Question about the formal definition of a polynomial in relation to $sin(x)$ not being a polynomial
$begingroup$
This question has been asked before, but none of the answers seems to satisfy what I'm asking here ..
So, in class, we introduced set of polynomials as a set of sequences of real numbers, where only finitely many elements of a sequence are non zero.
Now, I get it that, for example:
$p(x) = x^2 + x + 50$ fits the definition. The sequence in definition is actually a sequence of coefficients of polynomial. So, in my example, $p = (50, 1, 2, 0, 0 ...)$, right ?
But how does $sin(x)$ not fit the definition ?
So my guess is that this set contains all the polynomials, but is larger than that. I suppose that condition that is missing is that polynomial is actually linear combination of elements of given sequence and power functions, in natural order $dots$
Is that correct, or am I wrong? I might have misunderstood the definition or $sin(x)$ is polynomial ?
polynomials definition
$endgroup$
|
show 5 more comments
$begingroup$
This question has been asked before, but none of the answers seems to satisfy what I'm asking here ..
So, in class, we introduced set of polynomials as a set of sequences of real numbers, where only finitely many elements of a sequence are non zero.
Now, I get it that, for example:
$p(x) = x^2 + x + 50$ fits the definition. The sequence in definition is actually a sequence of coefficients of polynomial. So, in my example, $p = (50, 1, 2, 0, 0 ...)$, right ?
But how does $sin(x)$ not fit the definition ?
So my guess is that this set contains all the polynomials, but is larger than that. I suppose that condition that is missing is that polynomial is actually linear combination of elements of given sequence and power functions, in natural order $dots$
Is that correct, or am I wrong? I might have misunderstood the definition or $sin(x)$ is polynomial ?
polynomials definition
$endgroup$
2
$begingroup$
ask yourself the question, can I write $sin(x)$ as a FINITE combination of powers of the indeterminate X and coefficients $a_i$, such that: $$ sin(x) = a_0 + a_1 X + dots + a_n X^n $$
$endgroup$
– Wesley Strik
Dec 17 '18 at 10:58
2
$begingroup$
The expansion of $sin x$ has infinitely many nonzero terms. The given definition says that a polynomial is a sequence with finitely many nonzero terms.
$endgroup$
– Sujit Bhattacharyya
Dec 17 '18 at 11:01
$begingroup$
I would recommend taking some real-analysis and specifically a section on sequences of functions, this really cleared things up for me to not see Taylor series as an "infinite polynomial".
$endgroup$
– Wesley Strik
Dec 17 '18 at 11:10
2
$begingroup$
Just in case you also find this confusing. Looks like the risk for that is low, but still... something that confuses many at first.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:14
3
$begingroup$
Also, sine cannot be a polynomial. After all, a polynomial has limits $pminfty$ when $xtopminfty$ (think about the leading term), but sine is bounded. The only bounded (real) polynomial functions are the constants.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:30
|
show 5 more comments
$begingroup$
This question has been asked before, but none of the answers seems to satisfy what I'm asking here ..
So, in class, we introduced set of polynomials as a set of sequences of real numbers, where only finitely many elements of a sequence are non zero.
Now, I get it that, for example:
$p(x) = x^2 + x + 50$ fits the definition. The sequence in definition is actually a sequence of coefficients of polynomial. So, in my example, $p = (50, 1, 2, 0, 0 ...)$, right ?
But how does $sin(x)$ not fit the definition ?
So my guess is that this set contains all the polynomials, but is larger than that. I suppose that condition that is missing is that polynomial is actually linear combination of elements of given sequence and power functions, in natural order $dots$
Is that correct, or am I wrong? I might have misunderstood the definition or $sin(x)$ is polynomial ?
polynomials definition
$endgroup$
This question has been asked before, but none of the answers seems to satisfy what I'm asking here ..
So, in class, we introduced set of polynomials as a set of sequences of real numbers, where only finitely many elements of a sequence are non zero.
Now, I get it that, for example:
$p(x) = x^2 + x + 50$ fits the definition. The sequence in definition is actually a sequence of coefficients of polynomial. So, in my example, $p = (50, 1, 2, 0, 0 ...)$, right ?
But how does $sin(x)$ not fit the definition ?
