- #26

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try [itex]n-2[/itex] :tongue2:

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- Thread starter dnt
- Start date

- #26

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try [itex]n-2[/itex] :tongue2:

- #27

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Data said:Take a look at what I said in the second sentence. That expression is the product of the firstn+1odd numbers and you want the product of the firstn-1odd numbers. So what should you sub in for [itex]n[/itex]?

do i substitute n-2?

- #28

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try it! and if it works, then try to understand how

- #29

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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

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holy crap i think that works

- #31

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indeed

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

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

- #32

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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)

[(-1)^(n+1)(2n-2)!]/[(2^n)(n-1)!(2^(n-1))]*X^(.5-n)

- #33

0rthodontist

Science Advisor

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Data's expression is the same as HallsofIvy's.

- #34

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Data said:indeed

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

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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.

- #37

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Halls is HallsofIvy

- #38

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

- #39

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- #40

0rthodontist

Science Advisor

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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~

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

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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 ofeven integersis 2*4*6*...*2n= (2*1)(2*2)(2*3)...(2*n)= (2*2*2...2)(1*2*3*...n)= 2^{n}n!.

A product ofodd 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

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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]

[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]

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