Question

If $x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+....$ to $\infty =y$, then $y+\frac{y^2}{2!}+\frac{y^3}{3!}+....$ to $\infty$ is equal to

• A. $-x$
• B. $x$
• C. $x+1$
• D. $x-2$

Answer 1

The correct option is B $x$

$y=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.....\infty \therefore \frac{dy}{dx}=1-x+x^2-x^3+.....\infty$

As we can see the above series is a GP. Thus applying the formula of infinite GP, we get;

$\frac{dy}{dx}=\frac{1}{1+x}$

Now let there be a function of $x$, $f\left(x\right)$ such that,

$f\left(x\right)=y+\frac{y^2}{2!}+\frac{y^3}{3!}+\frac{y^4}{4!}+.....\infty$

Our job is to find $f$;

$\therefore f^{\prime }=[1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+.....\infty ]\frac{dy}{dx}\frac{df}{dx}=[1+f]\frac{dy}{dx}\frac{df}{1+f}=\frac{dy}{dx}\cdot dx=\frac{dx}{1+x}$

Now integrate,

$\int \frac{df}{1+f}=\int \frac{dx}{1+x}\log (1+f)=\log (1+x)+C$

To find $C$ we need to put $x=0$, which implies $y=0$ which implies $f=0$. or $f\left(0\right)=0$

$\log (1+f(0\left)\right)=\log \left(1\right)+C0=0+CC=0$

$\therefore \log (1+f)=\log (1+x)1+f=1+xf\left(x\right)=x$