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Determine whether S is a subspace of $P_n$, the vectorspace of all real polynomial
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Determine whether S is a subspace of $P_n$, the vector space of all real polynomials of degree $leq n-1 $ of this form:
beginequation
p = a_0 + a_1X+a_2X^2+ dots + a_n-1X^n-1
endequation
To determine if the set $S = left p in p_n(mathbbR) midforall alpha in mathbbR: p(alpha) = p(-alpha) right$ is a subspace, I need to check for these 3 things:
- let q, p $in S$, then $r = q+p in S$
- let $mathbbB in mathbbR$ and $p in S$, then $mathbbBp in S$
- $mathbf0 in S$
So for number 1, we let $r(a) = q(a) + p(a)$ and because $q(a) + p(a) = q(-a) + p(-a)$ then $r(a) = r(-a)$.
And for number 2, we let $w(a) = mathbbBp(a)$. and because $mathbbB$ it's just a scaler $w(-a) = mathbbBp(-a)$.
But how do I check for number 3, i.e showing that $mathbf0 in S$?
linear-algebra polynomials
edited Aug 6 at 17:52
zzuussee
1,535419
asked Aug 6 at 16:47
user32091
638
add a comment |Â
up vote
1
down vote
favorite
Determine whether S is a subspace of $P_n$, the vector space of all real polynomials of degree $leq n-1 $ of this form:
beginequation
p = a_0 + a_1X+a_2X^2+ dots + a_n-1X^n-1
endequation
To determine if the set $S = left p in p_n(mathbbR) midforall alpha in mathbbR: p(alpha) = p(-alpha) right$ is a subspace, I need to check for these 3 things:
- let q, p $in S$, then $r = q+p in S$
- let $mathbbB in mathbbR$ and $p in S$, then $mathbbBp in S$
- $mathbf0 in S$
So for number 1, we let $r(a) = q(a) + p(a)$ and because $q(a) + p(a) = q(-a) + p(-a)$ then $r(a) = r(-a)$.
And for number 2, we let $w(a) = mathbbBp(a)$. and because $mathbbB$ it's just a scaler $w(-a) = mathbbBp(-a)$.
But how do I check for number 3, i.e showing that $mathbf0 in S$?
linear-algebra polynomials
edited Aug 6 at 17:52
zzuussee
1,535419
asked Aug 6 at 16:47
user32091
638
Is the $0$ polynomial (whose value is always $0$) an even function?
– Ethan Bolker
Aug 6 at 16:49
If $p=0$, the zero polynomial, what is $p(a)$?
– David C. Ullrich
Aug 6 at 17:08
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Determine whether S is a subspace of $P_n$, the vector space of all real polynomials of degree $leq n-1 $ of this form:
beginequation
p = a_0 + a_1X+a_2X^2+ dots + a_n-1X^n-1
endequation
To determine if the set $S = left p in p_n(mathbbR) midforall alpha in mathbbR: p(alpha) = p(-alpha) right$ is a subspace, I need to check for these 3 things:
- let q, p $in S$, then $r = q+p in S$
- let $mathbbB in mathbbR$ and $p in S$, then $mathbbBp in S$
- $mathbf0 in S$
So for number 1, we let $r(a) = q(a) + p(a)$ and because $q(a) + p(a) = q(-a) + p(-a)$ then $r(a) = r(-a)$.
And for number 2, we let $w(a) = mathbbBp(a)$. and because $mathbbB$ it's just a scaler $w(-a) = mathbbBp(-a)$.
But how do I check for number 3, i.e showing that $mathbf0 in S$?
linear-algebra polynomials
edited Aug 6 at 17:52
zzuussee
1,535419
asked Aug 6 at 16:47
user32091
638
Determine whether S is a subspace of $P_n$, the vector space of all real polynomials of degree $leq n-1 $ of this form:
beginequation
p = a_0 + a_1X+a_2X^2+ dots + a_n-1X^n-1
endequation
To determine if the set $S = left p in p_n(mathbbR) midforall alpha in mathbbR: p(alpha) = p(-alpha) right$ is a subspace, I need to check for these 3 things:
- let q, p $in S$, then $r = q+p in S$
- let $mathbbB in mathbbR$ and $p in S$, then $mathbbBp in S$
- $mathbf0 in S$
So for number 1, we let $r(a) = q(a) + p(a)$ and because $q(a) + p(a) = q(-a) + p(-a)$ then $r(a) = r(-a)$.
And for number 2, we let $w(a) = mathbbBp(a)$. and because $mathbbB$ it's just a scaler $w(-a) = mathbbBp(-a)$.
