I am trying to run my function. it shows
[root, ni]=value1(xu,xl,acceptable)
Error using value1
Too many input arguments.
function[root, ni]=value1(xu,xl,acceptable)
fu=HW10B(xu);
fl=HW10B(xl);
Err=1;
N=0;
if fu*fl>=0
end
while Err>=acceptable;
m=(xu+xl)/2;
fm=HW10B(m)
if fm*fu<0;
fl=m;
else fu=m;
Err=abs(xu-xl)/xu*100;
end
N=N+1;
end
function [ y] = HW10B( x)
%equation of x
y=3*x^3-8*x^2-4*x+9;
end
root=m;
ni=N;
end
function[m, N]=value1(xu,xl,acceptable)
y=#(x)3*x.^3-8.*x.^2-4.*x+9;%//Used anonymous function instead of private
fu=y(xu);%//Used above definition
fl=y(xl);
Err=1;%//Initialise error
N=0;%//Initialise counter
while Err>=acceptable && N<1e4;%//check for convergence
m=(fu+fl)/2;%//Get new midpoint
fm=y(m);%//Get value at midpoint
if fm*fu<0;%//Get left or right value to move
fl=m;
else
fu=m;
Err=abs(fu-fl)/fu*100;%//Calculate the error
end
N=N+1;%//Update iteration counter
end
end
Call it from the command line:
xu=15;xl=2;acceptable=1e-3;
[root, ni]=value1(xu,xl,acceptable)
root =
2.7554
ni =
29
As you can see I cleaned up your code quite a bit. Using the two separate storage variables at the end of the code was just taking up more space than necessary. The if statement fu*fl>0 did not do anything, thus I chucked it out. Finally, you needed to update your values in your functions, thus using the fl, fx and fm, not the xu and xl.
If you call the function exactly as I showed you from the command line (with your own values of course), it should not throw any errors.
What happens in your original code is that you calculate everything once for the input variables, get an error which is larger than acceptable and therefore executes again, taking the same input arguments, returning the same error as before, which is still larger than acceptable. This is what we call an infinite loop. I suggest you check for it using a maximum number of iterations, i.e.
while Err>=acceptable && N<1e4
and change the 1e4 to whatever maximum number of iterations you want to have. If you accidentally end up going infinite, this counter will kill it without having to resort to crtl-c or equivalents.
Related
I wanted to plot the load voltage across the resistor in series with a diode using matlab. This is a rather simple example involving piecewise functions, however I ran into an unexpected error.
t=[0:0.001:10]
vs=4*sin(pi * t)
for i =1:length(vs)
if(vs(i)<=0.7)
v(i)=0;
else
v(i)=vs(i)-0.7;
end
end
plot(t,v)
t is the time, vs is the source voltage, and v is the load voltage. Now, running this gave me an error saying "error, t and v are of different sizes..". Using length() I found out that while t and vs are of lengths 10001, v is somehow of length 1000001.
This is truly baffling to me. How can v and vs possibly differ in size? every element of vs was mapped to an element of v, and yet the size of v comes out to be about 100 times the size of vs.
Being new to matlab, I still am not very comfortable with not explicitly declaring the array v before using it in the for loop. However, I went through with it, because the example I worked on prior to this, used the same thing and it worked without any problems. We simply had a plot a piecewise function:
x=[-2 : 0.00001 : 20];
for i=1: length(x)
if(x(i)>=-2 && x(i)<0)
y(i)=sqrt(x(i)^2+1);
else if(x(i)>=0 && x(i)<10)
y(i)=3*x(i)+1;
else
y(i)=9*sin(5*x(i)-50);
end
end
end
plot(x,y)
This code worked flawlessly, and in my opinion the initial code is fundamentally doing the same thing, so again, I'm clueless as to why it failed.
The original code works if you initialise v to be of the same size as t (and therefore, vs), but still, I want to know why the code involving x,y worked (where y wasn't initialised) and the code involving (t,v) failed.
Also, my friend copy pasted the entire code into the command window of matlab 2016, and it worked. (and I'm using the 2021 version).
