- #1

RChristenk

- 49

- 4

- Homework Statement
- Logarithmic Series question for finding ##\log_e2##

- Relevant Equations
- Binomial Theorem, Logariths

By definition:

##\log_e(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}- \cdots ## ##(1)##

Replacing ##x## by ##−x##, we have:

##\log_e(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}- \cdots##

By subtraction,

##\log_e(\dfrac{1+x}{1-x})=2(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+ \cdots)##

Put ## \dfrac{1+x}{1-x}=\dfrac{n+1}{n}##, so that ##x=\dfrac{1}{2n+1}##; we thus obtain:

##\log_e(n+1)-\log_en=2[\dfrac{1}{2n+1}+\dfrac{1}{3(2n+1)^3}+\dfrac{1}{5(2n+1)^5}+\cdots]##

By putting ##n=1##,##\log_e2## is found. My question is why can't I set ##x=1## in ##(1)##?

##\log_e(1+1)=\log_e2=1-\dfrac{1}{2}+\dfrac{1}{3}- \cdots##

I realize the numerical value is wrong, but why?

##\log_e(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}- \cdots ## ##(1)##

Replacing ##x## by ##−x##, we have:

##\log_e(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}- \cdots##

By subtraction,

##\log_e(\dfrac{1+x}{1-x})=2(x+\dfrac{x^3}{3}+\dfrac{x^5}{5}+ \cdots)##

Put ## \dfrac{1+x}{1-x}=\dfrac{n+1}{n}##, so that ##x=\dfrac{1}{2n+1}##; we thus obtain:

##\log_e(n+1)-\log_en=2[\dfrac{1}{2n+1}+\dfrac{1}{3(2n+1)^3}+\dfrac{1}{5(2n+1)^5}+\cdots]##

By putting ##n=1##,##\log_e2## is found. My question is why can't I set ##x=1## in ##(1)##?

##\log_e(1+1)=\log_e2=1-\dfrac{1}{2}+\dfrac{1}{3}- \cdots##

I realize the numerical value is wrong, but why?