In this textbook: https://www.stitz-zeager.com/szprecalculus07042013.pdf
On Page $236$, it says:
But I can easily say $h(x)=|a_nx^n+a_{n-1}x^{n-1}+...a_2x^2+a_1x|$
Isn't my $h(x)$ written as a combination of powers?
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6No, what you have written is the absolute value of a combination of powers. By that logic, why isn't $\sin(x^2)$ a polynomial? – Randall Feb 04 '25 at 02:13
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You have absolute values signs in $h(x)$. You have to prove that's a polynomial first, you can't assume it. – CyclotomicField Feb 04 '25 at 02:15
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It is a good question, though, as to how to prove that $|x|$ is not a polynomial in $x$. – Randall Feb 04 '25 at 02:15
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I understand now I have to handle the absolute value first before handling what's inside. But then I don't understand why an absolute value is not a combination of powers. – ronald christenkkson Feb 04 '25 at 02:18
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It is a combination of powers and another function. "Combination" is a very sloppy word - formally, it should be "linear combination." – Thomas Andrews Feb 04 '25 at 02:49
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1Powers of $x$ is also sloppy - it could include $x^{-1}$ or $x^{1/2}.$ – Thomas Andrews Feb 04 '25 at 02:51
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The real answer is, look at Definition 3.1 (the definition of "polynomial") on the previous page, and observe that there's no way to make an expression like that that is equal to $|x|$ everywhere. The formula in Definition 3.1 is the only kind of "combination of powers of $x$" that matters here. – David K Feb 04 '25 at 03:15
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Suppose $|x|$ is a polynomial function, i.e., there exists coefficients $a_0,a_1,\ldots,a_n$ such that $$|x|=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\tag{1}$$ for all $x$. For this to be true for $x=0$ in particular, we need $a_0=0$. If we now assume $x\neq 0$, we can divide both sides of the previous equation to obtain
$$\frac{|x|}{x}=a_{n}x^{n-1}+a_{n-1}x^{n-2}+\cdots+a_1 \tag{2}$$
If we subtract $1$ from both sides, then the RHS is clearly a polynomial and the LHS is equal to $|x|/x-1$. But this LHS is equal to $0$ for all $x>0$ and thus has an infinitude of real roots. Since every nonzero polynomial has finitely-many real roots, the only such polynomial would be the zero polynomial. (See here for a proof that doesn't use the fundamental theorem of algebra.) But then $|x|/x=1$ for all $x$, which is false for $x<0$. Hence no such polynomial exists and $|x|$ cannot be a polynomial function.
Semiclassical
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It really comes down to the following definition of a polynomial. To quote Wikipedia,
a polynomial is a mathematical expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication and exponentiation to nonnegative integer powers
The operation of taking the absolute value is not included, thus an expression in which the absolute value of a variable is taken cannot be a polynomial.
H. sapiens rex
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Thanks for the answer. I understand it as well. But I still don't get why the textbook I linked says an absolute value is not a combination of powers. What does that mean? – ronald christenkkson Feb 04 '25 at 02:23
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4Except that $f(x) = |x^2|$ is a polynomial, so it's not sufficient just to say that the presence of an absolute value is an automatic dealbreaker. – Randall Feb 04 '25 at 02:25
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To use $|x|$ as an example, it is saying that there are no unambiguous scalars $a_0,a_1,a_2,...a_n$ such that $|x|=a_0+a_1x+a_2x^2+...a_nx^n$ that will work for all values of $x$. – H. sapiens rex Feb 04 '25 at 02:28
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@ronald there does not exist any choices of $n$ and $a_0,a_1,a_2,\dots,a_n$ such that $|x| = a_0+a_1x+a_2x^2+a_3x^3+\dots+a_nx^n$ for all values of $x$. There does happen to be such choices so that they equal for all positive values of $x$ (namely $n=1,a_0=0,a_1=1$ giving $|x|=0+1x$) but this fails to work for the negative and positive values of $x$ simultaneously. There also exist ways of rewriting that involve additional symbols and functions, however we are not interested in those other ways of rewriting, and in particular are not interested in ways of rewriting that involve abs value. – JMoravitz Feb 04 '25 at 02:28
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@Randall you mean $|x^2|=x^2$? So in essence the absolute value of something squared is always equivalent to squaring it without needing the absolute value? – ronald christenkkson Feb 04 '25 at 02:34
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I would be careful in saying that $|x^2|$ is a polynomial. It happens to be equal to a polynomial function (at least on the real numbers), but it's not a polynomial expression and that's often what's important. – ConMan Feb 04 '25 at 05:30
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But why are absolute values not a combination of the powers of $x$? For example $|x^2+x+1|$ to me is a combination of the powers of $x$. – ronald christenkkson Feb 04 '25 at 05:42
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@ronaldchristenkkson Robert Israel, JMoravitz and I have already explained why: there does not exist a set of scalars by which $|p(x)|$ (where $p(x)$ is a polynomial) can be written as a sum of powers of $x$, which work for all values of $x$. That's the problem. – H. sapiens rex Feb 04 '25 at 06:10
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Here's one way to prove that $|x|$ is not a polynomial, i.e. that there is no choice of $n$ and $a_0, a_1, \ldots, a_n$ that makes $|x| = a_0 + a_1 x + \ldots + a_n x^n$ for all real numbers $x$.
Suppose such $n$ and $a_0, a_1, \ldots, a_n$ did exist. First note that $n =0$ and $n= 1$ do not work (I'll leave this as an exercise). So $n \ge 2$. Then $|x| - x = a_0 + (a_1 - 1) x + \ldots + a_n x^n$ would be a polynomial of degree $n$. By the Fundamental Theorem of Calculus, a polynomial of degree $n$ has at most $n$ real roots. But $|x| - x$ has infinitely many real roots, namely all positive real numbers: contradiction. Thus they do not exist, and $|x|$ is not a polynomial.
Robert Israel
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Also worth noting that this applies more generally: if $f(x)$ is identical to some polynomial $p(x)$ at an infinite number of points, then either $f(x)=p(x)$ everywhere or $f(x)$ is not a polynomial. – Semiclassical Feb 04 '25 at 20:55