Help with 'solve' function in MATLAB for numerical solution!

For example, lets say I have the expression "x=2^x*b+c" and I use solve function as,

d=solve('x=(2^x)*b+c');

Now if I were to assign 'b' and 'c' values prior to writing the solve statement, the solution in 'd' will still return a SYMBOLIC solution.

If b=2, c=3, for example and I write,

d=solve('x=(2^x)*2+3');

I will get the numerical solution, but INSTEAD

if i type

d=solve('x=(2^x)*b+c');

with the values of 'b' and 'c' declared prior to this statement, I get a symbolic solution.

What do I need to do to get a NUMERICAL answer each time? (This is because the values of 'b' and 'c' are changing within a loop) (Also note this is just an example to illustrate my problem, not the real function I wish to solve)

Thanks Ian

Answer 1

[Matt Fig]

d = solve('x=(2^x)*b+c');
n = subs(d,{'b','c'},{2,3})  % Put 2 for b, 3 for c

If you are doing the substitutions in a loop, this will probably be faster:

C = matlabFunction(d); % Convert symbolic into a func handle.
C(2,3)

[Ian]

Matt I tried it, but its giving me an [empty sym]

This is the actual function: d=solve('x=2^(1+(b-3)/(3*2^x))-c','x');

And I get d=[empty sym] Even if I carry out n=subs(d,{'b','c'},{2,3}) I still get []

But if the values were inserted into the solve equation like so, d=solve('x=2^(1+(2-3)/(3*2^x))-3','x') then I get d=-2.41....

Any way to work around this? Thanks Ian

[Matt Fig]

Yes, that is a different problem isn't it?!

When symbolics won't work, go numerical...

b = 2;
c = 3;
f = @(x)  x - (2.^(1+(b-3)./(3*2.^x))-c);
fzero(f,1)
ans =

   -2.4184

[Ian]

thank you!

Answer 2

[Babak]

After you get the symbolic solution "d", you may need to say:

 num_sol = vpa(d)

which computes d with numbers provided before. You should look at the list of the functions provided in Symbolic Math Toolbox.

Answer 3

[Heidi Miller]

%%

%

syms x y g

x=x==6

x=6

g = 3*pi*x.^2

gives the numerical answer too

[Walter Roberson]

The original question had x on both sides of the equation.

The code you post here does not have x on both sides of the equation. It also overwrites values in such a way that the code is completely equivalent to just the last two lines, assigning numeric 6 to x and then calculating g purely numerically. No equation solving is performed.