Are polynomials infinitely many times differentiable?

Are polynomials infinitely many times differentiable?

If so, does it only mean that at some point we reach 0 and then we keep on getting 0?

Thank you!

• That is exactly correct. Good thinking! – Jahan Claes Jun 29 '15 at 22:33
• Yes, polynomials are infinitely many times differentiable, and yes, after some finite number of derivatives (specifically $\deg f + 1$) we get $0$, and then we continue to get $0$ thereafter. – mweiss Jun 29 '15 at 22:33
• ... but that's not what it means, it's just what happens in this case when you keep differentiating. – Robert Israel Jun 29 '15 at 22:34
• For example, $\sin x$ is infinitely differentiable, but you never reach and stay at zero. – Mark Fischler Jun 29 '15 at 22:36

Yes.

$$\frac{d}{dx} 0 = 0,$$ so once you've taken $\deg(p) + 1$ derivatives, where $p$ is the polynomial you're considering, you will just keep getting zero.

• If in this logic, then constant function is also infinitely differentiable? – Userkkr Apr 5 '17 at 1:40
• Yes --- the constant functions are just a special case of this, being the degree zero polynomials. – qaphla Apr 10 '17 at 21:03

The other answers probably answered your question, but in case you wanted to learn more, the key word to look up is smoothness of the function (see here for instance: https://en.wikipedia.org/wiki/Smoothness).

So another, more fancy way of asking your question is whether or not polynomials are smooth functions.

• Only one of the two other answers is above. You may want to change your wording to make it more independent of surroundings. – Ruslan Jun 30 '15 at 10:28
• Changed the wording. Thanks! – Physics Enthusiast Jun 30 '15 at 13:32

To be only a little more formal. I would consider the standard definitions of "derivative" and "differentiable".

The definition of the derivative is typically defined for a function on an interval or at a point.

Derivative Let $I \subseteq \Bbb R$ be an interval, let $f:I\to \Bbb R$, and let $c \in I$. $L \in \Bbb R$ is a derivative of $f$ at $c$ if the following conditions hold.

compacted definition using undefined limit, is given by the limit

$L=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

alternatively $\forall \epsilon$, $\exists \delta(\epsilon):$ if $\epsilon \gt 0$ then $\delta (\epsilon)\gt 0$.

If $x \in I$ and $0<|x-c|< \delta(\epsilon)$ then $|\frac{f(x)-f(c)}{x-c}-L|<\epsilon$.

Differentiable simply means that the derivative exists at a point $c$.

So for the case of the polynomials in from $\Bbb R \to \Bbb R$ we will consider this interval $I$ to be equal to the reals itself. What I typically do is consider the two inequalities

$1)$ $|x-c|<\delta(\epsilon)$

$2)$ $|\frac{f(x)-f(c)}{x-c}-L|<\epsilon$

as a system of inequalities, however this is stronger then necessary because solving for the variables simultaneously will imply that $1$ implies $2$ and $2$ also implies $1$ or $1\iff 2$. The definition only requires that $1 \Rightarrow 2$. You can investigate that on your own. First consider $f(x) = x$ then $|\frac{x-c}{x-c}-L|=|1-L|<\epsilon$

$|x-c|<\delta(\epsilon)$

In this case $\delta$ is not a function of $\epsilon$ because if $\delta =|x-c|+1$ $\Rightarrow |x-c|<\delta$. Normally $2$ would constrain the possible x values and a more specific delta would satisfy this constrained so that $1 \Rightarrow 2$ which is what is actually needed. So we see for all $\epsilon$ $2$ is true because $L$ need only be an element of the real. So we can take $L=1$.

axiom 1 $\frac{d}{dx}x|_{x=c}=1$ As you may recall. Next consider $f(x)=a$ for any $a \in \Bbb R$. $1$, $2$ become

$|x-c|<\delta(\epsilon)$

$|\frac{a-a}{x-c}-L|=|L|<\epsilon$

We can take $\delta(\epsilon)=|x-c|+1$ as before and $L=0$.

axiom 2 $\frac{d}{dx}a|_{x=c}=0, \forall a \in \Bbb R$ a constant.

