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Equation Solving Wolfram Language

Solve not working

Neerav Kaushal

Posted 4 years ago

I am trying to solve the following equation for x, but it is not working at all. What I am doing wrong? c1 = 1.610^(-33) c2 = 5.2310^(-27) eqn = ExpandAll[(c2/2) + ((1/c1)*(((x^2)/4) + c1)^(1/2)) == Log[(x/2 - ((((x^2)/4) + c1)^(1/2)))/((((x^2)/4) + c1 - x/2)^(1/2))]] Solve[eqn, x]

POSTED BY: Neerav Kaushal

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Werner Geiger

Posted 4 years ago

You should add the domain Reals to Solve, i.e. Solve[eqn,x,Reals] Then Solve (as well as NSolve) ends quickly with the correct solution {}, i.e. there is no solution. Block[{c1, c2, x, eqn}, c1 = 1.610^(-33); c2 = 5.2310^(-27); eqn = ExpandAll[(c2/2) + ((1/c1)(((x^2)/4) + c1)^(1/2)) == Log[(x/2 - ((((x^2)/4) + c1)^(1/2)))/((((x^2)/4) + c1 - x/2)^(1/2))]]; Solve[eqn, x, Reals] ] {} You can see this when you take care of the legal real x-domains of the equations left and right side. From the Log-Expression on the right, one can see that: x/2 > Sqrt[c1 + x^2/4] must hold. But this is never True for non-negative c1. I have enhanced your code a bit. One can see there what happens. See the attachment. Set orgVals at will for your or my (somewhat nicer) c1-, c2-Values. BTW.: If you allow complex values, neither Solve nor DSolve can solve the problem. Attachments:* 200711 Solve not...nb

POSTED BY: Werner Geiger

Bill Nelson

Posted 4 years ago

Look at this c1 = 1.610^(-33); c2 = 5.2310^(-27); Table[{x, (c2/2) + ((1/c1)(((x^2)/4) + c1)^(1/2)), Log[(x/2 - ((((x^2)/4) + c1)^(1/2)))/((((x^2)/4) + c1 - x/2)^(1/2))]}, {x,-10,10}] which returns {{-10, 3.125^33, 0.601986402162968 + PiI}, {-9, 2.8125^33, 0.5928118328288697 + PiI}, {-8, 2.5^33, 0.5815754049028404 + PiI}, {-7, 2.1875^33, 0.5674899664194921 + PiI}, {-6, 1.875^33, 0.549306144334055 + PiI}, {-5, 1.5625^33, 0.5249110622493387 + PiI}, {-4, 1.25^33, 0.49041462650586326 + PiI}, {-3, 9.375^32, 0.43773436867694987 + PiI}, {-2, 6.25^32, 0.3465735902799726 + PiI}, {-1, 3.125^32, 0.1438410362258906 + PiI}, {0, 2.5^16, 0. + PiI}, {1, 3.125^32, Indeterminate}, {2, 6.25^32, Indeterminate}, {3, 9.375^32, Indeterminate}, {4, 1.25^33, Indeterminate}, {5, 1.5625^33, Indeterminate}, {6, 1.875^33, Indeterminate}, {7, 2.1875^33, Indeterminate}, {8, 2.5^33, Indeterminate}, {9, 2.8125^33, Indeterminate}, {10, 3.125^33, Indeterminate}} along with warnings about a division by zero and 0.ComplexInfinity

Look at this

c1 = 1.6*10^(-33);
c2 = 5.23*10^(-27);
Table[{x,
  (c2/2) + ((1/c1)*(((x^2)/4) + c1)^(1/2)),
  Log[(x/2 - ((((x^2)/4) + c1)^(1/2)))/((((x^2)/4) + c1 - x/2)^(1/2))]},
  {x,-10,10}]

which returns

{{-10, 3.125*^33, 0.601986402162968 + Pi*I}, 
 {-9, 2.8125*^33, 0.5928118328288697 + Pi*I}, 
 {-8, 2.5*^33, 0.5815754049028404 + Pi*I}, 
 {-7, 2.1875*^33, 0.5674899664194921 + Pi*I}, 
 {-6, 1.875*^33, 0.549306144334055 + Pi*I}, 
 {-5, 1.5625*^33, 0.5249110622493387 + Pi*I}, 
 {-4, 1.25*^33, 0.49041462650586326 + Pi*I}, 
 {-3, 9.375*^32, 0.43773436867694987 + Pi*I}, 
 {-2, 6.25*^32, 0.3465735902799726 + Pi*I}, 
 {-1, 3.125*^32, 0.1438410362258906 + Pi*I}, 
 {0, 2.5*^16, 0. + Pi*I}, 
 {1, 3.125*^32, Indeterminate},
 {2, 6.25*^32, Indeterminate},
 {3, 9.375*^32, Indeterminate}, 
 {4, 1.25*^33, Indeterminate},
 {5, 1.5625*^33, Indeterminate},
 {6, 1.875*^33, Indeterminate}, 
 {7, 2.1875*^33, Indeterminate},
 {8, 2.5*^33, Indeterminate},
 {9, 2.8125*^33, Indeterminate}, 
 {10, 3.125*^33, Indeterminate}}

along with warnings about a division by zero and 0.*ComplexInfinity

POSTED BY: Bill Nelson

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