I would like to solve one equation in Matlab with two unknown variables using the Newton raphson method.
The equation is
I(:,:,K) = IL(:,:,K)-Io(:,:,K)*(exp((V+I*Rs)/a(:,:,K))-1)-((V+I*Rs(:,:,K))/Rsh(:,:,K));
Can this be done in matlab and if so please guide me since I have not managed to find anything related to this equation form!
Thanks
No. In general, one equation in two unknowns has an infinite number of solutions. (Think of a contour plot. You are essentially looking for the level set, the locus of all points that yields zero for the dependent variable. It will be a curvilinear path in those variables.)
So you can't "solve" it. I would suggest a good solution to visualize the locus is indeed the function contour. ezplot will do it even better.
Related
I am trying to plot roots of a function that is composed of multiple bessel functions being added and multiplied in Matlab. The equation is Jm(omega)*Ik(omega)+Im(omega)*Jk(omega) where Jm is the bessel function of the first kind of order m (besselj). Im is the modified bessel function of the first kind of order m (besseli). For each mode m=o,1,2,...and n=1,2,3... The frequency omega(mn) is the corresponding root of the listed equation. m=0,1,2 n-1,2,3,4. I need to solve the equation for the 12 roots. I am new to Matlab and this is a little out of my league. So far I have this code but I wasn't sure if I needed the variable omega in the script or not. I have also looked at other people's questions on the suject but didn't see any quite like this. The plots I have seen look nothing like mine which tells me I am probably wrong. Thanks for any help.
m=(0:2); k=(1:3); n=(1:4);
Jm=besselj(m,n');
Ik=besseli(k,n');
Jk=besselj(k,n');
Im=besseli(m,n');
g=Jm.*Ik+Im.*Jk
plot(g)
Plotting
besselj and besseli take what you call omega as their second parameter, so to plot your function you should try something like
m=0; k=1; omega=0:0.02:10;
Jm=besselj(m,omega);
Ik=besseli(k,omega);
Jk=besselj(k,omega);
Im=besseli(m,omega);
g=Jm.*Ik+Im.*Jk;
plot(omega,g);
hold all;
plot(omega,0,'k');
axis([min(omega) max(omega) -100 100]);
This shows you that for m=1, k=1 the first zeros are around 3.2, 6.3 and 9.4:
Finding the roots numerically
You could implement Halley's method for your function g, similar to how the roots of besselj are determined in the MatlabCentral file linked by Cheery.
I would like to know if anybody knows how I can plot an integral calculated using quad/quadl, or if this is possible.
I read that I can set the trace parameter to be non-zero, and this results in the information of each iteration being provided, but I'm not sure how and if I can use the information to plot an integral.
Thanks.
quad and quadl do not compute an integral function anyway, i.e., an integral as a function of the parameter. And since tools like this work iteratively, refining their estimate until it satisfies a tolerance on the global value, they are not easily made to produce the plot you desire.
You can do what you desire by using a differential equation solver to generate the solution, ode45 for example.
This question has already confused me several days. While I referred to senior students, they also cannot give a reply.
We have ten ODEs, into which each a noise term should be added. The noise is defined as follows. since I always find that I cannot upload a picture, the formula below maybe not very clear. In order to understand, you can either read my explanation or go the this address: Plos one. You could find the description of the equations directly above the Support Information in this address
The white noise term epislon_i(t) is assumed with Gaussian distribution. epislon_i(t) means that for equation i, and at t timepoint, the value of the noise.
the auto-correlation of noise are given:
(EQ.1)
where delta(t) is the Dirac delta function and the diffusion matrix D is defined by
(EQ.2)
Our problem focuses on how to explain the Dirac delta function in the diffusion matrix. Since the property of Dirac delta function is delta(0) = Inf and delta(t) = 0 if t neq 0, we don't know how to calculate the epislonif we try to sqrt of 2D(x, t)delta(t-t'). So we simply assume that delta(0) = 1 and delta(t) = 0 if t neq 0; But we don't know whether or not this is right. Could you please tell me how to use Delta function of diffusion equation in MATLAB?
This question associates with the stochastic process in MATLAB. So we review different stochastic process to inspire our ideas. In MATLAB, the Wienner process is often defined as a = sqrt(dt) * rand(1, N). N is the number of steps, dt is the length of the steps. Correspondingly, the Brownian motion can be defined as: b = cumsum(a); All of these associate with stochastic process. However, they doesn't related to the white noise process which has a constraints on the matrix of auto-correlation, noted by D.
Then we consider that, we may simply use randn(1, 10) to generate a vector representing the noise. However, since the definition of the noise must satisfy the equation (2), this cannot enable noise term in different equation have the predefined partial correlation (D_ij). Then we try to use mvnrnd to generate a multiple variable normal distribution at each time step. Unfortunately, the function mvnrnd in MATLAB return a matrix. But we need to return a vector of length 10.
