Logarithmic Series question for finding ##\log_e2##

  • #1
RChristenk
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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?
 
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  • #2
RChristenk said:
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##.
RChristenk said:
I realize the numerical value is wrong, but why?
It isn't wrong.
 
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  • #3
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##.
 
  • #4
RChristenk said:
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 [itex]\ln(1 + x)[/itex] in a Taylor series about [itex]x = 0[/itex].

The series has radius of convergence 1 because [itex]\ln 0[/itex] diverges. But by the alternating series test [tex]\ln 2 = \ln (1 + 1) = 1 - \frac12 + \frac13 + \dots[/tex] converges.
 
  • #5
RChristenk said:
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.
 
  • #6
RChristenk said:
##\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.
 
  • #7
It should be noted that the standard notation for ##\log_e 2## is ##\ln 2##.
 
  • #8
FactChecker said:
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.
 
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  • #9
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.
 
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  • #10
RChristenk said:
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##.
 
  • #11
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