Functions that are always less than their derivatives

Question

I was wondering if there are functions for which $$f'(x) > f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c > 0$. Also, is there any significance in a function that is always less than its derivative?

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Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature... Are there more examples like - 1/x?

Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration is always greater than its velocity]

Answer 1

If $y'(x)>y(x)\quad\forall x\in\mathbb{R}$, we can define $f(x)=y'(x)-y(x)$ which is positive forall $x$. Suppose that $y'(x)$ is continuous function so that $f(x)$ is continuous too. Now with this element we can build the differential equation $$y'(x)=y(x)+f(x)$$ and its solutions are given by: $$y(x)=e^{x}\left(c+\int_{x_0}^{x}e^{-s}f(s)ds\right)$$

Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration is always greater than its velocity]

I don't know if there's application of this interesting property, but I'm sure that you can't compare velocity with the position because they are not homogeneous quantities.

Answer 2

Assuming $f(x)>0$, $f:\mathbb{R}\mapsto\mathbb{R}$

$f'(x) > f(x) \iff \frac{d}{dx}\ln(f(x))>1$

So you can turn any function $g$ where $g'(x)>1$ into this type of function by taking the exponential of it:

$\frac{d}{dx}g(x)>1 \implies \frac{d}{dx}\ln(e^{g(x)})>1 \implies \frac{d}{dx} e^{g(x)}>e^{g(x)}$

Answer 3

A simple example is $f(x)=-x^2-3$

Answer 4

A more interesting problem is to find a function $f:\mathbb{R}\rightarrow\mathbb{R}$, whose image is $\mathbb{R}$ and satisfies $f'(x)>f(x)$ for all $x\in\mathbb{R}$. One of those functions is

$$\sinh(x),$$

because

$$\frac{d}{dx}\sinh(x)=\cosh(x)>\sinh(x)$$ for all $x\in \mathbb{R}$.

Answer 5

$(e^{-x}f(x))'=e^{-x} (f'(x)-f(x)) >0$ so $e^{-x}f(x)$ is an increasing function. Since it is $0$ at $0$ we get $e^{-x}f(x)>0$ for all $x>0$. Hence $f(x)>0$ for all $x>0$.

Answer 6

Take $f(x)=e^{\alpha x}$. Then for $\alpha >1$ we have $f'(x)>f(x)$ and for $\alpha <1$ we have $f'(x)<f(x)$.

Answer 7

How about if you look at it as a differential equation. Say

$y' = y + 1$

which has solution $y=Ce^x -1$

Or $y'=y+x^2+1$

which has solution $y=Ce^x - (x^2+2x+3)$

Or $y'=y+2\sin x+3$

which has solution $y = Ce^x - \sin x - \cos x -3$

Answer 8

The inequality $$f'(x) > f(x)$$ is equivalent to $$\left[ f(x) e^{-x} \right]' > 0.$$

So the general solution is to take any differentiable function $g(x)$ with $g'(x) > 0$ and put $f(x) = g(x) e^x$.

Note that nothing is assumed about $f$ except differentiability, which is necessary to ask the question in the first place.

Answer 9

A very simple example is $f(x) = -1 < 0 = f'(x)$. Relevant to your edit: this isn't exponential at all.

Other examples that aren't immediately exponential:

• $\frac{-\pi}{2} + \arctan x$ is everywhere negative and everywhere strictly monotonically increasing, so is everywhere less than its derivative.
• $-1 + \mathrm{erf}(x)$ is also everywhere negative and everywhere strictly monotonically increasing. (These are very similar, since they are shifted copies of the CDFs of the (standard/normalized) Cauchy and Gaussian distributions.)
• $\frac{1}{2}\left( x - \sqrt{x^2 + 4} \right)$ is the lower branch of a hyperbola having the $x$-axis and the line $y = x$ as asymptotes. It is everywhere negative and everywhere strictly monotonically increasing.

Answer 10

Another simple example would be $f(x) = -e^{-x}$, $f'(x) = e^{-x}$

Answer 11

See, $-\frac{1}{x}, \frac{1}{x^{2}} \ in \ [0, \infty]$

Answer 12

For any differential function $f$ for which both $f(x)$ and $f'(x)$ are limited to finite ranges, $f'(x) - f(x)$ is also limited to a finite range, so there is a $c$ for which $f'(x) - f(x) > -c\ \forall\ x$. Therefore, a function $g(x) = f(x) - c$ can be formed for which $g'(x) - g(x) - c> -c\ \forall\ x$ or $g'(x) > g(x)\ \forall\ x$.

For example, this holds for many differential periodic functions.

Answer 13

Mike an answer to your additional question "Are there physical examples of this?" is enabled by dromastyx.

His example shows hyperbolic functions which describe accurately the physical phenomenon of 'solitons'.

Solitons are solitary waves such as sun flares, Tsunamis etc. An example of finding such waves hidden in known equations is :

http://rsos.royalsocietypublishing.org/content/2/7/140406.review-history

Answer 14

Here's a proof using the mean value theorem.

Let $m = \operatorname{inf} \{ x > 0 \colon f(x)\leq 0\}$. This infimum exists, since the reals are complete and the set is bounded below by $0$. Since $f'(0)>0$, there exists some $\varepsilon >0$ such that $f(x) > f(0)$ for $x\in (0,\varepsilon) $. This means that if there is some $x>0$ with $f(x) \leq 0$, then $\varepsilon$ is also a lower bound. (This is common lemma In the context of the mean value theorem).

Thus it suffices to show that $m=0$. Suppose $m>0$. By the mean value theorem, there is some point $y\in (0,m) $ with $f'(y) \leq 0$. By our assumptions, we then have $f(y) < f'(y) \leq 0$. Thus $y$ contradicts the minimality of $m$.