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How to derive nth derivative of x(t), when nth derivative of t(x) is known?

  1. Aug 21, 2012 #1
    I have a function, [itex]x(t)[/itex], where [itex]x[/itex] and [itex]t[/itex] are real scalars. I have been trying to derive an analytical expression for the nth derivative of [itex]x[/itex] with respect to [itex]t[/itex]:[tex]x^{(n)}(t) \triangleq \frac{d^n}{dt^n}x(t)= \ \ ?[/tex]However, getting a general expression is proving tricky.

    Meanwhile, I have managed to derive a method for recursively obtaining the nth derivative of [itex]t[/itex] with respect to [itex]x[/itex]:[tex]t^{(n)}(x) = f\{t^{(n-1)}(x), t^{(n-2)}(x), \dots\} = \ \ \mathrm{known}[/tex]

    Is there any way I can use this information to my advantage?
     
  2. jcsd
  3. Aug 21, 2012 #2
    Just to clarify, the reason I think this should be possible is as follows...

    I start with the first derivative:[tex]x^{ \prime }(t) = \frac{1}{t^{ \prime }(x)}[/tex](which is known). Then, differentiating, I get:[tex]x^{ \prime \prime}(t) = \frac{d}{dt}x^{ \prime }(t) = x^{ \prime }(t) \frac{d}{dx}x^{ \prime }(t) = x^{ \prime }(t) \frac{d}{dx}\frac{1}{t^{ \prime }(x)} \\ \ \ \ \ \ \ \ \ = \frac{-t^{ \prime \prime}(x)}{t^{ \prime }(x)^3}[/tex] ... and continuing, we get more functions of [itex]t^{ \prime }(x), t^{ \prime \prime}(x), t^{ \prime \prime \prime}(x), \dots[/itex] and so on.

    Since I know all these terms, I feel I should be able to get a general answer. I suppose if I could just get a general expression for the nth derivative of a function inverse, that would suffice (??)
     
  4. Aug 21, 2012 #3
    I may have found a solution in Equation 7 of [1] (link).

    It'll take me forever to interpret it, as I'm not a mathematician ... but perhaps it's on the right lines??

    If the link is dead:
    [1] J. F. Traub, "On the nth Derivative of the Inverse Function", American Mathematical Monthly, Vol. 69, No. 9, Nov. 1962.
     
  5. Aug 21, 2012 #4
    Yes, equation (9) seems to be what you're looking for. As you can see, it's a very nasty formula - which would explain why they don't teach it to you in most calculus courses.

    The reason being is that, in general, computing derivatives is easy - you will rarely ever need any fancy techniques in order to calculate some derivative. Implicit differentiation together with the chain rule and product rule will usually suffice.

    Of course, this is still an interesting problem, and I had never thought about it. Good read :)
     
  6. Aug 22, 2012 #5
    As with many maths papers, I'm completely stumped. In particular, I don't understand what the limits should be on the sum in Equation 7. The given explanation is: "with the sum taken over all [itex]j_i[/itex] such that:[tex]\sum_{i=2}^{n}(i-1)j_i = n - 1, \ \ \ j_i \geq 0 \ \ \ \ \ \ (8)[/tex]and where [itex]r=\sum_{i=2}^{n}j_i[/itex]."

    ... but I can't see how to get started. Presumably an issue is that [itex]j_i[/itex] must be integer valued (?). Any ideas how to get started for n=2,3,...?
    In my case, ultimately, it's a matter of practicality. I want a computer to generate large numbers of various derivatives quickly. However, for me, the fun lies in the mathematical steps needed to get there.
     
  7. Aug 22, 2012 #6
    Ok, so I think the point is that Eq. (8) produces one equation in multiple unknowns. Then, there exist many combinations of integers that will satisfy that equation. We have to find all possible valid combinations (presumably, by exhaustive search).

    Then, for each valid combination, we have a set of numbers [itex]\{j_2, j_3, \dots, j_n, r \}[/itex] which we can plug into Eq. (7), set by set (until we have summed over all the valid sets).

    For example...

    n = 2
    From Eq. 8:
    [itex]j_2 = 1[/itex]
    Therefore I find 1 valid set:
    [itex]\{j_2 = 1, r = 1\}[/itex]

    n = 3
    From Eq. 8:
    [itex]j_2 + 2j_3 = 2[/itex]
    Therefore I find 2 valid sets:
    [itex]\{j_2 = 0, j_3 = 1, r = 1\}[/itex]
    [itex]\{j_2 = 2, j_3 = 0, r = 2\}[/itex]

    n = 4
    From Eq. 8:
    [itex]j_2 + 2j_3 + 3j_4 = 3[/itex]
    Therefore I find 3 valid sets:
    [itex]\{j_2 = 3, j_3 = 0, j_4 = 0, r = 3\}[/itex]
    [itex]\{j_2 = 1, j_3 = 1, j_4 = 0, r = 2\}[/itex]
    [itex]\{j_2 = 0, j_3 = 0, j_4 = 1, r = 1\}[/itex]

    and so on...

    I have checked this up to n = 3, but will confirm also for n = 4.
     
  8. Aug 22, 2012 #7
    You might also be interested in the following

    http://arxiv.org/pdf/math/0501052.pdf

    Diego Dominici, Nested Derivatives: A Simple Method for Computing Series expansions of Inverse Functions, arXiv.math/0501052v2, IJMMS, 58, 2003
     
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