But how do I check for number 3, i.e showing that $mathbf0 in S$?
linear-algebra polynomials
edited Aug 6 at 17:52
zzuussee
1,535419
asked Aug 6 at 16:47
user32091
638
edited Aug 6 at 17:52
zzuussee
1,535419
edited Aug 6 at 17:52
zzuussee
1,535419
edited Aug 6 at 17:52
zzuussee
1,535419
1,535419
asked Aug 6 at 16:47
user32091
638
asked Aug 6 at 16:47
user32091
638
asked Aug 6 at 16:47
user32091
638
638
Is the $0$ polynomial (whose value is always $0$) an even function?
– Ethan Bolker
Aug 6 at 16:49
If $p=0$, the zero polynomial, what is $p(a)$?
– David C. Ullrich
Aug 6 at 17:08
add a comment |Â
Is the $0$ polynomial (whose value is always $0$) an even function?
– Ethan Bolker
Aug 6 at 16:49
If $p=0$, the zero polynomial, what is $p(a)$?
– David C. Ullrich
Aug 6 at 17:08
Is the $0$ polynomial (whose value is always $0$) an even function?
– Ethan Bolker
Aug 6 at 16:49
Is the $0$ polynomial (whose value is always $0$) an even function?
– Ethan Bolker
Aug 6 at 16:49
If $p=0$, the zero polynomial, what is $p(a)$?
– David C. Ullrich
Aug 6 at 17:08
If $p=0$, the zero polynomial, what is $p(a)$?
– David C. Ullrich
Aug 6 at 17:08
add a comment |Â
3 Answers
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$mathbf0$ is the zero polynomial, the null vector of the space $p_n(mathbbR)$, given with $mathbf0(alpha)=sum_i=0^n0cdotalpha^i=0$ f.a. $alpha$ as it works as the identity for pointwise addition of functions. Thus, especially $mathbf0(alpha)=0=mathbf0(-alpha)$ f.a. $alphainmathbbR$. Followingly, $mathbf0in S$.
Your work for the first two closure conditions look perfectly fine.
EDIT: A function $f$, i.e. in your case a polynomial of degree $n$(maximal), s.t. $f(x)=f(-x)$ is called an even function. Besides prominent examples like sine, especially all constant functions are even. (Can you see why?)
answered Aug 6 at 17:24
zzuussee
1,535419
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up vote
0
down vote
Other people have clarified to you what you are missing. I am presenting something that may be helpful to you in the future. After you learn about linear independence or linear maps, you can come back here and read this answer again.
If you already have some knowledge about linear maps, then you can show that $S$ is an $mathbbR$-vector subspace of $P_n$ of dimension $leftlceildfracn2rightrceil$ by exhibiting a surjective linear map $varphi:P_ntomathbbR^leftlfloorfracn2rightrfloor$ and showing that $S=ker(varphi)$. Since the kernel of a linear map from a vector space $V$ to another vector space $W$ is a vector subspace, the claim follows (using the Rank-Nullity Theorem).
Now, I shall give you a good map $varphi$. Define
$$varphi(p):=big(p(+1)-p(-1),p(+2)-p(-2),p(+3)-p(-3),ldots,p(+m)-p(-m)big),,$$
where $m:=leftlfloorfracn2rightrfloor$. It is easy to see that $varphi$ is linear. It is also surjective since
$$varphi(q_j)=e_j,,$$
where $e_1,e_2,ldots,e_minmathbbR^m$ are the usual standard basis vectors, and
$$q_j(x):=fracprodlimits_iin[m]setminusj,left(x^2-i^2right)prodlimits_iin[m]setminusj,left(j^2-i^2right)in mathbbR[x]text for j=1,2,ldots,m,.$$
Here, $[m]:=1,2,ldots,m$.
The next task is to show that $S=ker(varphi)$. Clearly, by the definition of $S$, we see that $Ssubseteq ker(varphi)$. We need to prove that $ker(varphi)subseteq S$ as well. Let $p$ be an arbitrary element of $ker(varphi)$. Define
$$f_p(x):=p(+x)-p(-x)inmathbbR[x],.$$
Thus, the roots of $f_p(x)$ are $0,pm1,pm2,ldots,pm m$. Consequently, $f_p$ has at least $$2m+1=2leftlfloorfracn2rightrfloor+1geq 2left(fracn-12right)+1=n$$ roots. As the degree of $f_p$ is at most the degree of $p$, which is $n-1$, we conclude that $f_p$ is identically zero. That is, $p(+x)=p(-x)$ identically, whence $p(+alpha)=p(-alpha)$ for all $alphainmathbbR$, so that $pin S$.
