Wednesday, November 9, 2011

Finding Multiple Derivatives


      When finding derivatives, it becomes tedious to continually derive the function to find further derivatives.  Although I'm not sure of the application of derivatives past the second, I came up with a formula for multiple derivations of polynomials.  To do this I observed the way each part of a term changes over each differentiation.

      Each time a new derivative is found, the coefficient is multiplied by the power.  Thus:
            C1 = C0.n(n - 1)(n - 2) ... (n - r) for all n > r

                 = C
0.nPr

      where C0 is the original coefficient, n gives the original power of x, and r gives the derivative to be found (e.g. For f'''(x), r = 3)

      Each time a derivative is found, the constant term is lost.  Thus:
           C1 = 0, for all 0 < n < r



      For negative indices, the sign of the coefficient depends on whether r is odd or even.  The magnitude of the coefficient (absolute value) is multiplied by a larger number each time a new derivative is found.  Therefore, beginning multiplication at the original power will give the new coefficient as follows:



           C
1 = C0.(-1)r.|n - r + 1|! / |n + 1|!, for all n < 0



      As the powers are simply lessened, in magnitude, by 1 for each derivative, and those for which n is positive and less than r should be eliminated, they are given by:
           n
1 = n0 - r, for all n0  > r
           n1 = 0, for all 0 < n0 < r     (as x0 = 1, and the coefficient is 0, thus term is 0 x 1 = 0)
           n1 = n0 + r, for all n0 < 0

      Collating these results gives a way of finding a term in the derivative of a polynomial function:

      Where T
k = C.xn for a polynomial function f(x).  tk in the rth derivative of f(x) is given by:
             tk = C.nPrxn - r, for all n > r

             t
k = 0, for all 0 < n < r
             t
k = C.(-1)r.|n - r + 1|! / |n + 1|!.xn + r, for all n < 0

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