Setting up vector function

I write a function SetUp to set up three vectors from an input vector k. Vector c = (-k(2); -k(3); ..... -k(n)) where I have n-1 elements. Vector d = (k(1)+k(2); k(2)+k(3); ...... k(n-1)+k(n); k(n)) where there are n elements And vector e = vector c.

Here is the code I wrote:

function [c,e,d] = SetUp

k = input('Enter k vector');
n = size(k,2);

for k1 = 1:n
    c = -k(k1)*ones(n-1,1);
    d = (k(k1)+k(k1+1))*ones(n,1); d(n) = k(n);
    e = c;
    end
end

By running this code I get an error: Index exceeds matrix dimensions.

Problem is in this line: d = (k(k1)+k(k1+1))*ones(n,1); d(n) = k(n);

In my opinion the problem is in indexing as the last term when Matlab runs d = (k(k1)+k(k1+1))*ones(n,1); is d = k(n)+k(n+1), where the program does not recognize k(n+1) as there is no such term in vector k.

So, the question is, how to overcome this problem as I don't even need the last term as the last term is d(n) = k(n)???

Thanks in advance!

[Guillaume]

You have to ask yourself several questions:

• What is the purpose of the loop in your code? Note that at each iteration, you're completely overwriting the previous value of c, d and e.
• Is a loop required to do what you want? (Hint: No)
• How can you have a different equation for the last term? (Hint: calculate all the other terms from 1 to n-1 instead of up to n)
• What happens with your code if the user enters a column vector instead of a row vector? (Hint: use numel instead of size(x, 2))

Answer 1

[Eric]

In your for loop, you are using k(k1+1). if k1 is the last element of k, then k(k1+1) will be exceeding the dimension of k. If you dont care about the last term, use

 for k1 = 1:n-1
     ... Stuff ...
 end
 d(n) = k(n);

or if you need the other vectors to go to n, stick it in a conditional

 for k1 = 1:n
     ... Other Stuff ...
     if k1~=n
         d = (k(k1)+k(k1+1))*ones(n,1);
     end
 end

[K O]

I see what I was missing.

if k1~=n
         d = (k(k1)+k(k1+1))*ones(n,1);
end

Thanks man!

Answer 2

[KL]

c = -k(2:end); %EDITED
d = k+[k(2:end);0];
e = c;

in case, k is row vector,

 c = -k(2:end); %EDITED
 d = k+[k(2:end),0];
 e = c

that's all you need. No need for a loop!

[Eric]

why not

 c = -k(2:end);

[K O]

I'll try without the for loop too.

Ty KL and Eric for discussion.

[KL]

Bam! I totally overlooked it! Thanks, I'll edit the answer.