Proofs for Taylors theorem and other forms

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 $f\in C^1(\Bbb R)$.
• @LutzL your first answer is for showing $f(x)=\sum_{j=0}^{k-1}\frac{f^{(j)}(x_0)}{j!}(x-x_0)^j+\frac{1}{(k-1)!}\int_{x_0}^x f^{(k)}(t)(x-t)^{k-1}\,dt$? – Legolas Sep 28 '15 at 17:00
• i had wrong values for the $i$ and $ii$ in sums.. – Legolas Sep 28 '15 at 17:04
• the second part holds for $j=0$-$k$? because i wrote wrong the values before – Legolas Sep 28 '15 at 17:39