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Topic: 10th Derivative... (Read 711 times) |
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Dimetri
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Find f(10)(0) for f(x) = sqrt(1 + 4x2 - 4x3 - 4x5 + 4x6 + x8)
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towr
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Re: 10th Derivative...
« Reply #1 on: Dec 8th, 2005, 7:00am » |
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Any better way than to grab the nearest mathematics program and have that work it out?
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Dimetri
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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.
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Michael Dagg
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Re: 10th Derivative...
« Reply #3 on: Dec 8th, 2005, 12:15pm » |
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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.
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| « Last Edit: Dec 8th, 2005, 12:17pm by Michael Dagg » |
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Michael Dagg
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Michael Dagg
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Re: 10th Derivative...
« Reply #4 on: Dec 8th, 2005, 12:36pm » |
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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.
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| « Last Edit: Dec 8th, 2005, 1:35pm by Michael Dagg » |
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Michael Dagg
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towr
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Re: 10th Derivative...
« Reply #5 on: Dec 8th, 2005, 12:47pm » |
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Oops, I wouldn't have removed my post about possibly using laplace, if I knew you were going to respond to it.. 
But tell me more about Leibnitz' rule. How would that work here?
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| « Last Edit: Dec 8th, 2005, 12:49pm by towr » |
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Michael Dagg
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Re: 10th Derivative...
« Reply #6 on: Dec 8th, 2005, 12:59pm » |
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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! ?
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| « Last Edit: Dec 8th, 2005, 12:59pm by Michael Dagg » |
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Icarus
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Re: 10th Derivative...
« Reply #7 on: Dec 8th, 2005, 4:00pm » |
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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.
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Demetri
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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.
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Michael Dagg
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Re: 10th Derivative...
« Reply #9 on: Dec 8th, 2005, 5:59pm » |
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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.
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| « Last Edit: Dec 8th, 2005, 6:38pm by Michael Dagg » |
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Michael Dagg
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Icarus
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Re: 10th Derivative...
« Reply #10 on: Dec 8th, 2005, 7:37pm » |
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on Dec 8th, 2005, 5:59pm, Michael_Dagg wrote:for n=10 it must be that
2pi i = [int] f^(n) (z) z^(-n-1) dz, contour as mentioned. |
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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).
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Michael Dagg
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Re: 10th Derivative...
« Reply #11 on: Dec 8th, 2005, 7:52pm » |
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Indeed, I inadvertently wrote ^(n).
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Michael Dagg
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Icarus
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Re: 10th Derivative...
« Reply #12 on: Dec 8th, 2005, 8:29pm » |
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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.
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| « Last Edit: Dec 8th, 2005, 8:29pm by Icarus » |
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Dimetri
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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!.
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Michael Dagg
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Re: 10th Derivative...
« Reply #14 on: Dec 21st, 2005, 12:47pm » |
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>> 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.
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Michael Dagg
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Icarus
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Re: 10th Derivative...
« Reply #15 on: Dec 21st, 2005, 7:12pm » |
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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).
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Icarus
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Re: 10th Derivative...
« Reply #16 on: Jan 9th, 2006, 7:17pm » |
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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!.
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| « Last Edit: Jan 10th, 2006, 5:18am by Icarus » |
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pcbouhid
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Re: 10th Derivative...
« Reply #17 on: Jan 10th, 2006, 4:20am » |
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Tks, Icarus.
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Sjoerd Job Postmus
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Re: 10th Derivative...
« Reply #18 on: Jan 10th, 2006, 6:52am » |
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Well, I let maple crunch, and it came to the solution that f[sup](10)[/sub](0) = 3628800... How? Just letting maple crunch.
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SWF
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Re: 10th Derivative...
« Reply #19 on: Jan 10th, 2006, 5:48pm » |
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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.
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Michael Dagg
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Re: 10th Derivative...
« Reply #20 on: Jan 11th, 2006, 11:35am » |
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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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| « Last Edit: Jan 11th, 2006, 12:48pm by Michael Dagg » |
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