I have the following matlab code. 1st line after while gives error. i am trying to make Newtons method to find roots. for that i need derivative f`(p0).
plz guide me what i am doing wrong and how can i get derivative of function f;
I also tried D(f(p0)) but that didn't work, gives error: Undefined function or method 'D' for input arguments of type 'double'.
format long;
f=#(x) cos(x)-x;
p0 = 0.5;
TOLL = 1e-4;
N = 100;
i = 1;
while (i <= N)
p = p0-f(p0)/diff(f(p0)); %Error, returns empty results this produced error
if ( abs(p-p0) < TOLL)
fprintf('Root of given equation is %f\n', p);
return;
end
i=i+1;
p0 = p;
end
fprintf('Method failed after %d iteration\n', i);
The error is because p0 is a scalar, so f(p0) is a scalar. Then taking diff(f(p0)) wont work.
To find the derivative at p0, you could use this definition of the derivative:
f'(p0) = limit as h->0 of (f(x+h)-f(x))/h.
Pick h as some small number (say 1e-3), using (f(p0+h) and f(p0)), you should be able to get an approximation of the derivative of f at p0.
Related
I'm working on my own Matlab code for the bisection method, and have defined an anonymous function I am trying to find the root for. It produces a simple graph and I know it could easily find the root if it would run properly.
When my code gets up to evaluating the function at a given x value, it returns:
Error using feval
Function to evaluate must be represented as a string scalar, character vector, or
function_handle object.
Error in bisection (line 3)
fa = feval(f, a);
My full code is:
function m=bisection(f,a,b,imax,tol)
fa = feval(f, a);
fb = feval(f, b);
i=0;
if fa*fb>0
disp('No root here: pick a new interval')
return
end
while abs(b-a) >= tol
i = i + 1;
m=(a+b)/2;
fm = feval(f, m);
if fa*fb<0
b = m;
else
a = m;
end
abs(fm);
end
% Show the last approximation considering the tolerance
w = feval(f, m);
fprintf('\n x = %f produces f(x) = %f \n %i iterations\n', m, fm, i-1);
fprintf(' Approximation with tolerance = %f \n', tol);
end
With my function being:
function [T] = freezing(x)
alpha = 0.138*10^-6;
Ti = 20;
Ts = -15;
t = 60*60*24*60;
tol = 10^-13;
%x = linspace(0,3);
T = #(x) (Ti-Ts)*erf(x./(2*sqrt(alpha*t)))+Ts;
dT = #(x) (Ti-Ts)*(1/sqrt(pi*alpha*t))*exp(-x.^2./(4*alpha*t));
T = T(x);
end
I'm really not sure what the issue is here – in my command window, I can easily input x values and get output. Any suggestions are much appreciated!
feval needs f to be a function handle. When you call bisection, I suspect you're passing the result of freezing to bisection instead of the function handle to freezing
You need to do bisection(#freezing, ...) instead of bisection(freezing(...), ...)
The #freezing creates a function handle that is passed to bisection. Doing the latter passes the result of freezing to bisection.
I am trying to compute the value of this integral using Matlab
Here the other parameters have been defined or computed in the earlier part of the program as follows
N = 2;
sigma = [0.01 0.1];
l = [15];
meu = 4*pi*10^(-7);
f = logspace ( 1, 6, 500);
w=2*pi.*f;
for j = 1 : length(f)
q2(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(2));
q1(j)= sqrt(sqrt(-1)*2*pi*f(j)*meu*sigma(1));
C2(j)= 1/(q2(j));
C1(j)= (q1(j)*C2(j) + tanh(q1(j)*l))/(q1(j)*(1+q1(j)*C2(j)*tanh(q1(j)*l)));
Z(j) = sqrt(-1)*2*pi*f(j)*C1(j);
Apprho(j) = meu*(1/(2*pi*f(j))*(abs(Z(j))^2));
Phi(j) = atan(imag(Z(j))/real(Z(j)));
end
%integration part
c1=w./(2*pi);
rho0=1;
fun = #(x) log(Apprho(x)/rho0)/(x.^2-w^2);
c2= integral(fun,0,Inf);
phin=pi/4-c1.*c2;
I am getting an error like this
could anyone help and tell me where i am going wrong.thanks in advance
Define Apprho in a separate *.m function file, instead of storing it in an array:
function [ result ] = Apprho(x)
%
% Calculate f and Z based on input argument x
%
% ...
%
meu = 4*pi*10^(-7);
result = meu*(1/(2*pi*f)*(abs(Z)^2));
end
How you calculate f and Z is up to you.
MATLAB's integral works by calling the function (in this case, Apprho) repeatedly at many different x values. The x values called by integral don't necessarily correspond to the 1: length(f) values used in your original code, which is why you received errors.
So, I'm trying to do the Gauss-Seidel method in Matlab and I found a code that does this but when I apply it to my matrices I get the Subscripted assignment dimension mismatch. error. I will show you my code in order to get a better idea.
%size of the matrix
n = 10;
%my matrices are empty in the beginning because my professor wants to run the algorithm for n = 100
and n = 1000. A's diagonal values are 3 and every other value is -1. b has the constants and the
first and last value will be 2,while every other value will be 1.
