vector uses info from itself to grow without for cycle

I need to solve this problem without a for loop:

B= (1:20)   
A = [];                     
A(1) = 1/(1+B(1));      
for k = 2:length(B)
    A(k,1) = (1-B(k)*sum(A))/(1+B(k));    
end

i.e. I need to know if it is possible to get information from prebvious calculation to create a vector, but without a for loop. Thanks.

[darova]

You code looks weird

[gabriele fadanelli]

thank you I modified

[gabriele fadanelli]

in particular this is a case where I need to use previous data to "build" my vector: is there a way to do it without for loop?

[Guillaume]

What is the final A that your code should be producing?

[gabriele fadanelli]

Let's say it is a vector which needs the sum of the previous values of itself to be computed up to a certain length set by vector B

[Guillaume]

Rather than making us guess what you mean, I'll repeat, what is the desired final (20x1) A for your code above? At least give us the first few elements of the sequence. Alternatively, please write the recursion mathematically.

[gabriele fadanelli]

B = (1:20)   
A = [];                     
A(1) = 1/(1+S(1));      

for k = 2:length(S)
    A(k,1) = (1-B(k)*sum(A))/(1+B(k));    
end

vector A should be

0.500000000000000
0
-0.125000000000000
-0.100000000000000
-0.0625000000000000
-0.0392857142857143
-0.0265625000000000
-0.0192460317460318
-0.0146651785714286
-0.0115823412698413
-0.00939504794973545
-0.00778140262515264
-0.00655451329837490
-0.00559878557013975
-0.00483920439085755
-0.00422520498138715
-0.00372162668292910
-0.00330337238480719
-0.00295212243142038
-0.00265424172588574

[Walter Roberson]

Your code talks about length(S) but it only uses S(1) and no other S element?

[gabriele fadanelli]

I am sorry. New right code:

B= (1:20)   
A = [];                     
A(1) = 1/(1+B(1));      
for k = 2:length(B)
    A(k,1) = (1-B(k)*sum(A))/(1+B(k));    
end

thanks

[Walter Roberson]

Are you looking for a closed form formula that can give A(K) without calculating A(K-1) ?

Are you looking for a recusive function that can calculate A(K) by calling itself, without an explicit loop?

[gabriele fadanelli]

yes that is exactly what I want to know if is possible to do somehow

[Walter Roberson]

Which of the two?

Recursive functions with no explicit loop are easy for this.

A closed form formula might be difficult.

[gabriele fadanelli]

even recursive function would be nice thank you. I just need an example.

Answer 1

[Walter Roberson]

function A = calculate_A(n)
   if n == 1
      A = 1/2;
   else
       A_before = calculate_A(n-1);
       A = [A_before; (1-n*sum(A_before))/(1+n)];
   end
end

Note: this will fail at roughly 75000, due to the recursion using up memory.

Answer 2

[Ameer Hamza]

For the original code in your question. Following is the simplified form.

k = 1:20;
A = 6.^(k-1);

[gabriele fadanelli]

you are right, but I am actually looking for above problem solution, sorry.