Math Problem Statement

Let U = { p in p_4(R) : p’’(6)=0. Find a basis of U

Solution

Given U={pP4(R):p(6)=0}U = \{ p \in P_4(\mathbb{R}) : p''(6) = 0 \}, we need to find a basis for UU. Here's the step-by-step breakdown:

Step 1: General form of a polynomial in P4(R)P_4(\mathbb{R})

A polynomial in P4(R)P_4(\mathbb{R}) is of degree at most 4, so we can express a general element of P4(R)P_4(\mathbb{R}) as: p(x)=ax4+bx3+cx2+dx+ep(x) = ax^4 + bx^3 + cx^2 + dx + e where a,b,c,d,eRa, b, c, d, e \in \mathbb{R} are constants.

Step 2: Second derivative of p(x)p(x)

To impose the condition p(6)=0p''(6) = 0, we first compute the second derivative of p(x)p(x).

First derivative: p(x)=4ax3+3bx2+2cx+dp'(x) = 4ax^3 + 3bx^2 + 2cx + d

Second derivative: p(x)=12ax2+6bx+2cp''(x) = 12ax^2 + 6bx + 2c

Step 3: Apply the condition p(6)=0p''(6) = 0

Substitute x=6x = 6 into the second derivative and set it equal to 0: 12a(6)2+6b(6)+2c=012a(6)^2 + 6b(6) + 2c = 0 Simplifying: 432a+36b+2c=0432a + 36b + 2c = 0 Dividing the entire equation by 2: 216a+18b+c=0216a + 18b + c = 0 Thus, the condition that must hold for p(x)Up(x) \in U is: c=216a18bc = -216a - 18b

Step 4: Express p(x)p(x) in terms of free variables

From the condition above, we can express cc in terms of aa and bb. So, the polynomial p(x)p(x) becomes: p(x)=ax4+bx3+(216a18b)x2+dx+ep(x) = ax^4 + bx^3 + (-216a - 18b)x^2 + dx + e

Step 5: Construct a basis

The free variables are a,b,d,ea, b, d, e, so we can generate a basis by assigning one of these variables to 1 and the others to 0 in turn:

  1. Set a=1a = 1, b=0b = 0, d=0d = 0, e=0e = 0: p(x)=x4216x2p(x) = x^4 - 216x^2

  2. Set a=0a = 0, b=1b = 1, d=0d = 0, e=0e = 0: p(x)=x318x2p(x) = x^3 - 18x^2

  3. Set a=0a = 0, b=0b = 0, d=1d = 1, e=0e = 0: p(x)=xp(x) = x

  4. Set a=0a = 0, b=0b = 0, d=0d = 0, e=1e = 1: p(x)=1p(x) = 1

Step 6: Final basis

Thus, a basis for UU is: {x4216x2,x318x2,x,1}\{ x^4 - 216x^2, x^3 - 18x^2, x, 1 \}

Would you like more details on any part of the solution or have any questions?

Here are 5 related questions to consider:

  1. What is the dimension of the space P4(R)P_4(\mathbb{R})?
  2. How would the basis change if the condition was p(6)=0p'(6) = 0 instead of p(6)=0p''(6) = 0?
  3. What does it mean geometrically for a polynomial to satisfy p(6)=0p''(6) = 0?
  4. How would you find a basis if the degree of the polynomial was higher, say P5(R)P_5(\mathbb{R})?
  5. How can this method be extended to polynomials over other fields?

Tip: When solving for a subspace of polynomials, always focus on the conditions imposed on the derivatives or values to reduce the degrees of freedom.

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Math Problem Analysis

Mathematical Concepts

Polynomials
Linear Algebra
Basis of a Vector Space
Differentiation

Formulas

General polynomial of degree 4: p(x) = ax^4 + bx^3 + cx^2 + dx + e
First derivative: p'(x) = 4ax^3 + 3bx^2 + 2cx + d
Second derivative: p''(x) = 12ax^2 + 6bx + 2c
Condition for second derivative at x=6: 216a + 18b + c = 0

Theorems

Basis Theorem: A basis is a set of linearly independent vectors that span the space.

Suitable Grade Level

Undergraduate Mathematics