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Even partial derivatives of a ratio

  1. Feb 12, 2009 #1
    Is there a general formula for the even partial derivatives of a ratio,
    where both A and B are functions of f?

    [tex]\frac{\partial ^{(2n)}}{\partial f^{(2n)}} \left( \frac{A}{B} \right)[/tex]

    Thanks
     
  2. jcsd
  3. Feb 13, 2009 #2

    CompuChip

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    I think not... if we let A' and B' denote the partial derivatives of A and B w.r.t. f, respectively, then for n = 1 you would get
    [tex]\frac{\partial^2}{\partial f^2} \frac{A}{B} = \frac{\partial}{\partial f} \left( \frac{A' B - A B'}{B^2} \right) = \frac{ (A'' B - A B'') B^2 - 2 B B' (A' B - A B') }{ B^4 }[/tex]
    which already looks really messy (and that's just n = 1).

    But you could try writing out some more terms and see if there is a pattern.
     
  4. Feb 13, 2009 #3

    arildno

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    It is fairly trivial to find an IMPLICIT equation for the n'th derivative of a fraction:

    Set [tex]y(x)=\frac{A(x)}{B(x)}\to{A}(x)=B(x)y(x)[/tex]
    Thus, we have for the n'th derivative:
    [tex]A^{(n)}(x)=\sum_{i=0}^{i=n}\binom{n}{i}B^{(n-i)}(x)y^{(i)}(x)[/tex]
    This is then readily solved for the n'th derivative for y:
    [tex]y^{(n)}(x)=\frac{1}{B(x)}(A^{(n)}(x)-\sum_{i=0}^{i=n-1}\binom{n}{i}B^{(n-i))(x}y^{(i)}(x))[/tex]
     
  5. Feb 13, 2009 #4
    Arildno, I have to disagree: that's not trivial, it's awesome! Thanks for your response. A couple of thoughts:

    1) How did you get from generic differentiation of a product to that binomial series? Simple pattern recognition? If not, could you suggest a specific reference so that I can follow your reasoning?

    2) There seems to be a tiny glitch in your final answer. Here's what I think it should be:

    [tex] y(x) = \frac{A(x)}{B(x)} \qquad (1)[/tex]

    [tex] A(x) = B(x) \cdot y(x) \qquad (2)[/tex]

    [tex] A^{(n)} (x) = \sum _{i = 0} ^{n} \left[ \binom{n}{i} B^{(n-i)} (x) \cdot y^{(i)}(x) \right] \qquad (3)[/tex]

    [tex] A^{(n)} (x) = \sum _{i = 0} ^{n-1} \left[ \binom{n-1}{i} B^{(n-1-i)} (x) \cdot y^{(i)}(x) \right] + B(x) \cdot y^{(n)}(x) \qquad (4)[/tex]

    [tex] y^{(n)} (x) = \frac{1}{B(x)} \left\{ A^{(n)} (x) - \sum _{i = 0} ^{n-1} \left[ \binom{n-1}{i} B^{(n-1-i)} (x) \cdot y^{(i)}(x) \right] \right\} \qquad (5)[/tex]

    Thanks again
     
  6. Feb 13, 2009 #5
    One more point. Unless I made a mistake, the expression for the n-th derivative of y

    [tex]
    y^{(n)} (x) = \frac{1}{B(x)} \left\{ A^{(n)} (x) - \sum _{i = 0} ^{n-1} \left[ \binom{n-1}{i} B^{(n-1-i)} (x) \cdot y^{(i)}(x) \right] \right\} \qquad (5)
    [/tex]

    does not seem to work. I tested it on MATLAB for this function

    [tex]
    y(x) = \frac{(2+4i)^x - (3-i)^x}{(6-2i)^x - (5+7i)^x}
    [/tex]

    where [tex]i = \sqrt{-1}[/tex], A is the numerator, and B is the denominator. According to the formula above, the first derivative is

    [tex]
    y^{\prime}(x) = \frac{1}{B(x)}\left( A^{\prime}(x) - B(x) \cdot y(x) \right)
    [/tex]

