- #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)! ?
The formula for finding the Nth derivative of the square root of x is:
dN/dxN √x = (-1)N (2N-1)!!/2N x-(2N-1)/2
To find the first derivative of the square root of x, you can use the power rule for derivatives. This means taking the exponent (1/2) and multiplying it by the coefficient (1), and then subtracting 1 from the exponent. This results in 1/2x-1/2, which simplifies to 1/2√x.
Yes, the Nth derivative of the square root of x can be negative. This is because the (-1)N term in the formula alternates between positive and negative values for even and odd values of N. So, for example, the 2nd derivative would be positive, but the 3rd derivative would be negative.
There is no known shortcut or easier way to find the Nth derivative of the square root of x. It is recommended to use the formula mentioned in question 1 to find the Nth derivative.
No, the Nth derivative of the square root of x cannot be simplified further. The formula mentioned in question 1 is the most simplified form for finding the Nth derivative of the square root of x.