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Boas Maclaurin series for ln(2)

  1. Sep 1, 2016 #1
    • Member warned that the homework template must be used
    Boas's Problem.png
     
  2. jcsd
  3. Sep 1, 2016 #2

    George Jones

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    $$2=\left( \frac{1}{2}\right)^{-1},$$
    so ##\ln 2 =##?
     
  4. Sep 2, 2016 #3
    Thank you George for your hint. I've seen clues like this in the past, but it's not ringing a bell for me. It just seems like an identity for ln(2)=-ln(1/2) ... I see no immediate series, other than the alternating series coming through.
     
  5. Sep 2, 2016 #4

    micromass

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    Series ##(2)## isn't always alternating. Maybe you can get the minus factors to drop out somehow?
     
  6. Sep 2, 2016 #5
    Thank you micromass! I've been through that already with no insight. For example the series we desire is [1]. Manipulating [3] we easily get the first term (1-1/2), but starting on the next term we need 1/8 to come out ... but to do this with a -1/8 in [3] it seems clear that stuff must add up to 1/4 and nothing seems to come up to this rational cleanly. After that we need a 1/24 and from the previous work, this might be a nightmare! :-)
     
  7. Sep 2, 2016 #6

    George Jones

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    Maybe, I am giving too much away, but .... What happens in (2) if x is negative?
     
  8. Sep 2, 2016 #7
    Well it is only allowed to be in the interval -1<x<=1 ... picking a negative value in this range takes [2] to all negative terms ... and in fact choosing x=-1/2 makes the series converge to -ln(2) ... the opposite sign. But the idea here is not to start from the answer and work backward ... although that gives insight at times ... but to start with the original sum [1] and arrive at ln(2) ... plus the alternating series ln(1+x) *requires* x=1 if we're looking at ln(2). So I'm still confused. I realize that once I 'get' the answer, I'll be shaking my head.

    Also my multiplication of the two series as shown might be in error as I'm multiplying by the harmonic series, a divergent series ... thus multiplying a divergent series by a convergent series ... which may be an invalid thing in some cases ... especially since the convergent series converges to 1.

    I love Mary Boas's book as she makes you think very hard with her problem sets. I've already finished Chapter 3 on Linear Algebra which was truly excellent. In fact it was so good, I decided to also work on Chapters 1 and 2.

    Thanks!
     
  9. Sep 2, 2016 #8

    micromass

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    What do you get if ##x=-1/2##?
     
  10. Sep 2, 2016 #9
    As I mentioned above ... -ln(2).
     
  11. Sep 2, 2016 #10

    micromass

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    And on the other side?
     
  12. Sep 2, 2016 #11
    If you mean $$x=1/2,$$ then we get $$ln(3/2)=ln(3)-ln(2)$$
     
  13. Sep 2, 2016 #12

    micromass

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    No, what happens to (2) if you take ##x=-1/2##.
     
  14. Sep 2, 2016 #13
    I get my desired series $$-1/2-1/8-1/24-1/64,$$ but with the signs negative ... the negative of what I need. Some progress ...
     
  15. Sep 2, 2016 #14

    micromass

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    So what about using the series associated to the function ##f(x) = -\text{ln}(1+x)##?
     
  16. Sep 2, 2016 #15
    I hear you, but [1] is $$\sum_{n=1}^{\infty}\frac{1}{n 2^n}$$ not $$-\sum_{n=1}^{\infty}\frac{1}{n 2^n}$$
     
  17. Sep 2, 2016 #16

    micromass

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    I know.
     
  18. Sep 2, 2016 #17
    Okay so then $$-\sum_{n=1}^{\infty}\frac{1}{n 2^n}=-1/2-1/8-1/24-1/64...=-ln(2),$$which makes$$\sum_{n=1}^{\infty}\frac{1}{n 2^n}=1/2+1/8+1/24+1/64...=ln(2).
    \hspace{1cm}\square$$

    Thank you all very much! I learned a lot from this including the term Mercator series to signify ln(1+x)
     
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