    Here's what I found:

    Code (Text):
    >> syms x
    >> syms y A B
    >> y = ((2+4*i)^x - (3-i)^x)/((6-2*i)^x - (5+7*i)^x);
    >> pretty(y)
     
                                          x          x
                                 (2 + 4 I)  - (3 - I)
                                -----------------------
                                         x            x
                                (6 - 2 I)  - (5 + 7 I)

    >> A = ((2+4*i)^x - (3-i)^x)
     
    A =
     
    (2+4*i)^x-(3-i)^x

    >> B = (6-2*i)^x - (5+7*i)^x
     
    B =
     
    (6-2*i)^x-(5+7*i)^x

    >> (1/B)*(diff(A,x,1) - B*y) == diff(y,x,1)

    ans =

         0
     
    Last edited: Feb 13, 2009
  7. Feb 14, 2009 #6
    I finally found the correct answer. Thank you for the hint, arildno!

    [tex]
    y= \frac{A}{B} \Longleftrightarrow A^{(0)} = B^{(0)} y^{(0)}
    [/tex]

    [tex]
    A^{(1)} = B^{(0)} y^{(1)} + B^{(1)} y^{(0)}
    [/tex]

    [tex]
    A^{(2)} = B^{(0)} y^{(2)} + 2 B^{(1)} y^{(1)} + B^{(2)} y^{(0)}
    [/tex]

    [tex]
    A^{(3)} = B^{(0)} y^{(3)} + 3 B^{(1)} y^{(2)} +3 B^{(2)} y^{(1)} + B^{(3)} y^{(0)}
    [/tex]

    [tex]
    A^{(4)} = B^{(0)} y^{(4)} + 4 B^{(1)} y^{(3)} + 6 B^{(2)} y^{(2)} + 4 B^{(3)} y^{(1)} + B^{(4)} y^{(0)}
    [/tex]

    [tex]
    A^{(5)} = B^{(0)} y^{(5)} + 5 B^{(1)} y^{(4)} + 10 B^{(2)} y^{(3)} + 10 B^{(3)} y^{(2)} + 5 B^{(4)} y^{(1)} + B^{(5)} y^{(0)}
    [/tex]

    These derivatives can be rephrased as follows.

    [tex]
    A^{(0)} = \binom{0}{0} B^{(0)} y^{(0)}
    [/tex]

    [tex]
    A^{(1)} = \binom{1}{0} B^{(0)} y^{(1)} + \binom{1}{1} B^{(1)} y^{(0)}
    [/tex]

    [tex]
    A^{(2)} = \binom{2}{0} B^{(0)} y^{(2)} + \binom{2}{1} B^{(1)} y^{(1)} + \binom{2}{2} B^{(2)} y^{(0)}
    [/tex]

    [tex]
    A^{(3)} = \binom{3}{0} B^{(0)} y^{(3)} + \binom{3}{1} B^{(1)} y^{(2)} +\binom{3}{2} B^{(2)} y^{(1)} + \binom{3}{3} B^{(3)} y^{(0)}
    [/tex]

    [tex]
    A^{(4)} = \binom{4}{0} B^{(0)} y^{(4)} + \binom{4}{1} B^{(1)} y^{(3)} + \binom{4}{2} B^{(2)} y^{(2)} + \binom{4}{3} B^{(3)} y^{(1)} + \binom{4}{4} B^{(4)} y^{(0)}
    [/tex]

    [tex]
    A^{(5)} = \binom{5}{0} B^{(0)} y^{(5)} + \binom{5}{1} B^{(1)} y^{(4)} + \binom{5}{2} B^{(2)} y^{(3)} + \binom{5}{3} B^{(3)} y^{(2)} + \binom{5}{4} B^{(4)} y^{(1)} + \binom{5}{5} B^{(5)} y^{(0)}
    [/tex]

