Answer to Question #263987 in Real Analysis for Ojugbele Daniel

Question #263987

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


1
Expert's answer
2021-11-11T12:16:12-0500

Maclaurin series expansion


11t=k=0tk,1<t<1\dfrac{1}{1-t}=\displaystyle\sum_{k=0}^\infin t^k, -1<t<1

Then


11+t2=11(t2)=k=0(t2)k\dfrac{1}{1+t^2}=\dfrac{1}{1-(-t^2)}=\displaystyle\sum_{k=0}^\infin (-t^2)^k

=k=0(1)kt2k,1<t<1=\displaystyle\sum_{k=0}^\infin (-1)^kt^{2k}, -1<t<1

Integrate both sides


0x11+t2dt=0xk=0(1)kt2kdx\displaystyle\int_{0}^x\dfrac{1}{1+t^2}dt=\displaystyle\int_{0}^x\displaystyle\sum_{k=0}^\infin (-1)^kt^{2k}dx

arctanx=k=0(1)kx2k+12k+1,1x1\arctan x=\displaystyle\sum_{k=0}^\infin \dfrac{(-1)^kx^{2k+1}}{2k+1}, -1\leq x\leq1

arctan1=k=0(1)k(1)2k+12k+1\arctan 1=\displaystyle\sum_{k=0}^\infin \dfrac{(-1)^k(1)^{2k+1}}{2k+1}π4=arctan1=k=0(1)k2k+1\dfrac{\pi}{4}=\arctan 1=\displaystyle\sum_{k=0}^\infin \dfrac{(-1)^k}{2k+1}

=113+1517+19111+...=1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+...

π=4(113+1517+19111+...)\pi=4(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+...)

π3.1416\pi\approx3.1416


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