Find (a) , (b) ||p||, (c) ||q||, and (d) d(p,q) for the polynomials in P2 using the inner product &lt;p,q≥a0b0+a1b1+a2b2. p(x)=1−3x+x2,q(x)=−x+2x2

We have given that. p(x)=1−3x+x2, q(x)=−x+2x2 a) The inner product of the polynomials will be: p(x)=1−3x+x2,q(x)=−x+2x2 &lt;p,q≥1×0+(−3)(−1)+1(2) &lt;p,q≥3+2 &lt;p,q≥5 b) ||p||=12+(−3)2+12 ||p||=1+9+1 ||p||=11 c) ||q||=(−1)2+22 ||q||=1+4 ||q||=5 d) d(p,q)=||p−q|| d(p,q)=||1−3x+x2−(−x+2x2)|| d(p,q)=||1−3x+x2+x−2x2|| d(p,q)=||1−2x−x2|| d(p,q)=12+(−2)2+(−1)2 d(p,q)=1+4+1 d(p,q)=6

Solved: "Find (a) , (b) ||p||, (c) ||q||, and..."

waigaK

Answered question

2021-09-04

Find (a) , (b) $| | p | |$ , (c) $| | q | |$ , and (d) $d (p, q)$ for the polynomials in $P_{2}$ using the inner product $< p, q \geq a_{0} b_{0} + a_{1} b_{1} + a_{2} b_{2}$ .
$p (x) = 1 - 3 x + x^{2}, q (x) = - x + 2 x^{2}$

Answer & Explanation

coffentw

Skilled2021-09-05Added 103 answers

We have given that.
$p (x) = 1 - 3 x + x^{2}, q (x) = - x + 2 x^{2}$
a) The inner product of the polynomials will be:
$p (x) = 1 - 3 x + x^{2}, q (x) = - x + 2 x^{2}$
$< p, q \geq 1 \times 0 + (- 3) (- 1) + 1 (2)$
$< p, q \geq 3 + 2$
$< p, q \geq 5$
b) $| | p | | = \sqrt{1^{2} + {(- 3)}^{2} + 1^{2}}$
$| | p | | = \sqrt{1 + 9 + 1}$
$| | p | | = \sqrt{11}$
c) $| | q | | = \sqrt{{(- 1)}^{2} + 2^{2}}$
$| | q | | = \sqrt{1 + 4}$
$| | q | | = \sqrt{5}$
d) $d (p, q) = | | p - q | |$
$d (p, q) = | | 1 - 3 x + x^{2} - (- x + 2 x^{2}) | |$
$d (p, q) = | | 1 - 3 x + x^{2} + x - 2 x^{2} | |$
$d (p, q) = | | 1 - 2 x - x^{2} | |$
$d (p, q) = \sqrt{1^{2} + {(- 2)}^{2} + {(- 1)}^{2}}$
$d (p, q) = \sqrt{1 + 4 + 1}$
$d (p, q) = \sqrt{6}$

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