Explicit functions with discontinous intervals?

**mfetch22** · June 22nd 2010, 10:24 PM

I understand that rational functions of certain forms will produce a "hole" or infinitismal break in the graph, like the following:

$g(x) = \frac{(x-2)(x+3)}{(x-2)}$

will obviously appear as the graph of $y = x+3$ with a hole at $x=2$ . But is there a way to explicitly define a function (supposedly composed of functions which are not defined for all values of x, or that may result in such a function) to explicitly define a function [without using domain restrictions, piecewise representations, or band-aid set theory definitions; (i.e. using set theory to to define the function with the desire discountinuities)] that would naturally have discountinuities over a specific interval? Say I wanted to explicitly define a function that had a complete discontinuity at all values

$2 \leq x \leq 6$

or more generally

$a \leq x \leq b$

is there a manner in which it is possible to construct a function that would produce this property?

Recap:

[1]Explicit, non-piece-wise function $f(x)$ . No using set theory directly "cut and delete". No given domain restrictions

[2] $f(x)$ is not simply discontinuos at specific values, or holes, of $x$ , but rather is discontinuos for
$a \leq x \leq b$

How would one go about creating such a function? Thanks in advance for any advice related to this topic

**Sudharaka** · June 22nd 2010, 10:56 PM

Originally Posted by mfetch22

I understand that rational functions of certain forms will produce a "hole" or infinitismal break in the graph, like the following:

$g(x) = \frac{(x-2)(x+3)}{(x-2)}$

will obviously appear as the graph of $y = x+3$ with a hole at $x=2$ . But is there a way to explicitly define a function (supposedly composed of functions which are not defined for all values of x, or that may result in such a function) to explicitly define a function [without using domain restrictions, piecewise representations, or band-aid set theory definitions; (i.e. using set theory to to define the function with the desire discountinuities)] that would naturally have discountinuities over a specific interval? Say I wanted to explicitly define a function that had a complete discontinuity at all values

$2 \leq x \leq 6$

or more generally

$a \leq x \leq b$

is there a manner in which it is possible to construct a function that would produce this property?

Recap:

[1]Explicit, non-piece-wise function $f(x)$ . No using set theory directly "cut and delete". No given domain restrictions

[2] $f(x)$ is not simply discontinuos at specific values, or holes, of $x$ , but rather is discontinuos for
$a \leq x \leq b$

How would one go about creating such a function? Thanks in advance for any advice related to this topic

Dear mfetch22,

Let me give you an idea. If we define a function such that $f:\Re\rightarrow{\Re}~and~f(x)=0~\forall~x\in[a,b]$ then, the function, $h(x)=\frac{f(x).g(x)}{f(x)}$ is a function which is not defined for $x\in[a,b]$ (here g(x) is another arbitary function)

I hope that the function h(x) satisfies all the conditions you have mentioned.

**Ackbeet** · June 23rd 2010, 02:13 AM

I don't think sudharaka's function works, because it's not defined anywhere, and therefore isn't really a function. We need a well-defined function that is discontinuous everywhere in an interval. So, mfetch, is the characteristic function out of bounds for your problem? You could easily define $f(x)=\chi_{\mathbb{Q}}(x)$ . There's a function that is discontinuous everywhere. Trouble is, the characteristic function is itself defined piecewise. So I don't know if this "once removed" type of function is allowed or not.

Your problem sounds like it's part of a bigger problem. Any chance you'd be willing to state this bigger problem?

**mfetch22** · June 23rd 2010, 10:45 AM

Originally Posted by Ackbeet

I don't think sudharaka's function works, because it's not defined anywhere, and therefore isn't really a function. We need a well-defined function that is discontinuous everywhere in an interval. So, mfetch, is the characteristic function out of bounds for your problem? You could easily define $f(x)=\chi_{\mathbb{Q}}(x)$ . There's a function that is discontinuous everywhere. Trouble is, the characteristic function is itself defined piecewise. So I don't know if this "once removed" type of function is allowed or not.

Your problem sounds like it's part of a bigger problem. Any chance you'd be willing to state this bigger problem?

Its not neccesarily a bigger problem, its just a conecpt that came to my mind while watching this science channell program about finding new planets by observing wobbling stars, or by ovserving tiny shawdows or gaps in which the sun is partially blocked out. Then I, for some reason that I am unaware, I pictured an ellipse tracing out the path of a planet. I pictured viewing this planet orbit a lonely star, from a view point for which the planet dissipears for some period of its orbit behind this star and then reappears as it comes back into view. This gap in the visibility of the planet made me consider dropping that section of the ellipse when I viewed it from above. Then, I wondered if there was a way to make this "cut" without simply just stating its there. Maybe this means that I should also allow implicit relationships. So add that to the list, any implicit or explicit relationship that may arise in "continous discontinuities"..... I like this as name for this problem, it might not be mathematically proper but I've always loved oxy-morons.

Sudharaka, thank you for your post:

SUDHARAKA:

I hope that the function h(x) satisfies all the conditions you have mentioned.

I dont mean to be spliting hairs here, but the manner in which you defined $f(x)$ falls under the use of set theory to me. I dont mean to make any general statements or negitive comments about set theory, I'm simply trying to find a way to solve this problem with out it. Let me give an explaination of why..

Say I want to find an equalitiy with a graphical property named $P$ . Lets, in addition call, this equality $S$ . Lets put some restriction on this equality:

[1] $S$ must be explicitally or implicitally expressed as an equality of the standard two rectangluar axis variables $x$ and $y$ , in $\mathbb{R}^2$

[2] $S$ must exhibit property $P$ as desribed or defined mathematically, but it must exhibit this propery as a natural consequence of the manner in which the equation was expressed. I would consider most all domain restrictions given as a supplement to the original equation to be beyond the "natural consequences" of the form of its expression.

