How do i loop a function inside another function until both sides equal each other ?
for example
after I find this "f"
f = (1/(-1.8*log(((epsilon/D)/3.7)^2 + 6.9/Re)))^2;
I want to use that value of f and input it here
f = (1/ -2.0*log((epsilon/D) + (2.51/Re*sqrt(***f***))))^2
the program is supposed to loop until both sides equal each other or are relatively close. the acceptable accuracy or error is 0.00001.
and how do I display that value of f that gives mme wha
It seems to me you are trying to solve an expression
f = somefun(f);
where the initial value for f is given by
f = (1/(-1.8*log(((epsilon/D)/3.7)^2 + 6.9/Re)))^2
Your best bet (if it's available to you) is to use Matlab's optimization toolbox, where you set the function to be minimized to
f - somefun(f)
and where you can set the tolerance using optimset('TolFun', 1e-5);
If you don't have the toolbox, then drN's suggestion in the comments of using Newton's method is probably as good as any -and you will learn more from doing it yourself.
I'm assuming that's fix point iteration you are attempting to use there, with your first code line being a starting estimate and your second code line being the actual fixed point iteration.
In this case you need to simply repeat that second statement while testing the difference between successive iterations. Something like this for example.
f = 1;
df = 1;
while abs(df) > 0.0001
fnew = log(20/f);
df = fnew - f;
f = fnew;
end;
BTW. The above is a simple example of fixed point iteration to solve f*exp(f)=20, [or equivalently f = ln(20/f)]. Apply the same logic to your particular equation for "f", but beware that not all equations are amenable to fixed point iteration.
Related
I am new to MatLab.
Write a function that takes as input a matrix D ∈ R^(N×2), D_i = (x_i,y_i), and the period ω and returns a plot showing a fit of the data without noise.
I need help with creating the function that takes the input as a matrix and period ω. Here is what I have so far. Am I on the right track?
function F = fftfuntion(D, omega)
check = 0;
x = D(:,1);
y = D(;,2);
You are on track but there are 3 problems:
First: you are using p variable inside your function which is not defined there. If it is defined in your main code you have to insert it to this function by adding an input to your function as p and when the function is called you have to put p there. Your other solution is to set p as a global variable which is not recommended.
function F = fftfuntion(D, omega,p)
Second: you have said that you need omega as an input and you are changing it with omega = 2*pi which is not right.
Finally I don't see any output which I think that's because you are not still done with the function.
Good luck
For my research, I needed to calculate some formula. I create a function for it. This is the very first time for me to write a function and also the first time to write a math formula in Matlab language. May I please ask you to check if where I do wrong?
Here is formula
And here is the function I wrote (P is x and P^ is y):
function biass = BIASS(x,y)
%This function calculates biass error
% Detailed explanation goes here
H = sum(y-x)/sum(y)
biass = H * 100
end
I'm not sure about results because I believe they are not reasonable.
If y is P^, then you should write H = sum(y-x)/sum(x), not H = sum(y-x)/sum(y).
A slightly more efficient way to compute this is
100 * (sum(y) / sum(x) - 1)
since your expression can be simplified.
I have N functions in MATLAB and I can define them using strcat, num2str and eval in a for loop. So without defining by hand I am able to define N functions. Let N=4 and let them be given as follows:
f1=#(x) a1*x+1;
f2=#(x) a2*x+1;
f3=#(x) a3*x+1;
f4=#(x) a4*x+1;
Now I add these four functions and I can do this by hand as follows:
f=#(x)(f1(x)+f2(x)+f3(x)+f4(x));
Here I can do it by hand because I know that N=4. However, in general I never know how many functions I will have. For all cases I cannot write a new function.
Is there any way to do this automatically? I mean if I give N=6 I am expecting to see MATLAB giving me this:
f=#(x)(f1(x)+f2(x)+f3(x)+f4(x)+f5(x)+f6(x));
Whenever I give N=2 then I must have the function f, defined as follows:
f=#(x)(f1(x)+f2(x));
How can we do this?
First of all, you should read this answer that gives a series of reasons to avoid the use of eval. There are very few occasions where eval is necessary, in all other cases it just complicates things. In this case, you use to dynamically generate variable names, which is considered a very bad practice. As detailed in the linked answer and in further writings linked in that answer, dynamic variable names make the code harder to read, harder to maintain, and slower to execute in MATLAB.
