I've wrote a rungekutta code and there is a specific x range that it should calculate up to it. The problem is that the prof. wants a code that keeps on adding up by itself till it reaches the required answer.
I tried using the 'if' function but it only works once and no more (thus, even creating a mismatch in array lengths.
Code:
h=0.5;% step size
z=0.5;
x = 0:h:z;% the range of x
y = zeros(1,length(x));
y(1) = 50;% initial condition
F_xy = #(t,x) (37.5-3.5*x);%function
for i=1:(length(x)-1)% calculation loop
k_1 = F_xy(x(i),y(i));
k_2 = F_xy(x(i)+h,y(i)+h*k_1);
y(i+1) = y(i) + (h/2)*(k_1+k_2);% main equation
if y(i+1)>11
x=0:h:z+1;
end
end
disp (y(i+1))
Error of arrays length (shows that if function only works once)
41.4063
Error using plot
Vectors must be the same length.
Error in code6rungekutta2ndorder (line 30)
plot(x,y), grid on
it should keep on increasing by +1 in the 'z' variable till the answer of y(i+1) is less than 11. (correct z should be 9.5)
As #medicine_man suggested, you need a while loop:
h=0.5;% step size
z=0.5;
x = 0;
y = 50;% initial condition
F_xy = #(t,x) (37.5-3.5*x);%function
m = 0;
while y(m+1) > 11
m = m+1;
x = x+h;
k_1 = F_xy(x,y(m));
k_2 = F_xy(x+h,y(m)+h*k_1);
y(m+1) = y(m) + (h/2)*(k_1+k_2);% main equation
end
figure;
plot(0:h:x, y);
Your terminate condition, which is y(m+1) > 11, is checked at the beginning of each iteration of the while loop. In the loop, you can increment the value of x and update your y array. The loop runs until the terminate condition is met.
The result of the above code:
Related
I am trying to use my second order Adams-Bashforth function here:
function [t,x] = Adams(f,t_max,x0,N)
h = t_max/N;
t = linspace(0,t_max,N+1);
x = zeros(2,N+1);
x(:,1) = x0;
x(:,2) = x0 + h.*(f(t(1),x(:,1)));
for i=2:N
x(:,i+1) = x(:,i) + h.*((3/2.*f(t(i),x(:,i))-(1/2).*f(t(i-1),x(:,i-1))));
end
end
In order to solve the Lorenz System Equation. However, whenever I try to call the function, I get an error.
sigma = 10;
beta = 8/3;
rho = 28;
f = #(t,a) [-sigma*a(1) + sigma*a(2); rho*a(1) - a(2) - a(1)*a(3); -beta*a(3) + a(1)*a(2)];
[t,a] = Adams(f,10,[1 1 1],100);
plot3(a(:,1),a(:,2),a(:,3))
Output:
"Unable to perform assignment because
the size of the left side is 2-by-1 and
the size of the right side is 1-by-3.
Error in Project>Adams (line 55)
x(:,1) = x0;"
Is the issue with my function, or with how I am calling my function? Any help would be appreciated.
in line 5 of your Adams function: x(:,1) = x0;
the input x0 is the initial conditions [1 1 1] a row vector of size [3x1], but x is an array of size zeros(2,N+1), or [2x101].
There are several mistakes here, the first you assign a row vector to a colon vector, not only that, but the size of the container in that first column of array x which is [2x1] , that is a different size than the size of x0. that is what the error tells you.
The function is buggy there , and it wont work unless you debug it according to the original intent of that method you are trying to implement.
note that also in the script f=#(t,... is expected to have an input t, but there is not t in the expression [-sigma*a(1) + sigma*a(2); rho*a(1) - a(2) - a(1)*a(3); -beta*a(3) + a(1)*a(2)];
One of the methods to compute sqrt(a), a>0 is
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1)), n = 1, 2, …,
with X0=1 and X1=a (That is, it is known that lim n-> infin of Xn = sqrt(a)
Write a function [sqa, nitr] = mySqrt(a) which implements this calculation. The function should use a while-loop, terminate when the difference between Xn+1 and Xn becomes smaller than eps(10*a), and output Xn+1 in sqa and the value of n at which the while-loop was terminated in nitr. Test your function for a = 103041.
I have written this but it does not work
function [sqa, nitr] = mySqrt (a)
%[sqa, nitr] = mySqrt (a)
% computes square root of a
% sqa = a;
sqaprev = a;
nitr = 0;
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1))); %find the second term
sqa= X(n+1)
while abs (sqaprev-sqa) >= eps (10*a)
sqaprev = sqa;
sqa = (1/2) *(sqaprev+ (a/sqaprev));
nitr = nitr + 1;
end %while
end
i get the error:
Unrecognized function or variable 'X'.
Error in mySqrt (line 7)
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1))); %find the second term
Could someone help me ?
You should start with declaring your variables and assigning them values
X(1)=1;
X(2)=a;
n=2;
Then in the loop you apply the given recursion formula, not the Heron/Babylonian formula that got from somewhere else.
