I'm trying to compute a rather ugly integral using MATLAB. What I'm having problem with though is a part where I multiply a very big number (>10^300) with a very small number (<10^-300). MATLAB returns 'inf' for this even though it should be in the range of 0-0.0005. This is what I have
besselFunction = #(u)besseli(qb,2*sqrt(lambda*(theta + mu)).*u);
exponentFuncion = #(u)exp(-u.*(lambda + theta + mu));
where qb = 5, lambda = 12, theta = 10, mu = 3. And what I want to find is
besselFunction(u)*exponentFunction(u)
for all real values of u. The problem is that whenever u>28 it will be evaluated as 'inf'. I've heared, and tried, to use MATLAB function 'vpa' but it doesn't seem to work well when I want to use functions...
Any tips will be appreciated at this point!
I'd use logarithms.
Let x = Bessel function of u and y = x*exp(-u) (simpler than your equation, but similar).
Since log(v*w) = log(v) + log(w), then log(y) = log(x) + log(exp(-u))
This simplifies to
log(y) = log(x) - u
This will be better behaved numerically.
The other key will be to not evaluate that Bessel function that turns into a large number and passing it to a math function to get the log. Better to write your own that returns the logarithm of the Bessel function directly. Look at a reference like Abramowitz and Stegun to try and find one.
If you are doing an integration, consider using Gauss–Laguerre quadrature instead. The basic idea is that for equations of the form exp(-x)*f(x), the integral from 0 to inf can be approximated as sum(w(X).*f(X)) where the values of X are the zeros of a Laguerre polynomial and W(X) are specific weights (see the Wikipedia article). Sort of like a very advanced Simpson's rule. Since your equation already has an exp(-x) part, it is particularly suited.
To find the roots of the polynomial, there is a function on MATLAB Central called LaguerrePoly, and from there it is pretty straightforward to compute the weights.
Related
I'm solving a pair of non-linear equations for each voxel in a dataset of a ~billion voxels using fsolve() in MATLAB 2016b.
I have done all the 'easy' optimizations that I'm aware of. Memory localization is OK, I'm using parfor, the equations are in fairly numerically simple form. All discontinuities of the integral are fed to integral(). I'm using the Levenberg-Marquardt algorithm with good starting values and a suitable starting damping constant, it converges on average with 6 iterations.
I'm now at ~6ms per voxel, which is good, but not good enough. I'd need a order of magnitude reduction to make the technique viable. There's only a few things that I can think of improving before starting to hammer away at accuracy:
The splines in the equation are for quick sampling of complex equations. There are two for each equation, one is inside the 'complicated nonlinear equation'. They represent two equations, one which is has a large amount of terms but is smooth and has no discontinuities and one which approximates a histogram drawn from a spectrum. I'm using griddedInterpolant() as the editor suggested.
Is there a faster way to sample points from pre-calculated distributions?
parfor i=1:numel(I1)
sols = fsolve(#(x) equationPair(x, input1, input2, ...
6 static inputs, fsolve options)
output1(i) = sols(1); output2(i) = sols(2)
end
When calling fsolve, I'm using the 'parametrization' suggested by Mathworks to input the variables. I have a nagging feeling that defining a anonymous function for each voxel is taking a large slice of the time at this point. Is this true, is there a relatively large overhead for defining the anonymous function again and again? Do I have any way to vectorize the call to fsolve?
There are two input variables which keep changing, all of the other input variables stay static. I need to solve one equation pair for each input pair so I can't make it a huge system and solve it at once. Do I have any other options than fsolve for solving pairs of nonlinear equations?
If not, some of the static inputs are the fairly large. Is there a way to keep the inputs as persistent variables using MATLAB's persistent, would that improve performance? I only saw examples of how to load persistent variables, how could I make it so that they would be input only once and future function calls would be spared from the assumedly largish overhead of the large inputs?
