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 $f$?

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    $\begingroup$ 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. $\endgroup$
    – Jam
    Commented Oct 2, 2022 at 0:48

1 Answer 1


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
$f(0) = f(x) - xf'(x) + \frac {x^2} {2!} f''(x) - ...$,
which is your formula.


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