My Math Forum Function whose derivative is inversely proportional to the function itself

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 December 7th, 2018, 12:38 PM #1 Newbie   Joined: Dec 2018 From: Canada Posts: 1 Thanks: 0 Function whose derivative is inversely proportional to the function itself Is there such a function that its derivative is one divided by the function itself? This would be analogous to how the derivative of an exponantial is directly proportional to the exponential function itself, but in this case I’m looking for it to be inversely proportional. Does such a function exist at all, and if so what is it?
 December 7th, 2018, 12:59 PM #2 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 670 $y=k\sqrt{x}$. Thanks from topsquark
 December 7th, 2018, 01:01 PM #3 Senior Member   Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 Consider equation $\displaystyle y’= \frac{C}{y}$ for C-constant Thanks from topsquark Last edited by idontknow; December 7th, 2018 at 01:04 PM.
 December 8th, 2018, 01:10 PM #4 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 670 $y'y=k$ leads to $\frac{y^2}{2}=kx$.
December 8th, 2018, 01:31 PM   #5
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Quote:
 Originally Posted by mathman $y'y=k$ leads to $\frac{y^2}{2}=kx$.
That is not the most general solution.

 December 8th, 2018, 03:13 PM #6 Senior Member   Joined: Aug 2012 Posts: 2,200 Thanks: 645 I had an idea that you could proceed analogously to the way you solve $f' = f$. Assume $f$ is complex analytic, apply the theorem on term-by-term differentiation of a power series, and set the coefficients of $f$ and $f'$ equal. The result gives the well-known power series for $e^x$ [with initial condition $f(0) = 1$]. Of course you can't just take a term-by-term multiplicative inverse, so that idea doesn't work directly. However it turns out there's a formula for the multiplicative inverse of a power series, but it's a little involved. I found two nice writeups: https://math.stackexchange.com/quest...a-power-series and http://www.math.caltech.edu/~2016-17...r%20series.pdf. It's a special case of Faà di Bruno's formula and it involves Bell numbers. Today I learned, right? There's an awesome amount of "elementary" math that nobody knows, let alone the higher abstract stuff. If someone is so inclined, they could set the coefficients equal and work out the general form of a complex analytic function that satisfies $f' = \frac{1}{f}$. When I was looking this up I found a lot of literature for the same question about the functional inverse rather than the multiplicative inverse. There's a formula for that too. Thanks from topsquark and SDK Last edited by Maschke; December 8th, 2018 at 03:21 PM.
December 8th, 2018, 05:55 PM   #7
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Quote:
Originally Posted by Micrm@ss
Quote:
 Originally Posted by mathman $y'y=k$ leads to $\frac{y^2}{2}=kx$.
That is not the most general solution.
$$y^2 = c_1x + c_2 \quad (c_1x + c_2 > 0)$$

December 9th, 2018, 02:39 AM   #8
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Quote:
 Originally Posted by Maschke I had an idea that you could proceed analogously to the way you solve $f' = f$. Assume $f$ is complex analytic, apply the theorem on term-by-term differentiation of a power series, and set the coefficients of $f$ and $f'$ equal. The result gives the well-known power series for $e^x$ [with initial condition $f(0) = 1$]. Of course you can't just take a term-by-term multiplicative inverse, so that idea doesn't work directly. However it turns out there's a formula for the multiplicative inverse of a power series, but it's a little involved. I found two nice writeups: https://math.stackexchange.com/quest...a-power-series and http://www.math.caltech.edu/~2016-17...r%20series.pdf. It's a special case of Faà di Bruno's formula and it involves Bell numbers. Today I learned, right? There's an awesome amount of "elementary" math that nobody knows, let alone the higher abstract stuff. If someone is so inclined, they could set the coefficients equal and work out the general form of a complex analytic function that satisfies $f' = \frac{1}{f}$. When I was looking this up I found a lot of literature for the same question about the functional inverse rather than the multiplicative inverse. There's a formula for that too.
I wouldn't say nobody knows this. Its actually extremely common to do this when solving differential equations. The technique (often called automatic differentiation) is even simpler than the formulas you found for the inverse due to the fact that they did not use the derivative information at all.

Starting from x' = 1/x, where x is assumed to be analytic, define a new variable y = 1/x. Take its derivative by the chain rule so you get
$y' = \frac{-x'}{x^2} = -y^3$
Thus, you are left to solve the 2-dimensional polynomial differential equation: $x' = f(x)$ where $x \in \mathbb{R}^2$ and $f(x) = (x_2, - x_2^3)$.

This specific example shows up for instance, when solving the restricted n-body problem. It works even more generally than this for sin/cos, and therefore periodic functions which can be expressed in Fourier series.

In general, it is used to convert an existing $N$-th dimensional vector field to a polynomial vector field in higher dimensions and it works whenever the non-polynomial part of the vector field satisfies an algebraic differential equation.

Last edited by SDK; December 9th, 2018 at 02:42 AM.

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