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Let B be the basis of P3 consisting of the Hermite polynomials 1, 2t, -2+4t^2, and -12t + 8t^3 and let p(t)= 2 - 4t^2 - 8t^3. Find the coordinate vector of p relative to B

[p]B = ?

User Seffix
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1 Answer

6 votes

polynomial coordinates

example

spline functions using hermite polynomials can help generate smooth curves without any sharp corners for the body panels in 3D car models

answer

[p]B = [2, 0, -4, -8]

steps

1. We have a set of 4 special polynomials called the Hermite polynomials. These 4 polynomials are:

- 1st polynomial: 1

- 2nd polynomial: 2t

- 3rd polynomial: -2 + 4t^2

- 4th polynomial: -12t + 8t^3

2. These 4 polynomials form a basis B for the vector space P3. This means any 3rd degree polynomial can be written as a combination of these 4 basis polynomials.

3. We are given one 3rd degree polynomial:

p(t) = 2 - 4t^2 - 8t^3

4. To write p(t) in terms of the basis B, we need to find numbers c1, c2, c3, c4 such that:

p(t) = c1(1st polynomial) + c2(2nd polynomial) + c3(3rd polynomial) + c4(4th polynomial)

5. By matching coefficients, we find:

c1 = 2

c2 = 0

c3 = -4

c4 = -8

6. Therefore, the coordinate vector of p(t) relative to the basis B is:

[2, 0, -4, -8]

So in very simple terms, we wrote the given polynomial p(t) as a combination of the 4 special Hermite polynomials, and the numbers [2, 0, -4, -8] tell us how much to use of each polynomial.

- The basis B of P3 consisting of the Hermite polynomials is:

B = {1, 2t, -2 + 4t^2, -12t + 8t^3}

- The polynomial p(t) is given as:

p(t) = 2 - 4t^2 - 8t^3

To find the coordinate vector of p(t) relative to the basis B, we express p(t) as a linear combination of the basis polynomials:

p(t) = c1(1) + c2(2t) + c3(-2 + 4t^2) + c4(-12t + 8t^3)

Matching coefficients, we get:

c1 = 2

c2 = 0

c3 = -4

c4 = -8

Therefore, the coordinate vector of p(t) relative to the basis B is:

\[p\]B = [2, 0, -4, -8]

So the final answer is:

\[p\]B = [2, 0, -4, -8]

claudeAI

User Shane Hou
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