Here is a Challenge Problem about the sum that converges to $\frac{\pi}{4}$.

## Proof

From class discussion we know that:

$\int_0^1 1-x^2+x^4-x^6+...+(-1)^Nx^{2(N+1)}dx=\frac{\pi}{4}-\int_0^1\frac{(-1)^{N+1}x^{2(N+1)}}{1+x^2}dx$

We call the part being subtracted from $\frac{\pi}{4}$ the Error since as n goes to infinity, the error becomes 0. Now we want to estimate the error.

We know that when x is between 0 and 1, $|\frac{(-1)^{N+1}x^{2(N+1)}}{1+x^2}|\le x^{2(N+1)}$

And therefore $\int_0^1|\frac{(-1)^{N+1}x^{2(N+1)}}{1+x^2}|dx\le\int_0^1x^{2(N+1)}dx=\frac{1}{2(N+1)+1}$

Now we want to find the average of the sum to N and the sum to N+1. For this, we simply add them together and divide by two. For the sake of space, let $S_N(x)=1-x^2+x^4-x^6+...+(-1)^Nx^{2(N+1)}$ and $S_{N+1}(x)=1-x^2+x^4-x^6+...+(-1)^Nx^{2(N+1)}+(-1)^{N+1}x^{2(N + 2)}$

We already know what $\int_0^1S_N(x)dx$ is, so now we have to figure out what $\int_0^1S_{N+1}(x)dx$ is.

By simply substituting N+1 for N in the first integration (since we're taking it to the N+1 terms now), we will see that $\int_0^1S_{N+1}(x)dx=\frac{\pi}{4}-\int_0^1\frac{(-1)^{N+2}x^{2(N+2)}}{1+x^2}dx$

Again we want to estimate the error for this integral on the interval x is between 0 and 1

$|\frac{(-1)^{N+2}x^{2(N+2)}}{1+x^2}|\le x^{2(N+2)}$

And therefore:

$\int_0^1|\frac{(-1)^{N+2}x^{2(N+2)}}{1+x^2}|dx\le\int_0^1x^{2(N+2)}dx=\frac{1}{2(N+2)+1}$

Again, for the sake of space, let $|E(N)|=\int_0^1\left |\frac{(-1)^{N+1}x^{2(N+1)}}{1+x^2} \right |dx\le\frac{1}{2(N+1)+1}$ and then we see that $|E(N+1)|=\int_0^1\left |\frac{(-1)^{N+2}x^{2(N+2)}}{1+x^2}\right |dx\le\frac{1}{2(N+2)+1}$

Now we're ready to average the two sums.

$\frac{\int_0^1S_N(x)dx+\int_0^1S_{N+1}(x)dx}{2}=\frac{\frac{\pi}{4}-E(N)+\frac{\pi}{4}-E(N+1)}{2}=\frac{\pi}{4}-\frac{E(N)+E(N+1)}{2}$

This shows us a new Error, which is in fact the average of the errors of the two sums. So now we want to estimate this error. Here you have to realize that one error is positive and one is negative. From this fact we realize that estimation of the abosolute value of the new error is half the difference of the absolute value of the other errors. Or,

$|ERROR|=\frac{|E(N)+E(N+1)|}{2}=\frac{|(|E(N)|)-(|E(N+1)|)|}{2}$

Using substitutions from estimations made earlier we see that:

$|\frac{(|E(N)|)-(|E(N+1)|)}{2}|\le|\frac{\frac{1}{2(N+1)+1}-\frac{1}{2(N+2)+1}}{2}|=|\frac{\frac{2N+5-2N-3}{(2N+3)(2N+5)}}{2}|=\frac{1}{4N^2+16N+15}$

Therefore, the error of averaging N and N+1 terms will be

$|ERROR|\le\frac{1}{4N^2+16N+15}$

So as can be seen, we can take the error from a first order polynomial of N to a second order, so it is therefore much, much more accurate to average the sums. In fact, in order to be guaranteed the same accuracy for averaging N = 1 and N = 2 series without averaging, you would have to go to the N = 16. That is a great deal more terms to add than simply averaging the sums of N = 1 and N = 2.

## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva