i have the data below
the columns are X,Y and Z
i want to compute this Integral
i don't know how to do it in matlab
i searched and didn't find any case like my problem with the integration limits are array values,the variables are also arrays.
and i really don't know how to solve it mathematically
i tried solving ,but the Y,Z and X being arrays with double values is not making things easy
You can do the following if you store X, Y and Z in MATLAB:
fyz = Y*(Z - 25);
% Compute the integral
I = trapz(fyz, X);
Related
Looking over some MATLAB code related to multivariate Gaussian distributions and I come across this line:
params.means(k, :) = mean(X(Y == y, :));
Looking at the MATLAB documentation http://www.mathworks.com/help/matlab/ref/mean.html, my assumption is that it calculates the mean over the matrix X in the first dimension (the column). What I don't see is the parentheses that comes after. Is this a conditional probability (where Y = y)? Can someone point me to some documentation where this is explained?
Unpacked, this single line might look like:
row_indices = find(Y==y);
new_X = X(row_indices,:);
params.means(k,:) = mean(new_X);
So, as you can see, the Y==y is simply being used to find a subset of X over which the mean is taken.
Given that you said that this was for computing multivariate Gaussian distributions, I bet that X and Y are paired sets of data. I bet that the code is looping (using the variable k) over different values y. So, it finds all of the Y equal to y and then calculates the mean of the X values that correspond to those Y values.
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).
I'm having some trouble trying to solve and plot an integral in matlab. In fact, I know that if a solve one, I'll solve all the integrals that I need now.
I have plot in x axis a value of a variable "d" and in y axis the value of a integral of a normalized gaussian function from -inf to ((40*log10(d)-112)/36) and I'm not finding out a way to do it correctly. D is between 0 and 1600
Can anybody here please help me?
In Matlab you can use the integral-function to evaluate integrals:
q = integral(fun,xmin,xmax)
fun needs to be a function handle, also called function functions, like these two examples:
square = #(x) x.^2;
plusone = #(x) x+1;
I know there must be a really simple answer to this question, but I just can't seem to find it. (Guess I'm probably Googling the wrong terms.)
I am plotting some data in Matlab using the plot(x, data) function.
I want to find the x-intercept(s) of the line, i.e. the point(s) where y = 0.
In some cases, it may be that the data vector doesn't actually contain values equal to zero, so it's not just a matter of finding the indexes of the elements in data which are equal to zero, and then finding the corresponding elements in the x vector.
Like I said, it's a really simple problem and I'd think there's already some in-built function in Matlab...
Thank you for your help.
If you want to find X-intercept as interpolate between 2 closest points around X axes you can use INTERP1 function:
x0 = interp1(y,x,0);
It will work if x and y are monotonically increasing/decreasing.
x=-1.999:0.001:1.999;
y=(x-1).*(x+1);
plot(x,y)
hold on
plot(x,zeros(length(x),1),'--r')
find(abs(y)<1e-3)
So the last part will guarantee that even there is not exact y-intercept, you will still get a close value. The result of this code are the indices that satisfy the condition.
You can make a linear fit (1st order polynomial) to your data, then, from the slope and Y intercept of the fitted line, you'll be able to find X intercept. Here is an example:
x1 = 1:10;
y1 = x1 + randn(1,10);
P = polyfit(x1,y1,1);
xint = -P(2)/P(1);
if you want to know what is the slope and y_int, here it is:
Slope = P(1); % if needed
yint = P(2); % if need
Is it possible to compute the numerical hessian matrix for this function with respect to W_i,C, epsilon_i easily Matlab? I have computed a hessian by manually take a derivative, but I want to verify if my result is correct.
W = Nx1;
X = NxM;
X_i = Nx1;
y = 1xM;
C = 1x1;
DERIVEST on the file exchange has a function for doing this. There are also tips for doing this eg in Section 18 of this tutorial, or many other places.