Alternatively, just observe that $S$ is spanned by the $leftlceildfracm2rightrceil$ linearly independent polynomials $$1,x^2,x^4,ldots,x^2,leftlceilfracn2rightrceil-2,.$$ That is, every nonzero term in a polynomial $pin S$ must have an even degree.
answered Aug 6 at 18:08
Batominovski
23.4k22779
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I claim that
$S = p(x) in P_n(x) mid forall alpha in Bbb R, ; p(alpha) = p(-alpha) tag 1$
is precisely the set
$E = left p(x) = displaystyle sum_0^n - 1 p_i x_i in P_n(x) mid p_i = 0, i ; textodd right ; tag 2$
that is,
$S = E, tag 3$
which means the polynomials in $P_n(x)$ which satisfy $p(alpha) = p(-alpha)$ are precisely those in which occur only even powers of $x$; for the present purposes we count the zero polynomial $mathbf 0$ as having only even-degree terms, which makes sense insofar as $mathbf 0$ is the constant polynomial whose value is $0$ for every $alpha$:
$mathbf 0(alpha) = 0, forall alpha in Bbb R; tag 4$
furthermore,
$mathbf 0(alpha) = 0 = mathbf 0(-alpha), ; forall alpha in Bbb R; tag 5$
thus we see that
$mathbf 0 in S. tag 6$
We may prove the assertion (3) as follows: first off, it is easy to see that
$E subset S, tag 7$
since even degree terms are of the form $ax^2i$, $a in Bbb R$; we have
$aalpha^2i = a(alpha^2)^i = a((-alpha)^2)^i = a(-alpha)^2i; tag 8$
since any element of $E$ is the sum of such expressions, we see that (7) must bind.
Now suppose
$p(x) in S, tag 9$
and write
$p(x) = eta(x) + omega(x), tag10$
where $eta(x)$ consists of the even-degree monomials comprising $p(x)$, and $omega(x)$ the odd; then for any $alpha in Bbb R$ we have
$eta(alpha) + omega(alpha) = p(alpha) = p(-alpha) = eta(-alpha) + omega(-alpha) = eta(alpha) - omega(alpha), tag11$
whence
$2 omega(alpha) = 0 Longrightarrow omega(alpha) = 0; tag12$
since $omega(alpha) = 0$ for every $alpha in Bbb R$, we conclude that
$omega(x) = 0, tag13$
whence
$p(x) = eta(x) tag14$
has only terms of even degree; thus we see that
$S subset E, tag15$
so in fact
$S = E tag16$
as claimed. Now $S = E$ is clearly a subspace of $P_n(x)$: the sum of two polynomials with only even-degree monomials obviously satisfies the same criterion, as does any product of such a polynomial with a scalar; and $mathbf 0 in E$ as we have seen.
And thats the $alpha$ and $omega$ of it!
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$mathbf0$ is the zero polynomial, the null vector of the space $p_n(mathbbR)$, given with $mathbf0(alpha)=sum_i=0^n0cdotalpha^i=0$ f.a. $alpha$ as it works as the identity for pointwise addition of functions. Thus, especially $mathbf0(alpha)=0=mathbf0(-alpha)$ f.a. $alphainmathbbR$. Followingly, $mathbf0in S$.
Your work for the first two closure conditions look perfectly fine.
EDIT: A function $f$, i.e. in your case a polynomial of degree $n$(maximal), s.t. $f(x)=f(-x)$ is called an even function. Besides prominent examples like sine, especially all constant functions are even. (Can you see why?)
answered Aug 6 at 17:24
zzuussee
1,535419
add a comment |Â
up vote
0
down vote
$mathbf0$ is the zero polynomial, the null vector of the space $p_n(mathbbR)$, given with $mathbf0(alpha)=sum_i=0^n0cdotalpha^i=0$ f.a. $alpha$ as it works as the identity for pointwise addition of functions. Thus, especially $mathbf0(alpha)=0=mathbf0(-alpha)$ f.a. $alphainmathbbR$. Followingly, $mathbf0in S$.
Your work for the first two closure conditions look perfectly fine.
EDIT: A function $f$, i.e. in your case a polynomial of degree $n$(maximal), s.t. $f(x)=f(-x)$ is called an even function. Besides prominent examples like sine, especially all constant functions are even. (Can you see why?)
answered Aug 6 at 17:24
zzuussee
1,535419
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$mathbf0$ is the zero polynomial, the null vector of the space $p_n(mathbbR)$, given with $mathbf0(alpha)=sum_i=0^n0cdotalpha^i=0$ f.a. $alpha$ as it works as the identity for pointwise addition of functions. Thus, especially $mathbf0(alpha)=0=mathbf0(-alpha)$ f.a. $alphainmathbbR$. Followingly, $mathbf0in S$.
