Question

If $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}$.....to ∞, then write $\frac{d^{2}y}{d{x}^2}$in terms of y.

Answer 1

Here,
$$\begin{gathered} y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}+...\infty \\ \mathrm{Thus}, \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=-1+\frac{2x}{2!}-\frac{3x^2}{3!}+\frac{4x^3}{4!}...\infty \\ =-1+x-\frac{x^2}{2!}+\frac{x^3}{3!}-...\infty \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=1-\frac{2x}{2!}+\frac{3x^2}{3!}-\frac{4x^3}{4!}+...\infty \\ =1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...\infty \\ =y \end{gathered}$$