So my guess is that this set contains all the polynomials, but is larger than that. I suppose that condition that is missing is that polynomial is actually linear combination of elements of given sequence and power functions, in natural order $dots$
Is that correct, or am I wrong? I might have misunderstood the definition or $sin(x)$ is polynomial ?
polynomials definition
polynomials definition
edited Dec 20 '18 at 13:32
someone
asked Dec 17 '18 at 10:36
someonesomeone
1188
1188
2
$begingroup$
ask yourself the question, can I write $sin(x)$ as a FINITE combination of powers of the indeterminate X and coefficients $a_i$, such that: $$ sin(x) = a_0 + a_1 X + dots + a_n X^n $$
$endgroup$
– Wesley Strik
Dec 17 '18 at 10:58
2
$begingroup$
The expansion of $sin x$ has infinitely many nonzero terms. The given definition says that a polynomial is a sequence with finitely many nonzero terms.
$endgroup$
– Sujit Bhattacharyya
Dec 17 '18 at 11:01
$begingroup$
I would recommend taking some real-analysis and specifically a section on sequences of functions, this really cleared things up for me to not see Taylor series as an "infinite polynomial".
$endgroup$
– Wesley Strik
Dec 17 '18 at 11:10
2
$begingroup$
Just in case you also find this confusing. Looks like the risk for that is low, but still... something that confuses many at first.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:14
3
$begingroup$
Also, sine cannot be a polynomial. After all, a polynomial has limits $pminfty$ when $xtopminfty$ (think about the leading term), but sine is bounded. The only bounded (real) polynomial functions are the constants.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:30
|
show 5 more comments
2
$begingroup$
ask yourself the question, can I write $sin(x)$ as a FINITE combination of powers of the indeterminate X and coefficients $a_i$, such that: $$ sin(x) = a_0 + a_1 X + dots + a_n X^n $$
$endgroup$
– Wesley Strik
Dec 17 '18 at 10:58
2
$begingroup$
The expansion of $sin x$ has infinitely many nonzero terms. The given definition says that a polynomial is a sequence with finitely many nonzero terms.
$endgroup$
– Sujit Bhattacharyya
Dec 17 '18 at 11:01
$begingroup$
I would recommend taking some real-analysis and specifically a section on sequences of functions, this really cleared things up for me to not see Taylor series as an "infinite polynomial".
$endgroup$
– Wesley Strik
Dec 17 '18 at 11:10
2
$begingroup$
Just in case you also find this confusing. Looks like the risk for that is low, but still... something that confuses many at first.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:14
3
$begingroup$
Also, sine cannot be a polynomial. After all, a polynomial has limits $pminfty$ when $xtopminfty$ (think about the leading term), but sine is bounded. The only bounded (real) polynomial functions are the constants.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:30
2
2
$begingroup$
ask yourself the question, can I write $sin(x)$ as a FINITE combination of powers of the indeterminate X and coefficients $a_i$, such that: $$ sin(x) = a_0 + a_1 X + dots + a_n X^n $$
$endgroup$
– Wesley Strik
Dec 17 '18 at 10:58
$begingroup$
ask yourself the question, can I write $sin(x)$ as a FINITE combination of powers of the indeterminate X and coefficients $a_i$, such that: $$ sin(x) = a_0 + a_1 X + dots + a_n X^n $$
$endgroup$
– Wesley Strik
Dec 17 '18 at 10:58
2
2
$begingroup$
The expansion of $sin x$ has infinitely many nonzero terms. The given definition says that a polynomial is a sequence with finitely many nonzero terms.
$endgroup$
– Sujit Bhattacharyya
Dec 17 '18 at 11:01
$begingroup$
The expansion of $sin x$ has infinitely many nonzero terms. The given definition says that a polynomial is a sequence with finitely many nonzero terms.
$endgroup$
– Sujit Bhattacharyya
Dec 17 '18 at 11:01
$begingroup$
I would recommend taking some real-analysis and specifically a section on sequences of functions, this really cleared things up for me to not see Taylor series as an "infinite polynomial".
$endgroup$
– Wesley Strik
Dec 17 '18 at 11:10
$begingroup$
I would recommend taking some real-analysis and specifically a section on sequences of functions, this really cleared things up for me to not see Taylor series as an "infinite polynomial".
