Sum and product of all linear combinations
Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$
I want to show that the set $mathcal{A}$ of all linear combinations of such functions is a $mathbb{R}$-subalgebra of $C(K)$, but I'm struggling specifically with proving that the sum and the product of elements of $mathcal{A}$ belongs to $mathcal{A}$ cause I've very much confused with manipulating the indexes.
$textbf{My attempt:}$
Consider the set
$$mathcal{A} = Big{ sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}: x_i, alpha_i in mathbb{R}, k_i in mathbb{Z}_+ Big}$$
1. $p, q in mathcal{A} Rightarrow pq in mathcal{A}$
Take $p, q in mathcal{A}$ such that $p = sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}$ and $q = sum_{j=1}^{N}tilde{alpha}_jprod_{j=1}^{N} x_j^{tilde{k}_j}$, then it follows:
begin{align*}
p(x)q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{2N}c_l prod_{i=1}^{l}x_i^{k_i}, \
&= (pq)(x)
end{align*}
with $c_l = sum_{m=1}^{l}alpha_mtilde{alpha}_{l-m}$.
- $p, q in mathcal{A} Rightarrow p+q in mathcal{A}$
Take the same $p, q in mathcal{A}$ written above, then
begin{align*}
p(x) + q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) + Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{N} (alpha_l + tilde{alpha}_l)prod_{l=1}^{N}x_l^{k_l} \
&= (p + q)(x)
end{align*}
real-analysis linear-algebra polynomials
asked Nov 30 at 19:47
user71487
948
add a comment |
Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$
I want to show that the set $mathcal{A}$ of all linear combinations of such functions is a $mathbb{R}$-subalgebra of $C(K)$, but I'm struggling specifically with proving that the sum and the product of elements of $mathcal{A}$ belongs to $mathcal{A}$ cause I've very much confused with manipulating the indexes.
$textbf{My attempt:}$
Consider the set
$$mathcal{A} = Big{ sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}: x_i, alpha_i in mathbb{R}, k_i in mathbb{Z}_+ Big}$$
1. $p, q in mathcal{A} Rightarrow pq in mathcal{A}$
Take $p, q in mathcal{A}$ such that $p = sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}$ and $q = sum_{j=1}^{N}tilde{alpha}_jprod_{j=1}^{N} x_j^{tilde{k}_j}$, then it follows:
begin{align*}
p(x)q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{2N}c_l prod_{i=1}^{l}x_i^{k_i}, \
&= (pq)(x)
end{align*}
with $c_l = sum_{m=1}^{l}alpha_mtilde{alpha}_{l-m}$.
- $p, q in mathcal{A} Rightarrow p+q in mathcal{A}$
Take the same $p, q in mathcal{A}$ written above, then
begin{align*}
p(x) + q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) + Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{N} (alpha_l + tilde{alpha}_l)prod_{l=1}^{N}x_l^{k_l} \
&= (p + q)(x)
end{align*}
real-analysis linear-algebra polynomials
asked Nov 30 at 19:47
user71487
948
add a comment |
Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$
I want to show that the set $mathcal{A}$ of all linear combinations of such functions is a $mathbb{R}$-subalgebra of $C(K)$, but I'm struggling specifically with proving that the sum and the product of elements of $mathcal{A}$ belongs to $mathcal{A}$ cause I've very much confused with manipulating the indexes.
$textbf{My attempt:}$
Consider the set
$$mathcal{A} = Big{ sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}: x_i, alpha_i in mathbb{R}, k_i in mathbb{Z}_+ Big}$$
1. $p, q in mathcal{A} Rightarrow pq in mathcal{A}$
Take $p, q in mathcal{A}$ such that $p = sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}$ and $q = sum_{j=1}^{N}tilde{alpha}_jprod_{j=1}^{N} x_j^{tilde{k}_j}$, then it follows:
begin{align*}
p(x)q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{2N}c_l prod_{i=1}^{l}x_i^{k_i}, \
&= (pq)(x)
end{align*}
with $c_l = sum_{m=1}^{l}alpha_mtilde{alpha}_{l-m}$.
- $p, q in mathcal{A} Rightarrow p+q in mathcal{A}$
Take the same $p, q in mathcal{A}$ written above, then
begin{align*}
p(x) + q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) + Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{N} (alpha_l + tilde{alpha}_l)prod_{l=1}^{N}x_l^{k_l} \
&= (p + q)(x)
end{align*}
real-analysis linear-algebra polynomials
asked Nov 30 at 19:47
user71487
948
Let $K$ be a compact non-empty subset of $mathbb{R}^N$. If $x in mathbb{R}^N$ and if $(k_1, k_2, dots, k_N), k_i in mathbb{Z}_+$, consider the function
$$x mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$
I want to show that the set $mathcal{A}$ of all linear combinations of such functions is a $mathbb{R}$-subalgebra of $C(K)$, but I'm struggling specifically with proving that the sum and the product of elements of $mathcal{A}$ belongs to $mathcal{A}$ cause I've very much confused with manipulating the indexes.
$textbf{My attempt:}$
Consider the set
$$mathcal{A} = Big{ sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}: x_i, alpha_i in mathbb{R}, k_i in mathbb{Z}_+ Big}$$
1. $p, q in mathcal{A} Rightarrow pq in mathcal{A}$
Take $p, q in mathcal{A}$ such that $p = sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i}$ and $q = sum_{j=1}^{N}tilde{alpha}_jprod_{j=1}^{N} x_j^{tilde{k}_j}$, then it follows:
begin{align*}
p(x)q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{2N}c_l prod_{i=1}^{l}x_i^{k_i}, \
&= (pq)(x)
end{align*}
with $c_l = sum_{m=1}^{l}alpha_mtilde{alpha}_{l-m}$.
- $p, q in mathcal{A} Rightarrow p+q in mathcal{A}$
Take the same $p, q in mathcal{A}$ written above, then
begin{align*}
p(x) + q(x) &= Big(sum_{i=1}^{N}alpha_iprod_{i=1}^{N} x_i^{k_i} Big) + Big(sum_{j=1}^{N}tilde{alpha_j}prod_{j=1}^{N} x_j^{tilde{k}_j} Big) \
&= sum_{l=1}^{N} (alpha_l + tilde{alpha}_l)prod_{l=1}^{N}x_l^{k_l} \
&= (p + q)(x)
end{align*}
real-analysis linear-algebra polynomials
real-analysis linear-algebra polynomials
asked Nov 30 at 19:47
user71487
948
asked Nov 30 at 19:47
user71487
948
edited Dec 5 at 19:03
asked Nov 30 at 19:47
user71487
948
asked Nov 30 at 19:47
user71487
948
asked Nov 30 at 19:47
user71487
948
948
add a comment |
add a comment |
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