Is there such a thing as an uncountable polynomial?

[cragar]

Is it possible to have a polynomial with an uncountable number of turns?
Like we could think of cos(x) as having a countable number of turns.
could I have $y=x^{\aleph_1}$
My question may not even make sense.

 

[DonAntonio]

Is it possible to have a polynomial with an uncountable number of turns?
Like we could think of cos(x) as having a countable number of turns.
could I have $y=x^{\aleph_1}$
My question may not even make sense.

If by "turns" you mean maximal/minimal points then the answer is no, as in any such point the derivative, which exists in all the reals, will

vanish, and the derivative of a polynomial is another polynomial, with only a finite number of roots.

DonAntonio

 

[Mentallic]

A polynomial must have a finite number of "turns" (with a turn you're probably thinking about its local min and max points which are found by the roots of the polynomial's derivative) so we can just consider the polynomial's roots instead.

Functions that aren't polynomials can often be described by its Taylor series which is just a 'polynomial' of infinite length. For example,

$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$

But clearly e^x isn't a polynomial, and in fact, polynomials of degree n must have n complex roots that aren't all necessarily distinct. But e^x=0 has no roots. If it were a polynomial it should have infinite roots, right? Because its Taylor series is a 'polynomial' of infinite degree.

 

[mathwonk]

the rolle theorem implies that a graph does not turn on any interval not containing a critical point. since a polynomial has only a finite number of critical points (zeroes of the derivative), it has only finitely many turns.

 

[cragar]

ok so if a polynomial won't work, could we do it with some other function? maybe that's what you guys were leading to above.

 

[Vorde]

If by turns you to mean max/min points than there are certain functions (sine, cosine) that have an infinite amount of them, but I'm pretty sure by definition a polynomial has to have a finite amount of terms, and in that case then the number of max/min points will always be countable.

 

[haruspex]

I think the question you're trying to ask is whether a continuous function can have uncountably many local minima. Anyway, it is an interesting question whether or not it's the one you had in mind.
My feeling is that it can't. If f has a local minimum at x then there exists some interval (x-ε_x, x+ε_x) s.t. if y ≠ x is in the interval then f(y) > f(x). (OK, you might also want to allow f(y) = f(x) so long as f is constant on the interval between.)
If we could find non-overlapping such intervals then the result would follow; every such interval would contain a rational not found in any of the other intervals, so there could only be countably many of them. But consider e.g. f(x) = x²sin²(1/x) + x⁴. This has a local minimum at 0, but every interval around zero contains infinitely many other local minima.
I'll give it some more thought.

 

[SteveL27]

Is it possible to have a polynomial with an uncountable number of turns?
Like we could think of cos(x) as having a countable number of turns.
could I have $y=x^{\aleph_1}$
My question may not even make sense.

It's interesting that Euler used the same type of reasoning to attack the Basel problem (what we'd call zeta(2)) in 1735. He took the power series for sine and regarded it as a "polynomial" with an infinite number of roots to produce a simple derivation that zeta(2) = pi²/6.

http://en.wikipedia.org/wiki/Basel_problem

So the moral of the story is that if a student treated sine as an infinite polynomial, we'd mark that wrong; but when Euler did it, he was a genius!

 

[Number Nine]

It's interesting that Euler used the same type of reasoning to attack the Basel problem (what we'd call zeta(2)) in 1735. He took the power series for sine and regarded it as a "polynomial" with an infinite number of roots to produce a simple derivation that zeta(2) = pi²/6.

http://en.wikipedia.org/wiki/Basel_problem

So the moral of the story is that if a student treated sine as an infinite polynomial, we'd mark that wrong; but when Euler did it, he was a genius!

That's sort of misleading. He considered the zeros of the Taylor series for sine, yes, but it really didn't have anything to do with polynomials.

OP: By definition, a polynomial has a finite degree. Since a polynomial can have only as many roots as it's degree, and the "turns" correspond to the zeros of the derivative, which has an even smaller degree, the number of turns must be finite.

 

[cragar]

what about the Weierstrass function

 

[Number Nine]

what about the Weierstrass function

What about it?

 

[Vorde]

what about the Weierstrass function

The Weierstrass function isn't a polynomial.

 

[DonAntonio]

The Weierstrass function isn't a polynomial.

Yes, it isn't (big time!) a polynomial...so? This is such a pathological function that I highly doubt any more or less "reasonable"

definition of "turn" can be attached to it...

DonAntonio

 

[Vorde]

My point was that I believed the OP might have been citing the Weierstrass function as an example to support his original claim and I responded that separate of concrete definitions of minima and maxima on the function (to which I cannot speak to), the Weierstrass function cannot be cited as evidence solely on the basis of it not being a polynomial.

 

[cragar]

I was wondering if the Weierstrass function had an uncountable number of local minima?
Who cares if it is a polynomial.

 

[Infinitum]

As a very simple example, sin(x) is a function with infinite number of local minima. Of course, not a polynomial.

Weierstrass function, I'm not so sure...It should have infinite local minima, I think.

 

[Millennial]

The necessary condition that we will have an infinite number of stationary points is that the derivative of the function has infinitely many real roots. For this to be achieved, its degree must be infinite; for a polynomial has as many roots as its degree. Hence it is impossible to achieve it with a finite polynomial.

