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How can one find if a function $f$ is infinitely differentiable?

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up vote 12 down vote accepted

By differentiating it an infinite number of times?

But seriously, that's what you do. Just that usually you can infer what the higher order derivatives will be, so you don't have to compute it one by one.

Example To see that $\sin(x)$ is infinitely differentiable, you realize the following: $\frac{d}{dx}\sin(x) = \cos(x)$, and $\frac{d}{dx} \cos(x) = -\sin(x)$. So you see that $(\frac{d}{dx})^4\sin(x) = \sin(x)$, and so the derivatives are periodic. Therefore by continuity of $\sin(x)$ and its first three derivatives, $\sin(x)$ must be infinitely differentiable.

Example To see that $(1 + x^2)^{-1}$ is infinitely differentiable, you realize that $\frac{d}{dx}(1+x^2)^{-n} = -2n x (1+x^2)^{-n-1}$. So therefore by induction you have the following statement: all derivatives of $(1+x^2)^{-1}$ can be written as a polynomial in $x$ multplied by $(1 +x^2)^{-1}$ to some power. Then you can use the fact that (a) polynomial functions are continuous and (b) quotients of polynomial functions are continuous away from where the denominator vanishes to conclude that all derivatives are continuous.

The general philosophy at work is that in order to show all derivatives are bounded and continuous, you can take advantage of some sort of recursive relationship between the various derivatives to inductively give a general form of the derivatives. Then you reduce the problem to showing that all functions of that general form are continuous.

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I should add that most everyday "special functions" are, in fact, solutions to ordinary differential equations. Therefore, for those special functions (such as $\ln$, $\sin$, $\exp$ etc.) the recursive relationship between derivatives is almost given to you by definition. –  Willie Wong Dec 10 '10 at 15:16
    
One last remark: using the fundamental properties of the derivative, finite sums, finite differences, finite products of infinitely differentiable functions are all going to be smooth. The quotient of two smooth functions will be smooth away from when the denominator vanishes. So you can also break an expression down to component parts and check them one by one. –  Willie Wong Dec 10 '10 at 15:18
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There is no magic recipe to tell whether a function is infinitely differentiable, or even once differentiable, or even continuous indeed. It really depends on the function itself. Of course there are really general and trivial statements like all polynomials are infinitely differentiable and so on.

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Depending on how your function is given, that can be very easy or very hard. For functions given by expressions in closed form, you can try deriving a general formula for the $n$-th derivative. E.g. this way you can prove that all polynomials are smooth (=infinitely differentiable), so are sin, cos and exponential functions and any concatenation, product and sum of smooth functions is smooth.

Often, you will have to work with the actual derivative as a limit and differentiate until you find either a non-smooth point or some closed form that you recognise, you maybe some periodicity (i.e. so that some $n$-derivative is equal to some $m$-the derivative, like with trigonometric functions) or you have to get really clever.

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As Willie Wong points out, that's the only way to do it. By the way $\sin{x},\cos{x}$ are examples of infinitely differentiable functions, and most famous example is the function $f(x)=e^{x}$. This function is infinitely differentiable and it's derivative, is $e^{x}$. To see this:

\begin{align*} e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots \\ \frac{\textrm{d}} {\textrm{dx}}e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots \end{align*}

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