Given polynomials $P$ and $Q$ of degree less than equal to $n,$ is $|PQ|leq K |P| cdot |Q|?$
Given polynomials $P$ and $Q$ of degree less than equal to $n,$ is $|PQ|leq K |P| cdot |Q|?$ Note that here the norm is the infinite norm and therefore $$|P| =sup_{1leq kleq n}a_k$$
where $P=sum_{i=0}^{n}a_{i}x^{i}.$
I am guessing that $|PQ|leq (n+1)(2n+1)|P|cdot |Q|$, because each term can be controlled by $|P| cdot |Q|.$ However I am not sure whether this is true. Perhaps someone could give a hint.
real-analysis
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
add a comment |
Given polynomials $P$ and $Q$ of degree less than equal to $n,$ is $|PQ|leq K |P| cdot |Q|?$ Note that here the norm is the infinite norm and therefore $$|P| =sup_{1leq kleq n}a_k$$
where $P=sum_{i=0}^{n}a_{i}x^{i}.$
I am guessing that $|PQ|leq (n+1)(2n+1)|P|cdot |Q|$, because each term can be controlled by $|P| cdot |Q|.$ However I am not sure whether this is true. Perhaps someone could give a hint.
real-analysis
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
1
Do you mean $|P| = max_{0 le i le n} |a_i|$?
– Hans Engler
Nov 26 '18 at 19:27
add a comment |
Given polynomials $P$ and $Q$ of degree less than equal to $n,$ is $|PQ|leq K |P| cdot |Q|?$ Note that here the norm is the infinite norm and therefore $$|P| =sup_{1leq kleq n}a_k$$
where $P=sum_{i=0}^{n}a_{i}x^{i}.$
I am guessing that $|PQ|leq (n+1)(2n+1)|P|cdot |Q|$, because each term can be controlled by $|P| cdot |Q|.$ However I am not sure whether this is true. Perhaps someone could give a hint.
real-analysis
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
Given polynomials $P$ and $Q$ of degree less than equal to $n,$ is $|PQ|leq K |P| cdot |Q|?$ Note that here the norm is the infinite norm and therefore $$|P| =sup_{1leq kleq n}a_k$$
where $P=sum_{i=0}^{n}a_{i}x^{i}.$
I am guessing that $|PQ|leq (n+1)(2n+1)|P|cdot |Q|$, because each term can be controlled by $|P| cdot |Q|.$ However I am not sure whether this is true. Perhaps someone could give a hint.
real-analysis
real-analysis
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
edited Nov 26 '18 at 18:40
Henning Makholm
238k16303537
238k16303537
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
asked Nov 26 '18 at 18:38
Hello_World
3,82021630
3,82021630
1
Do you mean $|P| = max_{0 le i le n} |a_i|$?
– Hans Engler
Nov 26 '18 at 19:27
add a comment |
1
Do you mean $|P| = max_{0 le i le n} |a_i|$?
– Hans Engler
Nov 26 '18 at 19:27
1
1
Do you mean $|P| = max_{0 le i le n} |a_i|$?
– Hans Engler
Nov 26 '18 at 19:27
Do you mean $|P| = max_{0 le i le n} |a_i|$?
– Hans Engler
Nov 26 '18 at 19:27
add a comment |
1 Answer
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Let $P=sum_{i=0}^{n}a_{i}x^{i}, Q=sum_{i=0}^{n}b_{i}x^{i}$
$PQ=sum_{i=0}^{2n}(sum_{j=max{0, i-n}}^{min{i, n}}a_{j}cdot b_{i-j})x^{i}=sum_{i=0}^{2n}c_{i}x^{i}$
$|PQ| =sup_{0leq kleq 2n}|c_k|=max_{0leq kleq 2n}|sum_{j=max{0, k-n}}^{min{k, n}}a_{j}cdot b_{k-j}|\ leq(n+1)max_{0leq ileq n}|a_i|cdot max_{0leq jleq n}|b_j|\ =(n+1)|P|cdot|Q|$
The $(n+1)$ factor arises because the coefficient of $x^{n}$ in $PQ$ contains $n+1$ addends of the form $(a_icdot b_j)$, which is the maximum number of addends of this form for any $c_k$. It can be easily seen that the number of such addends increases progressively from $1$ to $n+1$ while calculating $c_0$ through $c_n$, and then decreases progressively to $1$. In other words, the "lengthiest" coefficent is $c_n$.