Your work for the first two closure conditions look perfectly fine.
EDIT: A function $f$, i.e. in your case a polynomial of degree $n$(maximal), s.t. $f(x)=f(-x)$ is called an even function. Besides prominent examples like sine, especially all constant functions are even. (Can you see why?)
answered Aug 6 at 17:24
zzuussee
1,535419
$mathbf0$ is the zero polynomial, the null vector of the space $p_n(mathbbR)$, given with $mathbf0(alpha)=sum_i=0^n0cdotalpha^i=0$ f.a. $alpha$ as it works as the identity for pointwise addition of functions. Thus, especially $mathbf0(alpha)=0=mathbf0(-alpha)$ f.a. $alphainmathbbR$. Followingly, $mathbf0in S$.
Your work for the first two closure conditions look perfectly fine.
EDIT: A function $f$, i.e. in your case a polynomial of degree $n$(maximal), s.t. $f(x)=f(-x)$ is called an even function. Besides prominent examples like sine, especially all constant functions are even. (Can you see why?)
answered Aug 6 at 17:24
zzuussee
1,535419
edited Aug 6 at 17:42
answered Aug 6 at 17:24
zzuussee
1,535419
answered Aug 6 at 17:24
zzuussee
1,535419
answered Aug 6 at 17:24
zzuussee
1,535419
1,535419
add a comment |Â
add a comment |Â
up vote
0
down vote
Other people have clarified to you what you are missing. I am presenting something that may be helpful to you in the future. After you learn about linear independence or linear maps, you can come back here and read this answer again.
If you already have some knowledge about linear maps, then you can show that $S$ is an $mathbbR$-vector subspace of $P_n$ of dimension $leftlceildfracn2rightrceil$ by exhibiting a surjective linear map $varphi:P_ntomathbbR^leftlfloorfracn2rightrfloor$ and showing that $S=ker(varphi)$. Since the kernel of a linear map from a vector space $V$ to another vector space $W$ is a vector subspace, the claim follows (using the Rank-Nullity Theorem).
Now, I shall give you a good map $varphi$. Define
$$varphi(p):=big(p(+1)-p(-1),p(+2)-p(-2),p(+3)-p(-3),ldots,p(+m)-p(-m)big),,$$
where $m:=leftlfloorfracn2rightrfloor$. It is easy to see that $varphi$ is linear. It is also surjective since
$$varphi(q_j)=e_j,,$$
where $e_1,e_2,ldots,e_minmathbbR^m$ are the usual standard basis vectors, and
$$q_j(x):=fracprodlimits_iin[m]setminusj,left(x^2-i^2right)prodlimits_iin[m]setminusj,left(j^2-i^2right)in mathbbR[x]text for j=1,2,ldots,m,.$$
Here, $[m]:=1,2,ldots,m$.
The next task is to show that $S=ker(varphi)$. Clearly, by the definition of $S$, we see that $Ssubseteq ker(varphi)$. We need to prove that $ker(varphi)subseteq S$ as well. Let $p$ be an arbitrary element of $ker(varphi)$. Define
$$f_p(x):=p(+x)-p(-x)inmathbbR[x],.$$
Thus, the roots of $f_p(x)$ are $0,pm1,pm2,ldots,pm m$. Consequently, $f_p$ has at least $$2m+1=2leftlfloorfracn2rightrfloor+1geq 2left(fracn-12right)+1=n$$ roots. As the degree of $f_p$ is at most the degree of $p$, which is $n-1$, we conclude that $f_p$ is identically zero. That is, $p(+x)=p(-x)$ identically, whence $p(+alpha)=p(-alpha)$ for all $alphainmathbbR$, so that $pin S$.
Alternatively, just observe that $S$ is spanned by the $leftlceildfracm2rightrceil$ linearly independent polynomials $$1,x^2,x^4,ldots,x^2,leftlceilfracn2rightrceil-2,.$$ That is, every nonzero term in a polynomial $pin S$ must have an even degree.
answered Aug 6 at 18:08
Batominovski
23.4k22779
add a comment |Â
up vote
0
down vote
Other people have clarified to you what you are missing. I am presenting something that may be helpful to you in the future. After you learn about linear independence or linear maps, you can come back here and read this answer again.
