I wanted to solve the following equation in MATLAB R2013a using the Symbolic Math Toolbox.
(y/x)-(((1+r)^n)-1)/r=0 where y,x and n>3 are given and r is the dependent variable
I tried myself & coded as follows:
f=solve('(y/x)-(((1+r)^n)-1)/r','r')
but as the solution for r is not exact i.e. it is converging on successive iterations hence MATLAB is giving a warning output with the message
Warning: Explicit solution could not be found.
f =
[ empty sym ]
How do I code this?
There are an infinite number of solutions to this for an unspecified value of n > 3 and unknown r. I hope that it's pretty clear why – it's effectively asking for a greater and greater number of roots of (1+r)^n. You can find solutions for fixed values of n, however. Note that as n becomes larger there are more and more solutions and of course some of them are complex. I'm going to assume that you're only interested in real values of r. You can use solve and symbolic math for n = 4, n = 5, and n = 6 (for n = 6, the solution may not be in a convenient form):
y = 441361;
x = 66990;
n = 5;
syms r;
rsol = solve(y/x-((1+r)^n-1)/r==0,r,'IgnoreAnalyticConstraints',true)
double(rsol)
However, the question is "do you need all the solutions or just a particular solution for a given value of n"? If you just need a particular solution, you shouldn't be using symbolic math at all as it's slower and has practical issues like the ones you're experiencing. You can instead just use a numerical approach to find a zero of the equation that is near a specified initial guess. fzero is the standard function for solving this sort of problem in a single variable:
y = 441361;
x = 66990;
n = 5;
f = #(r)y/x-((1+r).^n-1)./r;
r0 = 1;
rsol = fzero(f,r0)
You'll see that the value returned is the same as one of the solutions from the symbolic solution above. If you adjust the initial guess r0 (say r0 = -3), it will return the other solution. When using numeric approaches in cases when there are multiple solutions, if you want specific solutions you'll need to know about the behavior of your function and you'll need to add some clever extra code to choose initial guesses.
I think you forgot to define n as well.
f=solve('(y/x)-(((1+r)^n)-1)/r=0','n-3>0','r','n')
Should solve your problem :)
Related
I am trying to compare two simple expressions using Matlab symbolic toolbox. For some reason, the code returns 0. Any idea ?
syms a b c
A = (a/b)^c
B = a^c/b^c
isequal(A,B)
It seems like MATLAB has a hard time telling that two expressions are the same when (potentially) fractional exponents are involved.
So, one solution, as suggested by Mikhail is to restrict the values of c to be only integers although, as discussed in the Math.SE question jodag posted, there is nothing wrong with fractional exponents in this case.
Hence, since this restriction to integers is not necessary for the statement to be true, another solution is to use simplify function on the expression for B but allowing it to run more simplification steps in order to get the most simplified expression.
syms a b c
A = (a/b)^c
B = a^c/b^c
isequal(A,simplify(B,'step',4))
Four steps is actually the smallest number that worked for me, but that could vary across versions of MATLAB I'm assuming. To be sure, I would include more, but for really large expressions, this could become computationally intensive, so some judgment is necessary. Note that, you could also use the 'Seconds' option to limit the amount of time allowed for simplification.
In general what you wrote isn't true, under the right "assumptions" it becomes true: for example, assuming c is an integer you can trick MATLAB into expanding A
clc; clear all;
syms a
syms b
syms c integer
A = (a/b)^c;
B = simplify((a^c)/(b^c));
disp(isequal(A,B));
disp(A);
disp(B);
1
a^c/b^c
a^c/b^c
I'm using Matlab 2014b. I've tried:
clear all
syms x real
assumeAlso(x>=5)
This returned:
ans =
[ 5 <= x, in(x, 'real')]
Then I tried:
int(sqrt(x^2-25)/x,x)
But this still returned a complex answer:
(x^2 - 25)^(1/2) - log(((x^2 - 25)^(1/2) + 5*i)/x)*5*i
I tried the simplify command, but still a complex answer. Now, this might be fixed in the latest version of Matlab. If so, can people let me know or offer a suggestion for getting the real answer?
The hand-calculated answer is sqrt(x^2-25)-5*asec(x/5)+C.
This behavior is present in R2017b, though when converted to floating point the imaginary components are different.
Why does this occur?
