frac{f^{(j)}(x_0)}{j!}(x-x_0)^j+\frac{1}{(k-1)!}\int_{x_0}^x f^{(k)}(t)(x-t)^{k-1}\,dt$? $\endgroup$ – Legolas Sep 28 '15 at 17:00
  • $\begingroup$ i had wrong values for the $i$ and $ii$ in sums.. $\endgroup$ – Legolas Sep 28 '15 at 17:04
  • $\begingroup$ the second part holds for $j=0$-$k$? because i wrote wrong the values before $\endgroup$ – Legolas Sep 28 '15 at 17:39

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