How to apply the sum function of a factorial equation without using the factorial function

How do I use SUM and nested FOR loops to calculate the sigma notation of (2^i)/(i!) when i=0..20 without using the function factorial(i)?
I've tried
A=2^i
SUM(A)=0
for i=1:20
SUM(A)=SUM(A)+i
but this already doesn't work, and I cannot add the nested for loop of the factorial until I have this. How would this nested for loop look without using the factorial() function?

 Accepted Answer

Dear Anj, do you need something like this:
i = 1:20;
val = sum(2.^i);

10 Comments

Yes, I've gotten this far but I need to use a FOR Loop to divide each of these values by i! without using the factorial() function on matlab.
In other words SIGMA[(2^i)/(i!)] when i goes from 1 to 20.
you need this:
2^i / i!
in each loop iteration? and then sum up all values?
Yes, but without explicitly using the factorial() function.
Here is the code:
sum_all = 0;
for i = 1:20
power_val = 2^i;
fact = 1;
for j = 1:i
fact = fact * j;
end
sum_all = sum_all + power_val / fact;
end
disp(sum_all)
Your inner for loop can be replaced simply by
fact = prod(1:i);
Also we recommend using k or some other variable than i (the imaginary variable) for a loop counter (though it will work). So then your entire program above (now hidden) becomes simply:
sum_all = 0;
for k = 1:20
sum_all = sum_all + 2^k / prod(1:k);
end
disp(sum_all)
Yes you are right. But I used nested for loop because it was insisted by Anj to use nested for loop. Thanks for suggestions about 'i' as loop variable. I will try to avoid it although it is my habit to use 'i' as loop variable. Thanks a lot nevertheless
You're right. I don't know why teachers have students ignore the best part of MATLAB (vectorization) when that is exactly the perfect opportunity to demonstrate it.

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Asked:

Anj
on 20 Oct 2013

Commented:

on 20 Oct 2013

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