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Simplify the mathematical expression in any desired form?

Isa Comez

Isa Comez, KTU

Posted 6 years ago

Dear All, I want to write the following expression f[x_] := (E^(-h k0 x) + E^( h k0 x)) (-(-1 + 1/4 E^(-h (n1 + n2) x) (1 + E^(2 h n1 x)) (1 + E^( 2 h n2 x))) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33) - 1/4 E^(-h (n1 + n2) x) (-1 + E^(2 h n1 x)) (-1 + E^( 2 h n2 x)) (m2^2 Z13 - m1 m2^2 n1 Z33 + m1^2 (Z13 - m2 n2 Z33))); grouping by the exponential parts, that is, f[x]=A1* Exp(B1x)+ A2 * Exp(B2x). +An * Exp(Bnx) + C0 How can I do it? Thanks Isa Attachments:* SimplifyFunction.nb

Dear All,

I want to write the following expression

f[x_] := (E^(-h k0 x) + E^(
     h k0 x)) (-(-1 + 
         1/4 E^(-h (n1 + n2) x) (1 + E^(2 h n1 x)) (1 + E^(
            2 h n2 x))) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33) - 
     1/4 E^(-h (n1 + n2) x) (-1 + E^(2 h n1 x)) (-1 + E^(
        2 h n2 x)) (m2^2 Z13 - m1 m2^2 n1 Z33 + 
        m1^2 (Z13 - m2 n2 Z33)));

grouping by the exponential parts, that is, f[x]=A1* Exp(B1x)+ A2 * Exp(B2x). +An * Exp(Bn*x) + C0

How can I do it?

Thanks

Isa

POSTED BY: Isa Comez

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Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

Maybe like so: Simplify[Expand[f[x]] /. Exp[a__] :> \[FormalE][a]] /. \[FormalE][a__] :> Exp[a]

POSTED BY: Henrik Schachner

Bill Nelson

Posted 6 years ago

Asking for "any desired form" probably doesn't have an answer. Asking it to group by exponents is more feasible. I believe this will accomplish that f[x_] := (E^(-h k0 x) + E^(h k0 x)) (-(-1 + 1/4 E^(-h (n1 + n2) x) (1 + E^(2 h n1 x)) (1 + E^( 2 h n2 x))) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33) - 1/4 E^(-h (n1 + n2) x) (-1 + E^(2 h n1 x) )(-1 + E^(2 h n2 x)) (m2^2 Z13 - m1 m2^2 n1 Z33 + m1^2 (Z13 - m2 n2 Z33))); Expand[f[x]] //. a_Exp[c_]+b_Exp[c_] :> (a+b)*Exp[c]

Asking for "any desired form" probably doesn't have an answer.

Asking it to group by exponents is more feasible.

I believe this will accomplish that

f[x_] := (E^(-h k0 x) + E^(h k0 x)) (-(-1 + 1/4 E^(-h (n1 + n2) x) (1 + E^(2 h n1 x)) (1 + E^(
2 h n2 x))) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33) - 1/4 E^(-h (n1 + n2) x) (-1 + E^(2 h n1 x)
)(-1 + E^(2 h n2 x)) (m2^2 Z13 - m1 m2^2 n1 Z33 + m1^2 (Z13 - m2 n2 Z33)));
Expand[f[x]] //. a_*Exp[c_]+b_*Exp[c_] :> (a+b)*Exp[c]

POSTED BY: Bill Nelson

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

POSTED BY: Henrik Schachner

Isa Comez

Isa Comez, KTU

Posted 6 years ago

@Bill Nelson Great ! Many thanks. @Henrik Schachner I also thank you so much. That is what i want.

