Question

Find the Maclaurin's expansion of (sin^-1 x)^2 by forming a second order differential equation and
using Leibnitz's rule.
​

Answer 1

Answer:

$\begin{aligned}& x^2-\frac{x^6}{3 !}+\frac{x^{10}}{5 !}-\ldots \\& \sum_{n=0}^{\infty} \frac{x^{4 n+2}}{(2 n+1) !} \cdot(-1)^n\end{aligned}$

Step-by-step explanation:

Step : 1  According to the Leibniz rule, if two functions f(x) and g(x) can be differentiated by n distinct ways, then their product f (x). Additionally differentiable n times is g(x). Leibniz's first law is (f (x). g(x))n=∑nCrf(n−r)(x).

Step : 2 First we must find the series for sin(x)

$\begin{aligned}& \text { let } f(x)=\sin (x) \\& f^{\prime}(0)=\sin (0)=0 \\& f^{\prime}(0)=\cos (0)=1 \\& f^{\prime \prime}(0)=-\sin (0)=0 \\& f^{\prime \prime}(0)=-\cos (0)=-1\end{aligned}$

Now we can apply to the macluarin series;

$f(x)=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0) x^2}{2 !}+\frac{f^{\prime \prime \prime}(0) x^3}{3 !}+\ldots$

$Hence \sin (x)=x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\ldots$

Step : 3 Hence for sin(x2) we replace each x by x2 in the series for sin(x)

$\begin{aligned}& \sin \left(x^2\right)=\left(x^2\right)-\frac{\left(x^2\right)^3}{3 !}+\frac{\left(x^2\right)^5}{5 !}-\ldots \\& =x^2-\frac{x^6}{3 !}+\frac{x^{10}}{5 !}-\ldots\end{aligned}$

What we can write in sigma summation notation as;

$\sum_{n=0}^{\infty} \frac{x^{4 n+2}}{(2 n+1) !} \cdot(-1)^n$

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