Consider an inner product on two P3 vectors as (f, g) = f(-1)g(-1) + f(0)g(0) + f(1)g(1) + f(2)g(2): It is known that that basis B that contains (2-x-2x² + x³), (x) is an orthogonal basis for W under this inner product. Then we have: 1) If we know that 2 + 2x + (-2) x² + 1x³ is inside W, then if we wish to write 2 + 2x + (-2) x² + 1x³ = c₁(2-x-2x² + x³) + c₂(x), the value of c₂ would be? 2) The orthogoanl decomposition (under this inner product) of y = 6+(-3) x + (-9) x² + 12x³ to get y = y + z where y EW and Z EW¹ would be: y = ,Z = 3) The closest point in W from y = 6+(-3) x + (-9) x² + 12x³ under this inner product would be?
Consider an inner product on two P3 vectors as (f, g) = f(-1)g(-1) + f(0)g(0) + f(1)g(1) + f(2)g(2): It is known that that basis B that contains (2-x-2x² + x³), (x) is an orthogonal basis for W under this inner product. Then we have: 1) If we know that 2 + 2x + (-2) x² + 1x³ is inside W, then if we wish to write 2 + 2x + (-2) x² + 1x³ = c₁(2-x-2x² + x³) + c₂(x), the value of c₂ would be? 2) The orthogoanl decomposition (under this inner product) of y = 6+(-3) x + (-9) x² + 12x³ to get y = y + z where y EW and Z EW¹ would be: y = ,Z = 3) The closest point in W from y = 6+(-3) x + (-9) x² + 12x³ under this inner product would be?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Note: To use square roots, you can simply write "a^0.5".
Consider an inner product on two P3 vectors as (f, g) = f(−1)g(−1) + ƒ(0)g(0) + ƒ(1)g(1) + ƒ(2)g(2):
It is known that that basis B that contains (2-x-2x² + x³), (x) is an orthogonal basis for W under this inner product. Then we have:
1) If we know that 2 + 2x + (-2) x² + 1x³ is inside W, then if we wish to write 2 + 2x + (-2) x² + 1x³ = c₁(2-x-2x² + x³) + c₂(x), the value
of c₂ would be?
2) The orthogoanl decomposition (under this inner product) of y = 6 + (−3) x + (-9) x² + 12x³ to get y = y + z where y EW and ZE W would
be:
y=
Z=
3) The closest point in W from y = 6 + (-3) x + (-9) x² + 12x³ under this inner product would be?
4) The shortest distance to W from y = 6 + (-3) x + (-9) x² + 12x³ under this inner product would be?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdef25b02-21c7-4f0b-87c4-78c9bf8df921%2F9bb0f0fc-5771-49b1-b85e-245651c3b6e9%2Ffaxmdn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Note: To use square roots, you can simply write "a^0.5".
Consider an inner product on two P3 vectors as (f, g) = f(−1)g(−1) + ƒ(0)g(0) + ƒ(1)g(1) + ƒ(2)g(2):
It is known that that basis B that contains (2-x-2x² + x³), (x) is an orthogonal basis for W under this inner product. Then we have:
1) If we know that 2 + 2x + (-2) x² + 1x³ is inside W, then if we wish to write 2 + 2x + (-2) x² + 1x³ = c₁(2-x-2x² + x³) + c₂(x), the value
of c₂ would be?
2) The orthogoanl decomposition (under this inner product) of y = 6 + (−3) x + (-9) x² + 12x³ to get y = y + z where y EW and ZE W would
be:
y=
Z=
3) The closest point in W from y = 6 + (-3) x + (-9) x² + 12x³ under this inner product would be?
4) The shortest distance to W from y = 6 + (-3) x + (-9) x² + 12x³ under this inner product would be?
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