Testing for a function

Agent Smith said:
@fresh_42 , muchas gracias for your help, but these concepts are beyond my ken.

I'll tell you what I know and maybe you could curate your comments based on that.
1. The fundamental theorem of algebra: A polynomial of the \(\displaystyle n^{th}\) degree has \(\displaystyle n\) roots.
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Yes, however, not necessarily distinct roots.

[imath] (x-2)^6 [/imath] has one root, [imath] x=2, [/imath] six times. It is called the multiplicity of a root. If you start to dig deeper here, you will be entering the rabbit hole of field theory. Fields are domains like the rational, real or complex numbers where we can add, subtract, multiply and divide. There are others. The light switch in your room is a field: ON = 1 and OFF = 0. We can add: ON + ON = OFF, multiply: ON times OFF = OFF, ON times ON = ON, OFF times OFF = OFF, and divide by all numbers different from zero, i.e. we can divide by ON = 1 which doesn't change anything.

Agent Smith said:
2. I can't see how \(\displaystyle y^6 = x\) is a \(\displaystyle 6^{\text{th}}\) degree polynomial, unless you're swapping the x and y.
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Not sure what you mean. I told you that the letters are problematic. Let's make an example: [imath] y^6=729. [/imath] You have asked whether this means that [imath] y=\pm \sqrt[6]{729}=\pm 3. [/imath] I said, no, because there are also another four complex roots. We have the problem to determine all solutions of the equation
[math] y^6=729 \Longleftrightarrow y^6-729=0 [/math]which is the question about all roots of the polynomial [imath] p(y)=y^6-729. [/imath]

The results are
[math] y^6=729 \Longleftrightarrow y^6-729=0 \Longleftrightarrow y\in \left\{ \pm 3 \, , \, \pm \left( \dfrac{3}{2} + i \dfrac{3\sqrt{3}}{2}\right)\, , \,\pm \left(\dfrac{3}{2} - i \dfrac{3\sqrt{3}}{2}\right)\right\}. [/math]
See, the factor [imath] \sqrt[6]{729}=3 [/imath] is in all roots. That's why we solve [imath] y^6=1 .[/imath] To multiply the results by [imath] \sqrt[6]{x}=\sqrt[6]{729}=3 [/imath] afterwards is easy. No need to carry it all the way through.
 
By the way @fresh_42 , how is all this you're posting related to an algebraic test for a function, which is what I want to know?
 
fresh_42 said:
(x−2)6 has one root, x=2, x=2, x=2, six times. It is called the multiplicity of a root.
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I need to pick your brains on this one.

I recall a poster once asked, \(\displaystyle y = x^2\) is a polynomial of the \(\displaystyle 2^{\text{nd}}\) degree and should have 2 roots, but only has one viz. \(\displaystyle x = 0\).

I told him the following ...
\(\displaystyle x^2 = 0\)
\(\displaystyle x = \pm 0\)
The \(\displaystyle 2\) roots are \(\displaystyle x = +0\) and \(\displaystyle x = -0\). It just happens to be the case that \(\displaystyle +0 = -0 = 0\)

For \(\displaystyle y = (x - 2)^ 6\), my argument fails. There's multiplicity of roots, but there's no way I can demonstrate that there actually are 6 roots, it's just that they're all identical (like with \(\displaystyle x^2 = 0\))

Maybe ...
\(\displaystyle (x-2)^6 = 0\) means that \(\displaystyle x - 2 = \text{?} \sqrt [6] 0\) and maybe \(\displaystyle \text{?} \sqrt [6] 0 = 0 = +0 = -0 =\otimes 0 = \oplus 0 = \square 0 = \triangle 0\)
 
Agent Smith said:
By the way @fresh_42 , how is all this you're posting related to an algebraic test for a function, which is what I want to know?
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You said ...
Agent Smith said:
One guess that I made is \(\displaystyle y^n = \text{Something}, n \in \text{Evens}\), can't be a function, because \(\displaystyle y = \pm \text{Something else}\).
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... which seduced me to speak about the roots in general and the ##\pm## notation in particular. I thought you wanted to discuss the specific equation [imath] y^n=x. [/imath] But you are right, sorry, my fault.

A function [imath] f [/imath] is a set of pairs [imath] (x,y) [/imath]
[math] f=\{(x,y)\,|\,y=f(x)\text{ AND }x_1=x_2 \Rightarrow f(x_1)=f(x_2)\} [/math]i.e. the same value of [imath] x [/imath] cannot be mapped to two different images. It is a special case of a relation which is simply any set of pairs without the mapping condition.

Now, what is your relation you want to test whether it is a function?

[imath] y=x^2 [/imath]1721582242274.pngis a function, [imath] y^2=x [/imath] 1721582272441.pngis only a relation, and not a function.

That means, [imath] y=x^2 [/imath] is a function since identical values of [imath] x [/imath] have the same function value. On the other side, [imath] y=\pm \sqrt{x} [/imath] or [imath] y^2=x [/imath] is only a relation because [imath] (1,1) [/imath] and [imath] (1,-1) [/imath] are both possible: same [imath]x[/imath], different [imath] y. [/imath]

The test should be: are there two or more [imath] y [/imath] for the same value [imath] x [/imath]? Yes, its a relation and no function; no, its a function.
 
Agent Smith said:
I need to pick your brains on this one.

I recall a poster once asked, \(\displaystyle y = x^2\) is a polynomial of the \(\displaystyle 2^{\text{nd}}\) degree and should have 2 roots, but only has one viz. \(\displaystyle x = 0\).

I told him the following ...
\(\displaystyle x^2 = 0\)
\(\displaystyle x = \pm 0\)
The \(\displaystyle 2\) roots are \(\displaystyle x = +0\) and \(\displaystyle x = -0\). It just happens to be the case that \(\displaystyle +0 = -0 = 0\)

For \(\displaystyle y = (x - 2)^ 6\), my argument fails. There's multiplicity of roots, but there's no way I can demonstrate that there actually are 6 roots, it's just that they're all identical (like with \(\displaystyle x^2 = 0\))

Maybe ...
\(\displaystyle (x-2)^6 = 0\) means that \(\displaystyle x - 2 = \text{?} \sqrt [6] 0\) and maybe \(\displaystyle \text{?} \sqrt [6] 0 = 0 = +0 = -0 =\otimes 0 = \oplus 0 = \square 0 = \triangle 0\)
Click to expand...

Every real polynomial is also a complex polynomial and every complex polynomial [imath] p(x) [/imath] of degree [imath] n [/imath] can be written as
[math] p(x)=a_nx^n+a_{n-1}x^{x-1}+a_{n-2}x^{n-2}+\ldots +a_2x^2+a_1x+a_0=a_n\cdot (x-c_1)\cdot \ldots \cdot (x-c_n) [/math]We can sort all roots [imath] c_k [/imath] that are equal and write this as
[math] p(x)=a_n\cdot (x-d_1)^{r_1}\cdot (x-d_2)^{r_2}\cdot \ldots \cdot (x-d_m)^{r_m}[/math]where the sets of zeros are equal
[math] \{c_1,c_2,\ldots\, , \,c_n\}=\{d_1,d_2,\ldots,d_m\} [/math]and
[math] n=r_1+r_2+\ldots+ r_m\,. [/math]The numbers [imath] r_k [/imath] are called the multiplicities of the pairwise distinct roots [imath] d_k [/imath].


You can investigate these by finding the greatest common divisor between [imath] p(x) [/imath] and its derivative [imath] p'(x) [/imath]. Roots [imath] d [/imath] that occur more than once divide both polynomials, which means [imath] (x-d)\,|\,p(x) [/imath] and [imath] (x-d)\,|\,p'(x) .[/imath]
 
Agent Smith said:
I need to pick your brains on this one.

I recall a poster once asked, \(\displaystyle y = x^2\) is a polynomial of the \(\displaystyle 2^{\text{nd}}\) degree and should have 2 roots, but only has one viz. \(\displaystyle x = 0\).

I told him the following ...
\(\displaystyle x^2 = 0\)
\(\displaystyle x = \pm 0\)
The \(\displaystyle 2\) roots are \(\displaystyle x = +0\) and \(\displaystyle x = -0\). It just happens to be the case that \(\displaystyle +0 = -0 = 0\)

For \(\displaystyle y = (x - 2)^ 6\), my argument fails. There's multiplicity of roots, but there's no way I can demonstrate that there actually are 6 roots, it's just that they're all identical (like with \(\displaystyle x^2 = 0\))

Maybe ...
\(\displaystyle (x-2)^6 = 0\) means that \(\displaystyle x - 2 = \text{?} \sqrt [6] 0\) and maybe \(\displaystyle \text{?} \sqrt [6] 0 = 0 = +0 = -0 =\otimes 0 = \oplus 0 = \square 0 = \triangle 0\)
Click to expand...
Have you studied "complex algebra" - i.e. algebra involving complex number (i =\(\displaystyle \pm\sqrt{-1}\)) yet ?
 
fresh_42 said:
You said ...

... which seduced me to speak about the roots in general and the ##\pm## notation in particular. I thought you wanted to discuss the specific equation [imath] y^n=x. [/imath] But you are right, sorry, my fault.

A function [imath] f [/imath] is a set of pairs [imath] (x,y) [/imath]
[math] f=\{(x,y)\,|\,y=f(x)\text{ AND }x_1=x_2 \Rightarrow f(x_1)=f(x_2)\} [/math]i.e. the same value of [imath] x [/imath] cannot be mapped to two different images. It is a special case of a relation which is simply any set of pairs without the mapping condition.

Now, what is your relation you want to test whether it is a function?

[imath] y=x^2 [/imath]View attachment 38381is a function, [imath] y^2=x [/imath] View attachment 38382is only a relation, and not a function.

That means, [imath] y=x^2 [/imath] is a function since identical values of [imath] x [/imath] have the same function value. On the other side, [imath] y=\pm \sqrt{x} [/imath] or [imath] y^2=x [/imath] is only a relation because [imath] (1,1) [/imath] and [imath] (1,-1) [/imath] are both possible: same [imath]x[/imath], different [imath] y. [/imath]

The test should be: are there two or more [imath] y [/imath] for the same value [imath] x [/imath]? Yes, its a relation and no function; no, its a function.
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Thank you. For this post and the other one.

If a member of the domain maps to 2 or more members in the range/codomain, we don't have a function. I was hoping for some kinda test we could use to determine that. Aren't there tests for primality, etc.?

I like this 👉 [imath]y = a_1x^n + a_2x^{n -1} + \dots + k = a(x - r_1)(x - r_2) \dots (x - r_n)[/imath] from your other post.
 
Agent Smith said:
Aren't there tests for primality, etc.?
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That heavily depends on the size of your prime. Here is one for the smaller primes
www.dcode.fr

Prime Number Checker - Online Primality Tester/Calculator

Tool to check if a number is a prime number. A primality test is a mathematical and algorithmic test that indicates whether a number is prime or compound and answers true or false.
www.dcode.fr www.dcode.fr
The RSA encryption scheme relies on the fact that huge numbers are hard to factorize.

If you accept functions as a subset of "Domain Times Range" then you can check whether the same element from the domain has a pairing with more than one element from the range. This is sometimes easy to see like in [imath] y^2=x [/imath] where we have the pairs [imath] (-1,1)\, , \,(1,1) [/imath] (one [imath] x=1 [/imath] and two [imath] y=-1,+1 [/imath]), and sometimes not as easy, e.g. whether the inverse of the function [imath] y=x\cdot e^x [/imath] is a function or not, and if so, where? For which values of [imath] y [/imath]? If we would require that [imath] x,y [/imath] are both real and both positive, then [imath] y^n=x [/imath] would be a function. Simply because we ruled out the values that cause the problems.

Tests of such kinds are generally difficult. We cannot even check whether some problems like finding the shortest route for a traveling salesman, or staffing airplanes with flight crews are intrinsically hard or whether we are too stupid to find a smart, will say fast solution.
 
fresh_42 said:
That heavily depends on the size of your prime. Here is one for the smaller primes
www.dcode.fr

Prime Number Checker - Online Primality Tester/Calculator

Tool to check if a number is a prime number. A primality test is a mathematical and algorithmic test that indicates whether a number is prime or compound and answers true or false.
www.dcode.fr www.dcode.fr
The RSA encryption scheme relies on the fact that huge numbers are hard to factorize.

If you accept functions as a subset of "Domain Times Range" then you can check whether the same element from the domain has a pairing with more than one element from the range. This is sometimes easy to see like in [imath] y^2=x [/imath] where we have the pairs [imath] (-1,1)\, , \,(1,1) [/imath] (one [imath] x=1 [/imath] and two [imath] y=-1,+1 [/imath]), and sometimes not as easy, e.g. whether the inverse of the function [imath] y=x\cdot e^x [/imath] is a function or not, and if so, where? For which values of [imath] y [/imath]? If we would require that [imath] x,y [/imath] are both real and both positive, then [imath] y^n=x [/imath] would be a function. Simply because we ruled out the values that cause the problems.

Tests of such kinds are generally difficult. We cannot even check whether some problems like finding the shortest route for a traveling salesman, or staffing airplanes with flight crews are intrinsically hard or whether we are too stupid to find a smart, will say fast solution.
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Domain times range, as in (a, x), (a, y), (b, x), (b, y) for {a, b} and {x, y}? Rows n columns kinda thing? True, a function is a subset.
 
@fresh_42 , so there's no algebraic test for a function? Weird. You seem to know a lot. Why don't you invent one? It looks like there are good-enough-for-government-work primality checkers; there should be at least one function checker as well, oui? You could win karma points, if not the Fields Medal, making the life of a student less painful. 🙂
 
Agent Smith said:
@fresh_42 , so there's no algebraic test for a function? Weird. You seem to know a lot. Why don't you invent one? It looks like there are good-enough-for-government-work primality checkers; there should be at least one function checker as well, oui? You could win karma points, if not the Fields Medal, making the life of a student less painful. 🙂
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Such a test would be trivial if the domain has only finitely many elements and close to impossible in the general case. There are simply too many functions, uncountably infinite many. E.g. look at the inverse of the function [imath] y=xe^x. [/imath]

800px-Lambert-w.svg.png

(Source: Wikipedia, Lambert's W-function)

The blue part and the red part are both functions, just not the combination of them. You can see it and that's the answer in most cases: you can see it. Or read it from the definition of a relation.

However, there are mapping rules that are difficult to decide. Let's define
[math]\begin{array}{lll} R&=\{(n,G)\,|\,n=|G|\in \mathbb{N}\wedge G \text{ is a (finite) group with property P}\} \subseteq \mathbb{N}\times \{G\,|\,G \text{ is a finite group}\} \end{array}[/math]In the case of solvable (P) groups, [imath] R [/imath] isn't a function. In the case of simple (P) groups, we can only decide it since the early 80s of the last century (and I don't know the answer because I don't know the order of all finite simple groups):

"The proof is spread over more than 500 scientific articles with a total of almost 15,000 printed pages. However, not all of the proofs have been published. More than 100 mathematicians were involved from the late 1920s to the early 1980s." (Wikipedia)

It would take a serious amount of time to decide whether [imath] R [/imath] is a function or not in case [imath] P [/imath] stands for simplicity. But it can be done. Now, wait until someone comes around and asks this for any other property [imath] P .[/imath] There are simply way too many functions.

Even purely real functions can be nasty. Let's define
[math]\begin{array}{lll} f(x)&=\begin{cases} 2&\text{ if there exists a polynomial }p\text{ such that }p(\mathrm{i}^{\mathrm{i}})=x\\ 1&\text{ if there exists a polynomial }p\text{ such that }p(e^{\sqrt{2}})=x\\ 0&\text{ otherwise} \end{cases} \end{array}[/math]Is this a function or not? And I can complicate it in many ways with additional values and polynomials.
 
Agent Smith said:
Domain times range, as in (a, x), (a, y), (b, x), (b, y) for {a, b} and {x, y}? Rows n columns kinda thing? True, a function is a subset.
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Yes, that would be a relation that is not a function because [imath] f(a)=x [/imath] and [imath] f(a)=y .[/imath] This is exactly the difference between the two: every function is also a relation but a relation isn't necessarily a function. Functions are rules for [imath] a [/imath] that determine a unique value [imath] f(a). [/imath] We can have [imath] f(a)=x [/imath] and [imath] f(b)=x [/imath] but not [imath] f(a)=x [/imath] and [imath] f(a)=y .[/imath].
 
A function \(\displaystyle f: f(a) = x \wedge f(b) = x\) would be noninvertible, oui?
 
khansaheb said:
এটা সত্যি। - আমি জানি - কিন্তু আপনি কিভাবে সেখান থেকে এগোবেন ?
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It doesn't go on since there is no inverse. The inverse relation (switching the two components) is merely a relation and no function anymore.
 
Is it that \(\displaystyle y = x^2\) has no inverse or that it has an inverse which is not a function?

The inverse being \(\displaystyle x = \pm \sqrt y\)
 
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