I am trying to evaluate a function which is an infinite cosine series at some input values.
EDIT: Posting an image to describe what the infinite series looks like 
I wrote the following code to describe it in MATLAB.
function func = cosfun_hat(a,i) syms m x; assume(m,'integer'); assumeAlso(m > 0); sum(x) = sqrt(1-a^2)*symsum(sqrt(2)*a^m*cos(i*sym(pi)*x*2^m+1),m,0,Inf); func(x) = sum(x); end I want to evaluate the returned 'function' func to get numerical values for some input range say x_in = 0:0.001:1.
%Trying to evaluate func at x = 2 %In the command window I write func = cosfun_hat(0.5,2); func(2) which returns the symbolic expression:
(2^(1/2)*3^(1/2)*sum((1/2)^m*(exp(- pi*exp(m*log(2))*4*i - i)/2 + exp(pi*exp(m*log(2))*4*i + i)/2), m == 0..Inf))/2 I tried using subs to evaluate the expression:
%In the command window syms y; w(y) = func(y); y = 2; subs(w); But that returns the same symbolic expression. I am quite new to symbolic MATLAB.
Thanks!
EDIT Based on the comment by @NickyMattsson I tried
vpa(func(2)) which returns the numerical value of the expression. However, vpa(func(0.1)) returns a symbolic expression:
ans = 1.2247448713915890490986420373529*numeric::sum((1/2)^m*(exp(- (pi*exp(m*log(2))*i)/5 - i)/2 + exp((pi*exp(m*log(2))*i)/5 + i)/2), m == 0..Inf) The same problem with using double(func(0.1)), double doesn't return anything and is stuck.
2 Answers
Answers 1
Figured out a way to do it without using symbolic MATLAB.
function func = cosfun_hat(a,i,x) %syms m; % assume(m,'integer'); % assumeAlso(m > 0); % m = 0; sum = zeros(1,length(x)); sum2 = Inf(1,length(x)); while max(sum2-sum) > 1e-16 disp(m); sum2 = sum; sum = sum + sqrt(1-a^2)*sqrt(2)*a^m*cos(i*pi*x*2^(m+1)); m = m+1; end func = sum; end The sum converges inside 100 iterations.
Now if I do,
%In command window x_in = -2:0.001:2; f = cosfun_hat(0.6,2,x_in); plot(x_in,f); I get the plot:
Thanks everyone for your help!
Answers 2
Use this command
double(func(2)) 
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