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How to decompose the following expression?

  1. Jan 18, 2015 #1
    This comes from one of the research papers that I'm reading. The authors decompose Eq. 1 and obtain Eq. 2. After several attempts, I have been fairly unsuccessful at obtaining Eq. 2 from Eq. 1.

    I have tried using partial fractions decomposition without much success. Any help would be greatly appreciated.

    Eq. 1


    20-%20%5Cfrac%7B%5Cprod_%7Bs2%7D%20%7D%7B1-%5Cdelta%20+%5Cdelta%20%5Cepsilon%20_%7Bs2%7D%7D.gif

    Eq. 2

    gif.gif


    Thanks!


     
  2. jcsd
  3. Jan 18, 2015 #2

    mfb

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    Try to find the other direction, that is easier (and all steps work both ways). You can simplify equation 2 and get equation 1.
     
  4. Jan 18, 2015 #3

    Stephen Tashi

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    You have to explain the relations that exist among the symbols to get help on this question.
     
  5. Jan 18, 2015 #4

    Mark44

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    Yes, especially ##\Pi_{s1}## and ##\Pi_{sS2}##. Are they just numbers or is something else going on? The symbol ##\Pi## is usually used to denote a product.
     
  6. Jan 18, 2015 #5
    Sorry, I should have clarified. ∏s1 is profit in country 1, ∏s2 is profit in country 2, ∈s1 is risk in country 1, ∈s2 is risk in country 2, and δ is the discount rate.

    Eq. 2 tries to separate risk elements (∈s1 and ∈s2) from the profit gap between countries. Hope this helps.
     
  7. Jan 18, 2015 #6

    mfb

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    ∏ as variable is a bit unusual, but as long as all symbols are just variables, their meaning does not matter. This just needs the basic rules to work with fractions.
     
  8. Jan 19, 2015 #7

    Stephen Tashi

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    The results of a simple numerical test did produce identical answers for EPV. (Of course someone should check my program.)


    The output

     
  9. Jan 19, 2015 #8

    Stephen Tashi

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    Getting rid of terms in a denominator that don't cancel with anything in the numerator is a frequent mathematical daydream. Suppose we approach the problem as the desire to express:

    [itex]( \frac{A}{W + F} - \frac{B}{W + G}) [/itex] as [itex]( \frac{A}{W} - \frac{B}{W}) \ + \ L [/itex]

    where [itex] L [/itex] is the leftover stuff.

    Then
    [itex] L = \frac{A}{W+F} - \frac{A}{W} - \frac{B}{W + G} + \frac{B}{W }[/itex]

    [itex] = \frac{ AW - A(W+F)} {W(W+F)} + \frac{-BW+B(W+G)}{W(W+G)} [/itex]

    [itex] = \frac{-AF}{W(W+F)} + \frac{BG}{W(W+G)} [/itex]

    [itex] = \frac{1}{W} ( \frac{BG}{W+G} - \frac{AF}{W+F}) [/itex]

    In the problem at hand, there are relations between [itex] W, F, G[/itex]

    [itex] W = 1 -\delta [/itex]
    [itex]F = \delta \epsilon_1 = (1-W)\epsilon_1 [/itex]
    [itex]G = \delta \epsilon_2 = (1-W)\epsilon_2 [/itex]

    Substituting in selected places for [itex] F [/itex] and [itex] G [/itex]

    [itex] L = \frac{1}{W}( \frac{B(1-W)\epsilon_2}{W+G} - \frac{A(1-W)\epsilon_1}{W+F} ) [/itex]

    [itex] = \frac{1-W}{W} ( \frac{ \epsilon_2 B}{W+G} - \frac{\epsilon_1 A}{W+F}) [/itex]

    [itex] = \frac{1-W}{W} ( \frac{ (W+F)\epsilon_2 B - (W+G)\epsilon_1A}{(W+G)(W+F)}) [/itex]

    In the problem at hand:
    [itex] A = \prod_{s1} [/itex]
    [itex] B = \prod_{s2} [/itex]
    [itex] W = (1 -\delta) [/itex]
    [itex] F = \delta \epsilon_{s1} [/itex]
    [itex] G = \delta \epsilon_{s2} [/itex]
    [itex] \epsilon_1 = \epsilon_{s1} [/itex]
    [itex] \epsilon_2 = \epsilon_{s2} [/itex].
     
  10. Jan 31, 2015 #9

    Thanks so much! The simplification of L makes sense. Could you please dwell on how to separate [itex]( \frac{A}{W} - \frac{B}{W}) [/itex] from [itex]( \frac{A}{W + F} - \frac{B}{W + G}) [/itex]?
     
  11. Jan 31, 2015 #10

    Stephen Tashi

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    I'm not sure what you are asking.
     
  12. Feb 1, 2015 #11
    So, we expressed [itex]( \frac{A}{W + F} - \frac{B}{W + G}) [/itex] as [itex]( \frac{A}{W} - \frac{B}{W}) \ + \ L [/itex]. You then proceed to further simplify the leftover stuff. However, how can we get from [itex]( \frac{A}{W + F} - \frac{B}{W + G}) [/itex] to [itex]( \frac{A}{W} - \frac{B}{W}) \ + \ L [/itex]?

    Thanks.
     
  13. Feb 1, 2015 #12

    Stephen Tashi

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    I just defined [itex] L [/itex] to be [itex] ( \frac{A}{W + F} - \frac{B}{W + G}) - ( \frac{A}{W} - \frac{B}{W}) [/itex]
     
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