So, instead of defining functions f1, f2, f3, ... fN, what you do is define functions f{1}, f{2}, f{3}, ... f{N}. That is, f is a cell array where each element is an anonymous function (or any other function handle).
For example, instead of
f1=#(x) a1*x+1;
f2=#(x) a2*x+1;
f3=#(x) a3*x+1;
f4=#(x) a4*x+1;
you do
N = 4;
a = [4.5, 3.4, 7.1, 2.1];
f = cell(N,1);
for ii=1:N
f{ii} = #(x) a(ii) * x + 1;
end
With these changes, we can easily answer the question. We can now write a function that outputs the sum of the functions in f:
function y = sum_of_functions(f,x)
y = 0;
for ii=1:numel(f)
y = y + f{ii}(x);
end
end
You can put this in a file called sum_of_functions.m, or you can put it at the end of your function file or script file, it doesn't matter. Now, in your code, when you want to evaluate y = f1(x) + f2(x) + f3(x)..., what you write is y = sum_of_functions(f,x).
Hello I have written this to determine a root using Newton's method. The algorithm works. I also tried to implement an Experimental order of convergence EOC. It also works but I get the result that the order of convergence for Newton's method is 1 when in fact it is 2.
function [x,y,eoc,k]=newnew(f,df,x0,xe,eps,kmax)
x = x0;
y = feval(f,x);
for m=1:kmax
z = -y/feval(df,x);
x = x + z;
y = feval(f,x);
k = m;
for n=m
Ek=abs(x-xe);
end
for n=m+1
Ekp=abs(x-xe);
end
eoc=log(Ek)/log(Ekp);
if abs(y)<eps
return
end
end
disp('no convergence');
end
what is wrong?
When you say Ek=abs(x-xe) and Exp=abs(x-xe), they are exactly the same thing! That's why eoc evaluates to 1 every time.
Notice that you have no n in those equations. In fact, you don't need those extra for n=m loops either. Inside the for m=1:kmax loop, m is a single value not an array.
eoc needs to be calculated by comparing the previous loop iteration to the current one (since it doesn't make much sense to compare to a future loop iteration which hasn't happened yet). Because this looks like homework, I won't give you any code.. but this is a very strong hint.
I used nlfilter for a test function of mine as follows:
function funct
clear all;
clc;
I = rand(11,11);
ld = input('Enter the lag = ') % prompt for lag distance
A = nlfilter(I, [7 7], #dirvar);
% Subfunction
function [h] = dirvar(I)
c = (size(I)+1)/2
EW = I(c(1),c(2):end)
h = length(EW) - ld
end
end
The function works fine but it is expected that nlfilter progresses element by element, but in first two iterations the values of EW will be same 0.2089 0.4162 0.9398 0.1058. But then onwards for all iterations the next element is selected, for 3rd it is 0.4162 0.9398 0.1058 0.1920, for 4th it is 0.9398 0.1058 0.1920 0.5201 and so on. Why is it so?
This is nothing to worry about. It happens because nlfilter needs to evaluate your function to know what kind of output to create. So it uses feval once before starting to move across the image. The output from this feval call is what you see the first time.
From the nlfilter code:
% Find out what output type to make.
rows = 0:(nhood(1)-1);
cols = 0:(nhood(2)-1);
b = mkconstarray(class(feval(fun,aa(1+rows,1+cols),params{:})), 0, size(a));
% Apply fun to each neighborhood of a
f = waitbar(0,'Applying neighborhood operation...');
for i=1:ma,
for j=1:na,
x = aa(i+rows,j+cols);
b(i,j) = feval(fun,x,params{:});
end
waitbar(i/ma)
end
The 4th line call to eval is what you observe as the first output from EW, but it is not used to anything other than making the b matrix the right class. All the proper iterations happen in the for loop below. This means that the "duplicate" values you observe does not affect your final output matrix, and you need not worry.
I hope you know what the length function does? It does not give you the Euclidean length of a vector, but rather the largest dimension of a vector (so in your case, that should be 4). If you want the Euclidean length (or 2-norm), use the function norm instead. If your code does the right thing, you might want to use something like:
sz = size(I,2);
h = sz - (sz+1)/2 - ld;
In your example, this means that depending on the lag you provide, the output should be constant. Also note that you might want to put semicolons after each line in your subfunction and that using clear all as the first line of a function is useless since a function will always be executed in its own workspace (that will however clear persistent or global variables, but you don't use them in your code).