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One reason why the iteration

$$x_{n+1}=x_n-\tan\;x_n$$

converges quickly for appropriate starting values is that this is nothing more than the Newton-Raphson iteration for $\sin\;x$.

This got me thinking: given some arbitrary function $g(x)$, is there always a function $f(x)$ such that

$$\frac{f(x)}{f^\prime(x)}=g(x)$$

or are there restrictions on the nature of $g(x)$ so that the differential equation has a solution?

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2 Answers 2

The differential equation you have is an example of a first-order separable differential equation.

If you write $y = f(x)$, then you have $$ \frac{y}{\frac{dy}{dx}} = g(x) $$ Writing this in terms of differentials (treat $\frac{dy}{dx}$ as a fraction) so that all the $y$'s are on one side and all the $x$'s are on the other gives $$ \frac{dx}{g(x)} = \frac{dy}{y} $$ Integrating both sides yields $$ \int \frac{dx}{g(x)} + C = \ln |y| $$ where $C$ is an arbitrary constant. Then $$ \exp(\int \frac{dx}{g(x)} + C) = |y| $$ and so $$ y = \pm \exp(\int \frac{dx}{g(x)} + C) $$ This works as long as $\frac{1}{g(x)}$ has an antiderivative. For this purpose, it is sufficient to have $g$ continuous and not everywhere zero.

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Certainly not for "arbitrary" functions, as functions of the form $\frac{f}{f'}$ satisfy properties that not all functions satisfy. In fact, in any open set where $g$ is nonzero, $\frac{1}{g}=\log(|f|)'$, so $g$ must be the reciprocal of a derivative of a function. This also leads to a way to find such $f$ in many cases, namely by integrating the last equation to get $\log(|f(x)|)=\int_a^x\frac{dt}{g(t)}+C$ provided the integral exists, in which case $f(x)=c\cdot\exp\left(\int_a^x\frac{dt}{g(t)}\right)$.

For example, attempting this with $g(x)=\tan(x)$ leads to integrating $\cot(x)$ and ending up with $f(x)=c\cdot\sin(x)$.

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