Problem 1790

### Part (a)
1. Definition of the Set: Consider the set $P$ of all polynomials of degree less than or equal to 4 with rational coefficients. This set can be written as:
$$P = \{ a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 \mid a_i \in \mathbb{Q}, \, 0 \leq i \leq 4 \}$$

2. Vector Space Axioms: To prove that $P$ has a vector space structure over the field of rational numbers $\mathbb{Q}$, we need to verify the following axioms:
- Closure under Addition: If $p(x)$ and $q(x)$ are in $P$, then $p(x) + q(x)$ is also in $P$.
- Closure under Scalar Multiplication: If $p(x)$ is in $P$ and $c \in \mathbb{Q}$, then $c \cdot p(x)$ is also in $P$.
- Existence of Additive Identity: The zero polynomial $0$ is in $P$.
- Existence of Additive Inverses: For every $p(x) \in P$, there exists $-p(x) \in P$ such that $p(x) + (-p(x)) = 0$.
- Associativity and Commutativity of Addition: Polynomial addition is associative and commutative.
- Distributivity of Scalar Multiplication: Scalar multiplication distributes over both polynomial addition and scalar addition.

3. Verification:
- Closure under Addition: If $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$ and $q(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + b_4 x^4$, then:
$$p(x) + q(x) = (a_0 + b_0) + (a_1 + b_1) x + (a_2 + b_2) x^2 + (a_3 + b_3) x^3 + (a_4 + b_4) x^4$$
Since $a_i, b_i \in \mathbb{Q}$, $a_i + b_i \in \mathbb{Q}$, so $p(x) + q(x) \in P$.

- Closure under Scalar Multiplication: If $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$ and $c \in \mathbb{Q}$, then:
$$c \cdot p(x) = c a_0 + c a_1 x + c a_2 x^2 + c a_3 x^3 + c a_4 x^4$$
Since $a_i \in \mathbb{Q}$ and $c \in \mathbb{Q}$, $c a_i \in \mathbb{Q}$, so $c \cdot p(x) \in P$.

- Existence of Additive Identity: The zero polynomial $0 = 0 + 0x + 0x^2 + 0x^3 + 0x^4$ is in $P$.

- Existence of Additive Inverses: For $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4$, the additive inverse is $-p(x) = -a_0 - a_1 x - a_2 x^2 - a_3 x^3 - a_4 x^4$, which is in $P$.

- Associativity and Commutativity of Addition: Polynomial addition is associative and commutative by the properties of rational numbers.

- Distributivity of Scalar Multiplication: Scalar multiplication distributes over both polynomial addition and scalar addition by the properties of rational numbers.

Thus, $P$ satisfies all the axioms of a vector space over $\mathbb{Q}$.

### Part (b)
1. Linear Independence: To show that the polynomials $1, x - 2, (x - 2)^2, (x - 2)^3, (x - 2)^4$ are linearly independent, assume:
$$a(x-2)^4 + b(x-2)^3 + c(x-2)^2 + d(x-2) + e = 0$$
for some $a, b, c, d, e \in \mathbb{Q}$.

2. Expand the Polynomial:
$$a(x-2)^4 + b(x-2)^3 + c(x-2)^2 + d(x-2) + e = 0$$
Expanding each term:
$$a(x-2)^4 = a(x^4 - 8x^3 + 24x^2 - 32x + 16)$$
$$b(x-2)^3 = b(x^3 - 6x^2 + 12x - 8)$$
$$c(x-2)^2 = c(x^2 - 4x + 4)$$
$$d(x-2) = d(x - 2)$$
$$e = e$$

3. Combine Like Terms:
$$a x^4 + (b - 8a) x^3 + (24a - 6b + c) x^2 + (-32a + 12b - 4c + d) x + (16a - 8b + 4c - 2d + e) = 0$$

4. Set Coefficients to Zero: For the polynomial to be identically zero, each coefficient must be zero:
$$\begin{cases} a = 0 \\ b - 8a = 0 \\ 24a - 6b + c = 0 \\ -32a + 12b - 4c + d = 0 \\ 16a - 8b + 4c - 2d + e = 0 \end{cases}$$
Solving these equations:
$$a = 0 \implies b = 0 \implies c = 0 \implies d = 0 \implies e = 0$$

Since the only solution is $a = b = c = d = e = 0$, the polynomials are linearly independent.

5. Spanning the Space: Any polynomial of degree less than or equal to 4 can be written as a linear combination of $1, x - 2, (x - 2)^2, (x - 2)^3, (x - 2)^4$. Therefore, these polynomials span the space.

Thus, $1, x - 2, (x - 2)^2, (x - 2)^3, (x - 2)^4$ form a basis for the vector space.

### Part (c)
1. Express the Polynomial: We need to express $7 + 2x - 45x^2 + 3x^4$ in terms of the basis $1, x - 2, (x - 2)^2, (x - 2)^3, (x - 2)^4$.

2. Use Taylor Series Expansion: Compute the Taylor series of $3x^4 - 45x^2 + 2x + 7$ at $x = 2$.

3. Compute the Coefficients:
$$3(x-2)^4 + 24(x-2)^3 + 27(x-2)^2 - 82(x-2) - 121$$

Therefore, the polynomial $7 + 2x - 45x^2 + 3x^4$ can be written as:
$$3(x-2)^4 + 24(x-2)^3 + 27(x-2)^2 - 82(x-2) - 121$$

The final answer is $\boxed{ 3(x-2)^4 + 24(x-2)^3 + 27(x-2)^2 - 82(x-2) - 121 }$