Lf y-x-x2-x3-x4- -x-1- then x-

The correct option is A y/1 y y=x x^2+x^3 x^4...∞ ...(i) xy= x^2 x^3+x^4...∞ ...(ii) nbsp; Since |x| lt;1 Adding i and ii we get ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Algebra of Derivatives

lf y=x-x2+x...

Question

lf $y = x - x^{2} + x^{3} - x^{4} + \dots . \infty, | x | < 1$ , then $x =$

\frac{y}{y + 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y - \frac{1}{y}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{y}{1 - y}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y + \frac{1}{y}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

y = x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots . \infty

y - \frac{1}{2} y^{2} + \frac{1}{3} y^{3} - \frac{1}{4} y^{4} + \dots . \infty =

y = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots \infty

y + \frac{y^{2}}{2!} + \frac{y^{3}}{3!} + \dots =

\frac{x - y}{x} + \frac{1}{2} (\frac{x - y}{x})^{2} + \frac{1}{3} (\frac{x - y}{x})^{3} + \dots . =

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

\frac{d y}{d x} = y

x = {tanh}^{- 1} y

{log}_{e} (\frac{1 + y}{1 - y}) =

Join BYJU'S Learning Program