Multiple values of x what will be code

%it is just a example i want to know if there are multiple number of say x with other constant say a and b. Is it possible to plot graph between x and y. I am working with similar code that i wrote below but graph is not showing. Clc; Close all; Clear all; A=5; B=6; X=4,9,56,33 Y=(a+b)/x Plot (X,Y)

Answer 1

[Star Strider]

Try this:

A=5; 
B=6; 
X = [4,9,56,33]; 
Y=(A+B)./X; 

figure
plot (X,Y)

Use element-wise operations (the (.) operator) in the division. See Array vs. Matrix Operations for an extended discussion.

[rajdeep singh]

Thank you sir, but it is working perfectly for scalar.

Here is the real code

A=31*10^6; B=7.625*10^-43; C=8.38*10^-34; D=700; f=[23.27,24.72,33.18,34.63,44.08,44.54,44.99] i= arrayfun(@(z) ((2B*z^3)/(A*(exp((B*z))-1))),f) plot (f,i)

I tried arrayfun but still not working.

[rajdeep singh]

I mean / instead of ./ Is also working perfectly. For single valued x.

[Star Strider]

The arrayfun call (essentially a slow for loop) is not necessary here. Vectorised MATLAB commands make explicit loops and other such operations unnecessary.

If I understand correctly what you want to do, this should work:

A=31E6; 
B=7.625E-43; 
C=8.38E-34; 
D=700; 
f=[23.27,24.72,33.18,34.63,44.08,44.54,44.99]; 
zfcn = @(z) (2*B*z.^3)./(A*exp(B*z)-1);

figure
plot(f, zfcn(f))
grid

This works for vector ‘f’. Again, note the use of element-wise operations in the exponentiation and division. You can assign the output of ‘zfcn(f)’ to an intermediate variable if you wish. I do not here, and simply use it as an argument to the plot function.

Answer 2

[fred ssemwogerere]

While your question may not be so clear, from my understanding, you are trying to find a vector,"Y" from a vector of "X" values. Try this:

A=5; B=6; X=[4,9,56,33]; 
% your vector "Y". From this you can proceed with your plot
Y=arrayfun(@(z) (A+B)/z, X);

[rajdeep singh]

clc; close all; clear all;

A=31*10^6; B=7.625*10^-43; C=8.38*10^-34; D=700; f=[23.27,24.72,33.18,34.63,44.08,44.54,44.99] i= arrayfun(@(z) ((2B*z^3)/(A*(exp((B*z))-1))),f) plot (f,i)

[rajdeep singh]

Not working