Logarithmic Series question for finding ##\log_e2##

[RChristenk]

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?

 

[PeroK]

My question is why can't I set $x=1$ in $(1)$?

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

Why do you think you can't do that? That's exactly how you would produce a series expansion for $\log_e 2$.

I realize the numerical value is wrong, but why?

It isn't wrong.

 

[RChristenk]

Using $\log_e(n+1)-\log_en=2[\dfrac{1}{2n+1}+\dfrac{1}{3(2n+1)^3}+\dfrac{1}{5(2n+1)^5}+\cdots]$ and setting $n=1$, the result is $\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+ \cdots$. I thought this was different from $1-\dfrac{1}{2}+\dfrac{1}{3}- \cdots$.

 

[pasmith]

Homework Statement: Logarithmic Series question for finding $\log_e2$
Relevant Equations: Binomial Theorem, Logariths

By definition:

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

That's not a definition; it's the result of expanding \ln(1 + x) in a Taylor series about x = 0.

The series has radius of convergence 1 because \ln 0 diverges. But by the alternating series test \ln 2 = \ln (1 + 1) = 1 - \frac12 + \frac13 + \dots converges.

 

[PeroK]

Using $\log_e(n+1)-\log_en=2[\dfrac{1}{2n+1}+\dfrac{1}{3(2n+1)^3}+\dfrac{1}{5(2n+1)^5}+\cdots]$ and setting $n=1$, the result is $\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+ \cdots$. I thought this was different from $1-\dfrac{1}{2}+\dfrac{1}{3}- \cdots$.

It's a different series. But, two different series may have the same sum. You could check quite quickly on a spreadsheet whether those two series appear to converge to the same limit.

 

[FactChecker]

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

My question is why can't I set $x=1$ in $(1)$?

A Taylor series is best understood in the complex plane. It has a radius of convergence, so any pole in the complex plane determines the radius that applies to all x (it's better to use z in the complex plane). The invalid value of x= -1 for $\log_e(1+x)$ sets the radius of convergence centered at $z=0$ that means you can not use the Taylor series for x=1.

 

[Mark44]

It should be noted that the standard notation for $\log_e 2$ is $\ln 2$.

 

[PeroK]

A Taylor series is best understood in the complex plane. It has a radius of convergence, so any pole in the complex plane determines the radius that applies to all x (it's better to use z in the complex plane). The invalid value of x= -1 for $\log_e(1+x)$ sets the radius of convergence centered at $z=0$ that means you can not use the Taylor series for x=1.

That's not quite true. The series at $x = 1$ converges by the alternating series test. It's true that, technically, you still need to justify that it converges to $\ln(2)$. Which it does.

 

[PeroK]

PS that would explain the more detailed calculation in the OP. The shortcut $\ln(1+1)$ is a valid expansion, but technically it doesn't follow immediately from the basic theory of Taylor series. So, there would still be something to prove.

 

[PeroK]

Using $\log_e(n+1)-\log_en=2[\dfrac{1}{2n+1}+\dfrac{1}{3(2n+1)^3}+\dfrac{1}{5(2n+1)^5}+\cdots]$ and setting $n=1$, the result is $\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+ \cdots$. I thought this was different from $1-\dfrac{1}{2}+\dfrac{1}{3}- \cdots$.

I looked at this more closely. Better check the specific expansion you got for $n = 1$.

 

[mathwonk]

here is a reference for Perok's observation, "Abel's theorem".
https://en.wikipedia.org/wiki/Abel's_theorem