Answered
Replacing sym with function handle
Do you mean something like this? phi = pi/3; [nr, wr] = fn(phi); disp(phi) disp(nr(4)+56*3/2) disp(wr(4)) function [nr...

4 meses ago | 0

Answered
I'm trying to solve this system of ODE's describing a mechanical spring model.
Might be better to forget about symbolics, treat each 2nd order ode as two first order ode's and do the following: %applied for...

4 meses ago | 0

Answered
Solving a first order ODE with Euler backwards method
Your y_true is only valid for t>= 5 (smaller values give imaginary results for y). So, try going from 5 to 8: % y_true = log(t...

4 meses ago | 0

| accepted

Answered
Need help to create a loop
How about: e=2.7183; gamma=0.1333; k = [1,9,23,7,23,9,20,29,63,102,73,59,27,130,75,185,70,92,326]; Rt=0:0.01:10; P = zero...

4 meses ago | 1

Answered
Newton's method iterations
I suggest you plot a graph of your function, then you can see where good initial estimates would be. For example f = @(x) 2*ex...

4 meses ago | 0

Answered
the variable appears to change size every loop iteration
Answered here: https://uk.mathworks.com/matlabcentral/answers/1464024-the-variable-appears-to-change-size-every-loop-iteration?s...

4 meses ago | 1

Answered
Dealing with NaN Values
Try find(isnan(data))

4 meses ago | 0

Answered
Writing the dirac function as a function handle
Try dirac_i = @(x) x==i; % This assumes i has been fixed before the function is defined

4 meses ago | 0

| accepted

Answered
How to plot 2 graphs with input as a range and join them together?
Like this p0=0; p1=6; p2=1; p3=3; p4=3; t=[0:0.01:1]; x1 = (1/6)*[((((-t).^3)+(3*(t.^2))-(3*t)+1)*p0)+(((3*(t.^3))-(6*(t....

4 meses ago | 0

Answered
the variable appears to change size every loop iteration
You don't need the loop: If = [0 0.5 1.0 1.5 2.0 2.5]; Ea = [0 75 150 205 242 270]; Ra = 0.14; RI = 2; la = 100; Ifield = ...

4 meses ago | 1

Answered
Unable to find solution to matrix using Gauss Seidal code. How should I proceed to get the solution?
You are dividing by A(i,i) some of which are zero. These will introduce NaNs.

4 meses ago | 0

Answered
Solving Eqn with Varying Variable (Ms)
Like this: a1a4 = 1./[1, 2, 4, 10]; n = 10000; p4p1 = 1:n; M = zeros(numel(a1a4),n); for j = 1:numel(a1a4) m = 1.01; ...

4 meses ago | 1

| accepted

Answered
To solve two 2nd order coupled differential equation using ODE45?
Replace each 2nd order ODE by two 1st order ODEs. e.g. set V1 = X1', V2 = X2', then V1' = (50*sin(f(t) - V1)*V2')/sin(f(t)), V...

4 meses ago | 0

| accepted

Answered
Using Ode45 to solve dynamics problem (ISA model)
Like this z0 = 39045; %const.h0; v0 = 0; %const.v0; t0 = 0; tf = 800; N = 60000; tspan = linspace(t0, tf, N); X ...

4 meses ago | 0

| accepted

Answered
Curve fitting a power law function
Like this? h0=0.654; %[m] This seems to be unused h_t=[0.654;0.628;0.604;0.582;0.56;0.54;0.52;0.501;0.482;0.465;0.447;0.43;0.4...

4 meses ago | 0

Answered
Finding approximate real solution to a equation
You don't need syms here. Try the roots function. help roots

4 meses ago | 0

| accepted

Answered
Two Step Adam Bashford Method
As follows f=@(t,y) 3*t+y/t; alpha=5; a=1; b=2; n=3; [t, w, h] = abs2(f, a, b, alpha, n); plot(t,w,'-o'),grid xlabel('t'...

4 meses ago | 0

| accepted

Answered
How to add iterations in Newton Raphson code of Kepler Equation
Like this M = 0.908; e = 0.725; tol = 10^-6; [E, its] = Kepler(M,e,tol); disp(['E = ' num2str(E) ' after ' num2str(its), ...

4 meses ago | 0

| accepted

Answered
Let x=[2 4 6 8 10] and y=[1 3 5 7 9]. Compute for the vector z whose elements are equal to z= (xy+ y/x)/[(x+y)]^((y-x)) + 10^(x⁄y)
You need z = (((x.*y)+(y./x))./((x+y).^(y-x)))+10.^(x./y) % another dot here ^

4 meses ago | 0

| accepted

Answered
how to find the root of function?
Like this? n_h0 = 0.1; %initial guess n_h = fzero(@n_h_fun, n_h0); disp(n_h) function a=n_h_fun(n_h) alpha=0.74; ...

4 meses ago | 1

| accepted

Answered
Newton's method for 2 dimension vectors
LIke this? % Functions f = @(XY) [XY(1).^3 - 3*XY(1).*XY(2).^2 - 1; 3*XY(1).^2.*XY(2) - XY(2).^3]; J = @(XY) [...

4 meses ago | 0

Answered
How to plot a (which is changing from 0-1 in 0.01 increments) vs x(2) (using a for loop and fsolve to find the solution of a nonlinear equation containing x(s) sol based on a
Your first equation is a simple quadratic in x(1); your second is a quadratic in x(2) that depends on x(1), so, assuming you are...

4 meses ago | 0

Answered
to plot alpha(a) vs diameter(D) in the given problem. where A = l*sin(b), B = l*cos(b), C = ( h + 0.5*D )*sin (b) − 0.5*D*tan(b) and E = ( h + 0.5*D)*cos(b) − 0.5*D.
Try replacing D within the j-loop by D(j). Also add something like alpha(j) = a after the end of the i-loop (but inside the j-l...

4 meses ago | 0

Answered
Solving factorial equation with integer variables: Empty sym 0-by-1
Set x-2 = 659 and y-x = 679 for one solution, and x-2 = 679 and y-x = 659 for another.

4 meses ago | 0

Answered
Unable to plot exponential graphs properly
One possibility t = linspace(-10,10,400); y1 = exp(t); y2 = exp(-t); y3 = exp(2*t); plot(t,y1,t,y2,t,y3) legend('y1','y2...

4 meses ago | 0

Answered
Not enough input arguments.
Like this?: global PT S DELH Q CPA CPB CPC CPD NA NC NB ND NE NT PT = 1.5; % PT is total pressure in atmospheres S = 0.05; %...

4 meses ago | 1

| accepted

Answered
speed up integrating the same function over many overlapping intervals?
How about using cumtrapz to do the cumulative integral from 0 to max(X) just once, then select the sections specified by X(i).

4 meses ago | 0

| accepted

Answered
Solving single variable equation where other variables depends on that single variable
Try using fzero. For example: f = @(x) x.^2 +1./x; % arbitrary function: replace with your own y = @(x) f(x); z = @(x)...

4 meses ago | 0

| accepted

Answered
last value of array
Do you mean something like this %plot n vs x n vs y for n=1:10 [t,r]=ode45(@fn,[0 14],[0.01+n 0.01 ]); xlast(n) = r(end,1);...

5 meses ago | 0

Answered
How to keep both readability and calculation speed when using anonymous functions?
How about something like hexp = @(x,y) exp(1i*(x+y)); x = ..; % set to whatever is required y = sqrt(x.^2 + A^2); h1 = h...

5 meses ago | 0

Load more