Logarithmic Series question for finding ##\log_e2##

  • #1
RChristenk
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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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FAQ: Logarithmic Series question for finding ##\log_e2##

What is a logarithmic series?

A logarithmic series is a type of infinite series that can be used to represent the natural logarithm of a number. It is often written in the form of a power series, such as the series for ##\log(1+x)##, which is ##\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots## for ##|x| < 1##.

How can I use a logarithmic series to find ##\log_e 2##?

To find ##\log_e 2## using a logarithmic series, you can use the series expansion for ##\log(1+x)##. By setting ##x = 1##, you get ##\log_e 2 = \log(1+1)##. This can be approximated using the series: ##\log(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots##. Summing enough terms of this series will give an approximation of ##\log_e 2##.

What is the convergence rate of the logarithmic series for ##\log_e 2##?

The convergence rate of the logarithmic series for ##\log_e 2## is relatively slow because the series alternates and the terms decrease in magnitude as ##1/n##. To achieve a high degree of accuracy, a large number of terms must be summed, making it computationally intensive for precise calculations.

Are there alternative methods to find ##\log_e 2## more efficiently?

Yes, there are more efficient methods to find ##\log_e 2##. For example, numerical integration, iterative algorithms such as Newton's method, or using precomputed logarithm tables can provide more accurate results with fewer computations compared to the logarithmic series expansion.

What is the approximate value of ##\log_e 2##?

The approximate value of ##\log_e 2## is about 0.69314718056. This value can be found using various methods including the logarithmic series, but for practical purposes, it is often obtained from scientific calculators or logarithm tables.

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