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Mathematica Algebra Wolfram Language

Why I cannot get the results again?

Jie Li

Posted 10 years ago

Here is my equations, and I solved it once in a few second. But it cannot solve again when I tried to make a change (I did not change yet), (where E^ is for exponentiation) Solve[{2Dnena1 == a4Dse, Dsea4 == Dpe4a6(n + s - l), a1n^2 + a2 == a4n + a5, a6(l - n - s)^2 + a7 == a4(n + s) + a5, -a1enn^3bE^(bt)/3 - a2nbE^(bt) == a + 2 Dnen(1 - E^(bt)), -a4esb((n + s)^2 - n^2)E^(bt)/2 - a5bessE^(b t) == 0, epa6bE^(bt)(n + s - l)/3 - a7ep(l - n - s)E^(bt) == -a - Dps4a6(1 - E^(bt))(n + s - l)}, {a1, a2, a4, a5, a6, a7, b}] below is the solution it give me at first time: {{a1 -> -(( a Dse (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( Dne ep n^2 (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a2 -> -((a (Dse n + Dne s) (-Dne Dse ep - 6 Dpe Dse n^2 + 2 Dpe Dse en n^2 - 6 Dne Dpe n s))/( Dne ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a4 -> -((2 a (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a5 -> -((a (2 n + s) (-Dne Dse ep - 6 Dpe Dse n^2 + 2 Dpe Dse en n^2 - 6 Dne Dpe n s))/( ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a6 -> -((a Dse (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( 2 Dpe ep n (-l + n + s)^2 (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), a7 -> (a (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s))/( 2 Dpe ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s)), b -> (3 Dne ep (l - n - s) (Dse l - Dse n + 2 Dpe s - Dse s))/( Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s)}} I have no idea why the program cannot solve the equation again. Thank you a lot!

Here is my equations, and I solved it once in a few second. But it cannot solve again when I tried to make a change (I did not change yet), (where E^ is for exponentiation)

Solve[{2*Dne*n*a1 == a4*Dse, Dse*a4 == Dpe*4*a6*(n + s - l), 
  a1*n^2 + a2 == a4*n + a5, 
  a6*(l - n - s)^2 + a7 == 
   a4*(n + s) + a5, -a1*en*n^3*b*E^(b*t)/3 - a2*n*b*E^(b*t) == 
   a + 2 Dne*n*(1 - E^(b*t)), -a4*es*b*((n + s)^2 - n^2)*E^(b*t)/2 - 
    a5*b*es*s*E^(b* t) == 0, 
  ep*a6*b*E^(b*t)*(n + s - l)/3 - a7*ep*(l - n - s)*E^(b*t) == -a - 
    Dps*4*a6*(1 - E^(b*t))*(n + s - l)}, {a1, a2, a4, a5, a6, a7, b}]

below is the solution it give me at first time:

{{a1 -> -((
    a Dse (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 
       6 Dne Dpe n s))/(
    Dne ep n^2 (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + 
       Dse n - 2 Dpe s + Dse s))), 
  a2 -> -((a (Dse n + Dne s) (-Dne Dse ep - 6 Dpe Dse n^2 + 
       2 Dpe Dse en n^2 - 6 Dne Dpe n s))/(
    Dne ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + 
       Dse n - 2 Dpe s + Dse s))), 
  a4 -> -((2 a (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 
       6 Dne Dpe n s))/(
    ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + 
       Dse n - 2 Dpe s + Dse s))), 
  a5 -> -((a (2 n + s) (-Dne Dse ep - 6 Dpe Dse n^2 + 
       2 Dpe Dse en n^2 - 6 Dne Dpe n s))/(
    ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s) (-Dse l + 
       Dse n - 2 Dpe s + Dse s))), 
  a6 -> -((a Dse (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 
       6 Dne Dpe n s))/(
    2 Dpe ep n (-l + n + s)^2 (-3 Dse n + Dse en n - 
       3 Dne s) (-Dse l + Dse n - 2 Dpe s + Dse s))), 
  a7 -> (a (Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 
      6 Dne Dpe n s))/(
   2 Dpe ep n (-l + n + s) (-3 Dse n + Dse en n - 3 Dne s)), 
  b -> (3 Dne ep (l - n - s) (Dse l - Dse n + 2 Dpe s - Dse s))/(
   Dne Dse ep + 6 Dpe Dse n^2 - 2 Dpe Dse en n^2 + 6 Dne Dpe n s)}}

I have no idea why the program cannot solve the equation again.

Thank you a lot!

POSTED BY: Jie Li

3 Replies

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Bill Simpson

Posted 10 years ago

The behavior you are seeing was a design decision in the forum software that it was more important to make the default behavior of * be to toggle italic on or off, and hide the * in the process, than it was to display the * as typed. An * separated by white space is exempt from this. Many new users do not realize this.

POSTED BY: Bill Simpson

Eric Johnstone

Eric Johnstone, McGill University

Posted 10 years ago

Hi Jie Li, I think something has happened to your original equation set. There is no separation between the variables, so the multiplications have been lost. For example, if you copy out the equations list and then do a Shift/Return, you get {2 Dnena1 == a4Dse, Dsea4 == Dpe4a6 (-l + n + s), a1n^2 + a2 == a4n + a5, a7 + a6 (l - n - s)^2 == a5 + a4 (n + s), -a2nbE^bt - (a1enn^3 bE^bt)/3 == a + 2 Dnen (1 - E^bt), -a5bessE^bt - 1/2 a4esb E^bt (-n^2 + (n + s)^2) == 0, -a7ep E^bt (l - n - s) + 1/3 epa6bE^bt (-l + n + s) == -a - Dps4a6 (1 - E^bt) (-l + n + s)} Eric

Hi Jie Li,

I think something has happened to your original equation set. There is no separation between the variables, so the multiplications have been lost. For example, if you copy out the equations list and then do a Shift/Return, you get

{2 Dnena1 == a4Dse, Dsea4 == Dpe4a6 (-l + n + s), 
 a1n^2 + a2 == a4n + a5, 
 a7 + a6 (l - n - s)^2 == 
  a5 + a4 (n + s), -a2nbE^bt - (a1enn^3 bE^bt)/3 == 
  a + 2 Dnen (1 - E^bt), -a5bessE^bt - 
   1/2 a4esb E^bt (-n^2 + (n + s)^2) == 
  0, -a7ep E^bt (l - n - s) + 1/3 epa6bE^bt (-l + n + s) == -a - 
   Dps4a6 (1 - E^bt) (-l + n + s)}

Eric

POSTED BY: Eric Johnstone

Jie Li

Posted 10 years ago

Thank your reply, I edited the code but it still not work.

POSTED BY: Jie Li

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