Use a power series to approximate the definite integral to 6 decimal places, help

Question

Use a power series to approximate the definite integral to 6 decimal places

$$\int_0^{0.3} \frac{x^2}{1+x^4} dx$$

Answer 1

There is actually a fairly quick way to compute the power series. Remember the geometric sum: $$\sum_{n=0}^\infty r^n = \frac{1}{1-r}$$ if $|r| < 1$. This may not seem, at first, to be what we want, but do the following: $$\frac{x^2}{1+x^4} = x^2\frac{1}{1 - (-x^4)} = x^2 \sum_{n=0}^\infty (-x^4)^n = \sum_{n=0}^\infty (-1)^n x^{4n+2}$$ Integrate the series termwise. You'll get: $$\int_0^{0.3} \frac{x^2}{1+x^4}\, dx = \left. \sum_{n=0}^\infty (-1)^n \frac{x^{4n+3}}{4n+3}\right|_{0}^{0.3} = \sum_{n=0}^\infty (-1)^n \frac{(0.3)^{4n+3}}{4n+3}$$

Now, suppose $$S = \sum_{n=0}^\infty (-1)^n b_n$$ is an absolutely convergent sum, and then it is a fact that if we let $$S_N = \sum_{n=0}^N (-1)^n b_n$$ then the error $$|E_N| = |S - S_N| \leq b_{N+1}$$

So to finish the problem, just look for the smallest value of $N$ so that $\frac{(0.3)^{4N+3}}{4N+3} \leq 10^{-7}$. This would guarantee that the $N$th partial sum would be accurate to within 6 decimal places.

For the sake of completeness, you likely only need to add up the first 2 terms, but to be safe (and since the 3rd term is not terribly difficult to compute), you may want to go ahead and sum up the first 3 terms to be certain.

Answer 2

We need a power series representation of $$f(x)={x^2\over1+x^4}$$ that is valid on the interval $[0,.3]$. In fact we will find one that's valid on the interval $(-1,1)$.

We start with something familiar and similar to the function of interest: For $|x|<1$ $$\tag{1} {1\over 1-x} =1+x+x^2+\cdots. $$ Substituting $x=-u^4$ into $(1)$ gives $$\tag{2} {1\over 1+u^4}=1-u^4+u^8-u^{12}-\cdots; $$ which is valid for $|u|<1$.

Almost there. Multiplying both sides of $(2)$ by $u^2$ gives $$\tag{3} {u^2\over 1+u^4}=u^2-u^6+u^{10}-u^{14}-\cdots $$ which is valid for $|u|<1$.

Now that we have the series, we evaluate the integral by integrating the series representation term by term; this is valid since the radius of convergence of the series is 1, and since the region of integration is $[0,.3]\subset(-1,1)$: $$\eqalign{ \int_0^{.3} {u^2\over 1+u^4}\,du &= \int_0^{.3} u^2\,du- \int_0^{.3} u^6\,du+\int_0^{.3} u^{10}\,du- \cdots\cr &={u^3\over 3}\Bigl|_0^{.3} -{u^7\over 7}\Bigl|_0^{.3} +{u^{11}\over 11}\Bigl|_0^{.3} -\cdots\cr &= {(.3)^3\over 3} -{(.3)^7\over 7} +{(.3)^{11}\over 11} -\cdots } $$

But, what is the approximation? Where may we stop? Note that the series obtained after integrating is an alternating series and that$(.3)^{11}/11<1.62\cdot 10^{-7}$. By a standard result on estimating a convergent alternating series with a partial sum, the desired approximation is $$ \int_0^{.3} {u^2\over 1+u^4}\,du \approx {(.3)^3\over 3} -{(.3)^7\over 7} . $$