Expansion of log(1+x)

[uppaladhadium]

We know that log(1+x) = x+((x^2)/2)+((x^3)/3)+....((x^n)/n)+...
Could anybody please tell me the proof

 

[micromass]

It's just the Taylor series expansion:

f(x)=f(a)+f^\prime(a)(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+...

See http://en.wikipedia.org/wiki/Taylor_series

 

[mathman]

log(1+x)=x - x²/2 + x³/3 - x⁴/4 + ...
(Alternate signs)

The easiest way to see it is by using an integral representation.

log(1+x) = ∫dx/(1+x)

Since 1/(1+x) = 1 - x + x² - x³ + ...,

integrating term by term gives the series for log(1+x), where the integration limits are [0,x].

 

[uppaladhadium]

thankyou for the answers i am grateful to you

 

[uppaladhadium]

In taylor series if f(x)=log(1+x) then is f'(a)=0?
and is f(a)=log(1+a)

 

[gb7nash]

In taylor series if f(x)=log(1+x) then is f'(a)=0?
and is f(a)=log(1+a)

Assuming a > -1:

What do you get when you take the derivative of log(1+x)? What do you get when you plug a in?