If you already have some knowledge about linear maps, then you can show that $S$ is an $mathbbR$-vector subspace of $P_n$ of dimension $leftlceildfracn2rightrceil$ by exhibiting a surjective linear map $varphi:P_ntomathbbR^leftlfloorfracn2rightrfloor$ and showing that $S=ker(varphi)$. Since the kernel of a linear map from a vector space $V$ to another vector space $W$ is a vector subspace, the claim follows (using the Rank-Nullity Theorem).
Now, I shall give you a good map $varphi$. Define
$$varphi(p):=big(p(+1)-p(-1),p(+2)-p(-2),p(+3)-p(-3),ldots,p(+m)-p(-m)big),,$$
where $m:=leftlfloorfracn2rightrfloor$. It is easy to see that $varphi$ is linear. It is also surjective since
$$varphi(q_j)=e_j,,$$
where $e_1,e_2,ldots,e_minmathbbR^m$ are the usual standard basis vectors, and
$$q_j(x):=fracprodlimits_iin[m]setminusj,left(x^2-i^2right)prodlimits_iin[m]setminusj,left(j^2-i^2right)in mathbbR[x]text for j=1,2,ldots,m,.$$
Here, $[m]:=1,2,ldots,m$.
The next task is to show that $S=ker(varphi)$. Clearly, by the definition of $S$, we see that $Ssubseteq ker(varphi)$. We need to prove that $ker(varphi)subseteq S$ as well. Let $p$ be an arbitrary element of $ker(varphi)$. Define
$$f_p(x):=p(+x)-p(-x)inmathbbR[x],.$$
Thus, the roots of $f_p(x)$ are $0,pm1,pm2,ldots,pm m$. Consequently, $f_p$ has at least $$2m+1=2leftlfloorfracn2rightrfloor+1geq 2left(fracn-12right)+1=n$$ roots. As the degree of $f_p$ is at most the degree of $p$, which is $n-1$, we conclude that $f_p$ is identically zero. That is, $p(+x)=p(-x)$ identically, whence $p(+alpha)=p(-alpha)$ for all $alphainmathbbR$, so that $pin S$.
Alternatively, just observe that $S$ is spanned by the $leftlceildfracm2rightrceil$ linearly independent polynomials $$1,x^2,x^4,ldots,x^2,leftlceilfracn2rightrceil-2,.$$ That is, every nonzero term in a polynomial $pin S$ must have an even degree.
answered Aug 6 at 18:08
Batominovski
23.4k22779
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Other people have clarified to you what you are missing. I am presenting something that may be helpful to you in the future. After you learn about linear independence or linear maps, you can come back here and read this answer again.
If you already have some knowledge about linear maps, then you can show that $S$ is an $mathbbR$-vector subspace of $P_n$ of dimension $leftlceildfracn2rightrceil$ by exhibiting a surjective linear map $varphi:P_ntomathbbR^leftlfloorfracn2rightrfloor$ and showing that $S=ker(varphi)$. Since the kernel of a linear map from a vector space $V$ to another vector space $W$ is a vector subspace, the claim follows (using the Rank-Nullity Theorem).
Now, I shall give you a good map $varphi$. Define
$$varphi(p):=big(p(+1)-p(-1),p(+2)-p(-2),p(+3)-p(-3),ldots,p(+m)-p(-m)big),,$$
where $m:=leftlfloorfracn2rightrfloor$. It is easy to see that $varphi$ is linear. It is also surjective since
$$varphi(q_j)=e_j,,$$
where $e_1,e_2,ldots,e_minmathbbR^m$ are the usual standard basis vectors, and
$$q_j(x):=fracprodlimits_iin[m]setminusj,left(x^2-i^2right)prodlimits_iin[m]setminusj,left(j^2-i^2right)in mathbbR[x]text for j=1,2,ldots,m,.$$
Here, $[m]:=1,2,ldots,m$.
The next task is to show that $S=ker(varphi)$. Clearly, by the definition of $S$, we see that $Ssubseteq ker(varphi)$. We need to prove that $ker(varphi)subseteq S$ as well. Let $p$ be an arbitrary element of $ker(varphi)$. Define
$$f_p(x):=p(+x)-p(-x)inmathbbR[x],.$$
Thus, the roots of $f_p(x)$ are $0,pm1,pm2,ldots,pm m$. Consequently, $f_p$ has at least $$2m+1=2leftlfloorfracn2rightrfloor+1geq 2left(fracn-12right)+1=n$$ roots. As the degree of $f_p$ is at most the degree of $p$, which is $n-1$, we conclude that $f_p$ is identically zero. That is, $p(+x)=p(-x)$ identically, whence $p(+alpha)=p(-alpha)$ for all $alphainmathbbR$, so that $pin S$.
