Find Nth Derivative of Square Root of X

[dnt]

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.

 

[Data]

Halls is HallsofIvy

 

[dnt]

lol. i was thinking halls was some calc theorem :)

 

[Data]

occasionally an apparently useless thing (like, say, multiplying by 1, which is all I did) can allow you to avoid splitting things up into ugly "special cases"

 

[0rthodontist]

yes, it's the same except that you can still evaluate it for $n= -1$ (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~

Ah, good idea.
I do think it would have been easier to generalize from example (continuing from before, at the fourth derivative):
$$(-1)^3 \frac{6!}{(3!)(2^7)} \cdot x^{-\frac{7}{2}}$$
Now you know that you're going to change the sign every time, so the exponent on -1 is going to increase by 1 every time. The 6! came from 5 * 3 * 1 so that is going to increase by 2 every time. The 3! will increase by 1 every time, and the exponent on the 2 will increase by 1 every time from 5 * 3 * 1 and by another 1 every time from the denominator. So you can then write
$$(-1)^{n - 1} \frac{(2 * (n - 1))!}{((n-1)!)2^{2n - 1}} \cdot x^{\frac{1}{2} - n}$$
where all I did was make sure the various terms increase as they should, and otherwise adjust them by constants so they match the example. This saves you some fiddling with off-by-one errors and then you can verify it with more general notation.

 

[dnt]

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:
$$\frac{1*2*3*4*5*...*(2n)*(2n+1)}{2*4*6*...*(2n)}= \frac{(2n+1)!}{2^n(n!)}$$

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]

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.

can anyone explain it?
thanks.

 

[Data]

in regards to that numerator (product of the first n integers) how does it equal (2n+1)! ?

it doesn't. (2n+1)! is the product of the first 2n+1 integers, of course, not the first n integers.

But $(2n+1)!/(n!2^n)$ is the product of the first $n+1$ odd integers, because you're multiplying together the first 2n+1 integers and then dividing out the even ones. Here $n!2^n$ is the product of the first n even integers:

$$n!2^n = n(n-1)...(1)*2^n = (2n)(2n-2)...(2).$$

So in $(2n+1)!/(n!2^n)$, you're multiplying together the first $2n+1$ integers in the top, then you divide out the first $n$ even integers in the bottom, leaving you with the first $(2n+1)-n = n+1$ odd integers!