Question

Try for fun. Q. Use calculus to find the sum of [1/(x+1)+2/(x^2+1)+4/(x^4+1)+⋯+2^n/(x^2n+1)]

Answer 1

(I may be mistaken in typing, pl cooperate)
Ans. Let the sum of the series be P.
Now, Observe the pattern,

(1-x)(1+x)=1-x^2
(1-x)(1+x)(1+x^2)=1-x^(2^2 )
(1-x)(1+x)(1+x^2 )(1+x^4)=1-x^(2^3 )
---------------------------------
(1-x)(1+x)(1+x^2 )(1+x^4 )….(1+x^(2^n ) )=1-x^(2^(n+1) )
Taking log,
log(1-x)+log(1+x)+log(1+x^2 )+log(1+x^4 )+⋯.+log(1+x^(2^n ) )=log⁡(1-x^(2^(n+1) ))

Differentiating, w.r.t. x;
(-1)/(1-x)+1/(1+x)+2x/(1+x^2 )+(4x^3)/(1+x^4 )+⋯.+2^n x^(2^(n-1) )/(1+x^(2^n ) )=〖-2〗^(n+1) (x^(2^(n+1) )-1)/(1-x^(2^(n+1) ) )

Or,
1/x [x/(1+x)+(2x^2)/(1+x^2 )+(4x^4)/(1+x^4 )+⋯.+2^n x^(2^n )/(1+x^(2^n ) )]=1/(1-x) 〖-2〗^(n+1) (x^(2^(n+1) )-1)/(1-x^(2^(n+1) ) )

Or,
1/x [((x+1)-1)/(1+x)+(2〖(x〗^2+1)-2)/(1+x^2 )+(4(x^4+1)-4)/(1+x^4 )+⋯.+(〖2^n (x〗^(2^n )+1)-2^n)/(1+x^(2^n ) )]=1/(1-x) 〖-2〗^(n+1) (x^(2^(n+1) )-1)/(1-x^(2^(n+1) ) )
Or,
1/x [(1+2+4+⋯2^n )-P]=1/(1-x) 〖-2〗^(n+1) (x^(2^(n+1) )-1)/(1-x^(2^(n+1) ) )
Or,
[(2^(n+1)-1)/(2-1)-P]=x/(1-x) 〖-2〗^(n+1) x^(2^(n+1) )/(1-x^(2^(n+1) ) )
Or,
P =[2^(n+1)-1]- x/(1-x) 〖+2〗^(n+1) x^(2^(n+1) )/(1-x^(2^(n+1) ) )
=2^(n+1) [1+x^(2^(n+1) )/(1-x^(2^(n+1) ) )]-[1+x/(1-x)]
=2^(n+1) [1/(1-x^(2^(n+1) ) )]-[1/(1-x)]

Answer 2

this q was asked three days back.

Answer 3

i think i might be using slightly different definition for the word fun

Answer 4

hehe is that a tutorial or a question?

Answer 5

\[\text{ Q. Use calculus to find the sum of }\]
\[\left[\frac{1}{(x+1)}+\frac{2}{(x^2+1)}+\frac{4}{(x^4+1)}+⋯+\frac{2^n}{(x^2n+1)}\right]\]
(I may be mistaken in typing, pl cooperate)
Ans. Let the sum of the series be P.
Now, Observe the pattern,
\[(1-x)(1+x)=1-x^2\]
\[(1-x)(1+x)(1+x^2)=1-x^{2^2} \]
\[(1-x)(1+x)(1+x^2 )(1+x^4)=1-x^{2^3 }\]
\[\text{---------------------------------}\]
\[(1-x)(1+x)(1+x^2 )(1+x^4 )\cdots(1+x^{2^n } )=1-x^{2^{(n+1)} }\]
Taking log,
\[\log(1-x)+\log(1+x)+\log(1+x^2 )+\log(1+x^4 )+\cdots+\log(1+x^{2^n } )=\log(1-x^{2^{(n+1)}} )\]
\[\text{Differentiating, w.r.t. x;}\]
\[\frac{(-1)}{(1-x)}+\frac{1}{(1+x)}+\frac{2x}{(1+x^2 )}+\frac{4x^3}{(1+x^4 )}+\cdots.+\frac{2^n x^{2^{(n-1)}}}{(1+x^{2^n} )}=(-2)^{(n+1)} (x^{2^{(n+1)} })-\frac{1}{(1-x^{2^{(n+1)} } )}\]
Or,
\[ \frac 1x \left[\frac{x}{(1+x)}+\frac{(2x^2)}{(1+x^2 )}+\frac{(4x^4)}{(1+x^4 )}+\cdots+ \frac{2^nx^{(2^n )}}{(1+x^{2^n} )}\right]=\frac{1}{(1-x)} (-2)^{(n+1)} \frac{(x^{2^{(n+1)} }-1)}{(1-x^{2^{(n+1)}})}\]
Or,
\[\frac{1}{x} \left[\frac{((x+1)-1)}{(1+x)}+\frac{(2(x^2+1)-2)}{(1+x^2 )}+\frac{(4(x^4+1)-4)}{(1+x^4 )}+\cdots+\frac{((2^n x^{(2^n )}+1)-2^n)}{(1+x^{2^n})}\right]=\frac{1}{(1-x)}(-2)^{(n+1)} \frac{(x^{2^{(n+1)} }-1)}{(1-x^{2^{(n+1)} } )}\]
Or,
\[\frac{1}{x} \left[(1+2+4+\cdots+2^n )-P\right]=\frac{1}{(1-x)} (-2)^{(n+1)} \frac{(x^{2^{(n+1)}}-1)}{(1-x^{2^{(n+1)} } )}\]
Or,
\[\left[\frac{(2^(n+1)-1)}{(2-1)}-P\right]=\frac{x}{(1-x)} (-2)^{(n+1)} \frac{{x^{2^(n+1) }}}{(1-x^{2^{(n+1)} } )}\]
Or,
\[P =\left[2^{(n+1)}-1\right]- \frac{x}{(1-x)} (+2)^{(n+1)} \frac{x^{(2^{(n+1)} )}}{(1-x^{2^{(n+1)} } )}\]
\[=2^{(n+1)} \left[1+\frac{x^{2^{(n+1)} )}}{(1-x^{2^{(n+1)} } )}\right]-\left[1+\frac x{(1-x)}\right]\]
\[=2^{(n+1)} \left[\frac 1{(1-x^{2^{(n+1)} }) }\right]-\left[\frac{1}{(1-x)}\right]\]

Answer 6

@SMISHRA pls use the equation editor

Answer 7

that took me a while .

Answer 8

@UnkleRhaukus hw do u manage to write such a long latex in a small time?

Answer 9

the question has been up for over an hour,

Answer 10

:)

Answer 11

bt still u need so much patience to type that out in latex

Answer 12

latex is just a language , you get get more fluent with it , , the more you practice