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,

\[y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + . . . \infty \]
\[\text { Thus }, \]
\[ \Rightarrow \frac{d y}{d x} = - 1 + \frac{2x}{2!} - \frac{3 x^2}{3!} + \frac{4 x^3}{4!} . . . \infty \]
\[ = - 1 + x - \frac{x^2}{2!} + \frac{x^3}{3!} - . . . \infty \]
\[ \Rightarrow \frac{d^2 y}{d x^2} = 1 - \frac{2x}{2!} + \frac{3 x^2}{3!} - \frac{4 x^3}{4!} + . . . \infty \]
\[ = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + . . . \infty \]