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Suppose I have a generic polynomial:

$$f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0$$

If I continually differentiate $f(x)$, when will it end up as $f^{(z)}(x)=0$?

For example, given $f(x)=16x^3-x^2-3x+10$:

$$f'(x)=48x^2-2x-3\\f''(x)=96x-2\\f'''(x)=96\\f''''(x)=\color{red}{0}$$

It ends up as $f^{(4)}=0$.


I know that for the sine function this is not the case:

$$f(x)=\sin(x)\\f'(x)=\cos(x)\\f''(x)=-\sin(x)\\f'''(x)=-\cos(x)\\f''''(x)=\sin(x)\\\text{loop!}$$

Edit: This isn't a polynomial. $x$ is not in the base, among other things.

And if I define

$$f(x)=\sin(x)=\frac{e^{-ix}}{2}+\frac{e^{ix}}{2}$$

Edit: Then we have sine still not as a polynomial. When differentiating it, we get:

$$f'(x)=\frac{1}{2}ie^{ix}\left(-1+e^{2ix}\right)\\f''(x)=\frac{1}{2}e^{-ix}\left(1+e^{2ix}\right)\\f'''(x)=-\frac{1}{2}ie^{-ix}\left(-1+e^{2ix}\right)\\f''''(x)=\frac{e^{-ix}}{2}+\frac{e^{ix}}{2}\\\text{loop 2.0}!$$

What is "special" about the sine function that it does not eventually differentiate to $0$?

As I investigated this question, I started with $i$ as a possible culprit:

$$f(x)=10ix\\f'(x)=10i\\f''(x)=0\\\text{nope}$$


$$f(x)=10x^i\\f'(x)=10ix^{-1+i}\\f''(x)=(-10-10i)x^{-2+i}\\f^{(13)}(x)=(2716272000 - 8395946000i)x^{-13+i}\\\text{aha!}$$

I found that using $i$ in the exponent led to a polynomial that did not eventually differentiate to $0$.


When do polynomials eventually differentiate to $f^{(z)}(x)=0$? Is using $i$ in the exponent the only case where it does not?

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    $\begingroup$ using $i$ as an exponent is weird, have you learned about complex powers? Because I think it answers your question. $\endgroup$
    – mdave16
    Commented Aug 30, 2017 at 1:13
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    $\begingroup$ $e^{ix}$ is not a polynomial in $x$. $\endgroup$
    – MathMajor
    Commented Aug 30, 2017 at 1:13
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    $\begingroup$ $10x^i$ is not a polynomial. A polynomial of degree at most $n$ differentiates to zero after $n + 1$ differentiations, and this characterizes polynomials of degree at most $n$. $\endgroup$ Commented Aug 30, 2017 at 1:13
  • $\begingroup$ @MathMajor Dang, my bad. You're right, $x$ is not the base. $\endgroup$
    – esote
    Commented Aug 30, 2017 at 1:17
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    $\begingroup$ @esote The point doesn't stand because $x^i$ is not a polynomial. Polynomials are on the form $$a_n x^n + x_{n-1}x^{n-1} + ... + a_1x_{1} + a_0.$$ $\endgroup$
    – Eff
    Commented Aug 30, 2017 at 1:20

5 Answers 5

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A finite number of derivatives get to constant zero if and only if the original really is a polynomial. For degree $n,$ the $n+1$ derivative gives zero.

This is also the test, given a sequence of integers, for detecting whether it is given by a polynomial; this is the very simplest application of "finite differences." If I begin with $$ 1, \; 8, \; 27, \; 64, \; 125, \; 216, \; 343, $$ first difference sequence $$ 7, \; 19, \; 37, \; 61, \; 91, \; 127, $$ second differences $$ 12, \; 18, \; 24, \; 30, \; 36, $$ third $$ 6, 6,6,6,6, $$ fourth $$ 0,0,0,0 $$

