TITLE: Is this derivation right?
QUESTION [5 upvotes]: I was toying around with integration by parts and I found this formula. It works for all examples I have tried but I just want someone besides me to check the derivation: $$f(x)+C=\int f'(x)dx=xf'(x)-\int f''(x)xdx=xf'(x)-\frac{x^{2}}{2}f''(x)+\int f'''(x)\frac{x^{2}}{2}dx=xf'(x)-\frac{x^{2}}{2}f''(x)+\frac{x^{3}}{6}f'''(x)-\int f''''(x)\frac{x^{3}}{6}=\sum_{n=1}^{\infty}\frac{f^{(n)}(x)x^{n}(-1)^{n-1}}{n!}$$
Now use this formula for the inegral of f(x) and take the derivative of the sum to get rid of the constant:
$$\sum_{n=1}^{\infty}\frac{\frac{d}{dx}f^{(n-1)}(x)x^{n}(-1)^{n-1}}{n!}=\sum_{n=1}^{\infty}\frac{f^{(n)}(x)x^{n}(-1)^{n-1}+nf^{(n-1)}(x)x^{n-1}(-1)^{n-1}}{n!}$$
so:
$$f(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}(x)x^{n}(-1)^{n-1}+nf^{(n-1)}(x)x^{n-1}(-1)^{n-1}}{n!}$$
You can also solve for the constant denoted $C[f(x)]$:
$$C[f(x)]=\sum_{n=1}^{\infty}\frac{f^{(n-1)}(x)x^{n-1}(-1)^{n}}{(n-1)!}$$ Take the limit as x approaches 0 and you can see that: $C[f(x)]=-f(0)$

REPLY [5 votes]: What you have found is basically an inverted form of the Taylor expansion. Instead of expanding $f(x)$ in terms of $f^{(n)}(0)$, you can get this expression by writing $f(0)$ as an expansion in terms of $f^{(n)}(x)$.
\begin{align}
    f(0)&=f(x)+f'(x)(0-x)+\frac{f''(x)}{2}(0-x)^2+...\\
    f(0)&=f(x)-f'(x)x+\frac{f''(x)}{2}x^2+...\\
    -f(x)&+f(0)=-f'(x)x+\frac{f''(x)}{2}x^2+...\\
    f(x)&-f(0)=f'(x)x-\frac{f''(x)}{2}x^2+...
\end{align}