Power Series Problem?
Got 2 power series problems but can't get a solution to the answer
1) Ln(1+x/1-x)=2x+2/3 x^3
2)Sqrt (1+x/1-x) = approx 1+x+(x^2)/2
need the solution to these problems so I can see how it is done for both cases
Got 2 power series problems but can't get a solution to the answer
1) Ln(1+x/1-x)=2x+2/3 x^3
2)Sqrt (1+x/1-x) = approx 1+x+(x^2)/2
need the solution to these problems so I can see how it is done for both cases
Comments
1) By the geometric series, 1/(1-x) = 1 + x + x^2 + x^3 + ...
Integrate both sides from 0 to x:
- ln(1 - x) = x + x^2/2 + x^3/3 + x^4/4 + ...
Replace x with -x:
- ln(1 + x) = -x + x^2/2 - x^3/3 + x^4/4 + ...
==> ln(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ...
Therefore, ln((1+x)/(1-x))
= ln(1+x) - ln(1 - x)
= [x - x^2/2 + x^3/3 - x^4/4 + ...] + [x + x^2/2 + x^3/3 + x^4/4 + ...]
= 2x + (2/3)x^3 + ...
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2) Using the Binomial Series, (1 + x)^a = 1 + ax + (a(a-1)) x^2/2! + ...
sqrt [(1+x)/(1-x)]
= (1+x)^(1/2) * (1-x)^(-1/2)
= (1 + (1/2)x + (1/2)(-1/2) x^2/2! + ...) (1 + (-1/2)(-x) + (-1/2)(-3/2) (-x)^2/2! + ...)
= (1 + (1/2)x - (1/8) x^2 + ...) (1 + (1/2)x + (3/8) x^2 + ...)
= 1 + (1/2 + 1/2)x + (3/8 + 1/4 - 1/8) x^2 + ... by distributive property
= 1 + x + (1/2) x^2 + ...
I hope this helps!