For symbolic transformation of integrals and series I occasionally use this formula: $$\int_0^1f(x)\,dx=-\sum_{n=1}^\infty\sum_{m=1}^{2^n-1}\frac{(-1)^m}{2^n}f\left(\frac m{2^n}\right)\tag{$\diamond$}$$ I believe it holds for all piecewise-smooth functions $f$ of bounded variation defined on $(0,1).$ Is it correct? I also think this condition might be too tight and can be relaxed. Could you please suggest a wider natural class of functions for which $(\diamond)$ holds? Is there a known name for this formula? Could you provide some reference for it?
Asked
Active
Viewed 393 times
8
-
1A curious particular case of this formula is $\displaystyle,,\sum _{n=1}^\infty,\frac{2+2^{2^{-n}}}{\big(1+2^{2^{-n}}\big),2^n}=\frac1{\log2}.$ – Vladimir Reshetnikov Jun 17 '19 at 16:20
-
An obvious question would be: how or where did you find this formula? – Yuriy S Jun 17 '19 at 21:22
-
I can not recall. The formula is pretty intuitive — the partial sums are just increasingly precise results of the rectangle method and should converge to the correct value for non-pathological functions. – Vladimir Reshetnikov Jun 17 '19 at 23:22
-
3Intuitive? Is that you, Ramanujan? – Lázaro Albuquerque Jun 17 '19 at 23:31
-
@Izralbu Just try to write down a few first partial sums... – Vladimir Reshetnikov Jun 18 '19 at 00:36
-
1Welcome back Vladimir. – FDP Jun 18 '19 at 05:45
-
I wonder if the powers of $2$ present in this formula could be beneficial for binary (machine) implementation? – Yuriy S Jul 03 '19 at 06:15
-
2This formula is called "the generalized method of exhaustion" in this paper. – Paul Enta Jul 20 '19 at 20:18
-
@PaulEnta Thanks. Actually, you comment is a satisfactory answer to my question. – Vladimir Reshetnikov Jul 21 '19 at 17:08
-
Another link to the same paper: https://arxiv.org/abs/math/0011078 – Vladimir Reshetnikov Jul 21 '19 at 17:12
3 Answers
6
In this paper by A.A. Ruffa, this formula is shown to derive from the method of exhaustion of the ancient Greeks which was used to find "the area of a shape by inscribing inside it a sequence of polygons whose areas converge to the area of the containing shape".
Accordingly the author call it the generalized metthod of exhaustion. Two proofs of the formula are given and the case of an infinite boundary for the integral is discussed. Several decompositions of elementary functions are also given.
Paul Enta
- 14,113
2
Hint:
For $n=2$ $$\frac{-f(\frac12)}2+\frac{-f(\frac14)+f(\frac24)-f(\frac34)}4=- \frac{f(\frac14)+f(\frac24)+f(\frac34)}4$$ and $n=3$ $$\frac{-f(\frac12)}2+\frac{-f(\frac14)+f(\frac24)-f(\frac34)}4+\frac{-f(\frac18)+f(\frac28)-f(\frac38)+f(\frac48)-f(\frac58)+f(\frac68)-f(\frac78)}8\\= -\frac{f(\frac18)+f(\frac28)+f(\frac38)+f(\frac48)+f(\frac58)+f(\frac68)+f(\frac78)}8. $$
You see the pattern. The formula essentially computes the arithmetic mean on equidistant points.
The early form of the Romberg method ?
1
This is a neat formula that I had never seen before. I found another proof which I think is a little more straightforward than those given by Ruffa.
Proposition Let $f(x)$ be Riemann integrable on $[0,1]$, then
\begin{equation} \int_{0}^{1}f(x)\mathrm{d}x = \sum_{n=1}^{\infty}\sum_{k=1}^{2^{n}-1}\frac{(-1)^{k+1}}{2^{n}}f\left(\frac{k}{2^{n}}\right) \end{equation}
Proof:
Since $f(x)$ is an integrable function on $[0,1]$, then by definition we may write the integral of $f(x)$ as a Riemann sum. In particular partioning the interval $[0,1]$ into subintervals of length $1/2^n$, we may write
\begin{equation} \int_{0}^{1}f(x)\mathrm{d}x = \lim_{n\to\infty}\frac{1}{2^n}\sum_{k=1}^{2^n}f\left(\frac{k}{2^n}\right) \end{equation}
Letting
\begin{equation} g_n := \frac{1}{2^n}\sum_{k=1}^{2^n}f\left(\frac{k}{2^n}\right) \end{equation}
by telescoping we have
\begin{align} \int_{0}^{1}f(x)\mathrm{d}x= \lim_{n\to\infty}g_n &= g_0 + \sum_{n=0}^{\infty}\left(g_{n+1}-g_n\right) \\ &= f(1) + \sum_{n=0}^{\infty}\frac{1}{2^{n+1}}\left(\sum_{k=1}^{2^{n+1}}f\left(\frac{k}{2^{n+1}}\right)-2\sum_{k=1}^{2^n}f\left(\frac{k}{2^{n}}\right)\right) \end{align}
But this can be simplified using the fact that
\begin{equation} \sum_{k=1}^{2^{n+1}}f\left(\frac{k}{2^{n+1}}\right)-2\sum_{k=1}^{2^n}f\left(\frac{k}{2^{n}}\right) = \sum_{k=1}^{2^{n+1}}(-1)^{k+1}f\left(\frac{k}{2^{n+1}}\right) \end{equation}
Therefore
\begin{align} \int_{0}^{1}f(x)\mathrm{d}x &= f(1) + \sum_{n=1}^{\infty}\frac{1}{2^{n}}\sum_{k=1}^{2^{n}}(-1)^{k+1}f\left(\frac{k}{2^{n}}\right) \\ &= \sum_{n=1}^{\infty}\sum_{k=1}^{2^{n}-1}\frac{(-1)^{k+1}}{2^{n}}f\left(\frac{k}{2^{n}}\right) \end{align}
as desired.
FofX
- 1,041