Cumulative sum with a for loop

I need to create a program for 1^2 +2^2 +...+1000^2

Both vectorised and with for loops.

I have managed the vector one I think:

 x=1:1000
 xsums=cumsum(x.^2)
 y=xsums(1000)

however for the the for loop version of the program I can't seem to get it, what I have made is :

 x=1:1000
 for n = 1:length(x)
     y=sum(n.^2)
 end

I'm also not even sure if that is the right idea.

any help would be great thanks

Answer 1

[Andrei Bobrov]

"I need to create a program for 1^2 +2^2 +...+1000^2"

sum((1:1000).^2)

or

s = 0;
for ii = 1:1000
    s = s + ii^2;
end

[KL]

@Phil Whitfield: The difference between using cumsum (you don't even need it!) and Andrei's solution is worth noting.

>> s = sum((1:1000).^2);
>> x=1:1000;
>> xsums=cumsum(x.^2);
>> whos s xsums
  Name       Size              Bytes  Class     Attributes

    s          1x1                   8  double              
    xsums      1x1000             8000  double

the size difference is 1000 times! Imagine you're doing it for a much more bigger number.

[Jan]

This is a homework question. Phil has shown his own effort and posted almost running code. Therefore I consider it as okay and useful to provide working solutions. +1

The loop has the advantage here, that it does not use a lot of temporary memory. It would run with n=1e12 also on a 8GB machine, in opposite to the vectorized version.

[Phil Whitfield]

how would I do the same thing for 1- (1/2) + (1/3) -(1/4)... +(1/999)?

I have managed this so far:

 i=0
 z=0
 b=0
 x=-1;
 for a=1:999;
     if mod(a,2)==0
         i=x./a
     elseif mod(a,2)~=0
         z=1./a
     end
 b= b+(i+z)
 end
         disp(b)

and

sum((1./(1:2:1000))-(1./(2:2:1000)))

I'm pretty sure the second one is right but I cant get the first to give the same answer.

I really appreciate it, thanks

[Stephen23]

Going up to 1000 gives the wrong answer. Try something like these:

>> 1+sum((1./(3:2:999))-(1./(2:2:999)))
ans =    0.693647430559821
>> sum(1./(1:2:999))-sum(1./(2:2:999))
ans =    0.693647430559813

loop, gives same output:

>> b = 0;
>> for k=2:2:999, b=b-1/k; end
>> for k=1:2:999, b=b+1/k; end
>> b
b =    0.693647430559823

Note that these only differ at the 14th significant figure.

[Jan]

@Phil:

S = 0;
for a = 1:999
  S = S + (-1)^(a-1) / a;
end

Or without the expensive power operation:

S = 0;
m = 1;
for a = 1:999
  S = S + m / a;
  m = -m;
end

Answer 2

[Jan]

Further solutions:

• DOT product:

v = 1:n;
s = v * v.';

This uses one temporary vector only, while sum(v .^ 2) needs to create two of them: v and v.^2 .

• Avoid the squaring: The elements of 1^2, 2^2, 3^2, ... are:

1, 4, 9, 16, 25

The difference is

3, 5, 7, 9

with an obvious pattern. Then:

s = sum(cumsum(1:2:2*n))

This is cheaper as squaring the elements. As loop:

s = 0;
c = 1;
d = 1;
for ii = 1:n
  s = s + d;
  c = c + 2;
  d = d + c;
end

Only additions, but s = s + ii * ii is nicer and slightly faster.

• Finally remember C.F. Gauss, who provided some methods to process the results of sums efficiently:

s = n * (n+1) * (2*n+1) / 6

Nice!