We are rather confused, so could you please give me just a light? Thanks so much!
NOTE: I see two hazy questions in here: 1) how to deal with a stochastic term in a DE and 2) how to deal with a delta function in a DE. Both of these are math related questions and http://www.math.stackexchange.com will be a better place for this. If you had a question pertaining to MATLAB, I haven't been able to pin it down, and you should perhaps add code examples to better illustrate your point. That said, I'll answer the two questions briefly, just to put you on the right track.
What you have here are not ODEs, but Stochastic differential equations (SDE). I'm not sure how you're using MATLAB to work with this, but routines like ode45 or ode23 will not be of any help. For SDEs, your usual mathematical tools of separation of variables/method of characteristics etc don't work and you'll need to use Itô calculus and Itô integrals to work with them. The solutions, as you might have guessed, will be stochastic. To learn more about SDEs and working with them, you can consider Stochastic Differential Equations: An Introduction with Applications by Bernt Øksendal and for numerical solutions, Numerical Solution of Stochastic Differential Equations by Peter E. Kloeden and Eckhard Platen.
Coming to the delta function part, you can easily deal with it by taking the Fourier transform of the ODE. Recall that the Fourier transform of a delta function is 1. This greatly simplifies the DE and you can take an inverse transform in the very end to return to the original domain.
I have a function of two variables of the type: y = f(x1,x2) to be approximated and I would like to use least squares method to do it.
Polyval and Polyfit work with two-dimensional function, here I need to solve a three-dimensional function.
Thanks in advance.
G.B.
I've solved it in this way
A = [x1.^2,x1.*x2,x2.^2,x1,x2,ones(length(x1),1)];
c=A\y;
yEval = c(1)*x1.^2+c(2)*x1.*x2+c(3)*x2.^2+c(4)*x1+c(5)*x2+c(6);
Thanks anyway for your help.
Regards,
G.B.
Looking at your function, it looks like you're using a full quadratic response surface to fit against.
You could use x2fx function to generate all the terms. It's nothing groundbreaking here, but it might be a little cleaner. You could also use it to do not just OLS fitting, but also use the robust methods.
Here's some code I've written:
% set up terms for the variables, linear, quadratic, interactive, and constant
paramVEcomponents= x2fx([MAPkpa,RPM],'quadratic');
% robust fit using a Talwar weighting function
[coefs,robuststats]= robustfit(paramVEcomponents(2:6),(CAM2.*TEMPd./MAPkpa),'talwar');
% generating points for all the data we have based on the new parameters of the response surface
GMVEhat= paramVEcomponents * coefs;
how can one obtain coordinates of intersections of two line diagrams with given expression or equation?
for example:
L1= sin(2x) , L2= Ln(x); or anything else.
Amazingly, nobody has yet suggested using the function designed to do this in matlab. Use fzero here. Fzero is a better choice than fsolve anyway, which requires the optimization toolbox. And, yes, you could do this with Newton's method, or even bisection or the secant method. But reinventing the wheel is the wrong thing to do in general. Use functionality that already exists when it is there.
The problem at hand is to find a point where
sin(2*x) == log(x)
Here log(x) refers to the natural log. Do this by subtracting one from the other, then looking for a zero of the result.
fun = #(x) sin(2*x) - log(x);
Before you do so, ALWAYS plot it. ezplot can do that for you.
ezplot(fun)
The plot will show a single root that lies between 1 and 2.
fzero(fun,2)
ans =
1.3994
Since you tagged with matlab, you can do it with fsolve(#(x)sin(2*x)-log(x),1) which gives 1.3994 (1 is the initial starting point or guess). The y-coordinate is log(1.3994) = 0.3361.
That is, you use fsolve, pass it the function you want to solve for the zero of, in this case sin(2*x) == log(x) so you want sin(2*x) - log(x) == 0 (log is the natural log in matlab).
If you already have functions set up like, e.g. L1 = #(x)sin(2*x) and L2 = #(x)log(x) (or in functions L1.m and L2.m) you can use fsolve(#(x)L1(x)-L2(x),1).
In general, you have to solve the equation L1(x) = L2(x). If you don't know from the beginning what L1 and L2 are (linear, polynominal...) then the only solution is numeric solving for example with Netwon algorithm. The problem is then reduced to finding roots (zeros) of function f(x) = L1(X) - L2(X).
This is not a trivial question: what you're asking for is a general method for solving any mathematical equation.
For instance, you could consider using the bisection method, or Newton's method.
There is no general answer.
As a general non-analytic solution, when you have any 2 curves described by 2 sets of points, there is great submission at File Exchange - Fast and Robust Curve Intersections.