1. Jun 6, 2012

### 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.

2. Jun 6, 2012

### DonAntonio

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

3. Jun 6, 2012

### 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 ex isn't a polynomial, and in fact, polynomials of degree n must have n complex roots that aren't all necessarily distinct. But ex=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.

4. Jun 6, 2012

### 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.

5. Jun 7, 2012

### cragar

ok so if a polynomial wont work, could we do it with some other function? maybe thats what you guys were leading to above.

6. Jun 7, 2012

### 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.

7. Jun 8, 2012

### 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-εx, x+εx) 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.

8. Jun 8, 2012

### SteveL27

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!

Last edited: Jun 8, 2012
9. Jun 8, 2012

### Number Nine

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.

10. Jun 9, 2012

### cragar

11. Jun 9, 2012

### Number Nine

12. Jun 9, 2012

### Vorde

The Weierstrass function isn't a polynomial.

13. Jun 9, 2012

### DonAntonio

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

14. Jun 9, 2012

### 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.

15. Jun 10, 2012

### cragar

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

16. Jun 10, 2012

### 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.

Last edited: Jun 10, 2012
17. Jun 10, 2012

### 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.)

18. Jun 10, 2012

### 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.

19. Jun 10, 2012

### 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.

20. Jun 10, 2012

### 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.