EDIT:
The original equations in full form look like:
Where:
and:
Everything else is known, solving for x_1 and x_2. f_KN was approximated by a spline. S_low (E) and S_high(E) are splines, the histograms they are from look like:
So, there's a few things I thought of:
Lookup table
Because the integrals in your function do not depend on any of the parameters other than x, you could make a simple 2D-lookup table from them:
% assuming simple (square) range here, adjust as needed
[x1,x2] = meshgrid( linspace(0, xmax, N) );
LUT_high = zeros(size(x1));
LUT_low = zeros(size(x1));
for ii = 1:N
LUT_high(:,ii) = integral(#(E) Fhi(E, x1(1,ii), x2(:,ii)), ...
0, E_high, ...
'ArrayValued', true);
LUT_low(:,ii) = integral(#(E) Flo(E, x1(1,ii), x2(:,ii)), ...
0, E_low, ...
'ArrayValued', true);
end
where Fhi and Flo are helper functions to compute those integrals, vectorized with scalar x1 and vector x2 in this example. Set N as high as memory will allow.
Those lookup tables you then pass as parameters to equationPair() (which allows parfor to distribute the data). Then just use interp2 in equationPair():
F(1) = I_high - interp2(x1,x2,LUT_high, x(1), x(2));
F(2) = I_low - interp2(x1,x2,LUT_low , x(1), x(2));
So, instead of recomputing the whole integral every time, you evaluate it once for the expected range of x, and reuse the outcomes.
You can specify the interpolation method used, which is linear by default. Specify cubic if you're really concerned about accuracy.
Coarse/Fine
Should the lookup table method not be possible for some reason (memory limitations, in case the possible range of x is too big), here's another thing you could do: split up the whole procedure in 2 parts, which I'll call coarse and fine.
The intent of the coarse method is to improve your initial estimates really quickly, but perhaps not so accurately. The quickest way to approximate that integral by far is via the rectangle method:
do not approximate S with a spline, just use the original tabulated data (so S_high/low = [S_high/low#E0, S_high/low#E1, ..., S_high/low#E_high/low]
At the same values for E as used by the S data (E0, E1, ...), evaluate the exponential at x:
Elo = linspace(0, E_low, numel(S_low)).';
integrand_exp_low = exp(x(1)./Elo.^3 + x(2)*fKN(Elo));
Ehi = linspace(0, E_high, numel(S_high)).';
integrand_exp_high = exp(x(1)./Ehi.^3 + x(2)*fKN(Ehi));
then use the rectangle method:
F(1) = I_low - (S_low * Elo) * (Elo(2) - Elo(1));
F(2) = I_high - (S_high * Ehi) * (Ehi(2) - Ehi(1));
Running fsolve like this for all I_low and I_high will then have improved your initial estimates x0 probably to a point close to "actual" convergence.
Alternatively, instead of the rectangle method, you use trapz (trapezoidal method). A tad slower, but possibly a bit more accurate.
Note that if (Elo(2) - Elo(1)) == (Ehi(2) - Ehi(1)) (step sizes are equal), you can further reduce the number of computations. In that case, the first N_low elements of the two integrands are identical, so the values of the exponentials will only differ in the N_low + 1 : N_high elements. So then just compute integrand_exp_high, and set integrand_exp_low equal to the first N_low elements of integrand_exp_high.
The fine method then uses your original implementation (with the actual integral()s), but then starting at the updated initial estimates from the coarse step.
The whole objective here is to try and bring the total number of iterations needed down from about 6 to less than 2. Perhaps you'll even find that the trapz method already provides enough accuracy, rendering the whole fine step unnecessary.