    Therefore,

    [tex]
    A^{(n)} = \sum _{i=0} ^{n} \binom{n}{i} B^{(i)} y^{(n-i)} = B^{(0)} y^{(n)} + \sum _{i=1} ^{n} \binom{n}{i} B^{(i)} y^{(n-i)}
    [/tex]

    which means

    [tex]
    y^{(n)} = \frac{1}{B ^{(0)} } \left[ A ^{(n)} - \sum _{i=1} ^{n} \binom{n}{i} B^{(i)} y^{(n-i)} \right] \qquad \mbox{for } n \geq 0
    [/tex]
     
    Last edited: Feb 14, 2009
  8. Feb 14, 2009 #7

    arildno

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    This is dead wrong. The upper argument in the binomial coefficient IS n, not n-1!
     
  9. Feb 18, 2009 #8
    Derivatives of a ratio of products

    I thought I was done with this thread, but I decided to expand it to a more general case:

    [tex]R = \frac{A\cdot C}{B\cdot D}[/tex]

    where R, A, B, C, and D are functions of x. I've found an expression for the n-th derivative, but I'm not sure it correct. If anyone finds a mistake, please let me know. Thanks.

    Here it is:

    [tex] R^{\displaystyle (n)} = \frac{1}{B^{\displaystyle (0)} \cdot C^{\displaystyle (0)}} \left\{ \sum _{\displaystyle i = 0} ^{\displaystyle n} \left[ \binom{n}{i} A^{\displaystyle (i)} C^{\displaystyle (n-i)} \right] + \mathop{\sum _{\displaystyle j=0} ^{\displaystyle j=n-1}} _{\displaystyle k = 0} ^{\displaystyle k = n} \left[ \binom{n}{j,k,n-j-k} R^{\displaystyle (j)}} B ^{\displaystyle (k)}} D^{\displaystyle (n-j-k)}} \right] \right\} [/tex]

    [tex]\mbox{for } n-(j+k) \geq 0[/tex]

    [tex]\mbox{where } \binom{n}{i} = \frac{n!}{i!(n-i)!} \mbox{ and } \binom{n}{j,k,n-j-k} = \frac{n!}{j! k! (n-j-k)!} [/tex]

    And this is how I got it:

    [tex] A^{(0)} C^{(0)} = R^{(0)} B^{(0)} D^{(0)}[/tex]

    [tex] A^{(1)} C^{(0)} + A^{(0)} C^{(1)} = R^{(1)} B^{(0)} D^{(0)} + R^{(0)} B^{(1)} D^{(0)} + R^{(0)} B^{(0)} D^{(1)}[/tex]

    [tex] A^{(2)} C^{(0)} + 2 A^{(1)} C^{(1)} + A^{(0)} C^{(2)} = R^{(2)} B^{(0)} D^{(0)} + 2\left( R^{(1)} B^{(1)} D^{(0)} + R^{(1)} B^{(0)} D^{(1)} + R^{(0)} B^{(1)} D^{(1)}\right) + \ldots[/tex]

    [tex]\ldots R^{(0)} B^{(2)} D^{(0)} + R^{(0)} B^{(0)} D^{(2)}[/tex]

    [tex] A^{(3)} C^{(0)} + 3\left( A^{(2)} C^{(1)} + A^{(1)} C^{(2)} \right) + A^{(0)} C^{(3)} = R^{(3)} B^{(0)} D^{(0)} + 3 \Big( R^{(2)} B^{(1)} D^{(0)} + R^{(2)} B^{(0)} D^{(1)} + \ldots [/tex]

    [tex]\ldots R^{(1)} B^{(2)} D^{(0)} + R^{(1)} B^{(0)} D^{(2)} + R^{(0)} B^{(2)} D^{(1)} + R^{(0)} B^{(1)} D^{(2)} \Big) + 6 R^{(1)} B^{(1)} D^{(1)} + R^{(0)} B^{(3)} D^{(0)} + R^{(0)} B^{(0)} D^{(3)}[/tex]

    and so forth.

    Any help is highly appreciated.
     
    Last edited: Feb 19, 2009
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