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        riddles >> easy >> 10th Derivative...
        
(Message started by: Dimetri on Dec 8th, 2005, 6:46am)
Title: 10th Derivative...
Post by Dimetri on Dec 8th, 2005, 6:46am
Find f(10)(0) for f(x) = sqrt(1 + 4x2 - 4x3 - 4x5 + 4x6 + x8)
Title: Re: 10th Derivative...
Post by towr on Dec 8th, 2005, 7:00am
Any better way than to grab the nearest mathematics program and have that work it out?
Title: Re: 10th Derivative...
Post by Dimetri on Dec 8th, 2005, 7:21am
There are probably other ways but I know of a way to do it which does not require diff'ing up to 10 times. Diff'ing up to 10 produces an enormous result.
Title: Re: 10th Derivative...
Post by Michael_Dagg on Dec 8th, 2005, 12:15pm
A less obvious way is to use Leibnitz' rule.

That is, let g = f^2 =  1 + 4x^2 - 4x^3 - 4x^5 + 4x^6 + x^8.
Title: Re: 10th Derivative...
Post by Michael_Dagg on Dec 8th, 2005, 12:36pm
If you are curious compute a few f^(n) (0) and you will find that there are a few n for which it is true that f^(n) (0) = n!. It is no accident that this f^(n) (0) and n! are parts of the n-th term in the Taylor series for f about x=0. In fact, it tells you that some coefficients equal 1, i.e, f^(n) (0) = n! -> f^(n) (0)/n! = 1 for various n.
Title: Re: 10th Derivative...
Post by towr on Dec 8th, 2005, 12:47pm
Oops, I wouldn't have removed my post about possibly using laplace, if I knew you were going to respond to it.. ;D

But tell me more about Leibnitz' rule. How would that work here?
Title: Re: 10th Derivative...
Post by Michael_Dagg on Dec 8th, 2005, 12:59pm
Towr and I are posting at the same time and that is why this thread has changed like it has.

This discussion has lead us to another question: For how many n is it true that f^(n) (0) = n!  ?
Title: Re: 10th Derivative...
Post by Icarus on Dec 8th, 2005, 4:00pm
By Liebnitz's rule, are you refering to dy/dx = dy/du * du/dx?

Another alternative is the generalized Cauchy formula:

f(n)(0) =(n!/2[pi]i) [int] f(z)z-n-1dz

where the integral is around a circle about z=0 small enough to be inside all roots of the polynomial.
Title: Re: 10th Derivative...
Post by Demetri on Dec 8th, 2005, 4:43pm
Ah. From what Icarus said I see the what is with Leibniz. Implicit diff on the left side it looks like

f^2 = 1 + 4x^2 - 4x^3 - 4x^5 + 4x^6 + x^8 then
(f*f)' = f*f' + f'*f = 2f*f' = d/dx(1 + 4x^2 - 4x^3 - 4x^5 + 4x^6 + x^8 ) = 8x - 12x^2 - 20x^4 + 24x^5 + 8x^7

f*f' = 4x - 6x^2 - 10x^2 + 12x^5 + 4x^7,

and then keep going with implicit diff on the left side. Looks like there are a couple of unknowns to solve for at the end since the right hand side eventually vanishes.

But, this is not the way I did it.
Title: Re: 10th Derivative...
Post by Michael_Dagg on Dec 8th, 2005, 5:59pm
Demetri you are right - I  thought someone would take off with it when I wrote g = f^2 = ...

Regarding the complex integral that Icarus mentioned, for n=10 it must be that

2pi i = [int] f^(n) (z) z^(-n-1) dz, contour as mentioned.
Title: Re: 10th Derivative...
Post by Icarus on Dec 8th, 2005, 7:37pm

on 12/08/05 at 17:59:25, Michael_Dagg wrote:
for n=10 it must be that

2pi i = [int] f^(n) (z) z^(-n-1) dz, contour as mentioned.


Only if f(20)(0) = 10!. The formula I gave is correct.

The full integral formula is: If f is analytic inside a simple closed contour C, then for all w inside C, and for all n >= 0,

f(n)(w) = (n!/2[pi]i) [int]C f(z)z-n-1 dz.

It's the formula that proves that for complex functions, C1(D) = Coo(D).
Title: Re: 10th Derivative...
Post by Michael_Dagg on Dec 8th, 2005, 7:52pm
Indeed, I inadvertently wrote ^(n).
Title: Re: 10th Derivative...
Post by Icarus on Dec 8th, 2005, 8:29pm
Then I see your point.

Another approach is to note that sqrt(1+x) = 1 + (1/2)x - (1/8)x2 + (1/16)x3 - (5/128)x4 + (7/256)x5 + o(x6).

If one replaces 1+x in the above with the polynomial expression 1 + 4x2 - 4x3 - 4x5 + 4x6 + x8, and simplifies, discarding anything with a degree of 11 or higher (which keeps the calculation from being quite so bad as it first sounds), then the coefficent of x10 = f(10)(0)/10!.

Admittedly, this is still more work than you want to do.
Title: Re: 10th Derivative...
Post by Dimetri on Dec 8th, 2005, 8:57pm
I did not think I would get a dialog going like this with that is problem. This problem was on one of my finals this week. The answer is definitely 3628800 = 10! but not up until the posts earlier did I now that the number was 10!.
Title: Re: 10th Derivative...
Post by Michael_Dagg on Dec 21st, 2005, 12:47pm
>> Demetri you are right - I  thought someone would take off with it when I wrote g = f^2 = ...

