Question

For every \(k \in \mathbb{N} \cup \{0\}\), let \(y_k(x)\) be a polynomial of degree k with \(y_k(1) = 5\). Further, let \(y_k(x)\) satisfy the Legendre equation \[ (1-x^2)y'' - 2xy' + k(k+1)y = 0. \] If \[ \frac{1}{2} \sum_{k=1}^{n} \int_{-1}^{1} (y_k(x) - y_{k-1}(x))^2 dx - \sum_{k=1}^{n} \int_{-1}^{1} (y_k(x))^2 dx = 24, \] for some positive integer n, then the value of n is ..................

Hint:

When you see an integral of a product of solutions to a Sturm-Liouville equation (like Legendre's), immediately think of the orthogonality property. This is almost always the key to simplifying the problem.

Answer 1

Step 1: Understanding the Concept:
The problem involves properties of Legendre polynomials, which are solutions to the Legendre differential equation. The key properties are the value at \(x=1\) and their orthogonality over the interval \([-1, 1]\).
Step 2: Key Formula or Approach:
1. The polynomial solution to the Legendre equation is the Legendre polynomial, \(P_k(x)\). So, \(y_k(x)\) must be a constant multiple of \(P_k(x)\), i.e., \(y_k(x) = c_k P_k(x)\).
2. Use the condition \(y_k(1)=5\) and the property \(P_k(1)=1\) to find the constant \(c_k\).
3. Use the orthogonality property of Legendre polynomials: \[ \int_{-1}^{1} P_m(x) P_n(x) dx = \begin{cases} 0 & \text{if } m \neq n
\frac{2}{2n+1} & \text{if } m = n \end{cases} \] 4. Substitute these properties into the given integral equation and solve for n.
Step 3: Detailed Explanation or Calculation:
Determine \(y_k(x)\):
We have \(y_k(x) = c_k P_k(x)\).
Given \(y_k(1) = 5\), we have \(c_k P_k(1) = 5\).
Since \(P_k(1) = 1\) for all \(k\), we get \(c_k = 5\).
Thus, \(y_k(x) = 5 P_k(x)\) for all \(k\).

Simplify the given equation:
Let the given equation be \(E\).
\[ E = \frac{1}{2} \sum_{k=1}^{n} \int_{-1}^{1} (y_k - y_{k-1})^2 dx - \sum_{k=1}^{n} \int_{-1}^{1} y_k^2 dx = 24 \] Expand the first term: \[ \int (y_k - y_{k-1})^2 dx = \int (y_k^2 - 2y_k y_{k-1} + y_{k-1}^2) dx = \int y_k^2 dx - 2\int y_k y_{k-1} dx + \int y_{k-1}^2 dx \] Using orthogonality: \[ \int_{-1}^{1} y_k y_{k-1} dx = \int_{-1}^{1} (5 P_k(x)) (5 P_{k-1}(x)) dx = 25 \int_{-1}^{1} P_k(x) P_{k-1}(x) dx = 25 \times 0 = 0 \] So the first term simplifies: \[ \frac{1}{2} \sum_{k=1}^{n} \left( \int y_k^2 dx + \int y_{k-1}^2 dx \right) - \sum_{k=1}^{n} \int y_k^2 dx = 24 \] \[ \frac{1}{2} \sum_{k=1}^{n} \int y_k^2 dx + \frac{1}{2} \sum_{k=1}^{n} \int y_{k-1}^2 dx - \sum_{k=1}^{n} \int y_k^2 dx = 24 \] \[ - \frac{1}{2} \sum_{k=1}^{n} \int y_k^2 dx + \frac{1}{2} \sum_{k=1}^{n} \int y_{k-1}^2 dx = 24 \] Now let's evaluate the integral \(\int y_k^2 dx\): \[ \int_{-1}^{1} y_k^2 dx = \int_{-1}^{1} (5 P_k(x))^2 dx = 25 \int_{-1}^{1} P_k(x)^2 dx = 25 \left( \frac{2}{2k+1} \right) = \frac{50}{2k+1} \] Substitute this back into the equation: \[ - \frac{1}{2} \sum_{k=1}^{n} \frac{50}{2k+1} + \frac{1}{2} \sum_{k=1}^{n} \frac{50}{2(k-1)+1} = 24 \] \[ -25 \sum_{k=1}^{n} \frac{1}{2k+1} + 25 \sum_{k=1}^{n} \frac{1}{2k-1} = 24 \] \[ 25 \left( \sum_{k=1}^{n} \frac{1}{2k-1} - \sum_{k=1}^{n} \frac{1}{2k+1} \right) = 24 \] This is a telescoping sum: \[ \sum_{k=1}^{n} \frac{1}{2k-1} = \frac{1}{1} + \frac{1}{3} + \frac{1}{5} + ...... + \frac{1}{2n-1} \] \[ \sum_{k=1}^{n} \frac{1}{2k+1} = \frac{1}{3} + \frac{1}{5} + ...... + \frac{1}{2n-1} + \frac{1}{2n+1} \] The difference is: \[ \left( \frac{1}{1} + \frac{1}{3} + ...... + \frac{1}{2n-1} \right) - \left( \frac{1}{3} + ...... + \frac{1}{2n-1} + \frac{1}{2n+1} \right) = 1 - \frac{1}{2n+1} \] So the equation becomes: \[ 25 \left( 1 - \frac{1}{2n+1} \right) = 24 \] \[ 25 \left( \frac{2n+1-1}{2n+1} \right) = 24 \implies 25 \left( \frac{2n}{2n+1} \right) = 24 \] \[ 50n = 24(2n+1) = 48n + 24 \] \[ 2n = 24 \] Step 4: Final Answer: \[ n = 12 \] Step 5: Why This is Correct: The solution correctly identifies the polynomials as multiples of Legendre polynomials and uses their orthogonality property to simplify the complex-looking equation. The simplification reveals a telescoping sum, which is evaluated correctly. The final algebraic steps to solve for n are accurate, leading to the correct result.