The exponential function has no real or complex roots. Since it is straightforward to show that it has no real roots, you can use Euler's formula to show it does not have any complex roots either. Hence, the exponential function does not have any stationary points.

The Gamma function is another example to a function that has infinitely many stationary points.

Functions with infinitely many stationary points involve periodic functions such as sine, cosine, and their sums (Fourier series.)

 

[Vorde]

The Weierstrass Function is differentiable nowhere, so based off of the definition we have all be using for minima/maxima (points where the derivative is zero), the Weierstrass function has no derivative, and therefore should have no minima/maxima.

That being said, its sort of tough to visualize a function like the weierstrass one, so its possible that definition of minima/maxima doesn't apply here, but I would assume it does and therefore believe it is safe to say that the Weierstrass function has no minima/maxima.

 

[slider142]

It is not necessarily true that a relative extremum of a function has a derivative of zero. The derivative may also be undefined there, as in f(x) = |x|.
Since the OP has moved the question onwards from polynomials to any function that has an uncountable number of relative minima and maxima, perhaps the OP may want to consider applying a precise definition of minimum and maximum. One definition that fits is that a point y = f(x) in the range of f is a relative minimum of f if there exists some finite interval I containing x such that y = f(x) is the minimum value of the set f(I). More fundamentally, given the set of all values of f over the finite interval I, the value y is less than or equal to each element of that set.
An analogous definition works for relative maximum.
The question can now be phrased as whether R can be partitioned into an uncountable number of finite intervals. If that can be done, we can then define our function with an uncountable number of relative minima and maxima. If not, no such function can be defined, as it would then define such a partition. This is a more tractable problem.

 

[Millennial]

Polynomials are differentiable everywhere...
He asked about the number of turns, so that is either the derivative is zero or undefined. Since polynomials cannot have undefined derivatives, considering the case when the derivative is zero suffices.

 

[slider142]

As of post #5 in this thread, the OP cragar was no longer interested in solely polynomials, but any function which may have an uncountable number of turning points. He expressed his disinterest in polynomials again in post #12, so I assume he means any function.

 

[Millennial]

Oh then, you are right.
But the definition of "turn" you make is also important. The Weierstrass function might have infinitely many turns or no turns at all, based on your definition. If you only take smooth turns, then the Weierstrass function has none (those are the ones yielded by differentiation.) Else, the absolute value function has a single turn, and the Weierstrass function has infinitely many.

 

[slider142]

Oh then, you are right.
But the definition of "turn" you make is also important. The Weierstrass function might have infinitely many turns or no turns at all, based on your definition. If you only take smooth turns, then the Weierstrass function has none (those are the ones yielded by differentiation.) Else, the absolute value function has a single turn, and the Weierstrass function has infinitely many.

The Weierstrass function has a countably infinite set of turns, so it does not satisfy the OP's question. You can show this by assuming that it has an uncountable number of relative extrema and then showing that this leads to a contradiction, using the interval argument.
Also, "my definition" is not proprietary. It is a restatement of the standard one: http://mathworld.wolfram.com/RelativeExtremum.html

 

[haruspex]

what about the Weierstrass function

Yes, if any continuous function is going to have uncountably many local minima then that would be a good candidate. I read somewhere that the Wiener function still only has countably many, so I still suspect the same will be true of Weierstrass.

 

[Number Nine]

It is not necessarily true that a relative extremum of a function has a derivative of zero. The derivative may also be undefined there, as in f(x) = |x|.

This is irrelevant. The OP asked specifically about polynomials, where the local extreme do correspond to the roots of the derivative. It follows immediately that the number of extrema is finite.

 

[haruspex]

This is irrelevant. The OP asked specifically about polynomials,

... but then changed the question. See post #5.

 

[Number Nine]

... but then changed the question. See post #5.

Ah...that he did.
In that case, we could just take a constant function. Then every point is a local minima/maxima, and there are uncountably many such points (mind you, there are no "turns"...)

 

[haruspex]

In that case, we could just take a constant function. .. (mind you, there are no "turns"...)

Quite. In fact one could consider an interval over which the function is constant as just one long local min/max/inflection, so that's not really in the spirit of the question. I think there's an interesting question lurking in there, but the first step might be to define it properly.

 

[Number Nine]

Quite. In fact one could consider an interval over which the function is constant as just one long local min/max/inflection, so that's not really in the spirit of the question. I think there's an interesting question lurking in there, but the first step might be to define it properly.

I think we can phrase the question as follows: Does there exist a real valued, continuous function who's derivative is zero at uncountably many points in its domain? Obviously, we have to reject the trivial answer of a derivative which is constant (and zero) on some interval; we could further specify that the derivative assumes a non-zero value in some neighbourhood of every point. My hunch is that no such function exists.

 

[haruspex]

Does there exist a real valued, continuous function who's derivative is zero at uncountably many points in its domain?

That adds the requirement that function is differentiable everywhere. Pretty sure you've no chance of finding a function like that. Need to rephrase it without recourse to differentiation.

 

[Number Nine]

That adds the requirement that function is differentiable everywhere. Pretty sure you've no chance of finding a function like that. Need to rephrase it without recourse to differentiation.

The only way I can think of, and it isn't very satisfying, is as follows: Does there exist a continuous, nowhere-constant function with uncountably many local extrema?

Elaborating on "nowhere-constant", let's say that a function f is constant at a point x if f assumes a single value within some neighbourhood of x.