Error: Invalid expression log equation

I am trying to solve an equation in Matlab mobile.

Following is the code -

 syms t a

 eqns =-a*log((3*t/4)*(3-t^2)/(2-t))-(1-a)*log((1/t^2)*((3-2*t)/(2-t)))+a*((1-t)^2)*(3-t)/((2-t)*(3-t^2)-(1-a)*(1+t)/(3-2*t))*((t*(-18+21*t-7*t^3+2*t^4)*(log((3-2*t)*(3-t^2)/(2*t(2-t)^2))))/(36-39*t+13*t^3-4*t^4+a*(-18+27*t-18*t^2+5*t^3)))+2*log((1/3)*(3-t^2)/(2-t))+t*log((4*(t^3)/27)*((3-t^2)/(3-2*t)))-2*t==0;

But I keep getting the error

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

Answer 1

[Walter Roberson]

syms t a

 eqns =-a*log((3*t/4)*(3-t^2)/(2-t))-(1-a)*log((1/t^2)*((3-2*t)/(2-t)))+a*((1-t)^2)*(3-t)/((2-t)*(3-t^2)-(1-a)*(1+t)/(3-2*t))*((t*(-18+21*t-7*t^3+2*t^4)*(log((3-2*t)*(3-t^2)/(2*t(2-t)^2))))/(36-39*t+13*t^3-4*t^4+a*(-18+27*t-18*t^2+5*t^3)))+2*log((1/3)*(3-t^2)/(2-t))+t*log((4*(t^3)/27)*((3-t^2)/(3-2*t)))-2*t==0;

Error using sym/subsindex (line 864)
Invalid indexing or function definition. Indexing must follow MATLAB indexing. Function arguments must be symbolic variables, and function body must be sym expression.

Error in sym/subsref (line 909)
R_tilde = builtin('subsref',L_tilde,Idx);

%            12     1 2     1 2   10 1   0    12     1 23     2 3   210   12   1  0 1   0 12   1 2     1 2   1 2   1 2     10 12  3                    2 3   45     4 5     4 5   6   5  4321 2                       3                     210      12   1 2     1 2   10      12  3   2   1 23     2 3     210

The bracket count is okay, it all balances out, and the error message you indicate does not occur. A different error occurs instead.

Have a look at the place the bracket count reaches 6:

% (2*t(2-t)^2)
% 5   6   5  4

Notice that you are trying to index t by 2-t . That is an error. You probably want (2*t*(2-t)^2)

[Sabrina Garland]

Thanks @Walter. It helped!

I wrote this for solution but i got the following error -

Warning: Solutions are parameterized by the symbols: [z, z1, z2], z1, z2. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'. > In sym/solve>warnIfParams (line 475) In sym/solve (line 357) Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'. > In sym/solve>warnIfParams (line 478) In sym/solve (line 357)

Is there a problem with following syntax?

Soln1= solve(eqn, t, a)

And how can fix domain for t and a between 0 and 1 for solution?

Thanks!

[Walter Roberson]

In theory....

 syms t a real
 assumeAlso(t >= 0 & t <= 1)
 assumeAlso(a >= 0 & a <= 1)

 eqn =-a*log((3*t/4)*(3-t^2)/(2-t))-(1-a)*log((1/t^2)*((3-2*t)/(2-t)))+a*((1-t)^2)*(3-t)/((2-t)*(3-t^2)-(1-a)*(1+t)/(3-2*t))*((t*(-18+21*t-7*t^3+2*t^4)*(log((3-2*t)*(3-t^2)/(2*t*(2-t)^2))))/(36-39*t+13*t^3-4*t^4+a*(-18+27*t-18*t^2+5*t^3)))+2*log((1/3)*(3-t^2)/(2-t))+t*log((4*(t^3)/27)*((3-t^2)/(3-2*t)))-2*t==0;
 Soln1 = solve(eqn, t, a, 'returnconditions', true, 'maxdegree', 3)
 Soln1.a
 Soln1.t
 Soln1.conditions
 

In practice, this is going to take a long time to compute, and the answers are probably going to be difficult to understand.

I will see if I can figure out anything more useful.

