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Fortran Factorials 
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#1
Feb912, 06:18 PM

P: 20

I don't understand at all how you tell the computer to evaluate a complicated factorial expression such as the one given in in the infinite sum of binomial theorem as
Ʃ [n! / k!(nk)! ] * x^k where n is the final value of the sum and k is where you are in the loop. It's supposed to be INTEGER :: k, n REAL :: sum, fact, x ASK INPUT (what are x and n?) DO k = 0,n sum = sum + fact*x**k fact = fact * (nk)/(k+1) END DO Is there a procedure to figure out what the term multiplied by the fact variable should be? 


#2
Feb912, 07:32 PM

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Look at what the term are summing, for each value of k.
When k = 0, it is n! / (0! n!) so k0 = 1 When k = 1, it is n! / (1! (n1)! so k1 = n = k0 * n / 1 When k = 2, it is n! / (2! (n2)! so k2 = n(n1) / 2! = k1 * (n1) / 2 When k = 3, it is n! / (3! (n3)! so k3 = n(n1)(n2) / 3! = k2 * (n2) / 3 etc. That is what the program is doing when it updates "fact". 


#3
Feb912, 09:35 PM

P: 20

Hmmm... then fact * (nk)/(k+1) can't be right because it doesn't match the results of working out all the factorials like that. I checked it in command prompt and it said the sum with x = 2 and n = 2 is 5. This sum represents (1+x)^n which should be 9 in that case.
I thought fact * ( (nk+1) / k ) would work but I'm not getting the right answer with that either. 


#4
Feb912, 09:50 PM

P: 20

Fortran Factorials
disregard that. i worked it out on paper and i got 9 but for some reason the program is outputting 5...
If the k0,k1,k2 etc values are the "fact", how is (n2)/3 = (nk)/(k+1) for k = 3? That's why I thought it should be ( (nk+1) / k ). 


#5
Feb1012, 07:33 AM

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Each time through the loop, it uses the current value of fact, then calculates the next value. 


#6
Feb1012, 10:30 AM

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#7
Feb1012, 10:38 AM

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