Mixing
Proof of all polynomials with degree n can for a basis
$begingroup$
Prove that {$1, x, x^2, x^3, ... , x^n$} form a basis of $P_n$ ($n$ is a non-negative integer) , the space of the polynomials of degree $n$. Using a general proof.
linear-algebra polynomials
edited Jan 19 at 22:33
Peter
48.1k1139133
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
$endgroup$
add a comment |
$begingroup$
Prove that {$1, x, x^2, x^3, ... , x^n$} form a basis of $P_n$ ($n$ is a non-negative integer) , the space of the polynomials of degree $n$. Using a general proof.
linear-algebra polynomials
edited Jan 19 at 22:33
Peter
48.1k1139133
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
$endgroup$
1
$begingroup$
Hello and welcome to M.SE. In order to get more chances to get a helpful answer it is worth you showing your efforts with your problem. That helps us to clarify exactly the doubt you have.
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– Dog_69
Jan 19 at 22:30
1
$begingroup$
Polynomials of degree $n$ are not a vector space.
$endgroup$
– Bernard
Jan 19 at 22:46
add a comment |
$begingroup$
Prove that {$1, x, x^2, x^3, ... , x^n$} form a basis of $P_n$ ($n$ is a non-negative integer) , the space of the polynomials of degree $n$. Using a general proof.
linear-algebra polynomials
edited Jan 19 at 22:33
Peter
48.1k1139133
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
$endgroup$
Prove that {$1, x, x^2, x^3, ... , x^n$} form a basis of $P_n$ ($n$ is a non-negative integer) , the space of the polynomials of degree $n$. Using a general proof.
linear-algebra polynomials
linear-algebra polynomials
edited Jan 19 at 22:33
Peter
48.1k1139133
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
edited Jan 19 at 22:33
Peter
48.1k1139133
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
edited Jan 19 at 22:33
Peter
48.1k1139133
edited Jan 19 at 22:33
Peter
48.1k1139133
edited Jan 19 at 22:33
Peter
48.1k1139133
48.1k1139133
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
asked Jan 19 at 22:22
Devina ArtsDevina Arts
1
1
1
$begingroup$
Hello and welcome to M.SE. In order to get more chances to get a helpful answer it is worth you showing your efforts with your problem. That helps us to clarify exactly the doubt you have.
$endgroup$
– Dog_69
Jan 19 at 22:30
1
$begingroup$
Polynomials of degree $n$ are not a vector space.
$endgroup$
– Bernard
Jan 19 at 22:46
add a comment |
1
$begingroup$
Hello and welcome to M.SE. In order to get more chances to get a helpful answer it is worth you showing your efforts with your problem. That helps us to clarify exactly the doubt you have.
$endgroup$
– Dog_69
Jan 19 at 22:30
1
$begingroup$
Polynomials of degree $n$ are not a vector space.
$endgroup$
– Bernard
Jan 19 at 22:46
1
1
$begingroup$
Hello and welcome to M.SE. In order to get more chances to get a helpful answer it is worth you showing your efforts with your problem. That helps us to clarify exactly the doubt you have.
$endgroup$
– Dog_69
Jan 19 at 22:30
$begingroup$
Hello and welcome to M.SE. In order to get more chances to get a helpful answer it is worth you showing your efforts with your problem. That helps us to clarify exactly the doubt you have.
$endgroup$
– Dog_69
Jan 19 at 22:30
1
1
$begingroup$
Polynomials of degree $n$ are not a vector space.
$endgroup$
– Bernard
Jan 19 at 22:46
$begingroup$
Polynomials of degree $n$ are not a vector space.
$endgroup$
– Bernard
Jan 19 at 22:46
add a comment |
1 Answer
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oldest
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$begingroup$
You need to show two things, first that they are linearly independent and second that they generate the set of all polynomials.
I'll give you a hint for linear independence:
You want to show that if for all $x in mathbb{R} $
$$alpha_0 + alpha_1 x + alpha_2 x^2 + dots + a_nx^{n} = 0 tag{A}$$
then $alpha_0 = alpha_1 = dots = alpha_n = 0$.
To do that, note that the polynomial $(mathrm{A})$ is $0$ for all $x in mathbb{R}$. However the zero-polynomial
$$0 + 0 cdot x + 0 cdot x^2+ dots + 0 cdot x^n = 0 $$
is also $0$ for all $x in mathbb{R}$.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
You need to show two things, first that they are linearly independent and second that they generate the set of all polynomials.
I'll give you a hint for linear independence:
You want to show that if for all $x in mathbb{R} $
$$alpha_0 + alpha_1 x + alpha_2 x^2 + dots + a_nx^{n} = 0 tag{A}$$
then $alpha_0 = alpha_1 = dots = alpha_n = 0$.
To do that, note that the polynomial $(mathrm{A})$ is $0$ for all $x in mathbb{R}$. However the zero-polynomial
$$0 + 0 cdot x + 0 cdot x^2+ dots + 0 cdot x^n = 0 $$
is also $0$ for all $x in mathbb{R}$.
$endgroup$
add a comment |
$begingroup$
You need to show two things, first that they are linearly independent and second that they generate the set of all polynomials.
I'll give you a hint for linear independence:
You want to show that if for all $x in mathbb{R} $
$$alpha_0 + alpha_1 x + alpha_2 x^2 + dots + a_nx^{n} = 0 tag{A}$$
then $alpha_0 = alpha_1 = dots = alpha_n = 0$.
To do that, note that the polynomial $(mathrm{A})$ is $0$ for all $x in mathbb{R}$. However the zero-polynomial
$$0 + 0 cdot x + 0 cdot x^2+ dots + 0 cdot x^n = 0 $$
is also $0$ for all $x in mathbb{R}$.
$endgroup$
add a comment |
$begingroup$
You need to show two things, first that they are linearly independent and second that they generate the set of all polynomials.
I'll give you a hint for linear independence:
You want to show that if for all $x in mathbb{R} $
$$alpha_0 + alpha_1 x + alpha_2 x^2 + dots + a_nx^{n} = 0 tag{A}$$
then $alpha_0 = alpha_1 = dots = alpha_n = 0$.
To do that, note that the polynomial $(mathrm{A})$ is $0$ for all $x in mathbb{R}$. However the zero-polynomial
$$0 + 0 cdot x + 0 cdot x^2+ dots + 0 cdot x^n = 0 $$
is also $0$ for all $x in mathbb{R}$.
$endgroup$
You need to show two things, first that they are linearly independent and second that they generate the set of all polynomials.
I'll give you a hint for linear independence:
You want to show that if for all $x in mathbb{R} $
$$alpha_0 + alpha_1 x + alpha_2 x^2 + dots + a_nx^{n} = 0 tag{A}$$
then $alpha_0 = alpha_1 = dots = alpha_n = 0$.
To do that, note that the polynomial $(mathrm{A})$ is $0$ for all $x in mathbb{R}$. However the zero-polynomial
$$0 + 0 cdot x + 0 cdot x^2+ dots + 0 cdot x^n = 0 $$
is also $0$ for all $x in mathbb{R}$.
answered Jan 19 at 22:32
user635162
add a comment |
add a comment |
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$begingroup$
Hello and welcome to M.SE. In order to get more chances to get a helpful answer it is worth you showing your efforts with your problem. That helps us to clarify exactly the doubt you have.
$endgroup$
– Dog_69
Jan 19 at 22:30
1
$begingroup$
Polynomials of degree $n$ are not a vector space.
$endgroup$
– Bernard
Jan 19 at 22:46