Question

if Mod X is less than 1 Y is equal to X minus x square + X power 3 minus x power 4 so on infinity then the value of x in terms of y is​

Answer 1

SOLUTION

TO DETERMINE

The value of

$\sf{y = x - {x}^{2} + {x}^{3} - {x}^{4} + ...... \infty } \: \: \: \: where \: |x| < 1$

FORMULA TO BE IMPLEMENTED

In an infinite Geometric series with first term = a and common ratio = r with | r | < 1

Then

$\displaystyle \sf{a +ar + a {r}^{2} + ..... \: \infty = \frac{a}{1 - r} }$

EVALUATION

Here it is given that

$\sf{y = x - {x}^{2} + {x}^{3} - {x}^{4} + ...... \infty } \: \: \: \: where \: |x| < 1$

This is an infinite Geometric series with first term = x and common ratio = - x

Now

$\sf{ | - x| = |x| < 1}$

Hence the given infinite Geometric series is convergent

$\sf{y = x - {x}^{2} + {x}^{3} - {x}^{4} + ...... \infty }$

$\displaystyle \sf{ = \frac{ - x}{1 - ( - x)} }$

$\displaystyle \sf{ = \frac{ - x}{1 + x} }$

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Answer 2

$\huge{\boxed{\bold{Question}}}$

if Mod X is less than 1 Y is equal to X minus x square + X power 3 minus x power 4 so on infinity then the value of x in terms of y is

$\huge{\boxed{\bold{Añswer}}}$

Solution,

The value of

y-x-x²+x³-x⁴+.........∞ when |x|

Formula

First term =a

Common term =r with |x|<1

Then,

a+ar+ar²+.........∞=a/1-r

y=x-x²+x³-x⁴.......∞ when |x|

First term=x

common term =-x

Now,

|-x|=|x|<1

y=x-x²+x³-x⁴+.........∞

=-x/1-(-x)

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