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 Subject: SYSTEM OF EQUATIONS Category: Science > Math Asked by: baroqqque-ga List Price: \$50.00 Posted: 10 Sep 2006 23:40 PDT Expires: 19 Oct 2006 05:56 PDT Question ID: 764063
 ```SYSTEM OF EQUATIONS, ALGEBRA, DIFFERENTIAL EQUATIONS Thanks for taking the time to check this. I have been struggling to solve the following system of equations: (x^?)*((1+x)^(-2))*(((1+y^(1-?))/(1-?))-1-y)=k (y^?)*((1+y)^(-2))*(((1+x^(1-?))/(1-?))-1-x)=(1/k) where 0
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 Subject: Re: SYSTEM OF EQUATIONS From: maplekiwi-ga on 13 Sep 2006 18:53 PDT
 ```I doubt that there is an explicit solution. So, you will have to settle for solving numerically. See http://www.math.gatech.edu/~carlen/2507/notes/NewtonMethod.html for solving two simultaneous implicit equations numerically using the Newton Raphson method. Evidently, you must start with a good guess.```
 Subject: Re: SYSTEM OF EQUATIONS From: mathtalk-ga on 19 Sep 2006 21:27 PDT
 ```Some observations that may be useful for formulating an approximate solution and thus a numerical strategy: As I'm sure baroqqque-ga has already observed, the equations can be restated: f(x) * g(y) = k f(y) * g(x) = 1/k where the functions on [0,1] are defined: f(z) = (z^?)/(1+z)^2 g(z) = ((1+z^(1-?))/(1-?) - (1+z) Bearing in mind that 0 < ? < 1 as well, we find taking the derivative: g'(z) = 1/(z^?) - 1 Hence g(z) is strictly increasing on [0,1], with: g(0) = ?/(1-?) and g(1) = 2?/(1-?). Also g"(z) = -?/(z^(1+?)) < 0 on (0,1), ie. the curve is concave down. The picture is less simple for f(z), which is not monotonic: f'(z) = ?*z^(?-1)/(1+z)^2 - 2*(z^?)/(1+z)^3 = [?/z + ? - 2] * (z^?)/(1+z)^3 Since 0 < ? < 1, we have: f increasing for 0 < z < ?/(2-?), f decreasing for ?/(2-?) < z < 1. In other words f(z) reaches its maximum at z = ?/(2-?), while: f(0) = 0 and f(1) = 1/4 f(?/(2-?)) = (?/(2-?))^?/(1 + ?/(2-?))^2 = (?/(2-?))^?/(2/(2-?))^2 = (?^?)*((2-?)^(2-?))/4 Given these points it would be helpful to know more about the specific values of ? (and k) for which a solution is sought. In principle the invertibility of g(z) can be exploited to eliminate one of the two variables. Let G(g(z)) = z on [0,1] be such an inverse: y = G(k/f(x)) x = G(1/(k*f(y))) = G(1/(k*f(G(k/f(x))))) Thus one obtains x as the fixed point of the function given last in terms of G,f, and k. Conceivably a simple fixed point iteration will produce a satisfactory starting approximation for a more sophisticated Newton iteration. regards, mathtalk-ga```
 Subject: Re: SYSTEM OF EQUATIONS From: maplekiwi-ga on 21 Sep 2006 10:31 PDT
 ```Here are some more observations that constrain k and ?. Combining the bounds on f and g: 0 < f(x)g(y) < 1/2 * ?/(1-?) * ?^? * (2-?)^(2-?) = U(?) We must have 0 < k < U(?) and similarly 0 < 1/k < U(?) giving 1 / U(?) < k < U(?). This requires that U(?) > 1 Solving numerically, ? > 0.6358. Therefore, there are no solutions if ? < 0.6358. If ? > 0.6358 there are no solutions if k is outside the bounds 1/U(?) and U(?). Further investigation is required to determine if there are 0, 1 or more solutions otherwise. The following web site is useful for graphing some of these functions: http://wims.unice.fr/wims/en_home.html```
 Subject: Re: SYSTEM OF EQUATIONS From: maplekiwi-ga on 24 Sep 2006 16:05 PDT
 ```Number of solutions: The graph of G(1/(k*f(G(k/f(x))))) generally looks like a "W" shaped curve. Some parts may be missing when the inverse of g is not defined. Also, the "W" may be deformed, looking more like the normal (statistical) curve. My experiments show that the number of solutions is 0, 1, 2, 3 or 4. Here are some numerical results: 0 solutions: ? = 0.7 k = 0.8 1 solution: ? = 0.8 k = 1 (x=0.1123,y=0.1123) 2 solutions: ? = 0.7 k = 0.95 (0.1559,0.8957) (0.1850,0.2925) 3 solutions: ? = 0.68 k = 1.03 (0.2271,0.9666) (0.3441,0.2278) (0.6863,0.1956) 4 solutions: ? = 0.66 k = 1.005 (0.2802,0.6015) (0.3948,0.3138) (0.5045,0.2832) (0.8927,0.9222)```
 Subject: Re: SYSTEM OF EQUATIONS From: maplekiwi-ga on 30 Sep 2006 14:02 PDT
 ```Graphs Make obvious changes to the website name: cs*senecac*on*ca/~lew*baxter to see graphs of f(x)g(y)=k and f(y)g(x) = 1/k where ? and k can be set.```