Even partial derivatives of a ratio

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
 

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.
 

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]
 
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
 
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:
>> 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:
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:

arildno

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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
This is dead wrong. The upper argument in the binomial coefficient IS n, not n-1!
 
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:

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