According to the algorithm you presented for the square root, you can try the code below
function [sqa, n] = mySqrt(a)
n = 2; % start from 2
X = [1,a]; % initial value for sequence X
while true % repeat procedure until termination condition is reached
X(n+1) = (a + (X(n)*X(n-1)))/(X(n)+X(n-1)); % update sequence X by assigning new values
if abs(X(n+1)-X(n)) <= eps(10*a) % termination condition
break;
end
n = n + 1;
end
sqa = X(end); % return the last element of X as the square root
end
such that
>> [sqa,n] = mySqrt(a)
sqa = 321.00
n = 20
I have many sets of data over the same time period, with a timestep of 300 seconds. Sets that terminate before the end of the observation period (here I've truncated it to 0 to 3000 seconds) have NaNs in the remaining spaces:
x = [0;300;600;900;1200;1500;1800;2100;2400;2700;3000];
y(:,1) = [4.65;3.67;2.92;2.39;2.02;1.67;1.36;1.07;NaN;NaN;NaN];
y(:,2) = [4.65;2.65;2.33;2.18;2.03;1.89;1.75;1.61;1.48;1.36;1.24];
y(:,3) = [4.65;2.73;1.99;1.49;1.05;NaN;NaN;NaN;NaN;NaN;NaN];
I would like to know at what time each dataset would reach the point where y is equal to a specific value, in this case y = 2.5
I first tried finding the nearest y value to 2.5, and then using the associated time, but this isn't very accurate (the dots should all fall on the same horizontal line):
ybreak = 2.5;
for ii = 1:3
[~, index] = min(abs(y(:,ii)-ybreak));
yclosest(ii) = y(index,ii);
xbreak(ii) = x(index);
end
I then tried doing a linear interpolation between data points, and then solving for x at y=2.5, but wasn't able to make this work:
First I removed the NaNs (which it seems like there must be a simpler way of doing?):
for ii = 1:3
NaNs(:,ii) = isnan(y(:,ii));
for jj = 1:length(x);
if NaNs(jj,ii) == 0;
ycopy(jj,ii) = y(jj,ii);
end
end
end
Then tried fitting:
for ii = 1:3
f(ii) = fit(x(1:length(ycopy(:,ii))),ycopy(:,ii),'linearinterp');
end
And get the following error message:
Error using cfit/subsasgn (line 7)
Can't assign to an empty FIT.
When I try fitting outside the loop (for just one dataset), it works fine:
f = fit(x(1:length(ycopy(:,1))),ycopy(:,1),'linearinterp');
f =
Linear interpolant:
f(x) = piecewise polynomial computed from p
Coefficients:
p = coefficient structure
But I then still can't solve f(x)=2.5 to find the time at which y=2.5
syms x;
xbreak = solve(f(x) == 2.5,x);
Error using cfit/subsref>iParenthesesReference (line 45)
Cannot evaluate CFIT model for some reason.
Error in cfit/subsref (line 15)
out = iParenthesesReference( obj, currsubs );
Any advice or thoughts on other approaches to this would be much appreciated. I need to be able to do it for many many datasets, all of which have different numbers of NaN values.
As you mention y=2.5 is not in your data set so the value of x which corresponds to this depends on the interpolation method you use. For linear interpolation, you could use something like the following
x = [0;300;600;900;1200;1500;1800;2100;2400;2700;3000];
y(:,1) = [4.65;3.67;2.92;2.39;2.02;1.67;1.36;1.07;NaN;NaN;NaN];
y(:,2) = [4.65;2.65;2.33;2.18;2.03;1.89;1.75;1.61;1.48;1.36;1.24];
y(:,3) = [4.65;2.73;1.99;1.49;1.05;NaN;NaN;NaN;NaN;NaN;NaN];
N = size(y, 2);
x_interp = NaN(N, 1);
for i = 1:N
idx = find(y(:,i) >= 2.5, 1, 'last');
x_interp(i) = interp1(y(idx:idx+1, i), x(idx:idx+1), 2.5);
end
figure
hold on
plot(x, y)
scatter(x_interp, repmat(2.5, N, 1))
hold off
It's worth keeping in mind that the above code is assuming your data is monotonically decreasing (as your data is), but this solution could be adapted for monotonically increasing as well.
There is a two-dimensional random walk that one can find here which works perfectly in Octave. However, when I tried to write a one-dimensional random walk program, I got an error. Here is the program:
t=[];
x=[];
for i=1:100000
J=rand;
if J<0.5
x(i+1)=x(i)+1;
t(i+1)=t(i)+1;
else
x(i+1)=x(i)-1;
t(i+1)=t(i)+1;
end
end
plot(t,x)
Here is the error:
error: A(I): index out of bounds; value 1 out of bound 0
Thank you.
No need for a loop:
N = 100000;
t = 1:N;
x = cumsum(2*(rand(1,N)<.5)-1);
plot(t,x)
For the 2D case you could use the same approach:
N = 100000;
%// t = 1:N; it won't be used in the plot, so not needed
x = cumsum(2*(rand(1,N)<.5)-1);
y = cumsum(2*(rand(1,N)<.5)-1);
plot(x,y)
axis square
You get an error because you ask MATLAB to use x(1) in the first iteration when you actually defined x to be of length 0. So you need to either initialize x and t with the proper size:
x=zeros(1,100001);
t=zeros(1,100001);
or change your loop to add the new values at the end of the vectors:
x(i+1)=[x(i) x(i)+1];
Since t and x are empty, therefore, you cannot index them through x(i+1) and x(i).
I believe you should intialize x and t with all zeros.
In the first iteration, i = 1, you have x(2) = x(1) +or- 1 while x has dimension of zero. You should define the starting point for x and t, which is usually the origin, you can also change the code a little bit,
x = 0;
N = 100000;
t = 0 : N;
for i = 1 : N
x(i+1) = x(i) + 2 * round(rand) - 1;
end
plot(t,x)
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.