[See Solution] (a) Find the Taylor series expansion for y= sin ^-1 x where sin ^-1 x=∫_0^x (d t)/(√1-t^2), x ∈ [-1,1] (b) Let y= sin ^-1
Question: (a) Find the Taylor series expansion for \(y=\sin ^{-1} x\) where
\[\sin ^{-1} x=\int_{0}^{x} \frac{d t}{\sqrt{1-t^{2}}}, x \in[-1,1]\](b) Let \(y=\sin ^{-1} x\). Observe that at \(x=0, y=0\) and at \(x=1, y=\) \(\pi / 2\). Use the series expansion in part (a) to obtain an infinite series expansion of
\[S=\int_{0}^{1} \sin ^{-1} x \int_{0}^{x} \frac{d t}{\sqrt{1-t^{2}}} d x\]You will have to integrate by parts to obtain a recurrence formula for the integrals. You may check the values in Mathematica.
(c) Show that the sum of the series in part (b) is
\[S=\frac{\pi^{2}}{8}\]by evaluating \(S=\int_{0}^{\pi / 2} y d y\)
(d) Deduce Euler's result that
\[\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}\]
Deliverable: Word Document 
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![[See Solution] Show that K(x, t)=(1)/(√4 π](/images/solutions/MC-solution-library-38693.jpg "[See Solution] Show that K(x, t)=(1)/(√4 π t) e^-x^2 /")
[See Solution] Show that K(x, t)=(1)/(√4 π t) e^-x^2 /
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