$endgroup$
– Wesley Strik
Dec 17 '18 at 11:10
2
2
$begingroup$
Just in case you also find this confusing. Looks like the risk for that is low, but still... something that confuses many at first.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:14
$begingroup$
Just in case you also find this confusing. Looks like the risk for that is low, but still... something that confuses many at first.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:14
3
3
$begingroup$
Also, sine cannot be a polynomial. After all, a polynomial has limits $pminfty$ when $xtopminfty$ (think about the leading term), but sine is bounded. The only bounded (real) polynomial functions are the constants.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:30
$begingroup$
Also, sine cannot be a polynomial. After all, a polynomial has limits $pminfty$ when $xtopminfty$ (think about the leading term), but sine is bounded. The only bounded (real) polynomial functions are the constants.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:30
|
show 5 more comments
2 Answers
2
active
oldest
votes
$begingroup$
We cannot write $sin(X)$ as a finite combination of the indeterminate $X$ and coefficients $a_i$. The problem is that you probably know that you can express
$$ sin(x) = sum_{} ^{infty} (-1)^{n+1} frac{x^{2n+1}}{(2n+1)!}$$
In terms of its Taylor series. This is however an infinite series, furthermore it is actually the limit as $n to infty$ of the partial sum of the sequence of polynomial functions. That's a mouthful. Basically we need some more tools to properly express $sin(x)$ as an "infinite polynomial". Notice that my quotes express that this is not the proper way to talk about it, the proper way to talk about Taylor series is in terms of limits of sequences of functions.
If you were to fit a polynomial of finite degree to $sin(x)$ we get that this polynomial must, like $sin(x)$ have infinitely many roots/zeros. We get a contradiction and therefore $sin(x)$ cannot ever be expressed in terms of a polynomial of finite degree.
Interestingly, just like with Taylor series being infinite we can define so-called formal power series, which are a generalisation of polynomials as you know them, because they encode infinitely long sequences:
https://en.wikipedia.org/wiki/Formal_power_series
Notice that we cannot substitute values into formal power series, or as my combinatorics professor calls them, fops.
Also see the definition on wikipedia
https://en.wikipedia.org/wiki/Polynomial#Definition
"That is, a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms."
$endgroup$
1
$begingroup$
This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ?
$endgroup$
– someone
Dec 17 '18 at 12:46
1
$begingroup$
I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring.
$endgroup$
– Wesley Strik
Dec 17 '18 at 12:54
1
$begingroup$
Well yes, so the definition in question is not really complete and precise, right ?
$endgroup$
– someone
Dec 17 '18 at 13:01
1
$begingroup$
Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ lim_ {n to infty } sum_{i=0}^n a_i x^i= sum_{i=0}^infty a_i x^i$$
$endgroup$
– Wesley Strik
Dec 17 '18 at 13:03
1
$begingroup$
Oh, okay, thats the thing that was bothering me. Thanks!
$endgroup$
– someone
Dec 17 '18 at 13:08
|
show 4 more comments
$begingroup$
This is one of these things where context really matters.
Often the first concept of polynomial one deals with is in the case of real (or potentially complex) functions. A polynomial is a function $f : mathbb{R} to mathbb{R}$ that can be written in the form
$$f(x) equiv a_0 + a_1 x + ldots + a_n x^n,$$
for some $a_0, ldots, a_n in mathbb{R}$ and $n$ a non-negative integer.
It's a handy definition for proving a function is a polynomial, but not so handy for proving a function is not a polynomial. For example, $sin$ is not a polynomial, but it takes some proving to show that no choice of $n$ and $a_0, ldots, a_n$ will make the above expression identical to $sin(x)$. Fortunately, it's not too hard to show; you can examine limits to $infty$, count roots (polynomials have only finitely many), or take repeated derivatives (polynomials differentiate to $0$).
However, in certain contexts, defining polynomials as functions is not sufficient. For example, to construct $mathbb{F}_4$, the finite field of order $4$, you start with $mathbb{F}_2$, which is the ring of integers modulo $2$. You then consider the ring $mathbb{F}_2[x]$, the ring of "polynomials" whose coefficients are elements of $mathbb{F}_2$, and quotient out the maximal ideal generated by the irreducible polynomial $x^2 + x + 1$.
