Approximating Integrals within an error with Maclaurin Series

Question

"Use Maclaurin's Series to approximate the integral to 3dp accuracy"

$$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}} $$

I was wondering if it is possible to solve this question by identifying out the number of terms needed to obtain a 3dp accuracy. I thought of using the Taylor Series Remainder Theorem, but it seems quite difficult with this binomial expression.

Answer 1

Yes it is possible:

The MacLaurin expansion of $$(1+x^2)^{-1/4}=1-\frac{x^2}{4}+\frac{5}{32}x^4+...$$ The numerical value of $$I=\int_{0}^{1/2} \frac{dx}{(1+x^2)^{1/4}}=0.490447$$ and $$\int_{0}^{1/2} (1-\frac{x^2}{4}+\frac{5}{32}x^4) dx= 0.409056.$$

Answer 2

Using the binomial expansion, we have $$ \frac{1}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} x^{2 n}$$ which, integrated gives $$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \frac{2^{-(2 n+1)}}{2 n+1} \binom{-\frac{1}{4}}{n}$$

So, summing $p$ terms, you look for $p+1$ such that $$R_p=\Big| \frac{2^{-(2 p+3)}}{2 p+3} \binom{-\frac{1}{4}}{p+1}\Big| \leq \epsilon$$

Using the representation in terms of the gamma function we have $$R_p=\frac{2^{-(2 p+3)}}{2 p+3} \frac{\Gamma \left(\frac{3}{4}\right)}{\Gamma (p+2) \left|\Gamma \left(-p-\frac{1}{4}\right)\right|}$$ and using the reflection formula this becomes $$R_p=\frac{2^{-(2 p+3)} \Gamma \left(\frac{3}{4}\right) \Gamma \left(p+\frac{5}{4}\right) \left|\sin \left(\left(p+\frac{5}{4}\right) \pi \right)\right|}{\pi (2 p+3) \Gamma (p+2)}=\frac{2^{-(2 p+\frac{7}{2})} \Gamma \left(\frac{3}{4}\right) \Gamma \left(p+\frac{5}{4}\right)}{\pi (2 p+3) \Gamma (p+2)}$$ Now, using Stirling approximations and continuing with Taylor series to $O\left(\frac{1}{p}\right)$,we have to solve for $p$ $$-2 p \log (2)-\frac{7 \log (p)}{4}+\log \left(\frac{\Gamma \left(\frac{3}{4}\right)}{16 \sqrt{2} \pi }\right)=\log(\epsilon)$$ and the solution is given in terms of Lambert function $$p=\frac{7 }{8 \log (2)}W\left(k \right)\qquad \text{where} \qquad k=\frac{2^{3/7} \log (2)}{7} \left(\frac{\Gamma \left(\frac{3}{4}\right)}{16 \sqrt{2} \pi \epsilon }\right)^{4/7}$$ Numerically, this gives $$p\sim 1.26236 \, W\left(\frac{0.0130926 } { \epsilon^{4/7}}\right)$$ Computed for $\epsilon=10^{-10}$, this would give $p=8.70$ while the exact solution would be $p=8.76$.