Still no one has produced the left-hand side.
Title: Re: 10th Derivative...
Post by Icarus on Dec 21st, 2005, 7:12pm
A hint on how to do it easily:

If g = [sum]k=0n C(n,k) f(k)f(n-k), then what is g'?

Remember the recursion relation C(n+1,k) = C(n,k-1) + C(n,k).
Title: Re: 10th Derivative...
Post by Icarus on Jan 9th, 2006, 7:17pm
In response to a query from pcbouhid, I fleshed out the details of the Taylor series for sqrt(1+x) method I mentioned above. Since I have, I thought I would post it here for anyone who is interested. While it may or may not be easier than the methods that Dimetri and Michael_Dagg used, I think it is easier than the implicit differentiation method we have discussed above:

sqrt(1+z) = 1 + (1/2)z - (1/8)z2 + (1/16)z3 - (5/128)z4 + (7/256)z5 + o(z6)

Make the substitution z = 4x2 - 4x3 - 4x5 + 4x6 + x8. Then we calculate the coefficient of x10 in each of the terms. The 1, z, and o(z6) terms clearly contribute nothing to the coefficient of x10.

z2 contributes 2 combinations: x2x8 occurs twice, and (x5)2 occurs once. Coefficient = 2(4)(1) + 1(-4)(-4) = 24.

z3 also contributes 2 combinations: x2x3x5 occurs 6 times, while (x2)2x6 occurs 3 times. Coefficient = 6(4)(-4)(-4) + 3(4)(4)(4) = 576.

z4 contributes 1 combination: (x2x3)2 occurs 6 times. Coefficient = 6(4)(4)(-4)(-4) = 1536.

z5 contributes 1 combination: (x2)5 occurs 1 time. Coefficient = 45 = 1024.

Total coefficient of x10 is therefore -(24)/8 + (576)/16 - (1536)5/128 + (1024)7/256 = -3 + 36 - 60 + 28 = 1.

Hence f(10)(0)/10! = 1, and f(10)(0) = 10!.
Title: Re: 10th Derivative...
Post by pcbouhid on Jan 10th, 2006, 4:20am
Tks, Icarus.
Title: Re: 10th Derivative...
Post by Sjoerd Job Postmus on Jan 10th, 2006, 6:52am
Well, I let maple crunch, and it came to the solution that f[sup](10)[/sub](0) = 3628800... How? Just letting maple crunch.
Title: Re: 10th Derivative...
Post by SWF on Jan 10th, 2006, 5:48pm
That the question asks for 10th derivative at x=0 is a big clue that Taylor series is to be used instead of actually taking derivatives.

A simple way to get the answer to solve for the coefficents of the series one at a time. Start with a series

  [sum] an * xn   (sum on n from 0 to infinity)

When this is squared it should equal  1 + 4*x2 - 4*x3 - 4*x5 + 4*x6 + x8

Squaring the assumed series and equating coefficients for each power of x, gives simple expressions for each an in terms of the previously solved an. Just start at n=0 and work your way up to n=10.

That gives:  a0=1, a1=0, a2=2, a3=-2, a4=-2, a5=2, a6=4, a7=-8, a8=-11/2, a9=28, a10=1.

Since a10 is 1, the 10th derivative at x=0 is 10 factorial.
Title: Re: 10th Derivative...
Post by Michael_Dagg on Jan 11th, 2006, 11:35am
Icarus and SWF are definitely correct. Implicit differentialation is not that messy in this particular case. For primes higher than three I use Roman numerals:

Let g^2 = 1 + 4x^2 - 4x^3 - 4x^5 + 4x^6 + x^8, then

gg' = 4x - 6x^2 - 10x^4 + 12x^5 + 4x^7

(g')^2 + gg'' = 4 - 12x - 40x^3 + 60x^4 + 28x^6

3g'g'' + gg''' = - 12 - 120x^2 + 240x^3 + 168x^5

3(g'')^2 + 4g'g''' + gg^{IV} = - 240^x + 720x^2 + 840x^4

10g''g''' + 5g'g^{IV} + gg^{V} = - 240 + 1440x + 3360x^3

10(g''')^2 + 15g''g^{IV} + 6g'g^{V} + gg^{VI} = 1440 + 10080x^2

35g'''g^{IV} + 21g''g^{V} + 7g'g^{VI} + gg^{VII} = 20160x

35(g^{IV})^2 + 56g'''g^{V} + 28g''g^{VI} + 8g'g^{VII} + gg^{VIII} = 20160

126g^{IV}g^{V} + 84g'''g^{VI} + 36g''g^{VII} + 9g'g^{VIII} + gg^{IX} = 0

126(g^(V))^2 + 210g^{IV}g^{VI} + 120g'''g^{VII} + 45g''g^{VIII} + 10g'g^{IX}  + gg^{X} = 0

These equations imply g(0) = 1, g'(0) = 0, g''(0) = 4, g'''(0) = -12, g^{IV}(0) = -48,

g^{V}(0) = 240, g^{VI}(0) = 2880, g^{VII}(0) = -40320, g^{VIII}(0) = -221760, g^{IX}(0) = 10160640,

and finally g^{X}(0) = 3628800 = 10!.


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