However, if you think about this (and are with me so far), this makes no sense if you think of polynomials as functions from $mathbb{F}_2$ to $mathbb{F}_2$. As a function from $mathbb{F}_2$ to $mathbb{F}_2$, the polynomial $f(x) = x^2 + x + 1$ is equivalent to the constant function $1$, in that it sends both elements of $mathbb{F}_2$, $0$ and $1$, to $1$. So, thinking about polynomials in this way, in this context, is unhelpful and impedes our way to some vital mathematics.
So, all of this is a long-winded way to say, it depends on context. You can think of them as a sequence that's eventually $0$. You can fit $sin$ in terms of this definition by considering it as a sequence of coefficients of its Maclaurin Series (which makes it not a polynomial).
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2 Answers
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2 Answers
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$begingroup$
We cannot write $sin(X)$ as a finite combination of the indeterminate $X$ and coefficients $a_i$. The problem is that you probably know that you can express
$$ sin(x) = sum_{} ^{infty} (-1)^{n+1} frac{x^{2n+1}}{(2n+1)!}$$
In terms of its Taylor series. This is however an infinite series, furthermore it is actually the limit as $n to infty$ of the partial sum of the sequence of polynomial functions. That's a mouthful. Basically we need some more tools to properly express $sin(x)$ as an "infinite polynomial". Notice that my quotes express that this is not the proper way to talk about it, the proper way to talk about Taylor series is in terms of limits of sequences of functions.
If you were to fit a polynomial of finite degree to $sin(x)$ we get that this polynomial must, like $sin(x)$ have infinitely many roots/zeros. We get a contradiction and therefore $sin(x)$ cannot ever be expressed in terms of a polynomial of finite degree.
Interestingly, just like with Taylor series being infinite we can define so-called formal power series, which are a generalisation of polynomials as you know them, because they encode infinitely long sequences:
https://en.wikipedia.org/wiki/Formal_power_series
Notice that we cannot substitute values into formal power series, or as my combinatorics professor calls them, fops.
Also see the definition on wikipedia
https://en.wikipedia.org/wiki/Polynomial#Definition
"That is, a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms."
$endgroup$
1
$begingroup$
This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ?
$endgroup$
– someone
Dec 17 '18 at 12:46
1
$begingroup$
I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring.
$endgroup$
– Wesley Strik
Dec 17 '18 at 12:54
1
$begingroup$
Well yes, so the definition in question is not really complete and precise, right ?
$endgroup$
– someone
Dec 17 '18 at 13:01
1
$begingroup$
Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ lim_ {n to infty } sum_{i=0}^n a_i x^i= sum_{i=0}^infty a_i x^i$$
$endgroup$
– Wesley Strik
Dec 17 '18 at 13:03
1
$begingroup$
Oh, okay, thats the thing that was bothering me. Thanks!
$endgroup$
– someone
Dec 17 '18 at 13:08
|
show 4 more comments
$begingroup$
We cannot write $sin(X)$ as a finite combination of the indeterminate $X$ and coefficients $a_i$. The problem is that you probably know that you can express
$$ sin(x) = sum_{} ^{infty} (-1)^{n+1} frac{x^{2n+1}}{(2n+1)!}$$
In terms of its Taylor series. This is however an infinite series, furthermore it is actually the limit as $n to infty$ of the partial sum of the sequence of polynomial functions. That's a mouthful. Basically we need some more tools to properly express $sin(x)$ as an "infinite polynomial". Notice that my quotes express that this is not the proper way to talk about it, the proper way to talk about Taylor series is in terms of limits of sequences of functions.
If you were to fit a polynomial of finite degree to $sin(x)$ we get that this polynomial must, like $sin(x)$ have infinitely many roots/zeros. We get a contradiction and therefore $sin(x)$ cannot ever be expressed in terms of a polynomial of finite degree.
Interestingly, just like with Taylor series being infinite we can define so-called formal power series, which are a generalisation of polynomials as you know them, because they encode infinitely long sequences:
https://en.wikipedia.org/wiki/Formal_power_series
Notice that we cannot substitute values into formal power series, or as my combinatorics professor calls them, fops.
Also see the definition on wikipedia
https://en.wikipedia.org/wiki/Polynomial#Definition
"That is, a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms."
$endgroup$
1
$begingroup$
This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ?