POSTED BY: Isa Comez

Isa Comez

Isa Comez, KTU

Posted 6 years ago

Thanks Dr Schachner for your reply. However this code does not gather the terms that have the same exponential expressions. Please see the following and attached file 1/4 (-E^(h (k0 - n1 - n2) x) m1^2 Z13 + E^(-h (k0 + n1 - n2) x) m1^2 Z13 + E^(h (k0 + n1 - n2) x) m1^2 Z13 + E^(-h (k0 - n1 + n2) x) m1^2 Z13 + E^(h (k0 - n1 + n2) x) m1^2 Z13 - E^(h (-k0 + n1 + n2) x) m1^2 Z13 - E^(-h (k0 + n1 + n2) x) m1^2 Z13 + 2 E^(h (k0 - n1 - n2) x) m1 m2 Z13 + 2 E^(-h (k0 + n1 - n2) x) m1 m2 Z13 + 2 E^(h (k0 + n1 - n2) x) m1 m2 Z13 + 2 E^(-h (k0 - n1 + n2) x) m1 m2 Z13 + 2 E^(h (k0 - n1 + n2) x) m1 m2 Z13 + 2 E^(h (-k0 + n1 + n2) x) m1 m2 Z13 + 2 E^(-h (k0 + n1 + n2) x) m1 m2 Z13 - E^(h (k0 - n1 - n2) x) m2^2 Z13 + E^(-h (k0 + n1 - n2) x) m2^2 Z13 + E^(h (k0 + n1 - n2) x) m2^2 Z13 + E^(-h (k0 - n1 + n2) x) m2^2 Z13 + E^(h (k0 - n1 + n2) x) m2^2 Z13 - E^(h (-k0 + n1 + n2) x) m2^2 Z13 - E^(-h (k0 + n1 + n2) x) m2^2 Z13 - E^(h (k0 - n1 - n2) x) m1^2 m2 n1 Z33 - E^(-h (k0 + n1 - n2) x) m1^2 m2 n1 Z33 - E^(h (k0 + n1 - n2) x) m1^2 m2 n1 Z33 - E^(-h (k0 - n1 + n2) x) m1^2 m2 n1 Z33 - E^(h (k0 - n1 + n2) x) m1^2 m2 n1 Z33 - E^(h (-k0 + n1 + n2) x) m1^2 m2 n1 Z33 - E^(-h (k0 + n1 + n2) x) m1^2 m2 n1 Z33 + E^(h (k0 - n1 - n2) x) m1 m2^2 n1 Z33 - E^(-h (k0 + n1 - n2) x) m1 m2^2 n1 Z33 - E^(h (k0 + n1 - n2) x) m1 m2^2 n1 Z33 - E^(-h (k0 - n1 + n2) x) m1 m2^2 n1 Z33 - E^(h (k0 - n1 + n2) x) m1 m2^2 n1 Z33 + E^(h (-k0 + n1 + n2) x) m1 m2^2 n1 Z33 + E^(-h (k0 + n1 + n2) x) m1 m2^2 n1 Z33 + E^(h (k0 - n1 - n2) x) m1^2 m2 n2 Z33 - E^(-h (k0 + n1 - n2) x) m1^2 m2 n2 Z33 - E^(h (k0 + n1 - n2) x) m1^2 m2 n2 Z33 - E^(-h (k0 - n1 + n2) x) m1^2 m2 n2 Z33 - E^(h (k0 - n1 + n2) x) m1^2 m2 n2 Z33 + E^(h (-k0 + n1 + n2) x) m1^2 m2 n2 Z33 + E^(-h (k0 + n1 + n2) x) m1^2 m2 n2 Z33 - E^(h (k0 - n1 - n2) x) m1 m2^2 n2 Z33 - E^(-h (k0 + n1 - n2) x) m1 m2^2 n2 Z33 - E^(h (k0 + n1 - n2) x) m1 m2^2 n2 Z33 - E^(-h (k0 - n1 + n2) x) m1 m2^2 n2 Z33 - E^(h (k0 - n1 + n2) x) m1 m2^2 n2 Z33 - E^(h (-k0 + n1 + n2) x) m1 m2^2 n2 Z33 - E^(-h (k0 + n1 + n2) x) m1 m2^2 n2 Z33 - E^(h (k0 + n1 + n2) x) (m1 - m2) ((m1 - m2) Z13 + m1 m2 (n1 - n2) Z33) + 4 E^(-h k0 x) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33) + 4 E^(h k0 x) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33)) Attachments: SimplifyFunction_V2.nb

Thanks Dr Schachner for your reply. However this code does not gather the terms that have the same exponential expressions. Please see the following and attached file