Its good practice to initialize variables before entering a loop. It will help avoid undefinied behaviour when you run the script multiple times. If you run the script with different lengths for t, it would fail the second run. One solution would be:
t=0:0.001:10;
vs=4*sin(pi * t);
v=nan(size(t));
for i =1:length(vs)
if(vs(i)<=0.7)
v(i)=0;
else
v(i)=vs(i)-0.7;
end
end
figure;
plot(t,v);
You could also avoid the for loop and use matrix operations instead:
t=0:0.001:10;
vs=4*sin(pi * t);
v=vs-0.7;
v(vs<=0.7)=0;
figure;
plot(t,v);
I have been asked to obtain the first 15 triplets according to this series and this code ought to work. However, it does only produce a table (15*3) filled with zero rather than the 15 Pythagorean triplets? Any help will be welcome.
A = zeros(15, 3);
ii = 1;
for c = 5:120
c2=c^2;
for a=1:c-1
a2=a^2;
for b=a:c-1
if c2-(a2+b^2) == 0
A(ii,1) = a;
A(ii,2) = b;
A(ii,3) = c;
ii=ii+1;
if A(15, 1) ~= 0
flag = 1;
break
end
end
end
if flag == 1
break
end
end
if flag == 1
break
end
end
T1 = array2table(A);
disp(T1)
So, the code generated a correct table on application-restart before failing on all subsequent attempts. And, now I notice that the code runs successfully only for the first time after every relaunch of the application. (Resolved, thanks Dan Pollard.)
Also, interested in knowing if there is any way to not write an upper limit (120) into the code.
I don't think your if statement is ever satisfied. For example, for c=5, you'd expect a=3, b=4 to be a triplet. But you're only letting a and b go up to floor(sqrt(c-1)), which is 2.
Do you mean to let a and b go up to floor(sqrt(c2-1))?
Edit As the question has changed.
When you run the code, Matlab creates all the variables which you assign, and stores them in the workspace. This can be useful, but here it's hurting you as you have the variable flag which is stored as 1. This means that when the code runs, it checks if flag==1 after the first run through b, which it is, so the code ends. Resolve this by placing clear; at the beginning of your script.
There isn't a practical way to remove the upper limit on c. Matlab has the built-in variable Inf but at best Matlab won't let you use it in that context. Realistically you could just replace the 120 with a really large number, but this will take more time and more memory as the number gets bigger. Computers have a finite RAM to store matlab arrays in though, and there are infinitely many pythagorean triples, so doing the calculation without an upper limit will fail in some way.
In a MATLAB-function the following code is used:
function stuff()
if a == 2
do1();
else
do2();
end
end
This code is placed inside a simulation-loop and gets called 1000 times or more per second. The if-statement does only matter in the first call of the function, after that either do1 or do2 are used, the variable a will not change any more.
How do I prevent to waste processing time with this if-statement? Basically, how do I tell Matlab, to not check the if-statement any more, and just call the one function, that gets selected in the first call to stuff?
Contrary to your beliefs this is not a problem, the compiler (should) automatically does this optimization for you. See e.g. Loop-invariant code motion.
What you can do to help the compiler is to move the calculation of the check outside as a flag, e.g.
flag = a==2;
for i = 1:100
stuff(flag)
end
Then you only have to do the calculation once and it is clear to the compiler that the value does not change.
NOTE: Obviously, if your check really is a==2, this would not make much of a difference.
EDIT: I have not been able to definitely verify that MATLAB does this automatically. However, this is only the first layer of optimization that is done for you. All modern processors use what is called a Branch predictor, see e.g. this brilliant answer Why is processing a sorted array faster than processing an unsorted array?, or this wiki page. In short, the processor guesses the result of the if-statement, if it is correct, everything goes faster. I think it is fair to say that the processor guesses correctly in all of your cases.
TLDR: Do not wory about it.
Given the comments above, it seems what you are actually looking for is a way to dynamically chose the function to be run in your simulation. This choice should be dynamic (you do not know which function to use at runtime) but the choice should only be done once. This is easily achievable using function handles: https://www.mathworks.com/help/matlab/function-handles.html
Here is an example:
function dynamicSimulation()
if ( rand() > 0.5 ) % determine which function should be called dynamically
sim=#func1;
else
sim=#func2;
end
other_params = [];
for k = 1:5 % run the simulation
sim( k, other_params );
end
end
function func1( index, other_params )
fprintf( 'Index=%d: Simulating using function 1\n', index );
end
function func2( index, other_params )
fprintf( 'Index=%d: Simulating using function 2\n', index );
end
If you run this several times you will notice that the (random) choice of func1 or func2 will mean you do not get the same function being run each time, although the same one is used for the entire simulation.