Alternatively, just observe that $S$ is spanned by the $leftlceildfracm2rightrceil$ linearly independent polynomials $$1,x^2,x^4,ldots,x^2,leftlceilfracn2rightrceil-2,.$$ That is, every nonzero term in a polynomial $pin S$ must have an even degree.
answered Aug 6 at 18:08
Batominovski
23.4k22779
Other people have clarified to you what you are missing. I am presenting something that may be helpful to you in the future. After you learn about linear independence or linear maps, you can come back here and read this answer again.
If you already have some knowledge about linear maps, then you can show that $S$ is an $mathbbR$-vector subspace of $P_n$ of dimension $leftlceildfracn2rightrceil$ by exhibiting a surjective linear map $varphi:P_ntomathbbR^leftlfloorfracn2rightrfloor$ and showing that $S=ker(varphi)$. Since the kernel of a linear map from a vector space $V$ to another vector space $W$ is a vector subspace, the claim follows (using the Rank-Nullity Theorem).
Now, I shall give you a good map $varphi$. Define
$$varphi(p):=big(p(+1)-p(-1),p(+2)-p(-2),p(+3)-p(-3),ldots,p(+m)-p(-m)big),,$$
where $m:=leftlfloorfracn2rightrfloor$. It is easy to see that $varphi$ is linear. It is also surjective since
$$varphi(q_j)=e_j,,$$
where $e_1,e_2,ldots,e_minmathbbR^m$ are the usual standard basis vectors, and
$$q_j(x):=fracprodlimits_iin[m]setminusj,left(x^2-i^2right)prodlimits_iin[m]setminusj,left(j^2-i^2right)in mathbbR[x]text for j=1,2,ldots,m,.$$
Here, $[m]:=1,2,ldots,m$.
The next task is to show that $S=ker(varphi)$. Clearly, by the definition of $S$, we see that $Ssubseteq ker(varphi)$. We need to prove that $ker(varphi)subseteq S$ as well. Let $p$ be an arbitrary element of $ker(varphi)$. Define
$$f_p(x):=p(+x)-p(-x)inmathbbR[x],.$$
Thus, the roots of $f_p(x)$ are $0,pm1,pm2,ldots,pm m$. Consequently, $f_p$ has at least $$2m+1=2leftlfloorfracn2rightrfloor+1geq 2left(fracn-12right)+1=n$$ roots. As the degree of $f_p$ is at most the degree of $p$, which is $n-1$, we conclude that $f_p$ is identically zero. That is, $p(+x)=p(-x)$ identically, whence $p(+alpha)=p(-alpha)$ for all $alphainmathbbR$, so that $pin S$.
Alternatively, just observe that $S$ is spanned by the $leftlceildfracm2rightrceil$ linearly independent polynomials $$1,x^2,x^4,ldots,x^2,leftlceilfracn2rightrceil-2,.$$ That is, every nonzero term in a polynomial $pin S$ must have an even degree.
answered Aug 6 at 18:08
Batominovski
23.4k22779
answered Aug 6 at 18:08
Batominovski
23.4k22779
answered Aug 6 at 18:08
Batominovski
23.4k22779
answered Aug 6 at 18:08
Batominovski
23.4k22779
23.4k22779
add a comment |Â
add a comment |Â
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0
down vote
I claim that
$S = p(x) in P_n(x) mid forall alpha in Bbb R, ; p(alpha) = p(-alpha) tag 1$
is precisely the set
$E = left p(x) = displaystyle sum_0^n - 1 p_i x_i in P_n(x) mid p_i = 0, i ; textodd right ; tag 2$
that is,
$S = E, tag 3$
which means the polynomials in $P_n(x)$ which satisfy $p(alpha) = p(-alpha)$ are precisely those in which occur only even powers of $x$; for the present purposes we count the zero polynomial $mathbf 0$ as having only even-degree terms, which makes sense insofar as $mathbf 0$ is the constant polynomial whose value is $0$ for every $alpha$:
$mathbf 0(alpha) = 0, forall alpha in Bbb R; tag 4$
furthermore,
$mathbf 0(alpha) = 0 = mathbf 0(-alpha), ; forall alpha in Bbb R; tag 5$
thus we see that
$mathbf 0 in S. tag 6$
We may prove the assertion (3) as follows: first off, it is easy to see that
$E subset S, tag 7$
since even degree terms are of the form $ax^2i$, $a in Bbb R$; we have
$aalpha^2i = a(alpha^2)^i = a((-alpha)^2)^i = a(-alpha)^2i; tag 8$
since any element of $E$ is the sum of such expressions, we see that (7) must bind.