Vectorization
The rectangle method in the coarse step outlined above is easy to vectorize:
% (uses R2016b implicit expansion rules)
Elo = linspace(0, E_low, numel(S_low));
integrand_exp_low = exp(x(:,1)./Elo.^3 + x(:,2).*fKN(Elo));
Ehi = linspace(0, E_high, numel(S_high));
integrand_exp_high = exp(x(:,1)./Ehi.^3 + x(:,2).*fKN(Ehi));
F = [I_high_vector - (S_high * integrand_exp_high) * (Ehi(2) - Ehi(1))
I_low_vector - (S_low * integrand_exp_low ) * (Elo(2) - Elo(1))];
trapz also works on matrices; it will integrate over each column in the matrix.
You'd call equationPair() then using x0 = [x01; x02; ...; x0N], and fsolve will then converge to [x1; x2; ...; xN], where N is the number of voxels, and each x0 is 1×2 ([x(1) x(2)]), so x0 is N×2.
parfor should be able to slice all of this fairly easily over all the workers in your pool.
Similarly, vectorization of the fine method should also be possible; just use the 'ArrayValued' option to integral() as shown above:
F = [I_high_vector - integral(#(E) S_high(E) .* exp(x(:,1)./E.^3 + x(:,2).*fKN(E)),...
0, E_high,...
'ArrayValued', true);
I_low_vector - integral(#(E) S_low(E) .* exp(x(:,1)./E.^3 + x(:,2).*fKN(E)),...
0, E_low,...
'ArrayValued', true);
];
Jacobian
Taking derivatives of your function is quite easy. Here is the derivative w.r.t. x_1, and here w.r.t. x_2. Your Jacobian will then have to be a 2×2 matrix
J = [dF(1)/dx(1) dF(1)/dx(2)
dF(2)/dx(1) dF(2)/dx(2)];
Don't forget the leading minus sign (F = I_hi/lo - g(x) → dF/dx = -dg/dx)
Using one or both of the methods outlined above, you can implement a function to compute the Jacobian matrix and pass this on to fsolve via the 'SpecifyObjectiveGradient' option (via optimoptions). The 'CheckGradients' option will come in handy there.
Because fsolve usually spends the vast majority of its time computing the Jacobian via finite differences, manually computing a value for it manually will normally speed the algorithm up tremendously.
It will be faster, because
fsolve doesn't have to do extra function evaluations to do the finite differences
the convergence rate will increase due to the improved precision of the Jacobian
Especially if you use the rectangle method or trapz like above, you can reuse many of the computations you've already done for the function values themselves, meaning, even more speed-up.
Rody's answer was the correct one. Supplying the Jacobian was the single largest factor. Especially with the vectorized version, there were 3 orders of magnitude of difference in speed with the Jacobian supplied and not.
I had trouble finding information about this subject online so I'll spell it out here for future reference: It is possible to vectorize independant parallel equations with fsolve() with great gains.
I also did some work with inlining fsolve(). After supplying the Jacobian and being smarter about the equations, the serial version of my code was mostly overhead at ~1*10^-3 s per voxel. At that point most of the time inside the function was spent passing around a options -struct and creating error-messages which are never sent + lots of unused stuff assumedly for the other optimization functions inside the optimisation function (levenberg-marquardt for me). I succesfully butchered the function fsolve and some of the functions it calls, dropping the time to ~1*10^-4s per voxel on my machine. So if you are stuck with a serial implementation e.g. because of having to rely on the previous results it's quite possible to inline fsolve() with good results.
The vectorized version provided the best results in my case, with ~5*10^-5 s per voxel.
I'm using octave 3.8.1 which works like matlab.
