Higher derivatives

  • Thread starter dnt
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  • #1
dnt
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how do you find the general formula for the nth derivative of x^1/2 (square root of x)?

i can get the full formulas but i cannot put it together as a general formula:

n=1 (1st der): (1/2)(x^-1/2)
n=2 (2nd der): (1/2)(-1/2)(x^-3/2)
n=3 (3rd der): (1/2)(-1/2)(-3/2)(x^-5/2)
n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2)
...
nth der = ???

i have it right so far, correct? i just cant figure out how put it in terms of n (is it an arithmetic series or something)?

thanks.
 

Answers and Replies

  • #2
998
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Try to figure out a formula that works for the terms you have (think factorials). Then prove by induction that it works for all [itex]n[/itex].
 
  • #3
dnt
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i think the denominator will be 2^n

i think the exponent of x will be [-(2n-1)]

but i cant figure out the numerator part that goes 1,-1,-3,-5

would that be 3-2n?
 
  • #4
dnt
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(3-2n)/2^n * x^(1-2n)

does that work?
 
  • #5
0rthodontist
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n=4 (4th der): (1/2)(-1/2)(-3/2)(-5/2)(x^-7/2)
= [tex](-1)^3 \frac{(1)(3)(5)}{2^4} \cdot x^{-\frac{7}{2}}[/tex]
= [tex](-1)^3 \frac{(1)(2)(3)(4)(5)(6)}{(1)(2)(3)(2^7)} \cdot x^{-\frac{7}{2}}[/tex]
= [tex](-1)^3 \frac{6!}{(3!)(2^7)} \cdot x^{-\frac{7}{2}}[/tex]

By the way, instead of asking us if it works, why not find out yourself? Also the exponent of n is not what you said. It starts at 1/2 and decreases by 1 each time.
 
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  • #6
dnt
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but how does that work in terms of n?
 
  • #7
0rthodontist
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Do some guessing and checking to figure out how it is generalized.
 
  • #8
dnt
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so is my previous guess of (3-2n)/2^n * x^(1-2n) totally wrong?
 
  • #9
0rthodontist
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It is, but you should verify that yourself.
 
  • #10
998
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Well, do any of the derivatives you've found so far have integer powers of x (as your guess would predict)? Does your guess work to find the derivatives you've already calculated?

I'll give you a hint: yes you'll need powers of 2 in the denominator. Also look at the numerators of the coefficients. You're multiplying together a lot of numbers to get it in each case. Do you see a pattern in the numbers you're mulplying together?
 
  • #11
HallsofIvy
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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]
 
  • #12
dnt
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i didnt know that HoI. thanks. it also didnt help that our teacher told us not to multiply the stuff out to look for patterns...

however, that still doesnt explain how the numerator starts 1,-1...if i use what you did i start at 3/8 (which is the 3rd derivative, not the first)
 
  • #13
dnt
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is the 2nd part x^(n-1/2)? am i at least correct there?
 
  • #14
0rthodontist
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Your teacher gave you bad advice if he/she meant not to try any examples--it is good to multiply it out, check, look for a pattern, generalize. For one thing it's easier to deal with say the 4th derivative than with an arbitrary product notation or ellipsis. Though your final solution and derivation shoud be in the general form, it's often easier to work it out first for yourself on a specific example.

You should be able to tell yourself whether or not xn-1/2 is correct.
 
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  • #15
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Wouldn't it be ((3-2n)/2^n)*x^(1/2 - n)?
 
  • #16
dnt
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ok i tried using the equation for the product of odd in integers, which started me at 3 if i used n=1.

so i backed it up by substituting (n-1) for n to get:

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

but that only brings me to the 2nd derivative if i plug in n=1.

how can i back it up another spot so that n=1 gives me the 1st derivative? i tried substituting n-2 for n but that gives me a negative factorial on the numerator.

i just cannot figure out the pattern for 1,1,3,15,105,etc...i know its the product of odd integers but those first numbers screw me up. how do i get it to start 1,1... instead of 1,3...
 
  • #17
dnt
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0rthodontist said:
Your teacher gave you bad advice--it is good to multiply it out, check, look for a pattern, generalize. For one thing it's easier to deal with say the 4th derivative than with an arbitrary product notation or ellipsis.

You should be able to tell yourself whether or not xn-1/2 is correct.

oops i meant to make it negative:

x1/2-n

i think that part is correct but the rest is tough
 
  • #18
dnt
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lapo3399 said:
Wouldn't it be ((3-2n)/2^n)*x^(1/2 - n)?

dont think that works because eg, n=4 gives you:

(-5/16) for the first part when it should be -15/16
 
  • #19
0rthodontist
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dnt said:
ok i tried using the equation for the product of odd in integers, which started me at 3 if i used n=1.

so i backed it up by substituting (n-1) for n to get:

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

but that only brings me to the 2nd derivative if i plug in n=1.

how can i back it up another spot so that n=1 gives me the 1st derivative? i tried substituting n-2 for n but that gives me a negative factorial on the numerator.

i just cannot figure out the pattern for 1,1,3,15,105,etc...i know its the product of odd integers but those first numbers screw me up. how do i get it to start 1,1... instead of 1,3...
It looks okay the way it is, now you just need to include the powers of -1 (sign alternation), the additional denominator of 2^(n), and the power of x, which you already have correctly.
 
  • #20
dnt
238
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i still dont see why thats ok.

using this: (2n-1)!/[2^(n-1)*(n-1)!]

gives me 1,3,15,105 for n=1,2,3,4

which means it starts off with the 2nd derivative for n=1.

it should start with the first derivative for n=1
 
  • #21
0rthodontist
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Oh, I see what you mean. So just use n-2 and treat the first derivative as a separate case. That formula for product of odd integers is anyway only good when the integers are all positive.
 
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  • #22
998
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Halls' expression isn't quite what you need. Keep in mind that all three of these expressions are for the product of the first n+1 odd numbers, and for your derivatives you want the product of the first n-1 odd numbers.

[tex] (1)(3)(5)...(2n+1) = \frac{(2n+1)!}{n!2^n} = \frac{(2n+2)!}{(n+1)!2^{n+1}}[/tex]


Give the expression on the farthest right a try :)

(the reason that Halls' doesn't work is that it's only applicable for [itex]n \geq 0[/itex], whereas the one on the right works for [itex]n \geq -1[/itex] - in the case [itex]n=-1[/itex], you get 1 [instead of the "product of the first 0 odd numbers" which doesn't make much sense])
 
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  • #23
dnt
238
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still doesnt work.

n=1 gives 4!/(2!*4)=3

or am i suppose to substitute n-1 in for n?
 
  • #24
998
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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:
 
  • #25
dnt
238
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n-1 substituted in doenst work either! i still get 1,3,15,105 but i cannot get the first two terms to be 1,1.

this is impossible. (just kidding but its freaking hard!)
 

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