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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: $C[f(x)]=-f(0)$

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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}

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