I have an array of thousands of values I've only included three groupings as an example below:
(amp1=0.2; freq1=3; phase1=1; is an example of one grouping)
t=0;
amp1=0.2; freq1=3; phase1=1; %1st grouping
amp2=1.4; freq2=2; phase2=1.7; %2nd grouping
amp3=0.8; freq3=5; phase3=1.5; %3rd grouping
The Octave / Matlab code below solves for Y so I can plug it back into the equation to check values along with calculating values not located in the array.
clear all
t=0;
Y=0;
a1=[.2,3,1;1.4,2,1.7;.8,5,1.5]
for kk=1:1:length(a1)
Y=Y+a1(kk,1)*cos ((a1(kk,2))*t+a1(kk,3))
kk
end
Y
PS: I'm not trying to solve for Y since it's already solved for I'm trying to solve for Phase
The formulas located below are used to calculate Phase but I'm not sure how to put it into a for loop that will work in an array of n groupings:
How would I write the equation / for loop for finding the phase if I want to find freq=2.5 and amp=.23 and the phase is unknown I've looked online and it may require writing non linear equations which I'm not sure how to convert what I'm trying to do into such an equation.
phase1_test=acos(Y/amp1-amp3*cos(2*freq3*pi*t+phase3)/amp1-amp2*cos(2*freq2*pi*t+phase2)/amp1)-2*freq1*pi*t
phase2_test=acos(Y/amp2-amp3*cos(2*freq3*pi*t+phase3)/amp2-amp1*cos(2*freq1*pi*t+phase1)/amp2)-2*freq2*pi*t
phase3_test=acos(Y/amp3-amp2*cos(2*freq2*pi*t+phase2)/amp3-amp1*cos(2*freq1*pi*t+phase1)/amp3)-2*freq2*pi*t
Image of formula below:
I would like to do a check / calculate phases if given a freq and amp values.
I know I have to do a for loop but how do I convert the phase equation into a for loop so it will work on n groupings in an array and calculate different values not found in the array?
Basically I would be given an array of n groupings and freq=2.5 and amp=.23 and use the formula to calculate phase. Note: freq will not always be in the array hence why I'm trying to calculate the phase using a formula.
Ok, I think I finally understand your question:
you are trying to find a set of phase1, phase2,..., phaseN, such that equations like the ones you describe are satisfied
You know how to find y, and you supply values for freq and amp.
In Matlab, such a problem would be solved using, for example fsolve, but let's look at your problem step by step.
For simplicity, let me re-write your equations for phase1, phase2, and phase3. For example, your first equation, the one for phase1, would read
amp1*cos(phase1 + 2 freq1 pi t) + amp2*cos(2 freq2 pi t + phase2) + amp3*cos(2 freq3 pi t + phase3) - y = 0
Note that ampX (X is a placeholder for 1, 2, 3) are given, pi is a constant, t is given via Y (I think), freqX are given.
Hence, you are, in fact, dealing with a non-linear vector equation of the form
F(phase) = 0
where F is a multi-dimensional (vector) function taking a multi-dimensional (vector) input variable phase (comprised of phase1, phase2,..., phaseN). And you are looking for the set of phaseX, where all of the components of your vector function F are zero. N.B. F is a shorthand for your functions. Therefore, the first component of F, called f1, for example, is
f1 = amp1*cos(phase1+...) + amp2*cos(phase2+...) + amp3*cos(phase3+...) - y = 0.
Hence, f1 is a one-dimensional function of phase1, phase2, and phase3.
The technical term for what you are trying to do is find a zero of a non-linear vector function, or find a solution of a non-linear vector function. In Matlab, there are different approaches.
For a one-dimensional function, you can use fzero, which is explained at http://www.mathworks.com/help/matlab/ref/fzero.html?refresh=true
For a multi-dimensional (vector) function as yours, I would look into using fsolve, which is part of Matlab's optimization toolbox (which means I don't know how to do this in Octave). The function fsolve is explained at http://www.mathworks.com/help/optim/ug/fsolve.html
If you know an approximate solution for your phases, you may also look into iterative, local methods.
In particular, I would recommend you look into the Newton's Method, which allows you to find a solution to your system of equations F. Wikipedia has a good explanation of Newton's Method at https://en.wikipedia.org/wiki/Newton%27s_method . Newton iterations are very simple to implement and you should find a lot of resources online. You will have to compute the derivative of your function F with respect to each of your variables phaseX, which is very simple to compute since you're only dealing with cos() functions. For starters, have a look at the one-dimensional Newton iteration method in Matlab at http://www.math.colostate.edu/~gerhard/classes/331/lab/newton.html .