1/4 (-E^(h (k0 - n1 - n2) x) m1^2 Z13 + 
   E^(-h (k0 + n1 - n2) x) m1^2 Z13 + 
   E^(h (k0 + n1 - n2) x) m1^2 Z13 + 
   E^(-h (k0 - n1 + n2) x) m1^2 Z13 + 
   E^(h (k0 - n1 + n2) x) m1^2 Z13 - 
   E^(h (-k0 + n1 + n2) x) m1^2 Z13 - 
   E^(-h (k0 + n1 + n2) x) m1^2 Z13 + 
   2 E^(h (k0 - n1 - n2) x) m1 m2 Z13 + 
   2 E^(-h (k0 + n1 - n2) x) m1 m2 Z13 + 
   2 E^(h (k0 + n1 - n2) x) m1 m2 Z13 + 
   2 E^(-h (k0 - n1 + n2) x) m1 m2 Z13 + 
   2 E^(h (k0 - n1 + n2) x) m1 m2 Z13 + 
   2 E^(h (-k0 + n1 + n2) x) m1 m2 Z13 + 
   2 E^(-h (k0 + n1 + n2) x) m1 m2 Z13 - 
   E^(h (k0 - n1 - n2) x) m2^2 Z13 + 
   E^(-h (k0 + n1 - n2) x) m2^2 Z13 + 
   E^(h (k0 + n1 - n2) x) m2^2 Z13 + 
   E^(-h (k0 - n1 + n2) x) m2^2 Z13 + 
   E^(h (k0 - n1 + n2) x) m2^2 Z13 - 
   E^(h (-k0 + n1 + n2) x) m2^2 Z13 - 
   E^(-h (k0 + n1 + n2) x) m2^2 Z13 - 
   E^(h (k0 - n1 - n2) x) m1^2 m2 n1 Z33 - 
   E^(-h (k0 + n1 - n2) x) m1^2 m2 n1 Z33 - 
   E^(h (k0 + n1 - n2) x) m1^2 m2 n1 Z33 - 
   E^(-h (k0 - n1 + n2) x) m1^2 m2 n1 Z33 - 
   E^(h (k0 - n1 + n2) x) m1^2 m2 n1 Z33 - 
   E^(h (-k0 + n1 + n2) x) m1^2 m2 n1 Z33 - 
   E^(-h (k0 + n1 + n2) x) m1^2 m2 n1 Z33 + 
   E^(h (k0 - n1 - n2) x) m1 m2^2 n1 Z33 - 
   E^(-h (k0 + n1 - n2) x) m1 m2^2 n1 Z33 - 
   E^(h (k0 + n1 - n2) x) m1 m2^2 n1 Z33 - 
   E^(-h (k0 - n1 + n2) x) m1 m2^2 n1 Z33 - 
   E^(h (k0 - n1 + n2) x) m1 m2^2 n1 Z33 + 
   E^(h (-k0 + n1 + n2) x) m1 m2^2 n1 Z33 + 
   E^(-h (k0 + n1 + n2) x) m1 m2^2 n1 Z33 + 
   E^(h (k0 - n1 - n2) x) m1^2 m2 n2 Z33 - 
   E^(-h (k0 + n1 - n2) x) m1^2 m2 n2 Z33 - 
   E^(h (k0 + n1 - n2) x) m1^2 m2 n2 Z33 - 
   E^(-h (k0 - n1 + n2) x) m1^2 m2 n2 Z33 - 
   E^(h (k0 - n1 + n2) x) m1^2 m2 n2 Z33 + 
   E^(h (-k0 + n1 + n2) x) m1^2 m2 n2 Z33 + 
   E^(-h (k0 + n1 + n2) x) m1^2 m2 n2 Z33 - 
   E^(h (k0 - n1 - n2) x) m1 m2^2 n2 Z33 - 
   E^(-h (k0 + n1 - n2) x) m1 m2^2 n2 Z33 - 
   E^(h (k0 + n1 - n2) x) m1 m2^2 n2 Z33 - 
   E^(-h (k0 - n1 + n2) x) m1 m2^2 n2 Z33 - 
   E^(h (k0 - n1 + n2) x) m1 m2^2 n2 Z33 - 
   E^(h (-k0 + n1 + n2) x) m1 m2^2 n2 Z33 - 
   E^(-h (k0 + n1 + n2) x) m1 m2^2 n2 Z33 - 
   E^(h (k0 + n1 + n2) x) (m1 - m2) ((m1 - m2) Z13 + 
      m1 m2 (n1 - n2) Z33) + 
   4 E^(-h k0 x) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33) + 
   4 E^(h k0 x) m1 m2 (-2 Z13 + m1 n1 Z33 + m2 n2 Z33))

POSTED BY: Isa Comez

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