MATLAB: How to Plot y=sin(2pimx)+sin(2pinx) , where m and n are positive integers and x is a variable

equationMATLABplot

so this is the code I wrote:

syms x y real;
syms m n integer;
m>0;
n>0;
y=sin(2*pi*m*x)+sin(2*pi*n*x);

Now I do not know how to plot this function y. I tried using plot , eplot , fplot. Nothing works. I am new to Matlab so please bare with me if the question is too vague or else.

Best Answer

m = 0:1:10 ; 
n = 0:1:10 ;
[M,N] = meshgrid(m,n) ; 
x = pi/4 ; 
Y=sin(2*pi*M*x)+sin(2*pi*N*x); 
surf(M,N,Y)

If you want to vary x and fix m,n:

m = 2 ; 
n = 3 ; 
x = linspace(0,2*pi) ;  
Y=sin(2*pi*m*x)+sin(2*pi*n*x); 
plot(x,Y)

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MATLAB: How can i use L’hospital rule i.e sin(x)/x for x tends to zero

By convention, 0/0 evaluates to NaN.

To be absolutely formal (if a bit kludgy) about using L’Hospital's rule in this situation:

fderiv = @(f,x) (f(x+eps)-f(x))/eps;                    % Derivative of a function
xderiv = @(x) ((x+eps)-x)/eps;                          % Derivative of an expression
for n = 0:7
    x(n+1)=sin((pi*n)/2)./(pi*n);
    if isnan( x(n+1) )                                  % Test for ‘NaN’
        x(n+1) = fderiv(@sin, pi*n/2)./(xderiv(pi*n));  % L'Hospital's Rule
    end
end

There are better and more numerically stable derivative expressions (such as the five-point algorithm), but this will do for an illustration.

Also, using L'Hospital's rule,

x=sin((pi*n)/2)./(pi*n)

will be 1 at n=0, however:

x=(sin(pi*n)/2)./(pi*n)

with the fderiv call then becoming:

x(n+1) = (fderiv(@sin, pi*n)/2)./(xderiv(pi*n));

will be 0.5 for n=0.

Best Answer

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