Interesting functions thread

The identity function

Constant functions cuz the don't give a fuck what you plug in

>>8943795
how is it constructed my nigga?

>>8943795
Isn't that impossible? I remember that if a function's derivative is 0 for every x in R and is continuous than the function is constant.

>>8943904
>I remember that if a function's derivative is 0 for every x in R and is continuous than the function is constant.

Wrong. To see why, try to prove that.

>>8943904
That's why its an interesting function. It isn't really intuitive.

>>8943907
What? No, that's true. The Cantor function is *almost* everywhere differentiable, which is the catch.

>>8943907
Suppose f is continuous and f'(x)=0 for all x in R. Let a,b be any two distinct points in R, then f satisfies the mean value theorem on [a,b]. Then there exists some c such that f'(c) (b-a)= f(b)-f(a). But f'(c) = 0, and thus f(b) = f(a). It follows that f is constant on R

>>8943822
Not him but its putting x into base 3 then forcing it to become a binary number that is then read back into base 10

>>8943904
the derivative isn't 0 everywhere, it's only 0 where it's defined (which is *almost* everywhere).

>>8943822
f(0)=0, f(1)=1
split (0,1) into 3 equal length open intervals and let f(x)=1/2 for the middle one
split both remaining in 3 equal length open intervals and let f(x) for the middle one be arithmetic mean between already defined neighboring values
repeat ad infinitum

File: Cantor_function.gif (60KB, 983x966px) Image search: [Google]

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>>8943822
Better description than >>8943938 in the related pic.

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>>8943795
I like this thread idea OP, I'll start dumping some classic ones to get discussion rolling. Here's sin(1/x), which has a discontinuity at zero where the limit from either side fails to exist, and also doesn't diverge to plus or minus infinity.

File: 795px-WeierstrassFunction.svg.png (60KB, 795x505px) Image search: [Google]

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>>8943967
Here's the Weierstrass function, which is continuous everywhere but differentiable nowhere.

>continuous everywhere, derivative exists almost everywhere, but it always 0.

>interesting functions thread
>people only post gay ass satanic functions that only exist to bother people

>>8943967
Not to mention that its graph is connected, but not path-connected.

>>8943973
The function I'm going to define one has no picture, to accompany it, but I still think it's cool. It requires a bit of context first.

Theorem: Suppose [math]f: \mathbb{R} \to \mathbb{R}[/math] is monotonically increasing (meaning that if [math]x \leq y[/math], then [math]f(x) \leq f(y)[/math]). Then f only has simple discontinuities, meaning the left and right limit must always exist.

Proof: The left limit at x will be the supremum of [math]f((-\infty, x))[/math], and the right limit will be the infimum of [math]f((x, \infty))[/math]. It's very straightforward to check this works.

We of course have an analogous result for a monotonically decreasing function. A function which is either monotonically increasing or decreasing is monotonic.

Corollary: If f is monotonic, then f has at most countably many discontinuities.

Proof : By the above theorem, any discontinuity for f is a jump discontinuity. Then at each point of discontinuity, we can choose a rational point between the left and right limit at the discontinuity. This implies there are at most countably many points of discontinuity.

This corollary raises the question of whether there is a monotonic which is discontinuous at countably many points. The answer is yes, and here's an example!

Let [math]\sum_{n=1}^\infty x_n[/math] be a convergent sequence of positive numbers. We enumerate the rationals [math]\{q_n\}_{n=1}^\infty[/math]. Now define [math]f: \mathbb{R} \to \mathbb{R}[/math] as follows:

[eqn]
f(x) = \sum_{n : q_n < x} x_n.
[/eqn]

Because the [math]x_n[/math] are positive, it's easy to see this function is monotonically increasing. It's also easy to see by definition that for a rational point [math]q_m[/math], the left limit is [math]\sum_{n: q_n < q_m} x_n[/math], while the right limit is [math]\sum_{n: q_n \leq q_m} x_n[/math], so there is a jump discontinuity at this point.

In sum, we've constructed a monotonic function which is discontinuous at every rational point!

