Functions and Transcendental Numbers

Agent Smith

Agent Smith

Full Member
Joined
Oct 18, 2023
Messages
303
How can any function be continuous if there exist nonalgebraic numbers (transcendental?) in any given interval?
 
mario99 said:
Write an example.
Click to expand...
You made me check. The Wikipage lists \(\displaystyle \pi\) and \(\displaystyle e\) as transcendental numbers. So, first I apologize for not doing my homework, and the follow up question then must be ... are there numbers that are not solutions to any polynomial? I can more easily grasp that the domain of functions are restricted (division by 0), but is it possible that the range is too? Si, there are discontinuous functions, but from the lessons I took, point discontinuities always have a handy continuous function to replace them to find limits. I mean range restriction for any and all functions (these maybe between numbers or at either extremes of the number line).
 
Agent Smith said:
You made me check. The Wikipage lists \(\displaystyle \pi\) and \(\displaystyle e\) as transcendental numbers. So, first I apologize for not doing my homework, and the follow up question then must be ... are there numbers that are not solutions to any polynomial? I can more easily grasp that the domain of functions are restricted (division by 0), but is it possible that the range is too? Si, there are discontinuous functions, but from the lessons I took, point discontinuities always have a handy continuous function to replace them to find limits. I mean range restriction for any and all functions (these maybe between numbers or at either extremes of the number line).
Click to expand...
First, do you consider [imath]\pi[/imath] and [imath]e[/imath] as real numbers? Second, what is the definition of a continuous function?
 
Agent Smith said:
but from the lessons I took, point discontinuities always have a handy continuous function to replace them to find limits.
Click to expand...
Really ? How would it work for this function:
[math]f(x) = \begin{cases} 1 & \text{if} & x\in \mathbb Q \\ 0 & \text{if} & x \notin \mathbb Q \end{cases}[/math]where [imath]\mathbb Q[/imath] is a set of all rational numbers?
 
mario99 said:
First, do you consider [imath]\pi[/imath] and [imath]e[/imath] as real numbers? Second, what is the definition of a continuous function?
Click to expand...
Yes, \(\displaystyle \pi\) and \(\displaystyle e\) are real numbers. A function f(x) is continuous over an interval [a, b] IFF for any point p in that interval \(\displaystyle \displaystyle \lim_{x \to p^-} f(x) = \lim_{x \to p^+} f(x) = f(p)\)??
 
blamocur said:
Do you mean polynomials with rational coefficients ? Algebraic coefficients?
Click to expand...
Now that I think of it, yeah, (per Wiki) a polynomial with rational coefficients will never have as its output \(\displaystyle \pi\) or \(\displaystyle e\) (assuming I understood what transcendental numbers are) i.e. there's a discontinuity in the function, which could be remedied by using a substitute, another polynomial with irrational (?) coefficients. Is that ever done?
 
blamocur said:
Really ? How would it work for this function:
[math]f(x) = \begin{cases} 1 & \text{if} & x\in \mathbb Q \\ 0 & \text{if} & x \notin \mathbb Q \end{cases}[/math]where [imath]\mathbb Q[/imath] is a set of all rational numbers?
Click to expand...
I guess not all functions can be fixed in the way I described.
 
Agent Smith said:
Now that I think of it, yeah, (per Wiki) a polynomial with rational coefficients will never have as its output \(\displaystyle \pi\) or \(\displaystyle e\) (assuming I understood what transcendental numbers are) i.e. there's a discontinuity in the function, which could be remedied by using a substitute, another polynomial with irrational (?) coefficients. Is that ever done?
Click to expand...
No, you have it backwards. It is not that the output can't be say e, but rather that p(e) = 0 is not possible. Note that the input is e and the output is 0.

What @blamocur questioned you about is quite important. After all, if p(x) = x-e, then yes e is a root. So e is NOT transcendental? No, not at all.
A number t is said to transcendental if no polynomial with rational coefficients can have t as a root.
Personally, I prefer to state that t is said to transcendental if no polynomial with integer coefficients can have t as a root.
 
Steven G said:
No, you have it backwards. It is not that the output can't be say e, but rather that p(e) = 0 is not possible. Note that the input is e and the output is 0.

What @blamocur questioned you about is quite important. After all, if p(x) = x-e, then yes e is a root. So e is NOT transcendental? No, not at all.
A number t is said to transcendental if no polynomial with rational coefficients can have t as a root.
Personally, I prefer to state that t is said to transcendental if no polynomial with integer coefficients can have t as a root.
Click to expand...
Gracias for the clarification. I do have it backwards. 😊
Transcendental numbers can't be roots of polynomials with rational coefficients. Good example 👉 f(x) = (x - e)(x - \(\displaystyle \pi\)). 👍

So my question is, are there numbers that never appear in the range of any function, whatsoever?

Perhaps I should bring up applied math. Let's talk physics, with functions describing flight, fluid dynamics, nuclear reactions, etc. If it's possible to plot all of these functions (laws of nature) on a single graph, will every point in the cartesian plane be "touched" i.e. all points are input-output pairs to at least one of the functions, or will there be gaps/pockets of "white space". For example take any object and plot its speed, a function, f, of time, t, and distance covered, d. We know that [imath]f(t, d) < 186000 \text{ miles per second}[/imath] (the speed of light).

We probably need to state up front that we're using a set of rational units, Je ne sais pas 🤔. @Steven G
 
Last edited:
A quick question, @Steven G ... and @Otis
If I wanted to color/fill a given section of a plane as "efficiently" as possible with just [imath]1[/imath] function, would that function be some kinda spiral?
 
In the old days when we had those big, bulky TV sets with a CRT inside it, what was the function for the electron beam's path? I know, cogito I know, there's only 1 stream of electrons, zipping across the screen so fast that it creates an illusion that the entire screen is lit up all at once.
 
Agent Smith said:
A quick question, @Steven G ... and @Otis
If I wanted to color/fill a given section of a plane as "efficiently" as possible with just [imath]1[/imath] function, would that function be some kinda spiral?
Click to expand...
Can you define efficiently a little bit? I'm not sure how the color is being added.
 
Steven G said:
Can you define efficiently a little bit? I'm not sure how the color is being added.
Click to expand...
As you draw, you fill in the space. Efficiently as in maximize, for a given section of the Cartesian plane, the area covered by the function.
 
Hello there,

A function's continuity is not affected by the presence of transcendental numbers in an interval. Here’s why:

Key Points:

1. Continuity Definition:
  • A function f(x) is continuous at a point a if lim(x -> a) f(x) = f(a). This holds regardless of the types of numbers.
2. Transcendental Numbers:
  • These are just numbers that aren’t roots of any polynomial with rational coefficients. Examples include π and e.
3. Dense Subsets:
  • Both rational and transcendental numbers are densely packed in the real numbers, but this doesn’t affect continuity.
Conclusion:

Continuity depends on how the function behaves as it approaches any point, not on the specific type of numbers. Therefore, a function can be continuous on an interval despite the presence of transcendental numbers.

Hope this helps!
 
You must log in or register to reply here.
Top