Use sym to define a sum function, but i have a problem with indexing

% Define N(r)
c = [-1*ones(3,1); 0.3*ones(5,1); 0.6*ones(6,1)];
n = length(c);
  
syms k r
f = c(k)/(1+r)^(k-1);
V = subs(f, k, 1:n);
S_sum = sum(V);

Apparently Matlab gives me this problem when I tried to use elements in array c, e.g. c(1) to c(k)

Error using sym/subsindex (line 836)
Invalid indexing or function definition. Indexing must follow MATLAB indexing. Function arguments
must be symbolic variables, and function body must be sym expression.

How should I fix this index problem?

Thank you!

[Yihan Hu]

so I tried a different approach still using sym but it is a bit labor

c = [-1*ones(3,1); 0.3*ones(5,1); 0.6*ones(6,1)];
n = length(c);
 
syms N(r)
N(r) = c(1)/(1+r)^(1-1)+c(2)/(1+r)^(2-1)+c(3)/(1+r)^(3-1)+c(4)/(1+r)^(4-1)...
    +c(5)/(1+r)^(5-1)+c(6)/(1+r)^(6-1)+c(7)/(1+r)^(7-1)+c(8)/(1+r)^(8-1)+...
    c(9)/(1+r)^(9-1)+c(10)/(1+r)^(10-1)+c(11)/(1+r)^(11-1)+c(12)/(1+r)^(12-1)...
    +c(13)/(1+r)^(13-1)+c(14)/(1+r)^(14-1);

the following answers don't really give what I want here

Answer 1

[Walter Roberson]

it is never permitted to use aa symbolic variable as a subscript , not even for symsum.

You need to construct the definite equation

k = 1:N;
V = c./(1+r).^(k-1); 
S_sum = sum(V);

[Yihan Hu]

this code returns 14 terms, but the coefficient of each term is not really the individual element in vector c but the sum of all elements

[Walter Roberson]

c = [-1*ones(3,1); 0.3*ones(5,1); 0.6*ones(6,1)];
N = length(c);
k = (1:N).';
V = c./(1+r).^(k-1); 
S_sum = sum(V);

After that you might want to simplify(S_sum)

[Yihan Hu]

now it works!!

I will look into what difference the following makes!

k = (1:N).';

Thank you so much!

[madhan ravi]

+1 sir Walter but I don't understand why symsum produced 14 different vectors in this case ?

[Walter Roberson]

c is a column vector but k was a row vector, ./ was doing the equivalent of bsxfun(@divide, column_vector_expression, row_vector_expression) and so producing a 2D array.

[madhan ravi]

Thank you very much ! sir Walter makes perfect sense.

Answer 2

[madhan ravi]

syms r
k=1:n;
f = c(k)./(1+r).^(k-1);
S_sum=vpa(symsum(f),2);