Even partial derivatives of a ratio

In summary: B^{(3)} y^{(0)} 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)} A^{(5)} = \binom{5}{0} B^{(0)}
  • #1
DivGradCurl
372
0
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
 
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  • #2
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.
 
  • #3
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]
 
  • #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
 
  • #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:
>> 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:
  • #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:
  • #7
thiago_j said:
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!
 
  • #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:

Related to Even partial derivatives of a ratio

What is a partial derivative?

A partial derivative is a mathematical concept that measures the rate of change of a multivariable function with respect to one of its variables, while holding the other variables constant.

What is a ratio?

A ratio is a comparison of two quantities, typically expressed in the form of a fraction. It represents the relationship between the two quantities and can be used to make predictions or solve problems.

How do you calculate the partial derivative of a ratio?

To calculate the partial derivative of a ratio, you first need to express the ratio as a function with two variables. Then, you can use the quotient rule to find the partial derivatives of each variable, while treating the other variable as a constant.

What is the significance of taking partial derivatives of a ratio?

Taking partial derivatives of a ratio allows us to analyze the impact of one variable on the ratio, while holding the other variables constant. This can be useful in many fields, such as economics, physics, and engineering.

Can you give an example of finding the partial derivative of a ratio?

Sure, let's say we have the ratio y=x^2/z, where x and z are variables. To find the partial derivative of y with respect to x, we would use the quotient rule and get (2x*z-x^2)/z^2. This tells us how the ratio changes as x changes, while z remains constant.

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