Problem Breakdown:

We are tasked with finding a basis for the vector space of polynomials $p(x) \in \mathbb{R}[X]$ such that:

1. $p(-1) = 0$
2. $p''(-1) = 0$

Also, each polynomial in the basis must include the term $-2000x$.

Step-by-Step Approach:

1. General Form of the Polynomial:

We are dealing with polynomials in $\mathbb{R}[X]$, and the space is defined by the two conditions:

• $p(-1) = 0$
• $p''(-1) = 0$

Let's start by considering a general polynomial of degree at most 3 (since polynomials of higher degree can potentially introduce more terms and could still satisfy the given conditions). Thus, let the general polynomial $p(x)$ be:

$p(x) = ax^3 + bx^2 + cx + d$

2. Apply the Condition $p(-1) = 0$:

Substitute $x = -1$ into the polynomial:

$p(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = -a + b - c + d = 0$

Thus, we have the equation:

-a + b - c + d = 0 \tag{1}

3. Apply the Condition $p''(-1) = 0$:

To apply the second condition, we first need to compute the second derivative of $p(x)$. The first derivative is:

$p'(x) = 3ax^2 + 2bx + c$

The second derivative is:

$p''(x) = 6ax + 2b$

Now, substitute $x = -1$:

$p''(-1) = 6a(-1) + 2b = -6a + 2b = 0$

Thus, we have the second equation:

-6a + 2b = 0 \quad \Rightarrow \quad 3a = b \tag{2}

4. Incorporate the Term $-2000x$:

We are asked to ensure that each polynomial in the basis contains the term $-2000x$. This means that $c = -2000$.

5. Solve the System of Equations:

From equation (2), we know that $b = 3a$. Substituting this into equation (1):

$-a + 3a - (-2000) + d = 0$

This simplifies to:

$2a + 2000 + d = 0 \quad \Rightarrow \quad d = -2a - 2000$

Thus, the general polynomial $p(x)$ becomes:

$p(x) = ax^3 + 3ax^2 - 2000x - 2a - 2000$

Factor out $a$ to simplify:

$p(x) = a(x^3 + 3x^2 - 2) - 2000(x + 1)$

This is the general form of a polynomial in the vector space defined by the given conditions.

6. Basis of the Vector Space:

The two terms $x^3 + 3x^2 - 2$ and $x + 1$ are independent, so they form a basis for the vector space. Thus, the basis for the vector space of polynomials is:

$\{ x^3 + 3x^2 - 2, x + 1 \}$

Each polynomial in the basis contains the term $-2000x$ as required.

Conclusion:

The basis for the vector space of polynomials $p(x) \in \mathbb{R}[X]$ that satisfy $p(-1) = 0$ and $p''(-1) = 0$ and include the term $-2000x$ is:

$\boxed{\{ x^3 + 3x^2 - 2, x + 1 \}}$

Would you like further details or clarification on any of these steps?

Related Questions:

1. How can we extend the basis for polynomials of higher degree satisfying similar conditions?
2. What happens if we change the conditions to $p(1) = 0$ and $p''(1) = 0$?
3. How do we find the basis of a polynomial space with more constraints, such as $p(2) = 0$ and $p'''(-1) = 0$?
4. What other methods can be used to find a basis for a vector space of polynomials?
5. How do we compute the dimension of the polynomial space for higher degree polynomials?

Tip:

When solving for the coefficients of polynomials that satisfy specific conditions (like vanishing at certain points or having zero derivatives), writing out the system of equations and simplifying step-by-step is key to managing the algebra.