>>8943795

The Dirichlet function.

It maps all real numbers x to the set {0,1}, such that:

if x is rational, then f(x) is 1
if x is irrational, then f(x) is 0

Defined everywhere, discontinuous everywhere, differentiable nowhere.

And the area under the curve from x=0 to x=1 is..... ???

File: pc3ba4sh.gif (7KB, 592x301px) Image search: [Google]

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>>8944012
I was going to talk about the Dirichlet function in this post, but >>8944025 snipped me. Here's another function that's somewhat similar. If x is irrational, define f(x) = 0. If x is rational, write it in simplest form as x = p/q. Then define f(x) = 1/q. It's easy to see this function is discontinuous an every rational point. In fact, [math]\lim_{x \to a} f(x) = 0[/math] for every a, which implies that f is continuous at every irrational point. Pic related is a graph of the function.

>>8944025
Also, the Riemann integral of that function is undefined, and the Lebesgue integral is 0.

File: takagi10.gif (2KB, 372x265px) Image search: [Google]

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>>8944033
Last one I have for now is the Takagi function, also called the Blancmange Curve. It is, like the Weierstrass function, continuous and nowhere differentiable. Further, it is neither increasing nor decreasing on any interval of positive length.

The last thing I would like to point out are two other cool properties of the Cantor function (from the OP) which haven't been mentioned yet.
1) If you think about it, the length of a graph in the unit square can have maximum length 2. The length of the graph of the Cantor function is in fact 2.
2) Let f be the Cantor function and C the cantor set. We can see by definition f(C^c) is countable, and thus a measure zero set. This implies that the measure of f(C) is 1. In other words, the Cantor function maps a null set to a set of positive measure (this means f fails to satisfy the Luzin property).

File: Dirac_function_approximation.gif (89KB, 200x335px) Image search: [Google]

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The dirac delta function is a *generalized* function which is 0 everywhere but at x=0, yet has an area of 1 under the graph.
(Not actually a function, more of a limit to a series of functions)
https://en.m.wikipedia.org/wiki/Dirac_delta_function

>>8943967
>>8943974
How's high school treating ya?

>>8944426
>dirac delta function

PolarPlot[(1 + 0.9 Cos[8 t]) (1 + 0.1 Cos[24 t]) (0.9 + 0.05 Cos[200 t]) (1 + Sin[t]), {t, -Pi, Pi}]

Sign function

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A flat function - meaning at some point, all derivatives of the function are 0.

Example: y = exp(-x^(-2))

Since any derivative of y is 0 at x=0, it cannot be approximated by a Taylor series.

>>8945227
its not defined at origin

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>>8944426
>mfw a electrical engineer is being brainlet near me

>>8945234
its defined as 0 at the origin which is why it has the described property of derivatives of all orders 0 at the origin

>>8943967
>Babbys first analysis class

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hey guys. I am very bad at math. Is it possible to draw this function? and what is mathemathics behind it?

>>8943795
Conway base 13 function, not continuous but still has the IVT property of a continuous function.

>>8945252
i meant origo

and often piecewise defined functions tend to have not taylor series

>>8943904
>derivative is 0 for every x in R
>ignoring the phrase "almost everywhere"
found the Analysis D-student

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>>8944426
>sit in on a /sci/ thread to see how much of a joke they are
>anon says dirac delta "function"

>>8945709
>Is it possible to draw this function?
It looks like you just did.

Cool problem I did in first year maths:
Construct an explicit bijection between (0,1) and [0,1]

>>8946111
I used paint program: ( Is it possible to draw it via math? is there function behind it?

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>>8946108

File: zxfeC.gif (5KB, 600x225px) Image search: [Google]

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sine wave

>>8944033
>>8944012
this is my favorite i like this, thanks anons

File: integral function.png (28KB, 768x460px) Image search: [Google]

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the integral function

>>8945709
-exp(-x)/(x-1)

>>8943795
my teacher called it "The Devils function"

>>8947635
2+exp(-0.5x)/(x-1)

>>8947635
>>8947645
>exp