## Interesting examples

For the discussion of math. Duh.

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Zohar
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### Interesting examples

My friends and I thought of a simple function whose limit at infinity is zero but whose derivative explodes to negative and positive infinity. I thought this was pretty cool and I'm wondering what other examples you may have of counterintuitive things (peano curve, the Weierstrass function etc.).

The function we thought of is:
cos(e2x)e-x

Which obviously goes to zero at infinity. And the derivative, after a tiny bit of rearranging, is:
-sin(e2x)ex-cos(e2x)e-x

Second term goes to zero, first term explodes.
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schmiggen
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### Re: Interesting examples

Hrm. Is the limx->∞cos(x)e-x... something we can make sense of? I mean, I know we know the range of cosine is just [-1,1], but does that mean we can just disregard its limit and say "whatever it is in those bounds, if you multiply it by 0, you get 0"?

I mean, if you were just going to look at limx->∞cos(x), wouldn't we be stuck? (i.e. no limit)
As I type this, it seems less and less problematic. Still, I'm... oddly confused.
Last edited by schmiggen on Mon Apr 14, 2008 8:14 am UTC, edited 1 time in total.
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Cycle
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### Re: Interesting examples

My favorite examples of these sort of things come from point set topology. For example, the "Lakes of Wada": a compact subset C of R2 such that it's complement has 3 connected components, and the boundary of each component is C.

GBog
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### Re: Interesting examples

There are a bunch of interesting stuff in Counterexamples in Analysis. I don't have it with me right now, but I'll post a few examples in the evening. The only ones I can think of off* the top of my head are basic examples, like the weierstrass function and the function that's nowhere continuous, but whose absolute value is everywhere continuous.

Also, I simpler function that tends to zero, but whose derivative tends to [imath]\pm \infty[/imath] would be
$\frac 1x \sin (x^3)$

* Did I get those right?

Ended
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### Re: Interesting examples

One of my favourite counterexamples is the (non-constructive proof of the) existence of an infinitely differentiable function f:[0,1] -> R whose Taylor series is divergent in any open interval.
[edit: re-worded]

schmiggen wrote:Hrm. Is the limx->∞cos(x)e-x... something we can make sense of? I mean, I know we know the range of cosine is just [-1,1], but does that mean we can just disregard its limit and say "whatever it is in those bounds, if you multiply it by 0, you get 0"?

I mean, if you were just going to look at limx->∞cos(x), wouldn't we be stuck? (i.e. no limit)
As I type this, it seems less and less problematic. Still, I'm... oddly confused.

Well, limx->∞ cos(x)e-x = 0, since the absolute value of cos(x)e-x is less than e-x which tends 0. So yes, as you said, since cos(x) is bounded, you can essentially just ignore it.

And yes, limx->∞ cos(x) does not exist, since the absolute value of cos(x) oscillates and does not tend to anything. Similarly, something like sin(e2x)ex is unbounded, but doesn't tend to infinity (because it oscillates).
Last edited by Ended on Mon Apr 14, 2008 7:50 pm UTC, edited 1 time in total.
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stoke
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### Re: Interesting examples

Hrm. Is the limx->∞cos(x)e-x... something we can make sense of? I mean, I know we know the range of cosine is just [-1,1], but does that mean we can just disregard its limit and say "whatever it is in those bounds, if you multiply it by 0, you get 0"? I mean, if you were just going to look at limx->∞cos(x), wouldn't we be stuck? (i.e. no limit). As I type this, it seems less and less problematic. Still, I'm... oddly confused.

It's bounded above and below by the functions e^(-x) and -e^(-x). Just think about it like this:
For all epsilon greater than zero, there is some sufficiently large x s.t. for all y>x, y is between epsilon and negative epsilon.

Torn Apart By Dingos
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### Re: Interesting examples

Ended wrote:One of my favourite counterexamples is the existence of an infinitely differentiable function f:[0,1] -> R which does not have a power series expansion valid in any open interval.
i can has this function, plz?

Ended
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### Re: Interesting examples

Sry, cannot has. Non-constructive proof (It's a bit involved to type out, but the crucial step uses Baire category to show that there is some function whose derivatives at some point q are large enough s.t. the Taylor terms go to infinity at all points in [0,1]).
Generally I try to make myself do things I instinctively avoid, in case they are awesome.
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GBog
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### Re: Interesting examples

If you allow the function to have power series that does not converge to it, the example is less pathological, but still rather weird.

Let $g(x)=\begin{cases}e^{-\frac{1}{x^2}}&\text{if }x\neq 0,\\ 0&\text{if }x=0,\end{cases}$
and let [imath]\{q_n\}_{n=1}^\infty[/imath] be an enumeration of the rational numbers in [imath][0,1][/imath]

Then
$\displaystyle f(x) = \sum_{n=1}^\infty \frac {1}{n^2}g(x-q_n)$
is everywhere infinitely differentiable but does nowhere have a power series which converges to it on a non-zero interval.

Torn Apart By Dingos
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### Re: Interesting examples

Cool, that was really simple! That was, btw, what I interpreted Ended's function to be (a function with a Taylor series that converges but not to the function (at a dense set of points), not a function with a divergent Taylor series).

Another cool pathological function is the Devil's Staircase: a continuous monotonic function with derivative zero almost everywhere which isn't constant.

ThomasS
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### Re: Interesting examples

Let g be the continuous function on [0,1] with g(0)=g(1)=0 and g(1/2) = 1/2 and linear everywhere else. Then define fn(x) = g(10n x - ⌊10n x⌋)/10n
Notice that the |fn|C0 = 1/2 10-n. So g = ∑ fn is C0, which is to say, continuous.

However, it is an entertaining analysis exercise to show that g is not differentiable anywhere.
Edited: C0 not C

Ended
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### Re: Interesting examples

Torn Apart By Dingos wrote:That was, btw, what I interpreted Ended's function to be (a function with a Taylor series that converges but not to the function (at a dense set of points), not a function with a divergent Taylor series).

Good point, I'll edit my post to make it clearer.

Also, can a proof that {a counterexample exists} be counted as a counterexample? *mind asplodes*
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### Re: Interesting examples

Ended wrote:Also, can a proof that {a counterexample exists} be counted as a counterexample? *mind asplodes*

No. But it's still valid to say, "There is a counterexample." Saying "I have a counterexample" might be going a bit far...
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Owehn
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### Re: Interesting examples

"I know I have that counterexample somewhere around here... Oops, I left it in my other model of ZFC."

I remember in my reals class constructing a (Lebesgue) integrable function that was everywhere discontinuous and unbounded on every interval, and retained these properties after modification on any set with zero measure. It used the same trick that GBog did: take an enumeration of the rationals, and add up a convergent series of displaced spikes.
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