evaluate symbolic expression in MATLAB

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.

func

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:

subs

%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:

vpa(func(0.1))

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.

double(func(0.1))

double

This question has not received enough attention.

Does vpa(func(2)) solve your problem?
– Nicky Mattsson
Jun 27 at 18:12

vpa(func(2))

@NickyMattsson Thanks for your comment. vpa(func(2)) returns numerical value.However if I use vpa(func(0.1)), MATLAB returns 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)...
– Vizag
Jun 27 at 18:22

vpa(func(2))

vpa(func(0.1))

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)

2 Answers
2

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!

Use this command

double(func(2))


double(func(2)) returns a numerical value.double(func(0.1)) doesn't return anything. As I have said in my question, I need to evaluate on a range of values x_in = 0:0.01:1, so I need to be able to get result for fractional values as well.
– Vizag
Jun 27 at 18:32

double(func(2))

double(func(0.1))

x_in = 0:0.01:1

func(0.1) contains complex number and can be converted to numerical value.
– PyMatFlow
Jun 27 at 18:37

how can there be a complex part? Each term in the infinite series is real valued.
– Vizag
Jun 27 at 18:39

Complex number can be created in many ways. When i take a look at result of func(0.1) , I see complex symbol in the output. I do not generate it. MATLAB generates the result.
– PyMatFlow
Jun 28 at 6:36

No. the 'i' in the result is the one I gave as input and not iota. Its symbolically represented. Change cosfun_hat(a,i) to cosfun_hat(a,k) and you'll see the symbol 'k' in the result obtained
– Vizag
Jun 28 at 6:38

cosfun_hat(a,i)

cosfun_hat(a,k)

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