An actual non-answer from solve

The following code results in a an empty sym. Would anyone care to comment on (a) why no solution? and (b) how to actually get a solution from the symbolic math toolbox?

reset(symengine);clear all;
syms x lambda A k positive;
syms eta;

lambda = sym(1);
k = sym(2*pi);

eq1 = A*sin(eta)-10;
eq2 = A*sin(k*lambda/6 + eta)-20;
eq3 = A*sin(k*5*lambda/12 + eta);

% try to solve
b=solve(eq1,eq2,eq3,A,eta,k)

% --> answer is [ empty sym ]

subs(eq1,{A,eta,k},{20,pi/6,2*pi/lambda})
subs(eq2,{A,eta,k},{20,pi/6,2*pi/lambda})
subs(eq3,{A,eta,k},{20,pi/6,2*pi/lambda})

Answer 1

[Sean de Wolski]

You force k to equal 2*pi; how do you expect it to solve for any other value?

Answer 2

[Walter Roberson]

solve() is failing because you are attempting to solve for k, but k has a value already so eq1 eq2 and eq3 do not contain k anymore. Even if they did somehow still contain k, because k has a value, it would be the value that would be passed in to solve() where you list k at the end; you would need to quote k to get just the name passed in

solve(eq1,eq2,eq3,A,eta,'k')

but like I said this would fail because there is no k in eq* because eq* were constructed after k was defined.

Answer 3

[Susan]

Ok,I actually didn't mean to ask it to solve for k any more.

Now the following produces the right answer:

b=solve(eq1,eq2,eq3,A,eta)
>> b.A
ans =
20
>> b.eta
ans =
pi/6

Still, I'm wondering what it would take to solve for all three, which was my original quest:

reset(symengine);clear all;
syms x lambda A k positive;
syms eta;

eq1 = A*sin(eta)-10;
eq2 = A*sin(k*lambda/6 + eta)-20;
eq3 = A*sin(k*5*lambda/12 + eta);

% try to solve
b=solve(eq1,eq2,eq3,A,eta,k)b =
[ empty sym ]

[Walter Roberson]

Works for me in Maple:

solve([eq1, eq2, eq3], [A, eta, k],UseAssumptions);

[[A = 20, eta = (1/6)*Pi, k = 2*Pi/lambda],

[A = 20, eta = (5/6)*Pi, k = 10*Pi/lambda],

[A = 20, eta = (5/6)*Pi, k = -2*Pi/lambda],

[A = 20, eta = (1/6)*Pi, k = -10*Pi/lambda],

[A = (80/7)*14^(1/2), eta = arctan((1/22)*14^(1/2)*2^(1/2)), k = 12*arctan((1/2)*14^(1/2)*2^(1/2))/lambda],

[A = (80/7)*14^(1/2), eta = -arctan((1/22)*14^(1/2)*2^(1/2))+Pi, k = 12*(-arctan((1/2)*14^(1/2)*2^(1/2))+Pi)/lambda],

[A = (80/7)*14^(1/2), eta = -arctan((1/22)*14^(1/2)*2^(1/2))+Pi, k = -12*arctan((1/2)*14^(1/2)*2^(1/2))/lambda],

[A = (80/7)*14^(1/2), eta = arctan((1/22)*14^(1/2)*2^(1/2)), k = 12*(arctan((1/2)*14^(1/2)*2^(1/2))-Pi)/lambda]]

You have to eliminate the ones with negative k from this, as Maple does not seem to fully process the assumptions on k with the assumptions on lambda,