# Infinite number of Derivatives

Is there a kind of function (other than trigonometric) that you can take infinite amount of derivatives without it ever becoming 0. Algebraic functions now matter how long, or how many powers it has it can eventually be derived to 0. I am not including trigonometric functions because they are circular in nature. I mean a function that will not go on a circle like trigonometric do; But will have infinite derivatives.

• Exponentials, though these are closely related to trig functions. Another example is any hypergeometric function $_pF_q$ with $q \ge p$. – Cameron Williams Mar 29 '15 at 1:02
• The $n$-th derivative of a function $f$ is identically $0$ for some $n$ if and only if $f$ is a polynomial function. – André Nicolas Mar 29 '15 at 1:05
• @CameronWilliams Surely not any hypergeometric function... ${}_pF_q(-n,a_1,\dotsc;b_1,\dotsc;z)$ is a polynomial, after all! – Chappers Mar 29 '15 at 1:06
• @Chappers Oh true! I didn't think about that. – Cameron Williams Mar 29 '15 at 1:11

Certainly, $f(x) = Ce^{\pm x} \neq 0$ for a fixed non-zero $C$.

$\log{x}$ and its various derivatives and antiderivatives.

Really, polynomials are the exception here: it is easy enough to prove by induction that any smooth function $f$ for which $d^nf/dx^n=0$ is a polynomial of degree less than $n$.

• And it's a bit less easy to prove that any function which is infinitely differentiable and has some derivative equal to zero at each point is still a polynomial. See the MathOverflow question – Mark S. Mar 29 '15 at 15:37
• @MarkS. a lot of the posters here add the words 'smooth' or 'infinitely differentiable', but it unnecessarily complicates it -- if the $n$'th derivative of $f$ for some $n$ is $0$, then it trivially implies that $f$ is smooth and there is no need to add this condition. If the $n$'th derivative of $f$ is $0$ for some $n$, then $f$ is a polynomial function. – user26486 Mar 30 '15 at 12:53

The only smooth functions for which a derivative of some order is identically zero are polynomials.

This is essentially just computing $\int^{(n)}0\;dx$ for arbitrary $n$ (don't forget the additive constant at each stage).

• The condition 'smooth' is redundant, since $(0)^{(n)}=0$ for any $n$. – user26486 Mar 30 '15 at 12:57
• @user31415: The function is smooth a priori as OP speaks of taking derivatives of arbitrary order everywhere. I should point out that I have implicitly assumed that the domain is all of $\mathbb R$ (or at least connected) since the result does not follow otherwise: consider the function $\mathbb R\setminus\{0\}\to\mathbb R$ with $x\mapsto |x|$; its second derivative vanishes identically on its domain, but it isn't a polynomial (in fact, it is smooth as well). – MPW Mar 31 '15 at 7:19

In addition to what everyone else is saying: rational functions that aren't polynomials, such as $\dfrac1x$, $\dfrac1{1-x}$, $\dfrac x{x^2+2x+2}$, etc.

Also things like $x^x$, which is halfway between a power (e.g. $x^n$) and an exponential (e.g. $n^x$). It grows faster than exponentials, by the way. You might not have learned how to differentiate this, yet.

Honestly, anything other than polynomials.

• How is $x^x$ halfway between $n^x$ and $x^n$. It was my understanding that it is asymptotically greater than both... Am I missing something here? – k_g Mar 29 '15 at 6:08
• @k_g as I understand columbus, is conceptually somewhere in between, as it has both exponentiation and a power. – Davidmh Mar 29 '15 at 11:36
• @k_g: Note that expressions like $x^x$ are easy to differentiate if you know the generalized power rule $$\left(f^g\right)' =gf^{g-1}\cdot f'+ f^g\log f\cdot g'$$ So it really is "halfway in between" powers and exponentials. This generalizes both. If either $f$ or $g$ is constant, you recover the corresponding classical rule because the corresponding derivative vanishes. – MPW Mar 29 '15 at 12:55
• One way to differentiate $x^x$ is to rewrite it as $e^{x\ln x}$ and use the Chain and Product rules. – Akiva Weinberger Mar 29 '15 at 13:20
• To differentiate $x^x$ you can just use the two rules $(x^a)'=ax^{a-1}$ and $(a^x)'=a^x\log a$, they dont quite work on their own, but if you just sum them together it works fine. – Alice Ryhl Mar 30 '15 at 7:32

Further to other answers here, only polynomials of positive integer order eventually run to a zero derivative. The repeated derivative of any polynomial $x^{y}$ where $y$ is non-integer or where $y<0$ will never become zero. (The logarithm is a special starting point for the latter case.)

Of course there are, a simple example is $f(x)=e^x$. Since $\frac{d}{dx}e^x=e^x$, the $n$th derivative of $f(x)$ will also be $f(x)$, which will never be 0.

• I should have included that one, I figured this one out too. – Lucian09474 Mar 29 '15 at 1:45
• On the other hand, that particular example is technically circular in the same manner as the trig functions, just with a repetition interval of one. – hBy2Py Mar 30 '15 at 14:29

Let's suppose a function $f$ is given by a power series which converges within a open set $D$ (for simplicity, let's assume $0 \in D$). Then, we can write

$$f(x) = \sum_{n=0}^\infty a_n x^n$$

provided that infinitely many $a_n$'s are nonzero, $f^{(k)}(0) = k!a_k$ will be nonzero for infinitely many $k$. For example, $f(x)=e^x$ can be written as $$e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n$$ and the derivatives of $e^x$ never vanish.

Many special functions, such as $\text{erf} (x),\Gamma(x), J_0 (x)$ etc have derivatives such that it will never equal 0 after $n$ derivatives.