Higher derivatives

  • Thread starter dnt
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  • #26
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try [itex]n-2[/itex] :tongue2:
 
  • #27
dnt
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Data said:
Take a look at what I said in the second sentence. That expression is the product of the first n+1 odd numbers and you want the product of the first n-1 odd numbers. So what should you sub in for [itex]n[/itex]? :smile:

do i substitute n-2?
 
  • #28
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try it! and if it works, then try to understand how :smile:
 
  • #29
dnt
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Data said:
try [itex]n-2[/itex] :tongue2:

ill try it. the post before you i typed in before your newest post :)
 
  • #30
dnt
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holy crap i think that works
 
  • #31
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indeed :wink:

so can you explain why it works? (take a look at the bottom of the post where I suggested it)
 
  • #32
dnt
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ok so for my final answer:

[(-1)^(n+1)(2n-2)!]/[(2^n)(n-1)!(2^(n-1))]*X^(.5-n)
 
  • #33
0rthodontist
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Data's expression is the same as HallsofIvy's.
 
  • #34
dnt
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Data said:
indeed :wink:

so can you explain why it works? (take a look at the bottom of the post where I suggested it)

still trying to understand it. i dont know what halls is though.
 
  • #35
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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.
 
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  • #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? :confused:

dont know what i did. i think i substituted wrong or something.
 
  • #37
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Halls is HallsofIvy
 
  • #38
dnt
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lol. i was thinking halls was some calc theorem :)
 
  • #39
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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"
 
  • #40
0rthodontist
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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~
Ah, good idea.
I do think it would have been easier to generalize from example (continuing from before, at the fourth derivative):
[tex](-1)^3 \frac{6!}{(3!)(2^7)} \cdot x^{-\frac{7}{2}}[/tex]
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
[tex](-1)^{n - 1} \frac{(2 * (n - 1))!}{((n-1)!)2^{2n - 1}} \cdot x^{\frac{1}{2} - n}[/tex]
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.
 
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  • #41
dnt
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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]


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.
 
  • #42
dnt
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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.

can anyone explain it?
thanks.
 
  • #43
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dnt said:
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 [itex](2n+1)!/(n!2^n)[/itex] is the product of the first [itex]n+1[/itex] odd integers, because you're multiplying together the first 2n+1 integers and then dividing out the even ones. Here [itex]n!2^n[/itex] is the product of the first n even integers:

[tex]n!2^n = n(n-1)...(1)*2^n = (2n)(2n-2)...(2).[/tex]

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