Exactly expressing integral as a sum

Question

Apparently (i.e. according to my professor), the following holds:$$\int_a^b f(x) dx = (b-a)\sum_{n=1}^\infty \sum_{m=1}^{2^n-1} (-1)^{m+1}2^{-n}f(a+m(b-a)2^{-n}).$$How would one go about proving such a formula?

Answer 1

We can break an alternating sum into the difference of the non-alternating sum and twice the sum of the even terms: $$ \begin{align} &\sum_{m=1}^{2^n}(-1)^{m-1}(b-a)2^{-n}f(a+m(b-a)2^{-n})\\ &=\sum_{m=1}^{2^n}(b-a)2^{-n}f(a+m(b-a)2^{-n}) -2\sum_{m=1}^{2^{n-1}}(b-a)2^{-n}f(a+2m(b-a)2^{-n})\\ &=\underbrace{\sum_{m=1}^{2^n}(b-a)2^{-n}f(a+m(b-a)2^{-n})}_{\text{Riemann Sum with $2^n$ partitions}} -\underbrace{\sum_{m=1}^{2^{n-1}}(b-a)2^{-n+1}f(a+m(b-a)2^{-n+1})}_{\text{Riemann Sum with $2^{n-1}$ partition}}\tag{1} \end{align} $$ The sum of $(1)$ telescopes: $$ \begin{align} &\sum_{n=1}^\infty\sum_{m=1}^{2^n}(-1)^{m-1}(b-a)2^{-n}f(a+m(b-a)2^{-n})\\ &=\int_a^bf(x)\,\mathrm{d}x-(b-a)f(b)\tag{2} \end{align} $$ The terms of $(2)$ where $m=2^n$ sum to $$ -(b-a)\sum_{n=1}^\infty2^{-n}f(b)=-(b-a)f(b)\tag{3} $$ Subtracting $(3)$ from $(2)$ yields $$ \sum_{n=1}^\infty\sum_{m=1}^{2^n-1}(-1)^{m-1}(b-a)2^{-n}f(a+m(b-a)2^{-n}) =\int_a^bf(x)\,\mathrm{d}x\tag{4} $$

Answer 2

Let $l = b-a$, now put multiplier $l$ inside the inner series and let's consider members of outer series for different $n$.
$n = 1:$ $$\sum_{m=1}^{1} (-1)^{m+1}\frac l2 f(a+\frac{lm}{2}) = \frac l2f(a + \frac l2)$$ $n = 2:$ $$\sum_{m=1}^{3} (-1)^{m+1}\frac l4 f(a+\frac{lm}{4}) = \frac l4f(a + \frac l4) - \frac l4f(a + \frac {2l}{4}) + \frac l4f(a + \frac {3l}{4})$$ Now consider partiam sums $S_n$ of the outer series. $$S_1 = \frac l2 f(a+\frac l2)$$ $$S_2 = \frac l2 f(a+\frac l2) + \frac l4f(a + \frac l4) - \frac l4f(a + \frac {2l}{4}) + \frac l4f(a + \frac {3l}{4}) = \frac l4f(a + \frac l4) + \frac l4f(a + \frac {2l}{4}) + \frac l4f(a + \frac {3l}{4})$$ From this we can see the form of $S_n$: $$S_n = \sum_{k=1}^{2^n-1}\frac{l}{2^n}f(a + \frac{kl}{2^n})$$ You may plot a graph to understand this expression better and I will explain it analytically: for every $n$ we split $[a,b]$ into $2^n$ parts and count sum of values of $f(x)$ in the start of each segment multiplied by the segment length. This means that this expression is a particular case of integral partial sum $\sigma_n = \sum_{k=1}^{n-1}f(\xi_k)\Delta x_k$, where $x_k$ are some points from $[a,b]$ and $\xi_k \in [x_k, x_{k+1}]$. In our case $\Delta x_k$ is independent on k and equals $\frac{l}{2^n}$ and $\xi_k = x_k$.
As $$\lim_{n \to \infty}\sigma_n = \int_a^bf(x)dx$$ and $$\lim_{n \to \infty}S_n = \sum_{n=1}^{\infty}\sum_{m=1}^{2^n-1} (-1)^{m+1}\frac {l}{2^n} f(a+\frac{lm}{2^n}),$$ we have proved that the initial series converges to $\int_a^bf(x)dx$.