- #36
dnt
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0rthodontist said:Data's expression is the same as HallsofIvy's.
your right...why didnt it work the first time i tried it?
dont know what i did. i think i substituted wrong or something.
0rthodontist said:Data's expression is the same as HallsofIvy's.
Ah, good idea.Data said:yes, it's the same except that you can still evaluate it for [itex]n= -1[/itex] (even though the "product of the first 0 odd integers" doesn't make any sense). It happens to give dnt exactly what he needs though~
HallsofIvy said:Also, your statement that "the numerator part goes 1,-1,-3,-5" is wrong.
What you put in your very first post was:
n=1 (1st der): (1/2)(x^-1/2) = (1/2)(x^(-1/2))
n=2 (2nd der): (1/2)(-1/2)(x^-3/2) = (-1/4)(x^(-3/2))
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2) = (+3/8)(x^(-5/2))
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2) = (-15/16)(x^(-7/2))
(I've added the last in each line.)
so the numerators are 1, -1, 3= 1*3, -15= -1*3*15.
It might help you to realize this:
A product of even integers is 2*4*6*...*2n= (2*1)(2*2)(2*3)...(2*n)= (2*2*2...2)(1*2*3*...n)= 2nn!.
A product of odd integer, 1*3*5*7*...*(2n+1) is missing the even integers: multiply and divide by them:
[tex]\frac{1*2*3*4*5*...*(2n)*(2n+1)}{2*4*6*...*(2n)}= \frac{(2n+1)!}{2^n(n!)}[/tex]
dnt said:i konw the problem has been solved but i wanted to make sure i really understood it and one more question about the above equation popped into my head.
in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?
eg. if n=5 (product of first 5 integers = 1*2*3*4*5), well that should be equal to 120 which is n!
if you use (2n+1)! you get 11!, a much bigger number. can someone reexplain that? thanks.
dnt said:in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?