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Question

If $$\displaystyle f=\frac{x}{1+x^2} +\frac 1 3\left(\frac{x} {1+x^2}\right)^3+\frac1 5\left(\frac{x} {1+x^2}\right)^5 + \,...\,$$ to $$\infty$$ and $$\displaystyle g=x-\frac2 3 x^3 + \frac1 5 x^5 + \frac1 7 x^7 - \frac2 9 x^9 +...  $$, then  $$f= d \times g$$. Find 4d.


Solution

$$\displaystyle f=\frac{x}{1+x^2} +\frac 1 3\left(\frac{x} {1+x^2}\right)^3+\frac1 5\left(\frac{x} {1+x^2}\right)^5 + \,...\,$$
We know, $$\displaystyle Log(1+x)=x-\frac{x^2}{2}+\frac {x^3}{3}-\frac {x^4}{4}...$$

$$\therefore f\displaystyle=\frac { 1 }{ 2 } \left[ \log { (1+\frac { x }{ 1+x^{ 2 } } )- } \log { (1-\frac { x }{ 1+x^{ 2 } } ) }  \right] $$
$$\displaystyle =\frac { 1 }{ 2 } \left[ \log { (\frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } )- } \log { (\frac { 1+x^{ 2 }-x }{ 1+x^{ 2 } } ) }  \right] $$
$$\displaystyle \Rightarrow f=\frac { 1 }{ 2 } \log { (\frac { x^{ 2 }+x+1 }{ x^{ 2 }-x+1 } ) } $$
Now, $$\displaystyle g=x-\frac{2}{3} x^3 + \frac{1}{5} x^5 + \frac{1}{ 7} x^7 - \frac{2}{9} x^9 +...  $$
$$\displaystyle g=x-(1-\frac { 1 }{ 3 } )x^{ 3 }+\frac { 1 }{ 5 } x^{ 5 }+\frac { 1 }{ 7 } x^{ 7 }-\frac { 1 }{ 3 } (1-\frac { 1 }{ 3 } )x^{ 9 }+...$$
$$\displaystyle g=\left( x+\frac { 1 }{ 3 } x^{ 3 }+\frac { 1 }{ 5 } x^{ 5 }+\frac { 1 }{ 7 } x^{ 7 }+...... \right) -\left( x^{ 3 }+\frac { { (x^{ 3 }) }^{ 3 } }{ 3 } +\frac { { (x^{ 3 }) }^{ 5 } }{ 3 } +..... \right) $$
We know, $$\displaystyle \frac {1}{2}(Log (1+x)-Log (1-x))=x+\frac {x^3}{3}+\frac {x^5}{5}....$$
$$\therefore g\displaystyle =\frac { 1 }{ 2 } [\log { (1+x) } -\log { (1-x) } ]-\frac { 1 }{ 2 } [\log { (1+x^{ 3 }) } -\log { (1-x^{ 3 }) } ]$$
$$\displaystyle =\frac { 1 }{ 2 } \log { (\frac { 1+x }{ 1-x } ) } -\frac { 1 }{ 2 } \log { (\frac { 1+x^{ 3 } }{ 1-x^{ 3 } } ) } $$
$$\displaystyle \Rightarrow g=\frac { 1 }{ 2 } \log { \left( \frac { (1+x)(1-x^{ 3 }) }{ (1-x)(1+x^{ 3 }) }  \right)  } $$
$$\displaystyle \Rightarrow g=\frac { 1 }{ 2 } \log { (\frac { x^{ 2 }+x+1 }{ x^{ 2 }-x+1 } ) } $$
$$\Rightarrow g=f$$

Comparing with given value , 
$$d=1$$
$$\Rightarrow 4d=4$$

Maths

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