Question:
If $y=1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots .$ to $\infty$, then write $\frac{d^{2} y}{d x^{2}}$ in terms of $y$
Solution:
Given:
$y=1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\cdots \infty$
$\frac{d y}{d x}=0-1+\frac{2 x}{2 !}-\frac{3 x^{2}}{3 !}-\frac{4 x^{3}}{4 !}+\cdots \infty$
$\frac{d^{2} y}{d x^{2}}=0-0+1-\frac{2 x}{2 !}+\frac{3 x^{2}}{3 !}-\frac{4 x^{3}}{4 !}+\cdots \infty$
$=1-\frac{\mathrm{x}}{1 !}+\frac{\mathrm{x}^{2}}{2 !}-\frac{\mathrm{x}^{3}}{3 !}+\cdots \infty$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}$
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