Why is ##h(x)=|x|## not a polynomial?

[RChristenk]

Homework Statement

Why is $h(x)=|x|$ not a polynomial?

Relevant Equations

Definition of polynomial

According to this textbook: https://www.stitz-zeager.com/szprecalculus07042013.pdf

On Page $236$ it says:

But I can certainly write a combination of powers inside the absolute value: $|x+1|, |x^2+x+1|$...etc.

In fact I can put the definition of a polynomial inside the absolute value: $h(x)=|a_nx^n+a_{n-1}x^{n-1}+...a_2x^2+a_1x+a_0|$, then isn't $h(x)$ the absolute value written as a combination of powers?

 

[fresh_42]

Homework Statement: Why is $h(x)=|x|$ not a polynomial?
Relevant Equations: Definition of polynomial

According to this textbook: https://www.stitz-zeager.com/szprecalculus07042013.pdf

On Page $236$ it says:

View attachment 356812

But I can certainly write a combination of powers inside the absolute value: $|x+1|, |x^2+x+1|$...etc.

In fact I can put the definition of a polynomial inside the absolute value: $h(x)=|a_nx^n+a_{n-1}x^{n-1}+...a_2x^2+a_1x+a_0|$, then isn't $h(x)$ the absolute value written as a combination of powers?

If you argue like that, then you have two functions: a polynomial $p(x)$ and the absolute value function $\operatorname{abs}.$ Thus $h= \operatorname{abs}\circ p: \;h(x)=\operatorname{abs}(p(x))=|p(x)|.$

The combination isn't a polynomial. The absolute value function has a sharp bend where it is not differentiable. Polynomials are differentiable. They are defined as linear combinations of $x^n$ terms. You cannot express this sharp bend that $\operatorname{abs}$ has by terms of $x^n$.

 

[FactChecker]

You can't just ignore the absolute value as though it is not there.

 

[RChristenk]

If you argue like that, then you have two functions: a polynomial $p(x)$ and the absolute value function $\operatorname{abs}.$ Thus $h= \operatorname{abs}\circ p: \;h(x)=\operatorname{abs}(p(x))=|p(x)|.$

The combination isn't a polynomial. The absolute value function has a sharp bend where it is not differentiable. Polynomials are differentiable. They are defined as linear combinations of $x^n$ terms. You cannot express this sharp bend that $\operatorname{abs}$ has by terms of $x^n$.

I see. But what does it mean when the text says the absolute value "is not a combination of the powers of $x$"? Isn't $|x^2+x+1|$ for example a combination of the powers of $x$?

 

[fresh_42]

I see. But what does it mean when the text says the absolute value "is not a combination of the powers of $x$"? Isn't $|x^2+x+1|$ for example a combination of the powers of $x$?

$x^2+x+1$ is a combination of powers of $x$, $|x^2+x+1|$ is not because it has a second function "around" it, and that is no such combination.

is no polynomial.

 

[fresh_42]

I think I see the problem. You have chosen a function for which $|x^2+x+1|=x^2+x+1$ so it happens to be the same. But that is not generally the case. The absolute value is not part of the allowed tools to build a polynomial.

We define polynomials not only for the real numbers but also for domains that do not have an absolute value.

 

[RChristenk]

I think I see the problem. You have chosen a function for which $|x^2+x+1|=x^2+x+1$ so it happens to be the same. But that is not generally the case. The absolute value is not part of the allowed tools to build a polynomial.

I'm thinking if a polynomial is a person, then when that person puts on clothes (the absolute value), that person is still a person. Hence I couldn't understand why the text was saying absolute values can't be written as a combination of $x$. But you have explained clearly that by definition the absolute value is not a polynomial, so now I understand better. Thanks.

 

[fresh_42]

I'm thinking if a polynomial is a person, then when that person puts on clothes (the absolute value), that person is still a person. Hence I couldn't understand why the text was saying absolute values can't be written as a combination of $x$. But you have explained clearly that by definition the absolute value is not a polynomial, so now I understand better. Thanks.

A polynomial is not really a function. It is rather a sequence of its coefficients. E.g. $x^2+2x+3$ is the sequence $(3,2,1,0,0,0,\ldots).$ You can use it as a function by evaluating the polynomial at a certain location $x=a$ and get $a^2+2a+3$ and therewith a function. But the polynomial itself is just a sequence of numbers, the sequence of coefficients in front of each $x^n.$ But this gets us deeper into mathematics. For now, consider the polynomial as a function and the absolute value as another function. Both combined is no longer a polynomial because it cannot be written as such a sequence.

 

[pines-demon]

There has to be some condition of finiteness missing here $|x|$ is a polynmial in the sense that it can be written as a linear combination of powers of $x$, just not a finite sum:
$$|x|=\sqrt{1-(1-x^2)}=\sum_{m=0}^\infty \binom{1/2}{m} \left(1-x^2\right)^m\quad;\quad |x|\leq \sqrt{2}$$

 

[martinbn]

I'm thinking if a polynomial is a person, then when that person puts on clothes (the absolute value), that person is still a person.

May be you shouldn't think about polynomials this way.

Hence I couldn't understand why the text was saying absolute values can't be written as a combination of $x$.

You can try and see why it cannot. Say you have
$|x| = a_nx^n+a_{n-1}x^{n-1}+\cdots + a_0$
Then for positive $x$ you have
$x = a_nx^n+a_{n-1}x^{n-1}+\cdots + a_0$
Which means that all the $a_i$ are zero exept $a_1=1$.

For negative $x$ you have
$-x = a_nx^n+a_{n-1}x^{n-1}+\cdots + a_0$
Which means that all the $a_i$ are zero exept $a_1=-1$.

Clearly this is not possible.

 

[PeroK]

Homework Statement: Why is $h(x)=|x|$ not a polynomial?
Relevant Equations: Definition of polynomial

According to this textbook: https://www.stitz-zeager.com/szprecalculus07042013.pdf

On Page $236$ it says:

View attachment 356812

But I can certainly write a combination of powers inside the absolute value: $|x+1|, |x^2+x+1|$...etc.

In fact I can put the definition of a polynomial inside the absolute value: $h(x)=|a_nx^n+a_{n-1}x^{n-1}+...a_2x^2+a_1x+a_0|$, then isn't $h(x)$ the absolute value written as a combination of powers?

By your reasoning the number $1$ should classed as a negative number, because $1 = |-1|$.

 

[FactChecker]

Just because a part of the function is a polynomial does not mean the the entire thing is a polynomial. After all, $x$ is a polynomial so that would make all functions of $x$, like $|x|$, $\sin(x)$, $1/x$, etc. polynomials. That is clearly not true.

 

[vela]

I'm thinking if a polynomial is a person, then when that person puts on clothes (the absolute value), that person is still a person. Hence I couldn't understand why the text was saying absolute values can't be written as a combination of $x$. But you have explained clearly that by definition the absolute value is not a polynomial, so now I understand better. Thanks.

Definitions in math are very precise. It takes a little getting used to learning how to read them because in everyday language, we tend not to be painstakingly precise but instead rely on common sense when talking to others. So when you point to the person wearing clothes and ask "is that a person?" most people would ignore the clothing and answer yes. A mathematician, on the other hand, would answer no and explain that you're pointing to not just a person but clothes, which are not part of the person.