This occurs because Matlab's int function returns the full general solution when you ask for the indefinite integral. This solution is valid over the entire domain of of real values, including your restricted domain of x>=5.
With a bit of math you can show that the solution is always real for x>=5 (see complex logarithm). Or you can use more symbolic math via the isAlways function to show this:
syms x real
assume(x>=5)
y = int(sqrt(x^2-25)/x, x)
isAlways(imag(y)==0)
This returns true (logical 1). Unfortunately, Matlab's simplification routines appear to not be able to reduce this expression when assumptions are included. You might also submit this case to The MathWorks as a service request in case they'd consider improving the simplification for this and similar equations.
How can this be "fixed"?
If you want to get rid of the zero-valued imaginary part of the solution you can use sym/real:
real(y)
which returns 5*atan2(5, (x^2-25)^(1/2)) + (x^2-25)^(1/2).
Also, as #SardarUsama points out, when the full solution is converted to floating point (or variable precision) there will sometimes numeric imprecision when converting from exact symbolic form. Using the symbolic real form above should avoid this.
The answer is not really complex.
Take a look at this:
clear all; %To clear the conditions of x as real and >=5 (simple clear doesn't clear that)
syms x;
y = int(sqrt(x^2-25)/x, x)
which, as we know, gives:
y =
(x^2 - 25)^(1/2) - log(((x^2 - 25)^(1/2) + 5i)/x)*5i
Now put some real values of x≥5 to check what result it gives:
n = 1004; %We'll be putting 1000 values of x in y from 5 to 1004
yk = zeros(1000,1); %Preallocation
for k=5:n
yk(k-4) = subs(y,x,k); %Putting the value of x
end
Now let's check the imaginary part of the result we have:
>> imag(yk)
ans =
1.0e-70 *
0
0
0
0
0.028298997121333
0.028298997121333
0.028298997121333
%and so on...
Notice the multiplier 1e-70.
Let's check the maximum value of imaginary part in yk.
>> max(imag(yk))
ans =
1.131959884853339e-71
This implies that the imaginary part is extremely small and it is not a considerable amount to be worried about. Ideally it may be zero and it's coming due to imprecise calculations. Hence, it is safe to call your result real.
The following command
syms x real;
f = #(x) log(x^2)*exp(-1/(x^2));
fp(x) = diff(f(x),x);
fpp(x) = diff(fp(x),x);
and
solve(fpp(x)>0,x,'Real',true)
return the result
solve([0.0 < (8.0*exp(-1.0/x^2))/x^4 - (2.0*exp(-1.0/x^2))/x^2 -
(6.0*log(x^2)*exp(-1.0/x^2))/x^4 + (4.0*log(x^2)*exp(-1.0/x^2))/x^6],
[x == RD_NINF..RD_INF])
which is not what I expect.
The first question: Is it possible to force Matlab's solve to return the set of all solutions?
(This is related to this question.) Moreover, when I try to solve the equation
solve(fpp(x)==0,x,'Real',true)
which returns
ans =
-1.5056100417680902125994180096313
I am not satisfied since all solutions are not returned (they are approximately -1.5056, 1.5056, -0.5663 and 0.5663 obtained from WolframAlpha).
I know that vpasolve with some initial guess can handle this. But, I have no idea how I can generally find initial guessed values to obtain all solutions, which is my second question.
Other solutions or suggestions for solving these problems are welcomed.
As I indicated in my comment above, sym/solve is primarily meant to solve for analytic solutions of equations. When this fails, it tries to find a numeric solution. Some equations can have an infinite number of numeric solutions (e.g., periodic equations), and thus, as per the documentation: "The numeric solver does not try to find all numeric solutions for [the] equation. Instead, it returns only the first solution that it finds."
However, one can access the features of MuPAD from within Matlab. MuPAD's numeric::solve function has several additional capabilities. In particular is the 'AllRealRoots' option. In your case:
syms x real;
f = #(x)log(x^2)*exp(-1/(x^2));
fp(x) = diff(f(x),x);
fpp(x) = diff(fp(x),x);
s = feval(symengine,'numeric::solve',fpp(x)==0,x,'AllRealRoots')
which returns
s =
[ -1.5056102995536617698689500437312, -0.56633904710786569620564475006904, 0.56633904710786569620564475006904, 1.5056102995536617698689500437312]
as well as a warning message.