A = [];
b = [];
%assign the values to my matrices
for i=1:n
for j=1:n
if i == j
A(i,j) = 3;
else
A(i,j) = -1;
end
end
end
for i=2:n-1
b(i) = 1;
end
%here is the Gauss-Seidel algorithm
idx = 0;
while max(error) > 0.5 * 10^(-4)
idx = idx + 1;
Z = X;
for i = 1:n
j = 1:n; % define an array of the coefficients' elements
j(i) = []; % eliminate the unknow's coefficient from the remaining coefficients
Xtemp = X; % copy the unknows to a new variable
Xtemp(i) = []; % eliminate the unknown under question from the set of values
X(i) = (b(i) - sum(A(i,j) * Xtemp)) / A(i,i);
end
Xsolution(:,idx) = X;
error = abs(X - Z);
end
GaussSeidelTable = [1:idx;Xsolution]'
MaTrIx = [A X b]
I get the error for the Xsolution(:,idx) = X; part. I don't know what else to do. The code posted online works though, and the only difference is that the matrices are hardcoded in the m-file and A is a 5x5 matrix while b is a 5x1 matrix.
I am unable to run your code because some variables are not initialised, at least error and X. I assume the Problem is caused because Xsolution is already initialised from a previous run with a different size. Insert a Xsolution=[] to fix this.
Besides removing the error I have some suggestions to improve your code:
Use Functions, there are no "left over" variables from a previous run, causing errors like you got here.
Don't use the variable name error or i. error is a build-in function to throw errors and i is the imaginary unit. Both can cause hard to debug errors.
Initialise A with A=-1*ones(n,n);A(eye(size(A))==1)=3;, it's faster not to use a for loop in this case. To initialise b you can simply write b(1)=0;b(2:n-1)=1;
Use preallocation
the first time you run the code, Xsolution(:,idx) = X will create a Xsolution with the size of X.
the second time you run it, the existing Xsolution does not fit the size of new X.
this is another reason why you always want to allocate the array before using it.
I am new to Matlab. I am trying to solve a non-linear equation using this inbuilt Matlab function called fzero() but it's not giving me the results.
The main file goes like
A = 5;
B = 6;
C = 10;
eq = equation (A, B, C);
fzero(#(x)eq);
The other function file is:
function eq = equation (A, B, C)
syms x;
eq = A*x.^2 + B*x + C*(asinh(x)) ;
When I run this code, I get the following error:
Error using fzero (line 118)
The input should be either a structure with valid fields or at least two arguments to
FZERO.
Error in main (line 7)
fzero(#(x)eq);
Could someone help me with this?
EDIT:
WHen I specify the check point as 0, it returns me the following error.
Undefined function 'isfinite' for input arguments of type 'sym'.
Error in fzero (line 308)
elseif ~isfinite(fx) || ~isreal(fx)
Error in main (line 7)
fzero(#(x)eq, 0);
There are several mistakes in your code. For a start, fzero is for finding numerical roots of a non-linear equation, it is not for symbolic computations (check the documentation), so get rid of syms x. The correct way to call fzero in your case is a as follows:
A = 5;
B = 6;
C = 10;
eq = #(x) A*x^2 + B*x + C*(asinh(x));
x0 = 0; % or whatever starting point you want to specify
x = fzero(eq,x0)
You need to specify a guess, x0 point
fun = #sin; % function
x0 = 3; % initial point
x = fzero(fun,x0)
Hello I am relatively new to MATLAB and have received and assignment in which we could use any programming language. I would like to continue MATLAB and have decided to use it for this assignment. The questions has to do with the following formula:
x(t) = A[1+a1*E(t)]*sin{w[1+a2*E(t)]*t+y}(+/-)a3*E(t)
The first question we have is to develop an appropriate discretization of x(t) with a time step h. I think i understand how to do this using step but because there is a +/- in the end I am running into errors. Here is what I have (I have simplified the equation by assigning arbitrary values to each variable):
A = 1;
E = 1;
a1 = 1;
a2 = 2;
a3 = 3;
w = 1;
y = 0;
% ts = .1;
% t = 0:ts:10;
t = 1:1:10;
x1(t) = A*(1+a1*E)*sin(w*(1+a2*E)*t+y);
x2(t) = a3*E;
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
plot(y)
The problem is I keep getting the following error because of the +/-:
In an assignment A(I) = B, the number of elements in B and I must be the same.
Error in Try1 (line 21)
y(t) = [x1(t)+x2(t), x1(t)-x2(t)]
Any help?? Thanks!
You can remove the (t) from the left-hand side of all three assignments.
y = [x1+x2, x1-x2]
MATLAB knows what to do with vectors and matrices.
Or, if you want to write it out the long way, tell MATLAB there will be two columns:
y(t, 1:2) = [x1(t)'+x2(t)', x1(t)'-x2(t)']
or two rows:
y(1:2, t) = [x1(t)+x2(t); x1(t)-x2(t)]
But this won't work when you have fractional values of t. The value in parentheses is required to be the index, not a dependent variable. If you want the whole vector, just leave it out.