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

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #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 (??)
 
  • #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.
 
  • #4
123
0
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 :)
 
  • #5
As you can see, it's a very nasty formula
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 general, computing derivatives is easy - you will rarely ever need any fancy techniques in order to calculate some derivative.
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.
 
  • #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.
 
  • #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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