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Natural Log Proof

  1. Oct 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove the following statement:
    [tex] ln|1+\sigma x | = \frac{1}{2} ln|1-x^2| + \frac{\sigma}{2} ln| \frac{ |1+x|}{|1-x|}
    [/tex]
    2. Relevant equations


    3. The attempt at a solution
    Starting from right to left would be easier:
    [tex]
    = \frac{1}{2} ln|(1+x)(1-x)| + \frac{\sigma}{2} ln| 1+x| - \frac{\sigma}{2} ln(1-x) \\
    = \frac{1}{2} [ ln(1+x) + ln(1-x) + \sigma ln(1+x) - \sigma ln(1-x)] \\
    =\frac{1}{2} [ln|(1+x)(1+\sigma)| + ln|(1-x)(1-\sigma)|] \\
    = \frac{1}{2} [ ln |(1+x)^{1+\sigma} (1-x)^{1-\sigma} |]\\
    = \frac{1}{2} ln |( 1+x ) (1+x)^\sigma \frac{1+x}{(1-x)^\sigma}|]
    [/tex]
    Then I get stuck. Does anyone know what I'm missing?
     
  2. jcsd
  3. Oct 23, 2016 #2

    haruspex

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    The statement to be proved is clearly not true in general. Try σ=0. Are you sure you have stated the whole question, word for word?
     
  4. Oct 23, 2016 #3
    Do you know of any identities that would give me the left hand side? Any references? I cannot find anything online. Calculus textbooks just give integrals and not identities for logs.

    I need to take the natural log of the following expression.
    [tex] \frac{1}{2}( C + D ) + \frac{1}{2}( C - D ) \sigma
    [/tex]

    The idea was to use that identity, then rearranging above
    [tex] \frac{1}{2}( C + D ) [ 1 + \frac{(C-D)}{(C+D)} \sigma ]
    [/tex]

    The answer is

    [tex]
    ln2 + \frac{1}{2}ln|4CD|+\frac{\sigma }{2} ln|\frac{C}{D} |
    [/tex]

    Without that identity, It's pretty hard to achieve that result. Do you know of anything I can use?

    What's really important is that I at least get this form: [tex] \frac{\sigma }{2} ln|\frac{C}{D} | [/tex]
     
    Last edited: Oct 23, 2016
  5. Oct 23, 2016 #4

    SammyS

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    You appear to be a step or two away.

    Added in Edit:
    Never mind! I was looking at the wrong side of the equation.
     
    Last edited: Oct 23, 2016
  6. Oct 23, 2016 #5
    Last edited by a moderator: Oct 23, 2016
  7. Oct 23, 2016 #6

    Ray Vickson

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    Your desired result is false. For small ##|x|## we can expand both sides in a series in ##x##. For ease of writing, I will use the symbol ##a## instead of ##\sigma##. Let
    $$L(x) = \ln (1+ ax) , \;\; R(x) = \frac{1}{2} \ln(1-x^2) + \frac{a}{2} \ln \left(\frac{1+x}{1-x}\right).$$
    We have
    $$L(x) = ax - \frac{1}{2} a^2 x^2 + \frac{1}{3} a^3 x^3 - \frac{1}{4} a^4 x^4 + O(x^5),$$
    but
    $$R(x) = ax - \frac{1}{2} x^2 + \frac{1}{3} a^3 x^3 - \frac{1}{4} x^4 + O(x^5).$$
    Note that in ##L(x), R(x)## I have omitted the absolute-value signs; those signs are redundant when ##|x| < 1##, which is true for small ##|x|##.
     
  8. Oct 23, 2016 #7
    Was there a typo? Is there a way to determine if there is one?

    Or is there an easier way to simplify the natural log of the expression?

    [tex] \frac{1}{2}( C + D ) + \frac{1}{2}( C - D ) \sigma
    [/tex]

    with obtaining
    [tex] \frac{\sigma }{2} ln|\frac{C}{D} | [/tex] at least in the final answer?


    [tex]
    ln [ \frac{1}{2} D ( \frac{C}{D} +1 ) + \frac{\sigma}{2} D ( \frac{C}{D} -1) ]
    [/tex]
     
  9. Oct 23, 2016 #8

    haruspex

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    For it even to be approximately true you need to change the 1/2 factor in the first expression on the right to σ/2. That makes it more symmetric, so a lot more reasonable.
    It still is not going to be generally true. The roles of the σ on the left and the σ's on the right are too different. I would guess you are only supposed to show this as asymptotic behaviour for small x. (It is equivalent to (1+x)n approximating enx.)
     
  10. Oct 23, 2016 #9

    haruspex

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    a3?
     
  11. Oct 23, 2016 #10
    Is there a table of asymptotic expansions or reference I can refer to to get the desired result? I'm kind of stumped.
     
  12. Oct 23, 2016 #11

    haruspex

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    First step is to correct the RHS of the statement to be proved. See the first para of post #8.
    Next, as Ray points out, on the assumption that |x|<1 and |σx|<1, all those modulus signs can be thrown away.
    Then you can exponentiate both sides to get rid of the logarithms. Post what you get.
     
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