Finally, if you want to dig deeper, I found a textbook on this topic from the society for industrial and applied math: https://www.siam.org/books/textbooks/fr16_book.pdf .
As you can see, this is a very large field; Newton's method should be able to help you out, though.
Good luck!
I have a code that needs to evaluate the arc length equation below:
syms x
a = 10; b = 10; c = 10; d = 10;
fun = 4*a*x^3+3*b*x^2+2*c*x+d
int((1+(fun)^2)^.5)
but all that returns is below:
ans = int(((40*x^3 + 30*x^2 + 20*x + 10)^2 + 1)^(1/2), x)
Why wont matlab evaluate this integral? I added a line under to check if it would evaulate int(x) and it returned the desired result.
Problems involving square roots of functions may be tricky to intgrate. I am not sure whether the integral exists or not, but it if you look up the integral of a second order polynomial you will see that this one is already quite a mess. What you would have, would you expand the function inside the square root, would be a ninth order polynomial. If this integral actually would exist it may be too complex to calculate.
Anyway, if you think about it, would anyone really become any wiser by finding the analytical solution of this? If that is not the case a numerical solution should be sufficient.
EDIT
As thewaywewalk said in the comment, a general rule to calculate these kinds of integrals would be valuable, but to know the primitive function to the particular integral would probably be overkill (if a solution could be found).
Instead define the function as an anonymous function
fun = #(x) sqrt((4*a*x.^3+3*b*x.^2+2*c*x+d).^2+1);
and use integral to evaluate the function between some range, eg
integral(fun,0,100);
for evaluating the function in the closed interval [0,100].
My project require me to use Matlab to create a symbolic equation with square wave inside.
I tried to write it like this but to no avail:
syms t;
a=square(t);
Input arguments must be 'double'.
What can i do to solve this problem? Thanks in advance for the helps offered.
here are a couple of general options using floor and sign functions:
f=#(A,T,x0,x) A*sign(sin((2*pi*(x-x0))/T));
f=#(A,T,x0,x) A*(-1).^(floor(2*(x-x0)/T));
So for example using the floor function:
syms x
sqr=2*floor(x)-floor(2*x)+1;
ezplot(sqr, [-2, 2])
Here is something to get you started. Recall that we can express a square wave as a Fourier Series expansion. I won't bother you with the details, but you can represent any periodic function as a summation of cosines and sines (à la #RTL). Without going into the derivation, this is the closed-form equation for a square wave of frequency f, with a peak-to-peak amplitude of 2 (i.e. it goes from -1 to 1). Recall that the frequency is the amount of cycles per seconds. Therefore, f = 1 means that we repeat our square wave every second.