[3] $S$ may not be expressed in peice-wise form, or in the langue of set theory.

The reason I make this restriction is because without some of them, this question becomes extremely simple. Like the set theory restriction, say I wanted to define discontinuity from a to b. Without the set theory restriction, this can easiliy be done, like Sudharaka showed. One simply defines $P$ , a discontinuity in this case, for all values $f(x) \in [a, b]$ .

Does that clear things up? The only thing I could think of that would satisfy this definition (actually I think it doesnt fit into my bounds since its an infinite expression) is the following progression for a discontinuity for $[a,b]$ , I think you'll catch the pattern:

$f(x) = \frac{1}{(a-x)(b-x)(a+1-x)(b-1-x)(a+2-x)(b-2-x).......(a+d_n-x)(b-d_n-x)}$

$\;\;untill\;\; d_n=\frac{b-a}{2}\;\; and\;\; d_n = 0,1,2,3,4....\;\;\;for\;\;n=1,2,3,4.....$

next step:

$f(x) = \frac{1}{(a-x)(b-x)(a+0.5-x)(b-0.5-x)(a+1-x)(b-1-x)(a+1.5-x)(b-1.5-x)......<br /> .(a+d_n-x)(b-d_n-x)}$

$\;\;untill\;\; d_n = \frac{b-a}{2}\;\;and\;\;d_n = 0, 0.5, 1.0, 1.5,.....\;\;\;for\;\;n=1,2,3,4$

Can you see the pattern? This is the only manner in which I can concieve of a "continous discontinuity" from a to b. I dont think this fits into my restrictions, so I need help finding a much simpler form. As for the above, if the process is continued, would this be correct?

$\lim_{(d_{n+1}-d_n)\to 0}f(x) = (continous\;\;discontinuity\;\;on\;\;[a, b])???$

**earboth** · June 23rd 2010, 10:55 AM

Originally Posted by mfetch22

I understand that rational functions of certain forms will produce a "hole" or infinitismal break in the graph, like the following:

$g(x) = \frac{(x-2)(x+3)}{(x-2)}$

will obviously appear as the graph of $y = x+3$ with a hole at $x=2$ . But is there a way to explicitly define a function (supposedly composed of functions which are not defined for all values of x, or that may result in such a function) to explicitly define a function [without using domain restrictions, piecewise representations, or band-aid set theory definitions; (i.e. using set theory to to define the function with the desire discountinuities)] that would naturally have discountinuities over a specific interval? Say I wanted to explicitly define a function that had a complete discontinuity at all values

$2 \leq x \leq 6$

or more generally

$a \leq x \leq b$

is there a manner in which it is possible to construct a function that would produce this property?

Recap:

[1]Explicit, non-piece-wise function $f(x)$ . No using set theory directly "cut and delete". No given domain restrictions

[2] $f(x)$ is not simply discontinuos at specific values, or holes, of $x$ , but rather is discontinuos for
$a \leq x \leq b$

How would one go about creating such a function? Thanks in advance for any advice related to this topic

Not sure that I understand completely what you are looking for ...

Do you mean something like: $f(x)=\sqrt{x^2-4}$ ?

**mfetch22** · June 23rd 2010, 11:08 AM

Originally Posted by earboth

Not sure that I understand completely what you are looking for ...

Do you mean something like: $f(x)=\sqrt{x^2-4}$ ?

Of course! I didnt even consider the square root! Yes, this has a "continuous discontinuity", as I like to call it. Do you know if there is a way to use the property of this function you gave, to define some implict expression of an elipse that has a "continuous discontinuity" over some interval on the curve? Just asking, in the off chance that you know the answer.

**earboth** · June 26th 2010, 12:06 PM

Originally Posted by mfetch22

Of course! I didnt even consider the square root! Yes, this has a "continuous discontinuity", as I like to call it. Do you know if there is a way to use the property of this function you gave, to define some implict expression of an elipse that has a "continuous discontinuity" over some interval on the curve? Just asking, in the off chance that you know the answer.

First: I don't know the answer to your question ... but:

$\left\{\begin{array}{l}x=5 \cdot \cos\left(\sqrt{t^2-9}\right) \\ y= 3 \cdot \sin\left(\sqrt{t^2-9}\right)\end{array} , t\in [-2\pi, 2\pi]\right.$

will produce the graph in the attachment.

**Utherr** · June 26th 2010, 12:18 PM

Originally Posted by earboth

Not sure that I understand completely what you are looking for ...

Do you mean something like: $f(x)=\sqrt{x^2-4}$ ?

wow...that was easy rofl... here's another one although it has no values $f(x)=\sqrt{-(x^2+1)}$

**mfetch22** · June 27th 2010, 11:42 AM

Originally Posted by earboth

First: I don't know the answer to your question ... but:

$\left\{\begin{array}{l}x=5 \cdot \cos\left(\sqrt{t^2-9}\right) \\ y= 3 \cdot \sin\left(\sqrt{t^2-9}\right)\end{array} , t\in [-2\pi, 2\pi]\right.$

will produce the graph in the attachment.

Yes, but this has to do with a domain restriction. I was curious if there was a way to write the equation where the domain restrictions are inherent in the way it is written. Like division by zero, or a negitive square root? Who knows?

Any thoughts on the infinite construction of the rational function from my post above? The infinite rational expression? Is it even mathematically possible to denote such an expression as I did above? Or even mathematically valid?

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