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We know that there exist some functions $f(x)$ such that their derivative $f'(x)$ is strictly greater than the function itself. for example the function $5^x$ has a derivative $5^x\ln(5)$ which is greater than $5^x$. Exponential functions in general are known to be proportional to their derivatives. The question I have is whether it is possible for a function to grow "even faster" than this. To be more precise let's take the ratio $\frac{f'(x)}{f(x)}$ for exponential functions this ratio is a constant. For most elementary functions we care about, this ratio usually tends to $0$. But are there functions for which this ratio grows arbitrarily large? If so, is there an upper limit for how large the ratio $\frac{f'(x)}{f(x)}$ can grow? I also ask a similar question for integrals.

AllCatsAreGrey
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Ryan Cole
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  • "Arbitrarily large" where? Are you taking a limit as $x\to\infty$, or some finite interval, as in Ross Millikan's answer? – mr_e_man Jun 13 '18 at 04:52
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    By the chain rule, $f'(x)/f(x) = \bigl(\log f(x)\bigr)'$. Thus your question translates to "are there functions such that the derivative of the logarithm grows arbitrarily large?" – jochen Jun 13 '18 at 14:37
  • @jochen which is of course saying that there are functions whose derivative grows arbitrarily fast, by letting $\log(f) \to f$ – Brevan Ellefsen Jun 13 '18 at 22:29
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    Note the ratio you're talking about may not be a good way of measuring if "the derivative of a function grows arbitrarily faster than the function itself". Consider $f(x)\equiv x$. The derivative doesn't grow at all, the function itself grows in a constant rate. Still the ratio can be as large as you want if only you get close enough to $0$ from the positive side – not because the numerator gets large but because the denominator gets tiny. – Kamil Maciorowski Jun 13 '18 at 22:33
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    Take the derivative of a function that grows slower than $e^x$, and you get a function that grows even slower. Take a function that grows faster than $e^x$ and you get a function that grows even faster. Functions that grow like $e^x$ are the fixed point. – Nick Alger Jun 14 '18 at 04:41
  • $e^{x^2}$ is such an incredile simple example... – tired Jun 14 '18 at 06:32
  • 'For most elementary functions we care about, this ratio ($F'(x)/f(x)$) usually tends to 0' doesn't seem true to me. The sine and cosine functions are among those elementary functions which I care about, and the described ratio does not 'usually' tend to zero for them... – CiaPan Jun 14 '18 at 08:37

3 Answers3

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Consider the differential equation $$ \frac{f'}{f} = g $$ where $g$ is the fast-growing function you want. For instance, for $g(x) = e^x$ (and say the initial condition $f(0) =1$) you get $$f(x) = e^{e^x-1} $$ The ratio $f'/f$ grows arbitrarily large.

Clement C.
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    If $g$ is a continuous function defined on all of $\mathbb{R}$, will this differential equation also define a total function on $\mathbb{R}$? This reminds me of other differential equations which fail to define total functions, such as $y' = y^2$. – Sophie Swett Jun 13 '18 at 15:14
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    @TannerSwett $f(x) = Ae^{\int g(x)}$ for some constant $A$. So, yes, it is defined on all of $\Bbb R$ in that case. – Paul Sinclair Jun 13 '18 at 16:16
  • Isn't $f$ the function you want, rather than $g$? – Ovi Jun 14 '18 at 03:31
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    @Ovi g is the fast-growing function you want (of your choice) for the ratio. – Clement C. Jun 14 '18 at 03:56
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You can always try functions of the form $f(x) = e^{g(x)}$, where $g(x)$ is an antiderivative of a function with large growth. For example, if $f(x) = e^{e^x}$, then $$\frac{f'(x)}{f(x)} = e^x,$$ which is, of course, exponential.

Theo Bendit
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    +1, but would be nice and probably instructive if you could also show how to derive $f(x) = e^{g(x)}$ (how to start and separate the variables etc.), since right now it looks like you kind of pulled that rabbit out of a hat. – user541686 Jun 13 '18 at 22:24
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Even $f(x)=\frac 1x$ has derivative $\frac {-1}{x^2}$ and second derivative $\frac 2{x^3}$ which grow arbitrarily faster than $f(x)$ as $x \to 0$. The ratio $\frac {f'(x)}{f(x)}=-\frac 1x$ which is not bounded as $x \to 0$. The important message is that derivatives accentuate short range changes, so if you have a function that changes quickly in a short distance the derivative is large.

I don't understand what "a similar question for integrals" means. Integrals are smoothing functions, so the integral of a function can't grow faster than the function times the length of the integral.

Joonas Ilmavirta
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Ross Millikan
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