# Is this formula new and useful? [duplicate]

Recently, I have been doing some work relating to the Taylor Series on this site. I decided to integrate $$f'(x)$$ with integration by parts instead of using the obvious definition: $$\int f'(x)dx=xf'(x)-\int xf''(x)dx=xf'(x)-\frac{x^2}{2}f''(x)+\int\frac{x^2}{2}f'''(x)dx=\sum^N_{k=1}\frac{(-1)^{k+1}x^kf^{(k)}(x)}{k!}+\int\frac{(-1)^{N+1}x^Nf^{(N)}(x)}{N!}dx$$ Now this looks a lot like the Maclaurin series, but there is just that factor of $$(-1)^{k+1}$$.

I have a few questions about this formula. Can it be used alternatively to Taylor's Series? Is it novel? And finally, what are the limits for $$f$$?

• This is exactly equivalent to Taylor's theorem with the integral remainder, the function $f$, and the integration bounds $\int_x^0$. In general, I'm sorry to say, but you're extremely unlikely to discover anything new in calculus (especially not by repeatedly applying an existing rule) since the field is so well-trodden by centuries of students. But that's not to say it isn't worthwhile exploring for the sake of your own interest and practice.
– Jam
Commented Oct 2, 2022 at 0:48

Usually this formula is expressed between $$0$$ and $$x$$, with $$f(x)$$ being expressed with values of $$f$$ and its derivatives on $$0$$.
Here that's the other way round, i.e. you express $$f(x)$$ with values of derivatives of $$f$$ on $$x$$. But $$0$$ is still invisibly there: it is the lower bound of the integration interval.
So it is really the same as the usual formula, except the direction is reversed, we go from $$x$$ to $$0$$ - which explains the $$(-1)^{k+1}$$.
I.e from $$f(b) = f(a) + (b-a) f'(a) + \frac {(b-a)^2} {2!} f''(a) + ...$$
using $$a=x$$ and $$b=0$$, we get
$$f(0) = f(x) - xf'(x) + \frac {x^2} {2!} f''(x) - ...$$,