$endgroup$
– someone
Dec 17 '18 at 12:46
1
$begingroup$
I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring.
$endgroup$
– Wesley Strik
Dec 17 '18 at 12:54
1
$begingroup$
Well yes, so the definition in question is not really complete and precise, right ?
$endgroup$
– someone
Dec 17 '18 at 13:01
1
$begingroup$
Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ lim_ {n to infty } sum_{i=0}^n a_i x^i= sum_{i=0}^infty a_i x^i$$
$endgroup$
– Wesley Strik
Dec 17 '18 at 13:03
1
$begingroup$
Oh, okay, thats the thing that was bothering me. Thanks!
$endgroup$
– someone
Dec 17 '18 at 13:08
|
show 4 more comments
$begingroup$
We cannot write $sin(X)$ as a finite combination of the indeterminate $X$ and coefficients $a_i$. The problem is that you probably know that you can express
$$ sin(x) = sum_{} ^{infty} (-1)^{n+1} frac{x^{2n+1}}{(2n+1)!}$$
In terms of its Taylor series. This is however an infinite series, furthermore it is actually the limit as $n to infty$ of the partial sum of the sequence of polynomial functions. That's a mouthful. Basically we need some more tools to properly express $sin(x)$ as an "infinite polynomial". Notice that my quotes express that this is not the proper way to talk about it, the proper way to talk about Taylor series is in terms of limits of sequences of functions.
If you were to fit a polynomial of finite degree to $sin(x)$ we get that this polynomial must, like $sin(x)$ have infinitely many roots/zeros. We get a contradiction and therefore $sin(x)$ cannot ever be expressed in terms of a polynomial of finite degree.
Interestingly, just like with Taylor series being infinite we can define so-called formal power series, which are a generalisation of polynomials as you know them, because they encode infinitely long sequences:
https://en.wikipedia.org/wiki/Formal_power_series
Notice that we cannot substitute values into formal power series, or as my combinatorics professor calls them, fops.
Also see the definition on wikipedia
https://en.wikipedia.org/wiki/Polynomial#Definition
"That is, a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms."
$endgroup$
We cannot write $sin(X)$ as a finite combination of the indeterminate $X$ and coefficients $a_i$. The problem is that you probably know that you can express
$$ sin(x) = sum_{} ^{infty} (-1)^{n+1} frac{x^{2n+1}}{(2n+1)!}$$
In terms of its Taylor series. This is however an infinite series, furthermore it is actually the limit as $n to infty$ of the partial sum of the sequence of polynomial functions. That's a mouthful. Basically we need some more tools to properly express $sin(x)$ as an "infinite polynomial". Notice that my quotes express that this is not the proper way to talk about it, the proper way to talk about Taylor series is in terms of limits of sequences of functions.
If you were to fit a polynomial of finite degree to $sin(x)$ we get that this polynomial must, like $sin(x)$ have infinitely many roots/zeros. We get a contradiction and therefore $sin(x)$ cannot ever be expressed in terms of a polynomial of finite degree.
Interestingly, just like with Taylor series being infinite we can define so-called formal power series, which are a generalisation of polynomials as you know them, because they encode infinitely long sequences:
https://en.wikipedia.org/wiki/Formal_power_series
Notice that we cannot substitute values into formal power series, or as my combinatorics professor calls them, fops.
Also see the definition on wikipedia
https://en.wikipedia.org/wiki/Polynomial#Definition
"That is, a polynomial can either be zero or can be written as the sum of a finite number of non-zero terms."
edited Dec 17 '18 at 11:02
answered Dec 17 '18 at 10:50
Wesley StrikWesley Strik
1,665423
1,665423
1
$begingroup$
This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ?
$endgroup$
– someone
Dec 17 '18 at 12:46
1
$begingroup$
I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring.
$endgroup$
– Wesley Strik
Dec 17 '18 at 12:54
1
$begingroup$
Well yes, so the definition in question is not really complete and precise, right ?
$endgroup$
– someone
Dec 17 '18 at 13:01
1
$begingroup$
Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ lim_ {n to infty } sum_{i=0}^n a_i x^i= sum_{i=0}^infty a_i x^i$$
$endgroup$
– Wesley Strik
Dec 17 '18 at 13:03
1
$begingroup$
Oh, okay, thats the thing that was bothering me. Thanks!
$endgroup$
– someone
Dec 17 '18 at 13:08
|
show 4 more comments
1
$begingroup$
This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ?
$endgroup$
– someone
Dec 17 '18 at 12:46
1
$begingroup$
I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring.
$endgroup$
– Wesley Strik
Dec 17 '18 at 12:54
1
$begingroup$
Well yes, so the definition in question is not really complete and precise, right ?
$endgroup$
– someone
Dec 17 '18 at 13:01
1
$begingroup$
Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ lim_ {n to infty } sum_{i=0}^n a_i x^i= sum_{i=0}^infty a_i x^i$$
$endgroup$
– Wesley Strik
Dec 17 '18 at 13:03
1
$begingroup$
Oh, okay, thats the thing that was bothering me. Thanks!
$endgroup$
– someone
Dec 17 '18 at 13:08
1
1
$begingroup$
This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ?
$endgroup$
– someone
Dec 17 '18 at 12:46
$begingroup$
This one answers question precisely. Thanks .. Nice way to see it with infinitely many roots. But am I right about the additional condition for definition, though ? Of course, from definition one should intuitively get the idea that polynomial should be linear combination of a sequence with power functions, but person who never seen example of polynomial can't get a clear picture of one, right ? If definition is just a set of sequences alone, then many things are considered to be polynomial, right ?
$endgroup$
– someone
Dec 17 '18 at 12:46
1
1
$begingroup$
I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring.
$endgroup$
– Wesley Strik
Dec 17 '18 at 12:54
$begingroup$
I have only been taught that the coefficients of the polynomial of degree $n$ and a sequence of length $n$ of the form $(a_0, a_1 dots a_n )$ are isomorphic, so any of the two will do. We started out with the definition of a polynomial in $X$ over a field/ring $R$, where $a_i in R$. Then the elements of R[X] are represented by: $$P= a_0 +a_ 1 X dots a_n X^n$$ The structure where we consider all such expression we called a polynomial ring.
$endgroup$
– Wesley Strik
Dec 17 '18 at 12:54
1
1
$begingroup$
Well yes, so the definition in question is not really complete and precise, right ?
$endgroup$
– someone
Dec 17 '18 at 13:01
$begingroup$
Well yes, so the definition in question is not really complete and precise, right ?
$endgroup$
– someone
Dec 17 '18 at 13:01
1
1
$begingroup$
Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ lim_ {n to infty } sum_{i=0}^n a_i x^i= sum_{i=0}^infty a_i x^i$$
$endgroup$
– Wesley Strik
Dec 17 '18 at 13:03
$begingroup$
Well, yes. Your professor should really state something like "where any polynomial is a finite linear combination of power of the indeterminate and coefficients from the field/ring $R$" Notice that this sum runs up to $n$ and we are not saying something like : $$ lim_ {n to infty } sum_{i=0}^n a_i x^i= sum_{i=0}^infty a_i x^i$$
$endgroup$
– Wesley Strik
Dec 17 '18 at 13:03
1
1
$begingroup$
Oh, okay, thats the thing that was bothering me. Thanks!
$endgroup$
– someone
Dec 17 '18 at 13:08
$begingroup$
Oh, okay, thats the thing that was bothering me. Thanks!
$endgroup$
– someone
Dec 17 '18 at 13:08
|
show 4 more comments
$begingroup$
This is one of these things where context really matters.
Often the first concept of polynomial one deals with is in the case of real (or potentially complex) functions. A polynomial is a function $f : mathbb{R} to mathbb{R}$ that can be written in the form
$$f(x) equiv a_0 + a_1 x + ldots + a_n x^n,$$
for some $a_0, ldots, a_n in mathbb{R}$ and $n$ a non-negative integer.
It's a handy definition for proving a function is a polynomial, but not so handy for proving a function is not a polynomial. For example, $sin$ is not a polynomial, but it takes some proving to show that no choice of $n$ and $a_0, ldots, a_n$ will make the above expression identical to $sin(x)$. Fortunately, it's not too hard to show; you can examine limits to $infty$, count roots (polynomials have only finitely many), or take repeated derivatives (polynomials differentiate to $0$).
However, in certain contexts, defining polynomials as functions is not sufficient. For example, to construct $mathbb{F}_4$, the finite field of order $4$, you start with $mathbb{F}_2$, which is the ring of integers modulo $2$. You then consider the ring $mathbb{F}_2[x]$, the ring of "polynomials" whose coefficients are elements of $mathbb{F}_2$, and quotient out the maximal ideal generated by the irreducible polynomial $x^2 + x + 1$.
However, if you think about this (and are with me so far), this makes no sense if you think of polynomials as functions from $mathbb{F}_2$ to $mathbb{F}_2$. As a function from $mathbb{F}_2$ to $mathbb{F}_2$, the polynomial $f(x) = x^2 + x + 1$ is equivalent to the constant function $1$, in that it sends both elements of $mathbb{F}_2$, $0$ and $1$, to $1$. So, thinking about polynomials in this way, in this context, is unhelpful and impedes our way to some vital mathematics.
So, all of this is a long-winded way to say, it depends on context. You can think of them as a sequence that's eventually $0$. You can fit $sin$ in terms of this definition by considering it as a sequence of coefficients of its Maclaurin Series (which makes it not a polynomial).
$endgroup$
add a comment |
$begingroup$
This is one of these things where context really matters.
Often the first concept of polynomial one deals with is in the case of real (or potentially complex) functions. A polynomial is a function $f : mathbb{R} to mathbb{R}$ that can be written in the form
$$f(x) equiv a_0 + a_1 x + ldots + a_n x^n,$$
for some $a_0, ldots, a_n in mathbb{R}$ and $n$ a non-negative integer.
It's a handy definition for proving a function is a polynomial, but not so handy for proving a function is not a polynomial. For example, $sin$ is not a polynomial, but it takes some proving to show that no choice of $n$ and $a_0, ldots, a_n$ will make the above expression identical to $sin(x)$. Fortunately, it's not too hard to show; you can examine limits to $infty$, count roots (polynomials have only finitely many), or take repeated derivatives (polynomials differentiate to $0$).
However, in certain contexts, defining polynomials as functions is not sufficient. For example, to construct $mathbb{F}_4$, the finite field of order $4$, you start with $mathbb{F}_2$, which is the ring of integers modulo $2$. You then consider the ring $mathbb{F}_2[x]$, the ring of "polynomials" whose coefficients are elements of $mathbb{F}_2$, and quotient out the maximal ideal generated by the irreducible polynomial $x^2 + x + 1$.
However, if you think about this (and are with me so far), this makes no sense if you think of polynomials as functions from $mathbb{F}_2$ to $mathbb{F}_2$. As a function from $mathbb{F}_2$ to $mathbb{F}_2$, the polynomial $f(x) = x^2 + x + 1$ is equivalent to the constant function $1$, in that it sends both elements of $mathbb{F}_2$, $0$ and $1$, to $1$. So, thinking about polynomials in this way, in this context, is unhelpful and impedes our way to some vital mathematics.
So, all of this is a long-winded way to say, it depends on context. You can think of them as a sequence that's eventually $0$. You can fit $sin$ in terms of this definition by considering it as a sequence of coefficients of its Maclaurin Series (which makes it not a polynomial).
$endgroup$
add a comment |
$begingroup$
This is one of these things where context really matters.
Often the first concept of polynomial one deals with is in the case of real (or potentially complex) functions. A polynomial is a function $f : mathbb{R} to mathbb{R}$ that can be written in the form
$$f(x) equiv a_0 + a_1 x + ldots + a_n x^n,$$
for some $a_0, ldots, a_n in mathbb{R}$ and $n$ a non-negative integer.
It's a handy definition for proving a function is a polynomial, but not so handy for proving a function is not a polynomial. For example, $sin$ is not a polynomial, but it takes some proving to show that no choice of $n$ and $a_0, ldots, a_n$ will make the above expression identical to $sin(x)$. Fortunately, it's not too hard to show; you can examine limits to $infty$, count roots (polynomials have only finitely many), or take repeated derivatives (polynomials differentiate to $0$).
However, in certain contexts, defining polynomials as functions is not sufficient. For example, to construct $mathbb{F}_4$, the finite field of order $4$, you start with $mathbb{F}_2$, which is the ring of integers modulo $2$. You then consider the ring $mathbb{F}_2[x]$, the ring of "polynomials" whose coefficients are elements of $mathbb{F}_2$, and quotient out the maximal ideal generated by the irreducible polynomial $x^2 + x + 1$.
However, if you think about this (and are with me so far), this makes no sense if you think of polynomials as functions from $mathbb{F}_2$ to $mathbb{F}_2$. As a function from $mathbb{F}_2$ to $mathbb{F}_2$, the polynomial $f(x) = x^2 + x + 1$ is equivalent to the constant function $1$, in that it sends both elements of $mathbb{F}_2$, $0$ and $1$, to $1$. So, thinking about polynomials in this way, in this context, is unhelpful and impedes our way to some vital mathematics.
So, all of this is a long-winded way to say, it depends on context. You can think of them as a sequence that's eventually $0$. You can fit $sin$ in terms of this definition by considering it as a sequence of coefficients of its Maclaurin Series (which makes it not a polynomial).
$endgroup$
This is one of these things where context really matters.
Often the first concept of polynomial one deals with is in the case of real (or potentially complex) functions. A polynomial is a function $f : mathbb{R} to mathbb{R}$ that can be written in the form
$$f(x) equiv a_0 + a_1 x + ldots + a_n x^n,$$
for some $a_0, ldots, a_n in mathbb{R}$ and $n$ a non-negative integer.
It's a handy definition for proving a function is a polynomial, but not so handy for proving a function is not a polynomial. For example, $sin$ is not a polynomial, but it takes some proving to show that no choice of $n$ and $a_0, ldots, a_n$ will make the above expression identical to $sin(x)$. Fortunately, it's not too hard to show; you can examine limits to $infty$, count roots (polynomials have only finitely many), or take repeated derivatives (polynomials differentiate to $0$).
However, in certain contexts, defining polynomials as functions is not sufficient. For example, to construct $mathbb{F}_4$, the finite field of order $4$, you start with $mathbb{F}_2$, which is the ring of integers modulo $2$. You then consider the ring $mathbb{F}_2[x]$, the ring of "polynomials" whose coefficients are elements of $mathbb{F}_2$, and quotient out the maximal ideal generated by the irreducible polynomial $x^2 + x + 1$.
However, if you think about this (and are with me so far), this makes no sense if you think of polynomials as functions from $mathbb{F}_2$ to $mathbb{F}_2$. As a function from $mathbb{F}_2$ to $mathbb{F}_2$, the polynomial $f(x) = x^2 + x + 1$ is equivalent to the constant function $1$, in that it sends both elements of $mathbb{F}_2$, $0$ and $1$, to $1$. So, thinking about polynomials in this way, in this context, is unhelpful and impedes our way to some vital mathematics.
So, all of this is a long-winded way to say, it depends on context. You can think of them as a sequence that's eventually $0$. You can fit $sin$ in terms of this definition by considering it as a sequence of coefficients of its Maclaurin Series (which makes it not a polynomial).
edited Dec 18 '18 at 0:00
answered Dec 17 '18 at 11:03
Theo BenditTheo Bendit
17.3k12149
17.3k12149
add a comment |
add a comment |
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2
$begingroup$
ask yourself the question, can I write $sin(x)$ as a FINITE combination of powers of the indeterminate X and coefficients $a_i$, such that: $$ sin(x) = a_0 + a_1 X + dots + a_n X^n $$
$endgroup$
– Wesley Strik
Dec 17 '18 at 10:58
2
$begingroup$
The expansion of $sin x$ has infinitely many nonzero terms. The given definition says that a polynomial is a sequence with finitely many nonzero terms.
$endgroup$
– Sujit Bhattacharyya
Dec 17 '18 at 11:01
$begingroup$
I would recommend taking some real-analysis and specifically a section on sequences of functions, this really cleared things up for me to not see Taylor series as an "infinite polynomial".
$endgroup$
– Wesley Strik
Dec 17 '18 at 11:10
2
$begingroup$
Just in case you also find this confusing. Looks like the risk for that is low, but still... something that confuses many at first.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:14
3
$begingroup$
Also, sine cannot be a polynomial. After all, a polynomial has limits $pminfty$ when $xtopminfty$ (think about the leading term), but sine is bounded. The only bounded (real) polynomial functions are the constants.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 11:30