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    $\begingroup$ Also, OP, note that a polynomial is "an expression that can be built from constants and symbols called indeterminates or variables by means ofaddition, multiplication and exponentiation to a non-negative integer power." This of course excludes imaginary exponents. $\endgroup$ Commented Aug 30, 2017 at 1:19
  • $\begingroup$ @AndrewTawfeek Thanks, I was unaware that it did not extend to negative integer powers. Glad I got that straitened up. $\endgroup$
    – esote
    Commented Aug 30, 2017 at 1:23
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    $\begingroup$ @esote When some exponent are negative integers, we have a Laurent polynomial, but a Laurent polynomial is not considered a polynomial (a bit confusing). $\endgroup$ Commented Aug 30, 2017 at 11:14
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    $\begingroup$ @esote The deeper definition of "polynomial" is "an expression built up from addition, multiplication, constants and $x$", hence why $x^3=xxx$ is valid but $x^i=???$ is not. People didn't just wake up one day and decide to study sums of monomials with arbitrary restrictions on what the powers can be for no reason! $\endgroup$
    – Jack M
    Commented Aug 30, 2017 at 14:27
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Think of it the other way round: start from $0$, integrate it $n$ times... you'll end up with a polynomial of degree $n-1$. This means that if a function becomes $0$ (the function, not the number) after a finite number of differentiations, it can only be a polynomial.

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We can represent $\sin$ as sort of a polynomial, except with infinitely many terms, namely by power series expansion $$ \sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}=\sin x $$ which perhaps you have seen. What happens when you apply $$ \frac{d^k}{dx^k}\sin(x) $$ for any finite number $k$?

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    $\begingroup$ hi qbert, this is a good answer $\endgroup$
    – user217285
    Commented Aug 30, 2017 at 2:39
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    $\begingroup$ @Nitin There's a button called "upvote" to say that :) $\endgroup$
    – anderstood
    Commented Aug 30, 2017 at 3:18
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    $\begingroup$ You could call it a "sort of a polynomial", but not a polynomial, as that one, by definition, had a finite number of turns. Therefore the sine is not a polynomial. $\endgroup$
    – user365446
    Commented Aug 30, 2017 at 9:02
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    $\begingroup$ @Typhon, that's an experiment you can try. Find a general expression for the $n$-th derivative, and then see if the limit as $n\to\infty$ exists. $\endgroup$ Commented Aug 30, 2017 at 9:10
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    $\begingroup$ @Jeppe, that sounds like a good separate question to ask. :) $\endgroup$ Commented Aug 30, 2017 at 16:36
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Somewhat tongue-in-cheek answer: differentiating a function n times will result in a function identically equal to zero iff after n terms in the Taylor series, all the terms are zero. Which is just a roundabout way of saying "all the terms in the polynomial have exponent less than n". Sine's Taylor series goes on forever, so it derivatives go on forever.

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This is a consequence of the nature of "infinity". Sin/cos indeed seem to be polynomials , but of infinite order. So when you differentiate them, however often, the highest order term is still nonzero of infinite order. Therefore you never get to zero value as this would require a zero highest order term (and all other terms).

In fact (see comment below) once a 'polynomial' has an infinite number of terms, it isn't considered to be a polynomial. It is formally considered to be 'power series' instead. The introduction of infinitely many terms gives them a whole bunch of different properties (potentially) that polynomials don't have. As simple examples, polynomials are always well defined, smooth, continuous, non-periodic, even if they have a very large number of terms. But once you extend a polynomial to allow an infinite number of terms (= a power series) like those in the OP, or complex powers, or powers not in the natural numbers (0,1,2...), none of these properties of polynomials are guaranteed to still be the case.

You can't get round this by trying "infinite differentiation" either, as has been suggested. A simple reason is that an infinite set less an infinite set (each differentiation reduces the list of {ax^n} by one) can still leave an infinite set. Or almost any other size set of similar/lower cardinality (it's not always well defined and set theory isn't my strong point but you get the idea). Once you have infinitely many terms, then however much you differentiate, you can still have infinitely many terms.

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  • $\begingroup$ By definition a polynomial has a finite number of terms. The generalization where the order can be infinite is a formal power series. $\endgroup$
    – HBeel
    Commented Aug 30, 2017 at 12:36
  • $\begingroup$ True. However, to someone not used to the distinction, a power series just looks like an unending polynomial - the fact that their behaviour can fundamentally differ (smooth/continuous/differentiable/non-periodic/well defined at all points) may not be obvious, as in this question. I've clarified this in the answer, thanks. $\endgroup$
    – Stilez
    Commented Aug 30, 2017 at 16:32

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