Answer to Question #97155 in Linear Algebra for Nozipho

Question #97155
If V = P3 with the inner product < f, g >=
R 1
11
f(x)g(x)dx, apply the Gram-Schmidt algorithm
to obtain an orthogonal basis from B = {1, x, x2
, x3}
1
Expert's answer
2019-10-23T10:52:08-0400

Solution:

We need to find the Orthogonal Basis by applying Gram-Schmidt algorithm.

Given "\\beta ="{"{1, x, x^2, x^3}" } and < f, g > = "\\int_{-1}^ {1} f(x) \\space g(x) dx"


From this

"\\beta_1 = 1\\\\\n\\beta_2 = x \\\\\n\\beta_3 = x^2 \\\\\n\\beta_4 = x^3"

Next,

the Orthogonal vectors are

"\\alpha_1 = 1"


"||\\alpha_1||^2 = < \\alpha_1, \\alpha_1 > = <1, 1> = \\int _{-1}^ {1} 1 \\times 1 dx = [ x ]_{-1}^1 = 2"

"\\alpha_2 = \\beta_2 - \\frac {<\\beta_2, \\alpha_1>} {||\\alpha_1||^2 } \\space {\\alpha_1}"

"\\alpha_2 = x - \\frac {<x, 1>} {||1||^2 } \\space \\times {1}"

"\\alpha_2 = x - \\frac {\\int _{-1}^{1} x \\space dx} {\\int_{-1} ^{1}1 \\space dx } \\space \\times {1} = x - \\frac {0} {2} = x"


"(since \\int _{-1}^1 x \\space dx = 0 )"

Now,

"\\alpha_3 = \\beta_3 - \\frac {<\\beta_3, \\alpha_1>} {||\\alpha_1||^2 } \\times \\alpha_1 - \\frac {<\\beta_3, \\alpha_2>} {||\\alpha_2||^2} \\times \\alpha_2"

"\\alpha_3 = \\beta_3 - \\frac {<x^2, 1>} {||1||^2 } \\times 1 - \\frac {<x^3, x>} {||x||^2} \\times x"

"\\alpha_3 = \\beta_3 - \\frac {\\int_{-1}^1 x^2 \\space dx} {\\int _{-1}^1 1 \\space dx} \\times 1 - \\frac {\\int_{-1}^1 x^2 \\space \\times x \\space dx} {\\int_{-1}^1 x \\times x \\space dx } \\times x"

"\\alpha_3 = \\beta_3 - 2 \\times\\frac {2} {3} \\times 1 - 0 = x^2 - \\frac {1} {3}""\\alpha_4 = \\beta_4 - \\frac {<\\beta_4, \\alpha_1> } {||\\alpha_1||^2 } \\alpha_1 - \\frac {<\\beta_4, \\alpha_2> } {||\\alpha_2||^2 } \\alpha_2 - \\frac {<\\beta_4, \\alpha_3> } {||\\alpha_3||^2 } \\alpha_3"

"\\alpha_4 = x^3 - \\frac {< x^3, 1> } {||1||^2 } \\times1 - \\frac {<x^3, x> } {||x||^2 } x^2 - \\frac {<x^3, x^2 - \\frac {1}{3}> } {||x^2 - \\frac {1}{3}||^2 } \\times (x^2 - \\frac {1}{3})"


"= x^3 - \\frac {3}{5} x"



Answer: Orthogonal Basis = {"\\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4" } = {"1, x, x^2 - \\frac{1}{3}, x^3 - \\frac {3}{5} x" }


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS