Polynomial of finite degree actually infinite degree?

In summary: A polynomial function evaluated at a specific input is not a number. Rather, it is a value that can be added, subtracted, multiplied by other values in a field. It can be combined with other numbers to produce a new "number" in a field.If you have a finite degree polynomial and evaluate it for a particular input and the result is not a number, then the polynomial is not a polynomial function. In summary, the conversation discusses the contradiction between the degree of a polynomial and a rational polynomial function and the concept of evaluating a polynomial in different contexts, such as formal power series and real functions. The conversation
  • #1
Mr Davis 97
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##1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k##.

Isn't this a contradiction since the LHS has degree ##2## while the RHS has infinite degree?
 
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  • #2
Mr Davis 97 said:
##1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k##.

Isn't this a contradiction since the LHS has degree ##2## while the RHS has infinite degree?
Wherever your polynomials live in, ##\mathbb{F}(x)\; , \;\mathbb{F}[[x]]\,,## it only says that this field doesn't have a proper evaluation function given by the degree, or worse, that it does not included infinite (quotients of) polynomials.
 
  • #3
Mr Davis 97 said:
##1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k##.

Isn't this a contradiction since the LHS has degree ##2## while the RHS has infinite degree?

How do you define the degree of a polynomial in ##\mathbb{F}[[X]]##?
 
  • #4
Math_QED said:
How do you define the degree of a polynomial in ##\mathbb{F}[[X]]##?
To me the degree of a polynomial is the highest degree of its monomials with non-zero coefficients. I don't really know how to define degree precisely in the infinite case.

Actually, now that I expand the product, we get that all of the terms of degree higher than 2 cancel out, so I guess there is no problem.

But I do have another, slightly related question. We have that ##1+x+x^2 = \dfrac{1-x^3}{1-x}##. But the LHS can be evaluated at x=1 while the RHS cannot. How are they equal then?
 
  • #5
Mr Davis 97 said:
To me the degree of a polynomial is the highest degree of its monomials with non-zero coefficients. I don't really know how to define degree precisely in the infinite case.

Actually, now that I expand the product, we get that all of the terms of degree higher than 2 cancel out, so I guess there is no problem.

But I do have another, slightly related question. We have that ##1+x+x^2 = \dfrac{1-x^3}{1-x}##. But the LHS can be evaluated at x=1 while the RHS cannot. How are they equal then?
They are not equal. That's why I said: wherever your polynomials live. A rational polynomial and a rational polynomial function are not the same thing. But as just functions, they are not the same, because the quotient does not have the same range as the polynomial. The graph of the quotient has a hole!
 
  • #6
Let me ask something in return: Why is ##x^3-x = 0\,.##
 
  • #7
Mr Davis 97 said:
##1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k##.

Isn't this a contradiction since the LHS has degree ##2## while the RHS has infinite degree?

Hmm... aren't they the same thing? Assuming the series converge,

(1-x3)(1+x+x2+x3+x4+x5+...)
= (1+x+x2+x3+x4+x5+...)
- x3(1+x+x2+x3+x4+x5+...)
= (1+x+x2)+(x3+x4+x5+...)-(x3+x4+x5+...)
= 1+x+x2

The OP is similar to asking why the LHS is a 3-degree polymial and the RHS is a 10-degree polynomial... I don't think the RHS is a 10-degree polynomial:

1+x+x2 = 1+x+x2 + x10 - x10
 
  • #8
fbs7 said:
Hmm... aren't they the same thing? Assuming the series converge,

(1-x3)(1+x+x2+x3+x4+x5+...)
= (1+x+x2+x3+x4+x5+...)
- x3(1+x+x2+x3+x4+x5+...)
= (1+x+x2)+(x3+x4+x5+...)-(x3+x4+x5+...)
= 1+x+x2

The OP is similar to asking why the LHS is a 3-degree polymial and the RHS is a 10-degree polynomial... I don't think the RHS is a 10-degree polynomial:

1+x+x2 = 1+x+x2 + x10 - x10

These are formal power series. This is algebra, no analysis. Nothing converges here, these are just formal expressions that can be manipulated.
 
  • #9
Hmm... that's surprising...

Say that x = 2; then 1-x3 = -7, while (1+x+x2+...) = ∞, therefore (1-x3)*(1+x+x2+...) = -7 . ∞ = -∞

Meanwhile (1+x+x2+x3+x4+...) - (x3+x4+...) = ∞ - ∞ = indeterminate

I was trying to avoid assuming true a step that would be indeterminate if the series diverged, but if that's not necessary, then I stand corrected.
 
  • #10
fbs7 said:
Hmm... that's surprising...

Say that x = 2; then 1-x3 = -7, while (1+x+x2+...) = ∞, therefore (1-x3)*(1+x+x2+...) = -7 . ∞ = -∞

Meanwhile (1+x+x2+x3+x4+...) - (x3+x4+...) = ∞ - ∞ = indeterminate

I was trying to avoid assuming true a step that would be indeterminate if the series diverged, but if that's not necessary, then I stand corrected.
You first have to say what those polynomials mean:
  • formal power series
  • and if, invertible or not
  • real functions
Whether they allow the definition of a degree, a substitution map like ##x=2##, infinite length or whatever depends on this category. ##1+x+x^2+\ldots ## is without meaning if we do not attach one. As this meaning wasn't given, all we have are sequences:
\begin{align*}
1-x^3 &\triangleq (1,0,0,-1,0,0,\ldots) \\
1+x+x^2+\cdot & \triangleq (1,1,1,\ldots)
\end{align*}
and all they bring is a linear structure (addition and scalar multiplication) and perhaps a piecewise multiplication, which is not the usual polynomial multiplication. So neither polynomial multiplication, nor inversion, nor substitution, nor a degree, nor topological methods can be applied without further specifications.

It is the old fraud: start with something familiar, transfer it in a hidden world and draw false conclusions in the first environment.
Hint: This method is far better paid in Las Vegas, where it is called performances of magicians.
 
  • #11
Wow, I'm in deep and unchartered waters now! I thought that

##
1+x+x^2+x^3+...
##

always meant a limit

##
\lim_ {n \rightarrow +\infty} \sum_{i=0}^{n} x^i
##

but it may mean other things! I got to think about that over a hotdog. Living and learning!

Code:
int main(void)
{
    Dude fbs;
    fbs.ImproveMathSkill(0.00001,"%");
    fbs.GetFood(new HotDog());
}
 
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  • #12
fbs7 said:
Wow, I'm in deep and unchartered waters now! I thought that

##
1+x+x^2+x^3+...
##

always meant a limit
If you interpret the notation as a possibly divergent series then one can explore ways in which to evaluate the "sum" of a divergent series. https://en.wikipedia.org/wiki/Divergent_series

If you interpret the notation as a polynomial then you the advice of @fresh_42 is appropriate. One can distinguish between...

A formal polynomial: One can manipulate add, subtract and multiply formal polynomials by simply looking at the list of coefficients on each polynomial and coming up with a list of coefficients for the sum, difference or product. When looked at this way, there is no need to consider evaluating the polynomial or whether evaluation is even possible.

A polynomial function: One can treat a polynomial as a function produces a result, f(x) for each input, x. One can add, subtract or multiply polynomial functions by adding, subtracting or multiplying the evaluated results. When looked at this way, is is crucial to have a way of evaluating the polynomial for every possible input be aware of the input ranges for which evaluation is possible.

A polynomial function evaluated for a particular input. Again, one can add, subtract, multiply or divide. But one only needs to consider the selected input.

One thing that I find interesting is to contemplate polynomials over the boolean field GF(2) which contains just the integers 0 and 1. There are only four polynomial functions over this field: The constant 0 function, the constant 1 function, the identity function and the inverse function. But there are eight formal polynomials of degree 2 or less: x^2 + x + 1, x^2 + x, x^2 + 1, x^2, x + 1, x, 1 and 0.
 
  • #13
I think it is a widespread phenomenon at all levels to write down an expression and completely neglect the context, the definition or any other hint on how it should be interpreted. It is often clear by the circumstances, but as soon as it is written down, circumstances will be decoupled. We see this here quite often: members write down their problem and totally forget, that we readers are outstanders with respect to their circumstances. That's why I personally find section 2 of our template far more important than section 3.

I often try to fight these neglencies Don Quixote like as I think it is necessary to get used to a proper writing style, and actually do define what is meant - always. One of my famous objections is, if someone says ##2+2=4## or something. I'm literally forced to answer with a 'No'. If nothing else is said, we are in the naturals, the integers, the rationals, the reals or none of them? This might be nitpicking, but it is the mean seed of all later inaccuracies.
 
  • #14
I think the factorization doesn't go through##h: 1-x^3 =(1-x) (1+x+x^2)## so, if ##x\neq 1## then the ratio equals 1+x+x^2, but the main point remains. Sorry, please delete post.
 
  • #15
More simply, you should consider that ##\frac {1}{1-x} = \sum_{n=0}^{\infty} x^n## for ##0 \lt x \lt 1##. Would you say that the right-hand side is of infinite degree, when you know that the limit is a very smooth, simple, function? As @Math_QED indicated, that would be confusing algebra with analysis.
 
  • #16
Hmm... I'm kind of skimming on the almost-understanding this... So, let me try to organize my thoughts on this. This is a valid deduction

[a1] x=2 & y=x
[a2] y=2

The deduction below is false, of course.

[b1] x=2 & y=x
[b2] x=2 & y=x.(1-x)/(1-x)
[b3] x=2 & y=x.(1-x).(1+x+x2+x3+...) & 0<x<1
[b4] x=2 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...)) & 0<x<1
[b5] x=2 & y=x & 0<x<1
[b6] y=x & ( x=2 & 0<x<1 )

This leads to a contradiction, so it contains a mistake. I thought that the contradiction comes from a mistake on deducing [b3] from [b2], and that mistake is that I can't manipulate an infinite series when that series diverges. But, if I understood it right up so far:

(a) in Analysis fbs7 is correct, the contradiction arises because you cannot manipulate an infinite series if that series diverges
(b) in Algebra fbs7 is incorrect, and the contradiction arises because the clause 0<x<1 is artificial and not necessary, and without that clause this would be a perfectly fine deduction:

[c1] x=2 & y=x
[c2] x=2 & y=x.(1-x)/(1-x)
[c3] x=2 & y=x.(1-x).(1+x+x2+x3+...)
[c4] x=2 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...))
[c5] x=2 & y=x
[c6] y=2

So, tell me this... if [c6] is a valid deduction in Algebra, then is this also a valid deduction in Algebra?

[d1] x=2 & y=x
[d2] x=2 & y=x + log(0) - log(0)
[d3] x=2 & y=x
[d4] y=2

I can't believe you can deduce [d3] from [d2] above, because if you could, then you could also deduce

[e1] x=2 & y=x
[e2] x=2 & y=x + log(0) - log(0)
[e3] x=2 & y=x + log(2.0) - log(0)
[e4] x=2 & y=x + log(0) + log(0) - log(0)
[e5] x=2 & y=x + log(0)
[e6] y=2 + log(0)
[e7] y=log(102.0)
[e8] y=log(0)

So how in the world in Algebra I can deduce [c5] from [c4], but I cannot deduce [d3] from [d2]?
 
  • #17
A variant of the same theme... if in Algebra I can deduce

[c4] x=2 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...))
[c5] x=2 & y=x
[c6] y=2

then why I can't deduce

[cc4] x=1 & y=x.((1+x+x2+x3+...)-(x+x2+x3+...))
[cc5] x=1
[cc6] from [cc4], y=x.((1+x+x2+x3+...)-1*(x+x2+x3+...))
[cc7] from [cc5], y=x.((1+x+x2+x3+...)-x*(x+x2+x3+...))
[cc8] y=x.((1+x+x2+x3+...)-(x2+x3+x4+...))
[cc9] y=x.(1+x)
[cc10] y=2

The only difference between [c4] and [cc4] is that x=2 in the first, and x=1 in the second, and in the first I'm told that's valid, while the second it's certainly a mistake, and the only thing I did on [cc6] to [cc8] was to use the fact that x=1...
 
  • #18
I don't understand b3 of post #16. In the same statement, you are saying that x=2 and 0<x<1. That is wrong. You should not apply an infinite series with an x value where it does not converge. That is meaningless.
 
Last edited:
  • #19
FactChecker said:
I don't understand b3 of post #16. In the same statement, you are saying that x=2 and 0<x<1. That is wrong. You should not apply an infinite series with an x value where it does not converge. That is meaningless.

Precisely my point. I argued that when one replaces 1/(1-x) with an infinite series, that always adds the restriction 0<x<1, so you can't do that when x=2 and will lead to all kinds of jabberwoking. Ergo can't expand in series for values the series diverges.

But earlier in the thread I was told that the restriction 0<x<1 is not needed if the series is seen as a polynomial instead of a limit, and x*(1-x)*(.. power series...) still reduces to x even if the series diverges, ie, whatever the value of x. That I'm still trying to get in between my left ear and right ear.
 
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  • #20
fbs7 said:
But earlier in the thread I was told that the restriction 0<x<1 is not needed if the series is seen as a polynomial instead of a limit, and x*(1-x)*(.. power series...) still reduces to x even if the series diverges, ie, whatever the value of x. That I'm still trying to get in between my left ear and right ear.
I am also at a loss of what to think about that. But I see there are several good mathematicians who seem to accept it, so I will leave this discussion to them.
 
  • #21
Hmm... I couldn't help thinking about this while I was watching "Alita Battle Angel" today ($15 for popcorn and soda, holy choo-choo!).

I suspect everybody concurs that the expansion of 1/(1-x) in power series (an Analysis result) only makes sense if 0<x<1.

I think the point that Math_QED made on post #8 goes about this: if we set the series expansion aside, and only consider infinite polynomials, and say I define these purely Algebraic functions:

f:ℝ→ℝ, f(x) = 1+x+x2+x3+...
g:ℝ→ℝ, g(x) = x+x2+x3+...
h:ℝ→ℝ, h(x) = f(x) - g(x)

Then what's the value of h(2)? I think I empty-mindedly argued that, going by its definition, then h(2)=∞-∞ is undefined, (therefore h(x) only makes sense for 0<x<1), while Math_QED corrected me saying that if we replace f(x) and g(x) in the definition of h(x), and do the subtraction, then h(2) = 1 (therefore h(x) is valid for any x).

So, what's the truth of this matter? What's the value of h(2)?
 
  • #22
fbs7 said:
Hmm... I couldn't help thinking about this while I was watching "Alita Battle Angel" today ($15 for popcorn and soda, holy choo-choo!).

I suspect everybody concurs that the expansion of 1/(1-x) in power series (an Analysis result) only makes sense if 0<x<1.

I think the point that Math_QED made on post #8 goes about this: if we set the series expansion aside, and only consider infinite polynomials, and say I define these purely Algebraic functions:

f:ℝ→ℝ, f(x) = 1+x+x2+x3+...
g:ℝ→ℝ, g(x) = x+x2+x3+...
h:ℝ→ℝ, h(x) = f(x) - g(x)

Then what's the value of h(2)? I think I empty-mindedly argued that, going by its definition, then h(2)=∞-∞ is undefined, (therefore h(x) only makes sense for 0<x<1), while Math_QED corrected me saying that if we replace f(x) and g(x) in the definition of h(x), and do the subtraction, then h(2) = 1 (therefore h(x) is valid for any x).

So, what's the truth of this matter? What's the value of h(2)?

The reason why you are so confused is that you are mixing up two concepts.

Consider expressions of the form ##\sum_{n=1}^\infty a_n X^n##. These are called "formal power series".

Now, there are two ways to go: the algebraic way or the analytic way.

In algebra, we just define an addition and a multiplication with these formal power series: we consider formal series as elements on which we define some abstract operations which give another formal series, i.e. a series of the form ##\sum_{n=1}^\infty a_n X^n##.

In analysis, we consider ##\sum_{n=1}^\infty a_n X^n## as a limit of partial sums

##i.e. \sum_{n=1}^\infty a_n X^n = \lim_{k \to \infty }\sum_{n=1}^k a_n X^n## and we can ask ourselves for what values of ##X## this limit makes sense (= exists). If all the limits exist and some conditions are fulfilled, the algebraic operations that we can make with series in this way coincide with the ones that are defined in the algebraic way, but this is not always the case.
 
  • #23
So one can logically talk about the algebraic equality of expressions but should be careful about any implications regarding their numerical evaluations.
 
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  • #24
FactChecker said:
So one can logically talk about the algebraic equality of expressions but should be careful about any implications regarding their numerical evaluations.

That's a good TL:DR :)
 
  • #25
Math_QED said:
The reason why you are so confused is that you are mixing up two concepts.

Consider expressions of the form ##\sum_{n=1}^\infty a_n X^n##. These are called "formal power series".

Now, there are two ways to go: the algebraic way or the analytic way.

In algebra, we just define an addition and a multiplication with these formal power series: we consider formal series as elements on which we define some abstract operations which give another formal series, i.e. a series of the form ##\sum_{n=1}^\infty a_n X^n##.

In analysis, we consider ##\sum_{n=1}^\infty a_n X^n## as a limit of partial sums

##i.e. \sum_{n=1}^\infty a_n X^n = \lim_{k \to \infty }\sum_{n=1}^k a_n X^n## and we can ask ourselves for what values of ##X## this limit makes sense (= exists). If all the limits exist and some conditions are fulfilled, the algebraic operations that we can make with series in this way coincide with the ones that are defined in the algebraic way, but this is not always the case.

So, under the Algebraic understanding, is h(2)=1 or undefined?

I think that question is similar to this.. if we define f(x) = x/x, then is f(0)=1 or undefined?
 
  • #26
fbs7 said:
So, under the Algebraic understanding, is h(2)=1 or undefined?

I think that question is similar to this.. if we define f(x) = x/x, then is f(0)=1 or undefined?

I think your main confusion here is that you don't know the distinction between polynomials and polynomial functions.

A polynomial is a formal element ##\sum_{k=1}^n a_k X^k##.

A polynomial function is a map ##x \mapsto \sum_{k=1}^n a_k X^k##

You can evaluate a function in all points of its domain, so as for your question it does not make sense to ask what ##f(0)## is when ##f(x) = x/x## (here I regard ##f(x)## as a polynomial function), because ##0/0## is a meaningless expression. It is true however that ##f## has a continuous extension that is equal to ##1## everywhere.

Now, we can regard any polynomial as a polynomial function and then we have a notion of evaluating a polynomial. This can be generalised to general polynomial rings and should be treated in an algebraic geometry course.

Now, there is also the distinction between a formal power series and a power series.

A formal power series is a formal element ##\sum_{k=1}^\infty a_k X^k##.

A power series is a function ##x \mapsto \sum_{k=1}^\infty a_k x^k## with domain the elements ##x## where the limit of partial sums converges.

Unlike the polynomial part, I'm not aware of any algebraic evaluation of formal power series.

Thus, the question what is ##h(2)## makes no algebraic sense. Or at least not one that I'm aware of.
 
  • #27
I see; so that's what FactChecker referred to talking about equality of expressions, but being careful on the numerical evaluation of expressions!

That's a deep thought! I'll have to sleep on it! This is fascinating! This means that two formula F(x) and G(x) may be algebraically equal, say in F(x) = x/x and G(x) = 1, one might think they are the same thing, but F(x=0) and G(x=0) may actually be different, wow!

Thanks for the explanations!
 
  • #28
fbs7 said:
I see; so that's what FactChecker referred to talking about equality of expressions, but being careful on the numerical evaluation of expressions!

That's a deep thought! I'll have to sleep on it! This is fascinating! This means that two formula F(x) and G(x) may be algebraically equal, say in F(x) = x/x and G(x) = 1, one might think they are the same thing, but F(x=0) and G(x=0) may actually be different, wow!

Thanks for the explanations!

Yes, in the fraction field of the polynomials ##X/X =1##.
 
  • #29
fbs7 said:
Hmm... I couldn't help thinking about this while I was watching "Alita Battle Angel" today ($15 for popcorn and soda, holy choo-choo!).

I suspect everybody concurs that the expansion of 1/(1-x) in power series (an Analysis result) only makes sense if 0<x<1.

I think the point that Math_QED made on post #8 goes about this: if we set the series expansion aside, and only consider infinite polynomials, and say I define these purely Algebraic functions:

f:ℝ→ℝ, f(x) = 1+x+x2+x3+...
g:ℝ→ℝ, g(x) = x+x2+x3+...
h:ℝ→ℝ, h(x) = f(x) - g(x)

Then what's the value of h(2)? I think I empty-mindedly argued that, going by its definition, then h(2)=∞-∞ is undefined, (therefore h(x) only makes sense for 0<x<1), while Math_QED corrected me saying that if we replace f(x) and g(x) in the definition of h(x), and do the subtraction, then h(2) = 1 (therefore h(x) is valid for any x).

So, what's the truth of this matter? What's the value of h(2)?

You say "the expansion of 1/(1-x) in power series (an Analysis result) only makes sense if 0<x<1"

Actually, it makes sense if -1 < x < 1 (including x=0).
 
  • #30
thank you!
 
  • #31
Mr Davis 97 said:
##1+x+x^2 = \dfrac{1-x^3}{1-x} = (1-x^3)\cdot \dfrac{1}{1-x} = (1-x^3)\sum_{k=0}^\infty x^k##.

Isn't this a contradiction since the LHS has degree ##2## while the RHS has infinite degree?

By the usual definition of "polynomial" the RHS is not a polynomial. (The usual definition of polynomial doesn't permit a "polynomial" to be specified by the operation of taking a limit. Do your course materials actually define a type of "polynomial" that has "infinite degree"? )

A polynomial function f(x) may be equal to a function that is expressed in a way not permitted in the definition of "polynomial". For example f(x) = x = x + sin(x) - sin(x).

What qualifies a function to be a polynomial is that it may be expressed in a certain way. Alternative ways of expressing it don't involve any logical contradiction.
 

1. What is a polynomial of finite degree?

A polynomial of finite degree is a mathematical expression consisting of variables and constants, combined using addition, subtraction, and multiplication, with non-negative integer exponents. The degree of a polynomial is the highest exponent in the expression.

2. How is the degree of a polynomial related to its number of terms?

The degree of a polynomial is equal to the highest exponent in the expression, which is also equal to the number of terms in the polynomial. For example, a polynomial with 3 terms, each with an exponent of 2, has a degree of 2.

3. What does it mean for a polynomial to have infinite degree?

A polynomial with infinite degree means that its terms have increasingly larger exponents, without a finite highest exponent. This type of polynomial is also known as an "infinite polynomial" or "formal power series."

4. Can a polynomial of finite degree have an infinite number of terms?

No, a polynomial of finite degree can only have a finite number of terms. This is because the degree of a polynomial is equal to the highest exponent in the expression, and an infinite number of terms would result in an infinite degree.

5. What are some applications of polynomials of infinite degree?

Polynomials of infinite degree have various applications in mathematics, physics, and engineering. They are used to approximate functions, solve differential equations, and represent power series expansions. They also play a crucial role in the study of infinite series and convergence.

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