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.
Is it possible for the derivative of a function to grow arbitrarily faster than the function itself?
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$\begingroup$ "Arbitrarily large" where? Are you taking a limit as $x\to\infty$, or some finite interval, as in Ross Millikan's answer? $\endgroup$– mr_e_manCommented Jun 13, 2018 at 4:52
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4$\begingroup$ 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?" $\endgroup$– jochenCommented Jun 13, 2018 at 14:37
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$\begingroup$ @jochen which is of course saying that there are functions whose derivative grows arbitrarily fast, by letting $\log(f) \to f$ $\endgroup$– Brevan EllefsenCommented Jun 13, 2018 at 22:29
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1$\begingroup$ 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. $\endgroup$– Kamil MaciorowskiCommented Jun 13, 2018 at 22:33
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1$\begingroup$ 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. $\endgroup$– Nick AlgerCommented Jun 14, 2018 at 4:41
3 Answers
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.
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6$\begingroup$ 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 $y' = y^2$. $\endgroup$ Commented Jun 13, 2018 at 15:14
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4$\begingroup$ @TannerSwett $f(x) = Ae^{\int g(x)}$ for some constant $A$. So, yes, it is defined on all of $\Bbb R$ in that case. $\endgroup$ Commented Jun 13, 2018 at 16:16
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$\begingroup$ Isn't $f$ the function you want, rather than $g$? $\endgroup$– OviCommented Jun 14, 2018 at 3:31
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1$\begingroup$ @Ovi g is the fast-growing function you want (of your choice) for the ratio. $\endgroup$ Commented Jun 14, 2018 at 3:56
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.
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4$\begingroup$ +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. $\endgroup$ Commented Jun 13, 2018 at 22:24
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.
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1$\begingroup$ I think OP is talking about the antiderivative rather than a definite integral. $\endgroup$ Commented Jun 13, 2018 at 22:09