I reckon you don't waste much time on checking the validity that if statement. However, since you specifically mention it only checks for the first iteration: why not get that out? So instead of:
for ii = 1:10
if ii == 1
k = 1;
else
k = k + 1;
end
end
You could do
k = 1;
for ii = 2:10
k = k + 1;
end
Thus eliminating the check.
NB: this scales badly of course, but as it is only a single iteration here I consider it a good option.
When I have to display the variable value every n iterations of a for loop I always do something along these lines:
for ii=1:1000
if mod(ii,100)==0
display(num2str(ii))
end
end
I was wondering if there is a way to move the if condition outside the loop in order to speed up the code. Or also if there is something different I could do.
You can use nested loops:
N = 1000;
n = 100;
for ii = n:n:N
for k = ii-n+1:ii-1
thingsToDo(k);
end
disp(ii)
thingsToDo(ii);
end
where thingsToDo() get the relevant counter (if needed). This a little more messy, but can save a lot of if testing.
Unless the number of tested values is much larger than the number of printed values, I would not blame the if-statement. It may not seem this way at first, but printing is indeed a fairly complex task. A variable needs to be converted and sent to an output stream which is then printing in the terminal. In case you need to speed the code up, then reduce the amount of printed data.
Normally Matlab function takes vector inputs as well. This is the case for disp and display and does only take a single function call. Further, conversion to string is unnecessary before printing. Matlab should send the data to some kind of stream anyway (which may indeed take argument of type char but this is not the same char as Matlab uses), so this is probably just a waste of time. In addition to that num2str do a lot of things to ensure typesafe conversion. You already know that display is typesafe, so all these checks are redundant.
Try this instead,
q = (1:1000)'; % assuming q is some real data in your case
disp(q(mod(q,100)==0)) % this requires a single call to disp
I am trying to find use to Newton-Raphson method to find the roots. It does this by making a guess and then improving the guess after each iteration until you get one of the zeros.
Because the Newton-Raphson method quickly finds the zeros, it gives me a small error immediately and after two or three iterations max it should fail to meet the conditions of the while loop. However, the problem is that when I remove the semi-colon after "error" in my loop, I start getting fractions that should break the while loop, but its like Matlab doesn't know that 123/8328423 is less than 1. It continues to run until I manually force the program to stop running.
How do I fix this? I tried format long, format longe, and using double both in the command window, in the scrip file, and somewhere in the loop.
Thank you in advance for any tips, suggestions, or advice that may help!!
A = [1,2,-4;2,-2,-2;-4,-2,1;];
format longe
% syms x y z
% P = x^4 + 3*x^2*y^2-z^3+y+1;
% feval(symengine,'degree',P,x)
syms x
B = mateigenvalue(A);
f(x) = simplify(matdet(B));
x0 = 1;
error = 10;
while(error > .01)
x1 = x0 - f(x0)/(27*(x0)-3*(x0)^2);
error = abs(((f(x0)-f(x1))/f(x0))*100)
x0 = x1;
end
x0 = double(x0)
I reckon the main problem is with error.
It starts as double but inside the while-loop it turns into a symbolic variable and you can't easily compare symbolic variables with scalar values (the .01 in the while-loop condition).
Check in your workspace if error is symbolic (or type class(error) and check if sym is returned). I guess it is symbolic because a fraction is returned (123/8328423), as instead Matlab treats double values with decimals, not fractions.
If so, try doing (inside the while-loop) a conversion for error that is, under the line
error = abs(((f(x0)-f(x1))/f(x0))*100);
try putting
error=double(error);
So error will be temporarily converted in double and you can easily compare its value with .01 to check the while-loop condition.
Also, it is bad practice to call a variable error since error() is a built-in function in Matlab. By naming a variable error you cannot use the error() function. Same story goes for other built-in functions.