Basically, what you have to do is code up the first line of the equation... but how in the world would you do that? Welcome to the world of the Symbolic Math Toolbox. What we will need to do before hand is declare what our frequency is. Let's assume f = 1 for now. With the Symbolic Math Toolbox, you can define what are considered as mathematics variables within MATLAB. After, MATLAB has a whole suite of tools that you can use to evaluate functions that rely on these variables. A good example would be if you want to use this to define a closed-form solution of a function f(x). You can then use diff to differentiate and see what the derivative is. Try it yourself:
syms x;
f = x^4;
df = diff(f);
syms denotes that you are declaring anything coming after the statement to be a mathematical variable. In this case, x is just that. df should now give you 4x^3. Cool eh? In any case, let's get back to our problem at hand. We see that there are in fact two variables in the periodic square function that need to be defined: t and k. Once we do this, we need to create our function that is inside the summation first. We can do this by:
syms t k;
f = 1; %//Define frequency here
funcSum = (sin(2*pi*(2*k - 1)*f*t) / (2*k - 1));
That settles that problem... now how do we encapsulate this into an infinite sum!? The sum command in MATLAB assumes that we have a finite array to sum over. If you want to symbolically sum over a function, we must use the symsum function. We usually call it like this:
funcOut = symsum(func, v, start, finish);
func is the function we wish to sum over. v is the summation variable that we wish to use to index in the sum. In our case, that's k. start is the beginning of the sum, which is 1 in our case, and finish is where we wish to finish up our summation. In our case, that's infinity, and so MATLAB has a special keyword called Inf to denote that. Therefore:
xsquare = (4/pi) * symsum(funcSum, k, 1, Inf);
xquare now contains your representation of a square wave defined in terms of the Symbolic Math Toolbox. Now, if you want to plot your square wave and see if we have this right. We can do the following. Let's go between -3 <= t <= 3. As such, you would do something like this:
tVector = -3 : 0.01 : 3; %// Choose a step size of 0.01
yout = subs(xsquare, t, tVector);
You will notice though that there will be some values that are NaN. The reason why is because right at a multiple of the period (T = 1, 2, 3, ...), the behaviour is undefined as the derivative right at these points is undefined. As such, we can fill this in using either 1 or -1. Let's just choose 1 for now. Also, because the Fourier Series is generally a complex-valued function, and the square-wave is purely real, the output of this function will actually give you a complex-valued vector. As such, simply chop off the complex parts to get the real parts only:
yout = real(double(yout)); %// To cast back to double.
yout(isnan(yout)) = 1;
plot(tVector, yout);
You'll get something like:
You could also do this the ezplot way by doing: ezplot(xsquare). However, you'll see that at the points where the wave repeats itself, we get NaN values and so there is a disconnect between the high peak and low peak.
Note:
Natan's solution is much more elegant. I was still writing this post by the time he put something up. Either way, I wanted to give a more signal processing perspective to how to do this. Go Fourier!
A Fourier series for the square wave of unit amplitude is:
alpha + 2/Pi*sum(sin( n * Pi*alpha)/n*cos(n*theta),n=1..infinity)
Here is a handy trick:
cos(n*theta) = Re( exp( I * n * theta))
and
1/n*exp(I*n*theta) = I*anti-derivative(exp(I*n*theta),theta)
Put it all together: pull the anti-derivative ( or integral ) operator out of the sum, and you get a geometric series. Then integrate and finally take the real part.
Result:
squarewave=
alpha+ 1/Pi*Re(I*ln((1-exp(I*(theta+Pi*alpha)))/(1-exp(I*(theta-Pi*alpha)))))
I tried it in MAPLE and it works great! (probably not very practical though)
I have a symbolic function, whose zeros I am particular interested in knowing. I have searched through google, trying to find something related to my query, but was unsuccessful.
Could someone please help me?
EDIT:
T(x,t) = 72/((2*n+1)^2*pi^3)*(1 - (2*n+1)^2*pi^2*t/45 + (2*n+1)^4*pi^4*t^2/(2*45^2) - (2*n+1)^6*pi^6*t^3/(6*45^3))*(2*n+1)*pi*x/3;
for i=1:1:1000
T_new = 72/((2*i+1)^2*pi^3)*(1 - (2*i+1)^2*pi^2*t/45 + (2*i+1)^4*pi^4*t^2/(2*45^2) - (2*i+1)^6*pi^6*t^3/(6*45^3))*(2*i+1)*pi*x/3;
T = T + T_new;
end
T = T - 72/((2*n+1)^2*pi^3)*(1 - (2*n+1)^2*pi^2*t/45 + (2*n+1)^4*pi^4*t^2/(2*45^2) - (2*n+1)^6*pi^6*t^3/(6*45^3))*(2*n+1)*pi*x/3;
T = T(1.5,t);
T_EQ = 0.00001
S = solve(T - T_EQ == 0,t);
The problem that I get is that S is an a vector which contains imaginary numbers. I expected a real number, because I am trying to calculate a time.