Now suppose
$p(x) in S, tag 9$
and write
$p(x) = eta(x) + omega(x), tag10$
where $eta(x)$ consists of the even-degree monomials comprising $p(x)$, and $omega(x)$ the odd; then for any $alpha in Bbb R$ we have
$eta(alpha) + omega(alpha) = p(alpha) = p(-alpha) = eta(-alpha) + omega(-alpha) = eta(alpha) - omega(alpha), tag11$
whence
$2 omega(alpha) = 0 Longrightarrow omega(alpha) = 0; tag12$
since $omega(alpha) = 0$ for every $alpha in Bbb R$, we conclude that
$omega(x) = 0, tag13$
whence
$p(x) = eta(x) tag14$
has only terms of even degree; thus we see that
$S subset E, tag15$
so in fact
$S = E tag16$
as claimed. Now $S = E$ is clearly a subspace of $P_n(x)$: the sum of two polynomials with only even-degree monomials obviously satisfies the same criterion, as does any product of such a polynomial with a scalar; and $mathbf 0 in E$ as we have seen.
And thats the $alpha$ and $omega$ of it!
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
add a comment |Â
up vote
0
down vote
I claim that
$S = p(x) in P_n(x) mid forall alpha in Bbb R, ; p(alpha) = p(-alpha) tag 1$
is precisely the set
$E = left p(x) = displaystyle sum_0^n - 1 p_i x_i in P_n(x) mid p_i = 0, i ; textodd right ; tag 2$
that is,
$S = E, tag 3$
which means the polynomials in $P_n(x)$ which satisfy $p(alpha) = p(-alpha)$ are precisely those in which occur only even powers of $x$; for the present purposes we count the zero polynomial $mathbf 0$ as having only even-degree terms, which makes sense insofar as $mathbf 0$ is the constant polynomial whose value is $0$ for every $alpha$:
$mathbf 0(alpha) = 0, forall alpha in Bbb R; tag 4$
furthermore,
$mathbf 0(alpha) = 0 = mathbf 0(-alpha), ; forall alpha in Bbb R; tag 5$
thus we see that
$mathbf 0 in S. tag 6$
We may prove the assertion (3) as follows: first off, it is easy to see that
$E subset S, tag 7$
since even degree terms are of the form $ax^2i$, $a in Bbb R$; we have
$aalpha^2i = a(alpha^2)^i = a((-alpha)^2)^i = a(-alpha)^2i; tag 8$
since any element of $E$ is the sum of such expressions, we see that (7) must bind.
Now suppose
$p(x) in S, tag 9$
and write
$p(x) = eta(x) + omega(x), tag10$
where $eta(x)$ consists of the even-degree monomials comprising $p(x)$, and $omega(x)$ the odd; then for any $alpha in Bbb R$ we have
$eta(alpha) + omega(alpha) = p(alpha) = p(-alpha) = eta(-alpha) + omega(-alpha) = eta(alpha) - omega(alpha), tag11$
whence
$2 omega(alpha) = 0 Longrightarrow omega(alpha) = 0; tag12$
since $omega(alpha) = 0$ for every $alpha in Bbb R$, we conclude that
$omega(x) = 0, tag13$
whence
$p(x) = eta(x) tag14$
has only terms of even degree; thus we see that
$S subset E, tag15$
so in fact
$S = E tag16$
as claimed. Now $S = E$ is clearly a subspace of $P_n(x)$: the sum of two polynomials with only even-degree monomials obviously satisfies the same criterion, as does any product of such a polynomial with a scalar; and $mathbf 0 in E$ as we have seen.
And thats the $alpha$ and $omega$ of it!
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I claim that
$S = p(x) in P_n(x) mid forall alpha in Bbb R, ; p(alpha) = p(-alpha) tag 1$
is precisely the set
$E = left p(x) = displaystyle sum_0^n - 1 p_i x_i in P_n(x) mid p_i = 0, i ; textodd right ; tag 2$
that is,
$S = E, tag 3$
which means the polynomials in $P_n(x)$ which satisfy $p(alpha) = p(-alpha)$ are precisely those in which occur only even powers of $x$; for the present purposes we count the zero polynomial $mathbf 0$ as having only even-degree terms, which makes sense insofar as $mathbf 0$ is the constant polynomial whose value is $0$ for every $alpha$:
$mathbf 0(alpha) = 0, forall alpha in Bbb R; tag 4$
furthermore,
$mathbf 0(alpha) = 0 = mathbf 0(-alpha), ; forall alpha in Bbb R; tag 5$
thus we see that
$mathbf 0 in S. tag 6$
We may prove the assertion (3) as follows: first off, it is easy to see that
$E subset S, tag 7$
since even degree terms are of the form $ax^2i$, $a in Bbb R$; we have
$aalpha^2i = a(alpha^2)^i = a((-alpha)^2)^i = a(-alpha)^2i; tag 8$
since any element of $E$ is the sum of such expressions, we see that (7) must bind.
