I'm asked to estimate the integral $\int_0^{0.2} \frac {1}{1+x^4}$ to six decimal places.

I know that this can be represented by the power series $\sum_0^\infty (-1)^n x^{4n}$. I also know that the alternating series estimation theorem states that the |error| for a particular estimation will always be less than the term we did not include, which is $b_{n+1}$. In this case, $b_{n+1}$ is $(-1)^{n+1} x^{4n+4}$.

However, I'm having difficulty actually estimating this integral. I was told by my professor that if you want to estimate something to $n$ decimal places, such as 2, your $b_{n+1}$ should have $n$ zeros. For example, $b_{n+1} = 0.001$ if you wish to estimate to 2 decimal places.

This means that in our case, $b_{n+1} = 0.0000001$, with 6 zeros after the decimal point.

If this were not a power series, I would estimate this by simply trying arbitrary $n$ values until I achieved the desired $b_{n+1}$, but there is an $x$ involved because it is a power series. Thus, if I try n = 5, for instance, I will have $b_{n+1} = x^{24}$, which has a variable and is not 0.000something.

How do I find the value of n for which this estimation is accurate within 6 decimal places so that I can use the n terms to then calculate the definite integral?

  • $\begingroup$ Or do I have to calculate the definite integral for each "n" I select and see if that is less than 0.0000001? $\endgroup$ – AleksandrH Oct 29 '16 at 22:32

If $z\in(0,1)$, $$ \sum_{n\geq 0}(-1)^n \frac{z^{4n}}{4n+1} $$ is a converging series with alternating signs. In particular, $$ \left|\sum_{n\geq N}(-1)^n \frac{z^{4n}}{4n+1}\right|<\frac{z^{4N}}{4N+1} $$ and if $z=\frac{1}{5}$ and $N=2$, the RHS is less than $8\cdot 10^{-8}$, so the LHS is less than $8\cdot 10^{-8}$ as well.

  • $\begingroup$ A bit confusing, but I eventually figured it out. I'll give you a +1 and accept anyway, but I personally don't like all the "z" variables and whatnot :P $\endgroup$ – AleksandrH Oct 30 '16 at 17:36
  • $\begingroup$ Actually, no, I got n = 1, and the correct answer is 0.199936 $\endgroup$ – AleksandrH Oct 30 '16 at 18:18

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