How to decompose the following expression?

  • Thread starter Alec V
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  • #1
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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!


 

Answers and Replies

  • #2
35,261
11,512
Try to find the other direction, that is easier (and all steps work both ways). You can simplify equation 2 and get equation 1.
 
  • #3
Stephen Tashi
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Eq. 2

gif.gif

You have to explain the relations that exist among the symbols to get help on this question.
 
  • #4
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Eq. 2
gif.gif

Stephen Tashi said:
You have to explain the relations that exist among the symbols to get help on this question.
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.
 
  • #5
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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.
 
  • #6
35,261
11,512
∏ 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.
 
  • #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.)


#include <stdio.h>
#include <math.h>


void main()

{


double pi_1, pi_2, delta, ep_1, ep_2;
double frac1,frac2, numerator, denom1, denom2;
double epv_1, epv_2;
pi_1 = 10.0;
pi_2 = 30.0;
delta = 0.25;
ep_1 = 6.0;
ep_2 = 5.0;
printf("pi_1 = %6.3f pi_2 = %6.3f delta = %6.3f ep_1 = %6.3f ep_2 = %6.3f\n\n",pi_1,pi_2,delta,ep_1,ep_2);

denom1 = 1.0 - delta + delta * ep_1;
frac1 = pi_1/denom1;
printf("frac1 %6.4f\n", frac1);
denom2 = 1.0 - delta + delta * ep_2;
frac2 = pi_2/denom2;
printf("frac2 %6.4f\n",frac2);
epv_1 = frac1 - frac2;
printf("epv_1 = %6.3f\n\n",epv_1);

frac1 = (pi_1 - pi_2)/ (1.0 - delta);
printf("frac1 %6.4f\n", frac1);
frac2 = ( delta * ( denom1 * ep_2 * pi_2 - denom2 * ep_1 * pi_1) )/ ((1.0 - delta)*denom1* denom2 );
printf("frac2 %6.4f\n",frac2);
epv_2 = frac1 + frac2;
printf("epv_2 = %6.3f\n", epv_2);


}
The output

pi_1 = 10.000 pi_2 = 30.000 delta = 0.250 ep_1 = 6.000 ep_2 = 5.000

frac1 4.4444
frac2 15.0000
epv_1 = -10.556

frac1 -26.6667
frac2 16.1111
epv_2 = -10.556
 
  • #8
Stephen Tashi
Science Advisor
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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].
 
  • #9
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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].

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]?
 
  • #10
Stephen Tashi
Science Advisor
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. 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]?
I'm not sure what you are asking.
 
  • #11
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I'm not sure what you are asking.
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.
 
  • #12
Stephen Tashi
Science Advisor
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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]?
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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