Next consider $f(x)$, $g(x)$ defined similarly as functions from the reals to the reals and $h(x)=f(x)+g(x)$

$|x-c|<\delta(\epsilon)$

$|\frac{h(x)-h(c)}{x-c}|=|\frac{f(x)+g(x)-f(x)-g(x)}{x-c}-L|=|\frac{f(x)-f(c)}{x-c}-L_f+\frac{g(x)-g(c)}{x-c}-L_g|<\epsilon$, $L=L_f+L_g$

By the triangle inequality $|x+y|\le|x|+|y|$ which then implies

$|\frac{f(x)-f(c)}{x-c}-L_f+\frac{g(x)-g(c)}{x-c}-L_g|\le|\frac{f(x)-f(c)}{x-c}-L_f|+|\frac{g(x)-g(c)}{x-c}-L_g|$

But we see this is just the sum of two differentiable functions and if they are differentiable at $c$ then we know $L_f$, $L_g$ exist as well as these statements satisfied for all $\epsilon$ as well there will exist a $\delta$.

$|\frac{f(x)-f(c)}{x-c}-L_f|<\epsilon_f$, $|\frac{g(x)-g(c)}{x-c}-L_g|<\epsilon_g$

and $\epsilon_f+\epsilon_g=\epsilon$ so finally

axiom 3 $\frac{d}{dx}h(x)|_{x=c}=\frac{d}{dx}f(x)|_{x=c}+\frac{d}{dx}g(x)|_{x=c}$ if $h(x)=f(x)+g(x)$ where $f(x)$ and $g(x)$ are two differentiable functions on the interval considered.

Finally the product rule needs to be derived however as that is typically left as an exercise I shall do the same.

product rule if $f(x)$ and $g(x)$ are two differentiable functions on some interval $I$ and $h(x)=f(x) g(x)$then its derivative at all points on $I$ are $\frac{d}{dx}[f(x)]_{x=c}g(x)+f(x)\frac{d}{dx}[g(x)]_{x=c}$

Now we can start to move to the normal tools of calculus, for example if we consider the case were our functions are only as in axiom 1 and 2 and apply the product rule we get $1\cdot x+x\cdot 1=2x$. you can realize the pattern after a few iterations that $\frac{d}{dx}x^n|_{x=c}=nx^{n-1}$. Assume it is true for one $n \in \Bbb N$ then consider

$\frac{d}{dx}x^{n+1}|_{x=c}=x^n\frac{d}{dx}x|_{x=c}+x\frac{d}{dx}x^n|_{x=c}=x^n+x\frac{d}{dx}x^n|_{x=c}$

By our product rule then using our induction hypothesis we have

$x^n+x\cdot nx^{n-1}=x^n+nx^n=(n+1)x^n=(n+1)x^{(n+1)-1}$

which is what we have set out to show, all that is left is to evaluate the base case of $n=0$ which you can do at this point. We have actually shown it to be true for $n=2$ already. Also I will leave it to you to evaluate how the coefficients can be viewed as constant functions and use the product rule again to handle any term in a polynomial.

I will not prove but it is true that any polynomial can be witten in the form $\sum_{i=0}^n a_i x^i$ you can show again by induction that for a given $n$, $m \in \Bbb N$ assume $m \le n$

$\frac{d}{dx}\sum_{i=0}^m a_ix^i|_{x=c}=\sum_{i=0}^m a_i\frac{d}{dx}x^i|_{x=c}=\sum_{i=0}^m ia^ix^{i-1}$

then proceed to show true for m+1.

Finally to show that there are only finitely many derivatives of any polynomial greater than $0$ use contradiction and use the index in the coefficient $i$ with successive derivatives. This shows that there exists an n so that $\frac{d^n}{dx^n}\sum_{i=0}^la_ix^i|_{x=c}=0$ for all $l$. We have already seen that the derivative of a constant is zero.