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B Expressions of ##log(a+b), tan^{-1}(a+b),sin^{-1}(a+b)##,etc

  1. Feb 20, 2017 #1
    Hi, I got these:
    $$log(a+b)\approx \frac{b*logb-a*loga}{b-a} + log2 -1$$
    $$tan^{-1}(a+b)\approx \frac{b*tan^{-1}2b-a*tan^{-1}2a+\frac{1}{4}*ln\frac{1+4a^2}{1+4b^2}}{b-a}$$
    $$sin^{-1}(a+b)\approx \frac{b*sin^{-1}2b-a*sin^{-1}2a+\frac{1}{2}*(\sqrt{1-4b^2}-\sqrt{1-4a^2}}{b-a}$$

    And, similarly for ##sec^{-1}(a+b)##, ##cosec^{-1}(a+b)##, ##cot^{-1}(a+b)##, etc.
    So, you see that the RHS in each of these expressions is the average value of ##f(2x)## between x=a and x=b, i.e. $$\frac{\int_a^bf(2x)dx}{b-a}$$
    So, for what values of ##a## and ##b## do these approximations hold good? I checked that these had great accuracy for some pairs I put in my calculator but not so good for others.
     
    Last edited: Feb 20, 2017
  2. jcsd
  3. Feb 20, 2017 #2
    I think these formulas are good approximations if b and a are nearby in the real line. Cant exactly define how much nearby they should be.
     
  4. Feb 20, 2017 #3
    How does this work?
     
  5. Feb 22, 2017 #4
    For the log case, I 'd write
    b = a + x
    so
    b.logb = (a + x).log(a + x) =
    (a + x).log( a.(1 + x/a) )
    (a + x).( loga + log(1 + x/a) )
    ≈(a + x).(loga + x/a + . . . )
    where I've started an expansion of the second log assuming -1 <x/a <1 .
    In this way express your complete formula as a power series in x.

    Then do the same for log (a + b) = log(2a + x)
    = log( 2a.(1+x/2a) )
    ≅log(2a) + x/2a + . . .

    Compare the results, to see in which power of x = b - a the two expressions differ. A similar approach should work for your other formulae.
     
  6. Feb 26, 2017 #5

    Stephen Tashi

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    Here's an intuitive way to look at the situation. If ##A## and ##B## are nearly equal then ##A+B \approx 2A \approx 2B##. So a crude approximation is ##f(A+B) \approx f(2A) \approx f(2B)## Instead of using one of those crude approximations, we could use the "average" value of ##f## over the interval ##[2A,2B]##.

    That would be ## f(A+B) \approx m = \frac{ \int_{2A}^{2B} f(u) du} {2B- 2A} ##.

    Making the change of variable ##2x = u## the integration becomes ##m = \frac{ 2 \int_A^B f(2x) dx}{2B - 2A} = \frac{\int_A^B f(2x) dx}{B-A}##.
     
  7. Feb 26, 2017 #6
    ##A## and ##B## need not be nearly equal. For example, when ##A=0.4## and ##B=1.5##. Then ##A+B=1.9##, ##2A=0.8## and ##2B=3##. So, ##A+B## is neither approximately equal to ##2A## nor to ##2B##. The formula still gives a very accurate value of ##tan^{-1}1.9##.
     
  8. Feb 26, 2017 #7

    Stephen Tashi

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    However, A+B is still the average of 2A and 2B which leads to the justification for making an estimate by taking the average value of the function over the interval [2A,2B].
     
  9. Feb 26, 2017 #8
    2A and 2B will always have A+B as an average no matter what are A and B.
     
  10. Feb 26, 2017 #9

    Stephen Tashi

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    Yes.
     
  11. Feb 26, 2017 #10
    Well, that'd mean the formula would always work. You first said that it's supposed to work when ##2A\approx A+B\approx 2B##
     
  12. Feb 26, 2017 #11

    Stephen Tashi

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    No, it would suggest that the formula is worth trying. Whether the formula works or not depends on how the function we are approximating behaves.

    Yes, I did.
     
  13. Feb 28, 2017 #12
    I may have more to say later, but here are a few quick thoughts.

    I haven't got my notes in front of me but I believe that for large values of a and b (a, b > 1) your arctangent formula is good up to at least terms of order (a + b)-3. I think terms in (b -a) appear in higher orders.

    I expanded your formula about (a + b) = 1; i.e. arctan(a + b ) = π/4. The deviations seemed to be (b - a)2/3 (to be checked).

    The limiting case where a = 0 and b is small ( 4b2 < 1) is easy to analyse and seems instructive. I believe here your formula first deviates from the true formula in the cubic term.

    I do understand your interest in developing these formulae, but where do you think they would be useful?
     
  14. Mar 1, 2017 #13
    It's not about uses, I've just suddenly become interested in thinking about these things.
     
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