# Unable to solve a set of equations with Solve

I'm trying to solve the following equation using Mathematica 9.0.1.0 :

Solve[E^(-σ τ) k (1 + T σ)^-n == Subscript[M, 0] &&
2*E^(-σ τ) k (1 + T σ)^(-1 - n) (n T + τ + T σ τ) ==
Subscript[M, 1] &&
1/2 E^(-σ τ) k (1 + T σ)^(-2 - n) (n T^2 + n^2 T^2 + 2 n T τ +
2 n T^2 σ τ + τ^2 + 2 T σ τ^2 +
T^2 σ^2 τ^2)*6 == Subscript[M, 2] &&
k (-(1/6) E^(-σ τ) n (1 + n) (2 + n) T^3 (1 + T σ)^(-3 - n) -
1/2 E^(-σ τ) n (1 + n) T^2 (1 + T σ)^(-2 - n) τ -
1/2 E^(-σ τ) n T (1 + T σ)^(-1 - n) τ^2 -
1/6 E^(-σ τ) (1 + T σ)^-n τ^3) == Subscript[M, 3], {k, T,
n, τ}]


Mathematica outputs the following error:

Solve::nsmet: This system cannot be solved with the methods available to Solve. >>


And when I try to use "SolveAlwyas" instead of "Solve" I get the following errors :

Solve::incnst: Inconsistent or redundant transcendental equation. After reduction, the bad equation is 1-((1+T σ)^n)^(1/n) == 0. >>
Solve::incnst: Inconsistent or redundant transcendental equation. After reduction, the bad equation is -1+((1+T σ)^n)^(1/n) == 0. >>
Solve::incnst: Inconsistent or redundant transcendental equation. After reduction, the bad equation is 1-((1+T σ)^(-1-n))^(1/(-1-n)) == 0. >>
General::stop: Further output of Solve::incnst will be suppressed during this calculation. >>


This system can be solved.

Is there any way to solve this equations with Mathematica without giving numerical values to the parameters ?

• This is a vague question. – Artes Jan 12 '14 at 11:33
• How is it vague @Artes ? Try to run the code I provided, you won't get any solution. I can't find a way to solve this set of equations using Mathematica. Someone posted an answer where Reduce was used instead of Solve but it didn't work either. – James Dean Jan 12 '14 at 20:43
• I had answered your question but mistakenly I had evaluated n = 3 before, so it was a special solution for n == 3 only, when I noticed it I decided to delete my answer. I'm not going to work on your question since you haven't clarified your question. I'm guessinng n should be a natural number, and all parameters should be real and some of them positive. Then you have to decide what are the parameters and what are the unknowns. Unfortunately you haven't suggested what is your purpose, so I couldn't provide a viable answer. Moreover most likely SolveAlways isn't helpful. – Artes Jan 12 '14 at 21:26
• Indeed I am sorry. These are the unknowns : n is a positive integer, k, T and tau are positive real numbers. The other parameters Subscript[M, i] are known parameters and I want to express the unknowns in terms of them. – James Dean Jan 12 '14 at 21:34
• Still a bit vague. What about σ? Why do you need solutions for all natural n? How did you solve your equation since we have so many unclear assumptions? Are m0,...,m3 arbitrary or are they related to each other? – Artes Jan 12 '14 at 21:43

It can be solved in this way

From my observation, your are solving this system:

$$g(\sigma) = e^{-\sigma \tau}k(1+T\tau)^{-n}$$

$$g(\sigma)=A_0$$ $$g'(\sigma)=A_1$$ $$g''(\sigma)=A_2$$ $$g'''(\sigma)=A_3$$

My constants $A$ differ from your $M$ by some factor, you should be able to see it.

First, eliminate $k$

Reduce[g == A0 && g1 == A1 && g2 == A2 && g3 == A3, k]


you will get,

(1 + T \[Sigma] != 0 && A0 n != 0 && A1 + A0 \[Tau] != 0 &&
E^(\[Sigma] \[Tau]) (1 + T \[Sigma])^(1 + 2 n) != 0 &&
A1 == (-A0 n T - A0 \[Tau] - A0 T \[Sigma] \[Tau])/(
1 + T \[Sigma]) &&
A2 == (A1^2 + A1^2 n + 2 A0 A1 \[Tau] + A0^2 \[Tau]^2)/(A0 n) &&
A3 == (-A1^2 A2 + 2 A0 A2^2 - 2 A1^3 \[Tau] + 3 A0 A1 A2 \[Tau])/(
A0 (A1 + A0 \[Tau])))) &&
k == A0 E^(\[Sigma] \[Tau]) (1 + T \[Sigma])^n


And then solve for the variables without $k$

Solve[A1 == (-A0 n T - A0 \[Tau] - A0 T \[Sigma] \[Tau])/(
1 + T \[Sigma]) &&
A2 == (A1^2 + A1^2 n + 2 A0 A1 \[Tau] + A0^2 \[Tau]^2)/(A0 n) &&
A3 == (-A1^2 A2 + 2 A0 A2^2 - 2 A1^3 \[Tau] + 3 A0 A1 A2 \[Tau])/(
A0 (A1 + A0 \[Tau])), {T, \[Tau], n}]


By putting back the factors relating $A$ and $M$ , you will get the answer.

• Thank you for your answer it was very helpful. – James Dean Jan 14 '14 at 16:57