Math Problem Statement

For p \in P(\mathbb{R}), define Tp:\mathbb{R} \rightarrow \mathbb{R} by (Tp)(x) = \begin{cases} \frac{p(x) - p(3)}{x-3} & if x \neq 3 \ p'(3) & if x = 3 \end{cases} for each x \in \mathbb{R}. Show that Tp \in P(\mathbb{R}) for every polynomial p \in P(\mathbb{R}) and show that T:P(\mathbb{R}) \rightarrow P(\mathbb{R}) is a linear map.

Solution

To solve this problem, we need to demonstrate two key properties:

  1. That TpP(R)T_p \in P(\mathbb{R}) for every polynomial pP(R)p \in P(\mathbb{R}): This means Tp(x)T_p(x), as defined, is itself a polynomial.
  2. That T:P(R)P(R)T:P(\mathbb{R}) \rightarrow P(\mathbb{R}) is a linear map: This requires showing that TT satisfies linearity.

Step 1: TpP(R)T_p \in P(\mathbb{R})

Case 1: When x3x \neq 3

The definition of Tp(x)T_p(x) for x3x \neq 3 is: Tp(x)=p(x)p(3)x3.T_p(x) = \frac{p(x) - p(3)}{x - 3}. For p(x)P(R)p(x) \in P(\mathbb{R}), p(x)p(x) is a polynomial of degree nn. Subtracting p(3)p(3), a constant, does not change the degree of the polynomial. Dividing by (x3)(x - 3) reduces the degree by 1. Hence, p(x)p(3)x3\frac{p(x) - p(3)}{x - 3} is a polynomial of degree n1n - 1.

Case 2: When x=3x = 3

For x=3x = 3, Tp(3)=p(3)T_p(3) = p'(3). The derivative p(x)p'(x) of a polynomial p(x)p(x) is itself a polynomial (of degree n1n - 1). Therefore, Tp(3)=p(3)T_p(3) = p'(3) is well-defined as a specific value of this polynomial.

Combining the two cases

For all xRx \in \mathbb{R}, Tp(x)T_p(x) is defined piecewise. At x=3x = 3, Tp(x)T_p(x) matches the value of the polynomial Tp(x)T_p(x) for x3x \neq 3 due to the continuity of polynomials and derivatives. Hence, Tp(x)T_p(x) is a polynomial of degree n1n - 1.

Thus, TpP(R)T_p \in P(\mathbb{R}).


Step 2: Linearity of TT

To show TT is a linear map, we need to verify two properties:

  1. T(p1+p2)=T(p1)+T(p2)T(p_1 + p_2) = T(p_1) + T(p_2) for all p1,p2P(R)p_1, p_2 \in P(\mathbb{R}),
  2. T(cp)=cT(p)T(cp) = cT(p) for all pP(R)p \in P(\mathbb{R}) and cRc \in \mathbb{R}.

Property 1: T(p1+p2)=T(p1)+T(p2)T(p_1 + p_2) = T(p_1) + T(p_2)

For x3x \neq 3, T(p1+p2)(x)=(p1+p2)(x)(p1+p2)(3)x3.T(p_1 + p_2)(x) = \frac{(p_1 + p_2)(x) - (p_1 + p_2)(3)}{x - 3}. Expanding the numerator: (p1+p2)(x)(p1+p2)(3)=(p1(x)p1(3))+(p2(x)p2(3)).(p_1 + p_2)(x) - (p_1 + p_2)(3) = \big(p_1(x) - p_1(3)\big) + \big(p_2(x) - p_2(3)\big). Thus, T(p1+p2)(x)=p1(x)p1(3)x3+p2(x)p2(3)x3=T(p1)(x)+T(p2)(x).T(p_1 + p_2)(x) = \frac{p_1(x) - p_1(3)}{x - 3} + \frac{p_2(x) - p_2(3)}{x - 3} = T(p_1)(x) + T(p_2)(x). At x=3x = 3, using the definition: T(p1+p2)(3)=(p1+p2)(3)=p1(3)+p2(3)=T(p1)(3)+T(p2)(3).T(p_1 + p_2)(3) = (p_1 + p_2)'(3) = p_1'(3) + p_2'(3) = T(p_1)(3) + T(p_2)(3). Thus, T(p1+p2)=T(p1)+T(p2)T(p_1 + p_2) = T(p_1) + T(p_2) for all xRx \in \mathbb{R}.

Property 2: T(cp)=cT(p)T(cp) = cT(p)

For x3x \neq 3, T(cp)(x)=(cp)(x)(cp)(3)x3.T(cp)(x) = \frac{(cp)(x) - (cp)(3)}{x - 3}. Since (cp)(x)=cp(x)(cp)(x) = c \cdot p(x), T(cp)(x)=cp(x)cp(3)x3=cp(x)p(3)x3=cT(p)(x).T(cp)(x) = \frac{c \cdot p(x) - c \cdot p(3)}{x - 3} = c \cdot \frac{p(x) - p(3)}{x - 3} = c \cdot T(p)(x). At x=3x = 3, using the definition: T(cp)(3)=(cp)(3)=cp(3)=cT(p)(3).T(cp)(3) = (cp)'(3) = c \cdot p'(3) = c \cdot T(p)(3). Thus, T(cp)=cT(p)T(cp) = cT(p) for all xRx \in \mathbb{R}.


Conclusion

  1. TpP(R)T_p \in P(\mathbb{R}) for every polynomial pP(R)p \in P(\mathbb{R}).
  2. T:P(R)P(R)T:P(\mathbb{R}) \rightarrow P(\mathbb{R}) is a linear map.

Let me know if you'd like further clarifications or details on any part of the proof! Here are five follow-up questions:

  1. Why is the continuity of Tp(x)T_p(x) important at x=3x = 3?
  2. How does the degree reduction in Tp(x)T_p(x) affect the image of P(R)P(\mathbb{R}) under TT?
  3. Can TT be represented in terms of a matrix if P(R)P(\mathbb{R}) is finite-dimensional? How?
  4. What happens if we apply TT multiple times to the same polynomial p(x)p(x)?
  5. How does TT behave when applied to a constant polynomial?

Tip: Always verify that piecewise-defined functions are continuous to ensure smooth transitions across cases!

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Math Problem Analysis

Mathematical Concepts

Polynomial functions
Linear maps
Piecewise functions
Continuity and differentiability

Formulas

Tp(x) = \frac{p(x) - p(3)}{x - 3} \text{ for } x \neq 3
Tp(3) = p'(3)

Theorems

Linearity of differentiation
Degree reduction of polynomials

Suitable Grade Level

Undergraduate level