My answer to this question provides other way that various MuPAD solvers can be used, particularly if you can isolate and bracket your roots.
The above is not going to directly help with your inequalities other than telling you where the function changes sign. For those you could try:
s = feval(symengine,'solve',fpp(x)>0,x,'Real')
which returns
s =
(Dom::Interval(0, Inf) union Dom::Interval(-Inf, 0)) intersect solve(0 < 2*log(x^2) - 3*x^2*log(x^2) + 4*x^2 - x^4, x, Real)
Try plotting this function along with fpp.
While this is not a bug per se, The MathWorks still might be interested in this difference in behavior and poor performance of sym/solve (and the underlying symobj::solvefull) relative to MuPAD's solve. File a bug report if you like. For the life of me I don't understand why they can't better unify these parts of Matlab. The separation makes not sense from the perspective of a user.
I have the following script that defines a function and sets up and equation:
H = #(f) sum(log(f));
f = rand(1, 1);
syms a
H(f)-H(f-a)
I want to solve H(f)-H(f-a)=0 for a. I tried using fzero in the following manner, fzero('H(f)-H(f-a)', 0), but this doesn't yield me anything useful.
well the clear answer for this would be that a=0. This is of course the obvious answer, and may be why you are not getting "anything useful", especially when only considering one value for f.
However, you can actually get useful information when you add more values and use the solve function.
H = #(f) sum(log(f));
f = rand(5, 1);
syms a
temp = H(f)-H(f-a);
out = double(solve(temp == 0,a));
This will output a vector with all values of a that are solutions to your function. Just note that this may not result in the best answers in general, numerical solving of high order functions is rather difficult and unreliable. Because of this, it may take a long time to run as well.
I am trying to use solve() to solve a system of equations of the following form
eq1=a1x+a2y;
eq2=b1x+b2y;
where a1 = .05 for values of x<5, .1 for values of 5
Is there a way to solve for this using solve? As in sol = solve(eq1,eq2);
I'm not sure what you're trying to do here. Can you please post a real example (with numbers) and what you would like the output to be?
I think you're trying to solve linear simultaeneous equations. Assuming that is what you are trying to do:
I would suggest multiplying all of your equations by 20, so that your minimum quanta size of 0.05 becomes 1.00. Your problem then becomes the solution of linear equations for integer values.
Note that if the system is fully constrained (that is, if there are n independent constraints on the n equations you want to solve) then there will only be one solution and it may not necessarily be an integer solution. For example the system:
1 = 2a + 4b
3 = a + b
has the solution a = 5.5, b = -2.5. No other solution is possible.
For under-constrained systems, i.e.
0 = 3x + y
x > 0
Then there will be an infinite number of solutions, some of which may have both x and y being integer values. (Or there may be no integer solutions at all.)
Okay let me give you a quick rundown.
if you want to solve an equation or a system of equations and conditions then you need to define them as such, so let me explain.
so by example
clear all; %just to be safe
syms x y b
a=0.5;
somevalue=1;
someothervalue=3;
eq1= a*x+a*y == somevalue; %this is your first equation
eq2= b*x+b*y == someothervalue; %this is your 2nd equation
cond1= x<5; %this is a condition which matlab sees as an "equation"
eqs=[eq1,eq2,cond1]; %these are the equations and conditions you want to solve for, use this for solve
eqs=[eq1,eq2]; %use this for vpasolve and set your condition in range
vars=[x,y,b]; %these are the variable you want to solve for
range = [-Inf 5; NaN NaN; NaN NaN]; %NaN means you set no range
%you can use solve or vpasolve, second one being numeric, which is the one you'll probably want
n=5;
sol=zeros(n,numel(vars));
for i = 1:n
temp1 = vpasolve(eqs, vars, range, 'random', true);
temp = vpasolve(eqs, vars, 'random', true);
sol(i,1) = temp.x;
sol(i,2) = temp.y;
sol(i,3) = temp.b;
end
sol
Now when I run this myself I can't get the range to properly work for some reason, still trying to figure that out. When you don't set a range it works just fine, if you can use the solve function then there also isn't a problem.
In theory the range function should work fine like this so it might be a bug on my end.
If you use solve you have some nice options where you can use assume to set extra conditions that are a bit more advanced, like only checking for real solutions or only integers, etc.