Question

If $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x4}{4!}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_,\mathrm{then}\frac{d^{2}y}{d{x}^2}=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-...$

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=\frac{d}{dx}\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-...\right)$

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right)-\frac{d}{dx}\left(\frac{x^3}{3!}\right)+\frac{d}{dx}\left(\frac{x^4}{4!}\right)-...$

$\Rightarrow \frac{dy}{dx}=0-1+\frac{2x}{2!}-\frac{3x^2}{3!}+\frac{4x^3}{4!}-...$

$\Rightarrow \frac{dy}{dx}=-1+\frac{x}{1!}-\frac{x^2}{2!}+\frac{x^3}{3!}-...$

Again differentiating both sides with respect to x, we get

$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(-1+\frac{x}{1!}-\frac{x^2}{2!}+\frac{x^3}{3!}-...\right)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(-1\right)+\frac{d}{dx}\left(\frac{x}{1!}\right)-\frac{d}{dx}\left(\frac{x^2}{2!}\right)+\frac{d}{dx}\left(\frac{x^3}{3!}\right)-...$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=0+1-\frac{2x}{2!}+\frac{3x^2}{3!}-...$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...$


If $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_,\mathrm{then}\frac{d^{2}y}{d{x}^2}=\underline{1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...}.$