Here is a little background as to what I am trying to do:
http://hans.math.upenn.edu/~deturck/m241/solving_the_heat_eqn.pdf
In the given link is the heat equation solved for a particular one-dimensional case. The temperature distribution, that satisfies the prescribed boundary and initial conditions, is given on page 50, I believe.
What I would like to do is find the time at which the one-dimensional object equilibrates with the environment, which is held at a constant temperature of T=0. As far as I know, the easiest way to do this would be to use the Taylor expansion of the exponential function, using only the first few terms, because I expect the equilibrium time to be relatively short; and then use the small angle approximation for the sine function, because the rod has a relatively small length. Doing just this, I made a for loop to generate terms just as the summation function would--as you can see, I used 1000 terms.
Does what I am doing seem wrong to anyone? If there is a better method, could someone please recommend it?
You shouldn't be surprised to see imaginary roots provided that at least one root is real and positive, corresponding to your time. The question is if the time makes any sense due to the approximations that you're making. Have you plotted the the actual function to get a rough approximation for where the zero is?
I can't really comment on the particular problem you're trying to solve. You need to make sure that you're using enough Taylor expansion terms an that they are accurate for the domain. Have you tried this leaving in the exp and/or sin? Is there any reason that you can't just use zero? And have you checked that your summation has converged after 1,000 terms? Or does it converge much sooner or not at all?
The main question is why are you using symbolic math at all to solve this? This seems like a numeric problem unless you're experiencing overflow/underflow issues in your summation. You can find the zero using fzero in this case:
N = 32; % Number of terms in summation
x = 1.5;
T_EQ = 1e-5;
n = (2*(0:N)+1)*pi;
T = #(t)sum((72./n.^3).*exp(-n.^2*t/45).*sin(n*x/3))-T_EQ;
S = fzero(T,[0 1e3]) % Bounds around a root guarantees solution if function monotonic
which returns
S =
56.333877640358708
If you're going to use solve, I'd do something like the following to avoid for loops:
syms t
N = 32;
x = 1.5;
T_EQ = 1e-5;
n = (2*sym(0:N)+1)*sym(pi);
T(t) = sum((72./n.^3).*exp(-n.^2*t/45).*sin(n*x/3));
S = double(solve(T-T_EQ==0,t))
or, using symsum:
syms n t
N = 32;
x = 1.5;
T_EQ = 1e-5;
T(t) = symsum((72/(pi*(2*n+1))^3)*exp(-(pi*(2*n+1))^2*t/45)*sin(pi*(2*n+1)*x/3),n,0,N);
S = double(solve(T-T_EQ==0,t))
Lastly, your symbolic solutions are not even exact as some your pi variables are being converted to rational approximations. pi is floating point. Things like pi*t are generally safe if t is symbolic, because pi will be recognized as such. However, pi^2 is calculated in floating-point before being converted to symbolic due to order of operations. In general your should use sym('pi') or sym(pi) in symbolic expressions.
Assuming you have a polynomial or trigonometric function of x or y, and what you mean by "zeros" is the values where the function crosses the axis, i.e., either x or y is zero, you can call the value of the function when a variable is 0. An example:
syms x y
f=-cos(x)*exp(-(x^2)/40);
ezsurf(f,[-10,10])
F=matlabFunction(f,'vars',{[x]});
F([0])
The ezsurf just visualizes the plot. If you want a function of both x and y, you do something like the following:
syms x y
f=-cos(x)*cos(y)*exp(-(x^2+y^2)/40);
ezsurf(f,[-10,10])
F=matlabFunction(f,'vars',{[x,y]});
for y=0
solve(f)
end
This will give you the value of the function for which integer multiples of x correspond to zero points for y (values of the function that are on the y=0 plane).