Now suppose
$p(x) in S, tag 9$
and write
$p(x) = eta(x) + omega(x), tag10$
where $eta(x)$ consists of the even-degree monomials comprising $p(x)$, and $omega(x)$ the odd; then for any $alpha in Bbb R$ we have
$eta(alpha) + omega(alpha) = p(alpha) = p(-alpha) = eta(-alpha) + omega(-alpha) = eta(alpha) - omega(alpha), tag11$
whence
$2 omega(alpha) = 0 Longrightarrow omega(alpha) = 0; tag12$
since $omega(alpha) = 0$ for every $alpha in Bbb R$, we conclude that
$omega(x) = 0, tag13$
whence
$p(x) = eta(x) tag14$
has only terms of even degree; thus we see that
$S subset E, tag15$
so in fact
$S = E tag16$
as claimed. Now $S = E$ is clearly a subspace of $P_n(x)$: the sum of two polynomials with only even-degree monomials obviously satisfies the same criterion, as does any product of such a polynomial with a scalar; and $mathbf 0 in E$ as we have seen.
And thats the $alpha$ and $omega$ of it!
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
I claim that
$S = p(x) in P_n(x) mid forall alpha in Bbb R, ; p(alpha) = p(-alpha) tag 1$
is precisely the set
$E = left p(x) = displaystyle sum_0^n - 1 p_i x_i in P_n(x) mid p_i = 0, i ; textodd right ; tag 2$
that is,
$S = E, tag 3$
which means the polynomials in $P_n(x)$ which satisfy $p(alpha) = p(-alpha)$ are precisely those in which occur only even powers of $x$; for the present purposes we count the zero polynomial $mathbf 0$ as having only even-degree terms, which makes sense insofar as $mathbf 0$ is the constant polynomial whose value is $0$ for every $alpha$:
$mathbf 0(alpha) = 0, forall alpha in Bbb R; tag 4$
furthermore,
$mathbf 0(alpha) = 0 = mathbf 0(-alpha), ; forall alpha in Bbb R; tag 5$
thus we see that
$mathbf 0 in S. tag 6$
We may prove the assertion (3) as follows: first off, it is easy to see that
$E subset S, tag 7$
since even degree terms are of the form $ax^2i$, $a in Bbb R$; we have
$aalpha^2i = a(alpha^2)^i = a((-alpha)^2)^i = a(-alpha)^2i; tag 8$
since any element of $E$ is the sum of such expressions, we see that (7) must bind.
Now suppose
$p(x) in S, tag 9$
and write
$p(x) = eta(x) + omega(x), tag10$
where $eta(x)$ consists of the even-degree monomials comprising $p(x)$, and $omega(x)$ the odd; then for any $alpha in Bbb R$ we have
$eta(alpha) + omega(alpha) = p(alpha) = p(-alpha) = eta(-alpha) + omega(-alpha) = eta(alpha) - omega(alpha), tag11$
whence
$2 omega(alpha) = 0 Longrightarrow omega(alpha) = 0; tag12$
since $omega(alpha) = 0$ for every $alpha in Bbb R$, we conclude that
$omega(x) = 0, tag13$
whence
$p(x) = eta(x) tag14$
has only terms of even degree; thus we see that
$S subset E, tag15$
so in fact
$S = E tag16$
as claimed. Now $S = E$ is clearly a subspace of $P_n(x)$: the sum of two polynomials with only even-degree monomials obviously satisfies the same criterion, as does any product of such a polynomial with a scalar; and $mathbf 0 in E$ as we have seen.
And thats the $alpha$ and $omega$ of it!
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
edited Aug 7 at 16:25
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
answered Aug 7 at 1:40
Robert Lewis
37.2k22356
37.2k22356
add a comment |Â
add a comment |Â
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Is the $0$ polynomial (whose value is always $0$) an even function?
– Ethan Bolker
Aug 6 at 16:49
If $p=0$, the zero polynomial, what is $p(a)$?
– David C. Ullrich
Aug 6 at 17:08