[Walter Roberson]

Numerically, solutions include ones near

[
[0.80814734211076188233189764141161, 0.014157347318462409368745988278694]
[0.81774556051925063803415290145514, 0.017065493529859707439482572850095]
[0.82046261258401125142952904134027, 0.017914365002905370826371367553424]
[0.83165113247410330040663713139042, 0.021511344137270273069885390156103]
[0.86235181325356451804474711062975, 0.032006454691267775967023011531556]
[0.87728607400691739688823315300636, 0.037335270904563136299631940445499]
[0.89107163976355238503111378722198,  0.04233730158150518105730901834881]
[0.89922319441534142597544034193216, 0.045323008878958357652554141689136]
[0.92700631525224144351158031340986, 0.055600120484060467263409966499192]
[0.93050016660009881345983008776409, 0.056899680092455104708862984031671]
[0.94723947876611726467967480882978, 0.063136374364404389516159540203479]
[0.99119633032880843853815771634112, 0.079505614940861258639688187409334]
]

It is worth checking the accuracy of some of those: for example are the 2nd and 3rd really the same solution, or are they distinct solutions?

[Walter Roberson]

The computation took about 3/4 of an hour and produced 33 possible solutions. Now I am seeing if I can make any sense out of the conditions.

[Sabrina Garland]

Thank you so much! That was of great help. It solved my problem!!! I am not sure if i can thank you enough.

[Walter Roberson]

So... take

 syms t a real
 eqn =-a*log((3*t/4)*(3-t^2)/(2-t))-(1-a)*log((1/t^2)*((3-2*t)/(2-t)))+a*((1-t)^2)*(3-t)/((2-t)*(3-t^2)-(1-a)*(1+t)/(3-2*t))*((t*(-18+21*t-7*t^3+2*t^4)*(log((3-2*t)*(3-t^2)/(2*t*(2-t)^2))))/(36-39*t+13*t^3-4*t^4+a*(-18+27*t-18*t^2+5*t^3)))+2*log((1/3)*(3-t^2)/(2-t))+t*log((4*(t^3)/27)*((3-t^2)/(3-2*t)))-2*t;
 fsurf(eqn, [0 1 0 1]);
 view(3)

As you look closer in to this, it turns out that are there are an infinite number of solutions along a curved line. The limits are at approximately t < 0.1 and a > 0.7 and that otherwise you should be able to find a solution. For example,

 t1 = 0.0123;
 eqn1 = subs(eqn, t, t1)

eqn1 = 

 a1 = vpasolve(eqn1)

a1 = 

[Sabrina Garland]

@Walter Sorry to disturb you again but I keep getting solution in parameters. Why is that?

Soln1 =

struct with fields:

             t: [1×1 sym]
             a: [1×1 sym]
    parameters: [1×2 sym]
    conditions: [1×1 sym]

Solution for a is coming out to be z1. Why?

[Walter Roberson]

I demonstrated above that the solution is not a point or finite set of points, but is instead a line. As such, if MATLAB is going to be able to produce an answer at all, it would have to parameterize the line in some variable, and say "t has this relationship to the parameter, and a has this other relationship to the parameter."

For example:

syms x y real
sol = solve(x^2 + y^2 == 4, [x, y], 'returnconditions', true)

sol = struct with fields:
             x: [2×1 sym]
             y: [2×1 sym]
    parameters: [1×1 sym]
    conditions: [2×1 sym]

sol.x

ans = 

sol.y

ans = 

sol.conditions

ans = 

MATLAB had to parameterize the solution because there are an infinite number of solutions.

[Sabrina Garland]

Oh! Thanks at Walter! Then how did you get the vector of numerical solutions that you printed? (I am new to Matlab and don't know the exact syntax for this).

[Walter Roberson]

I used vpasolve with different random starting points, rand(2,1) and threw away results that did not meet the constraints.

[Sabrina Garland]

Okay! Thanks again!!!

[Tomas]

I'm also getting a similar errror in a code that I am trying to write. I have tried writing the equation in about three or four different ways and I'm still getting the same error. The expression and error are shown below.Again this is just one of the forms that I've tried for the equation. I've tried -1.*log(2-2u) and -log(2-2u). Any help is appreciated!

f_inv = @x log(-2(2-u))

f_inv = @x log(-2(2-u))

↑

Invalid expression. Check for missing multiplication operator,

missing or unbalanced delimiters, or other syntax error. To

construct matrices, use brackets instead of parentheses.

[Walter Roberson]

-2(2-u)

There is absolutely no where in MATLAB that supports implicit multiplication. -2(2-u) as far as MATLAB is concerned, is unary minus applied to the result of indexing the number 2 by the index 2-u -- which is an error because literals cannot be subscripted in MATLAB.

[Tomas]

the U in this case is a random variable. So, should I just change it to a different random variable to clear that issue?

[Walter Roberson]

 f_inv = @(x) log(-2*(2-x))

This will be complex-valued unless x >= 2.

It is not at all clear why you do not instead write

 f_inv = @(x) log(2*(x-2))