Answer to Question #263987 in Real Analysis for Ojugbele Daniel

Question

Find the maclaurin series expansion of f(x)= 1/1+x hence or otherwise find the expansion of 1/1+x^2 by integrating both sides of your result, find the series expansion of arctanx, by also putting x=1 find the expression π/4 hence show that π=3.142

Answer 1

Maclaurin series expansion

$$\dfrac{1}{1-t}=\displaystyle\sum_{k=0}^\infin t^k, -1<t<1$$

Then

$$\dfrac{1}{1+t^2}=\dfrac{1}{1-(-t^2)}=\displaystyle\sum_{k=0}^\infin (-t^2)^k$$

$$=\displaystyle\sum_{k=0}^\infin (-1)^kt^{2k}, -1<t<1$$

Integrate both sides

$$\displaystyle\int_{0}^x\dfrac{1}{1+t^2}dt=\displaystyle\int_{0}^x\displaystyle\sum_{k=0}^\infin (-1)^kt^{2k}dx$$

$$\arctan x=\displaystyle\sum_{k=0}^\infin \dfrac{(-1)^kx^{2k+1}}{2k+1}, -1\leq x\leq1$$

$$\arctan 1=\displaystyle\sum_{k=0}^\infin \dfrac{(-1)^k(1)^{2k+1}}{2k+1}$$ $$\dfrac{\pi}{4}=\arctan 1=\displaystyle\sum_{k=0}^\infin \dfrac{(-1)^k}{2k+1}$$

$$=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+...$$

$$\pi=4(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+...)$$

$$\pi\approx3.1416$$