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
add a comment |
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Let $P=sum_{i=0}^{n}a_{i}x^{i}, Q=sum_{i=0}^{n}b_{i}x^{i}$
$PQ=sum_{i=0}^{2n}(sum_{j=max{0, i-n}}^{min{i, n}}a_{j}cdot b_{i-j})x^{i}=sum_{i=0}^{2n}c_{i}x^{i}$
$|PQ| =sup_{0leq kleq 2n}|c_k|=max_{0leq kleq 2n}|sum_{j=max{0, k-n}}^{min{k, n}}a_{j}cdot b_{k-j}|\ leq(n+1)max_{0leq ileq n}|a_i|cdot max_{0leq jleq n}|b_j|\ =(n+1)|P|cdot|Q|$
The $(n+1)$ factor arises because the coefficient of $x^{n}$ in $PQ$ contains $n+1$ addends of the form $(a_icdot b_j)$, which is the maximum number of addends of this form for any $c_k$. It can be easily seen that the number of such addends increases progressively from $1$ to $n+1$ while calculating $c_0$ through $c_n$, and then decreases progressively to $1$. In other words, the "lengthiest" coefficent is $c_n$.
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
add a comment |
Let $P=sum_{i=0}^{n}a_{i}x^{i}, Q=sum_{i=0}^{n}b_{i}x^{i}$
$PQ=sum_{i=0}^{2n}(sum_{j=max{0, i-n}}^{min{i, n}}a_{j}cdot b_{i-j})x^{i}=sum_{i=0}^{2n}c_{i}x^{i}$
$|PQ| =sup_{0leq kleq 2n}|c_k|=max_{0leq kleq 2n}|sum_{j=max{0, k-n}}^{min{k, n}}a_{j}cdot b_{k-j}|\ leq(n+1)max_{0leq ileq n}|a_i|cdot max_{0leq jleq n}|b_j|\ =(n+1)|P|cdot|Q|$
The $(n+1)$ factor arises because the coefficient of $x^{n}$ in $PQ$ contains $n+1$ addends of the form $(a_icdot b_j)$, which is the maximum number of addends of this form for any $c_k$. It can be easily seen that the number of such addends increases progressively from $1$ to $n+1$ while calculating $c_0$ through $c_n$, and then decreases progressively to $1$. In other words, the "lengthiest" coefficent is $c_n$.
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
add a comment |
Let $P=sum_{i=0}^{n}a_{i}x^{i}, Q=sum_{i=0}^{n}b_{i}x^{i}$
$PQ=sum_{i=0}^{2n}(sum_{j=max{0, i-n}}^{min{i, n}}a_{j}cdot b_{i-j})x^{i}=sum_{i=0}^{2n}c_{i}x^{i}$
$|PQ| =sup_{0leq kleq 2n}|c_k|=max_{0leq kleq 2n}|sum_{j=max{0, k-n}}^{min{k, n}}a_{j}cdot b_{k-j}|\ leq(n+1)max_{0leq ileq n}|a_i|cdot max_{0leq jleq n}|b_j|\ =(n+1)|P|cdot|Q|$
The $(n+1)$ factor arises because the coefficient of $x^{n}$ in $PQ$ contains $n+1$ addends of the form $(a_icdot b_j)$, which is the maximum number of addends of this form for any $c_k$. It can be easily seen that the number of such addends increases progressively from $1$ to $n+1$ while calculating $c_0$ through $c_n$, and then decreases progressively to $1$. In other words, the "lengthiest" coefficent is $c_n$.
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
Let $P=sum_{i=0}^{n}a_{i}x^{i}, Q=sum_{i=0}^{n}b_{i}x^{i}$
$PQ=sum_{i=0}^{2n}(sum_{j=max{0, i-n}}^{min{i, n}}a_{j}cdot b_{i-j})x^{i}=sum_{i=0}^{2n}c_{i}x^{i}$
$|PQ| =sup_{0leq kleq 2n}|c_k|=max_{0leq kleq 2n}|sum_{j=max{0, k-n}}^{min{k, n}}a_{j}cdot b_{k-j}|\ leq(n+1)max_{0leq ileq n}|a_i|cdot max_{0leq jleq n}|b_j|\ =(n+1)|P|cdot|Q|$
The $(n+1)$ factor arises because the coefficient of $x^{n}$ in $PQ$ contains $n+1$ addends of the form $(a_icdot b_j)$, which is the maximum number of addends of this form for any $c_k$. It can be easily seen that the number of such addends increases progressively from $1$ to $n+1$ while calculating $c_0$ through $c_n$, and then decreases progressively to $1$. In other words, the "lengthiest" coefficent is $c_n$.
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
edited Nov 26 '18 at 20:12
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
answered Nov 26 '18 at 20:04
Shubham Johri
3,868716
3,868716
add a comment |
add a comment |
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Do you mean $|P| = max_{0 le i le n} |a_i|$?
– Hans Engler
Nov 26 '18 at 19:27