Question

Let \( l^2 = \{(x_1, x_2, x_3, \dots) : x_n \in \mathbb{R} \text{ for all } n \in \mathbb{N} \text{ and } \sum_{n=1}^\infty x_n^2 < \infty\} \).
For a sequence \( (x_1, x_2, x_3, \dots) \in l^2 \), define \( \|(x_1, x_2, x_3, \dots)\|_2 = \left( \sum_{n=1}^\infty x_n^2 \right)^{1/2} \).
Consider the subspace \( M = \{(x_1, x_2, x_3, \dots) \in l^2 : \sum_{n=1}^\infty \frac{x_n}{4^n} = 0\} \).
Let \( M^\perp \) denote the orthogonal complement of \( M \) in the Hilbert space \( (l^2, \|.\|_2) \).
Consider \( \left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\right) \in l^2 \).
If the orthogonal projection of \( \left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots\right) \) onto \( M^\perp \) is given by \( \alpha \left(\frac{1}{4}, \frac{1}{4^2}, \frac{1}{4^3}, \dots\right) \) for some \( \alpha \in \mathbb{R} \), then \( \alpha \) equals ..................

Hint:

In Hilbert space problems, when a subspace is defined as the set of vectors orthogonal to a given vector \( \mathbf{z} \) (i.e., \( M = \{\mathbf{x} | \langle \mathbf{x}, \mathbf{z} \rangle = 0\} \)), its orthogonal complement \( M^\perp \) is simply the span of that vector, \( \text{span}\{\mathbf{z}\} \). This simplifies projection problems significantly.

Answer 1

Step 1: Understanding the Concept:
This problem is set in the Hilbert space \( l^2 \). We are asked to find the scalar multiple in the orthogonal projection of a vector onto a one-dimensional subspace. The subspace \( M \) is defined as the set of vectors orthogonal to a specific vector, which means \( M^\perp \) will be the subspace spanned by that vector.
Step 2: Key Formula or Approach:
The orthogonal projection of a vector \( \mathbf{y} \) onto the subspace spanned by a non-zero vector \( \mathbf{z} \) is given by the formula: \[ \text{proj}_{\mathbf{z}}(\mathbf{y}) = \frac{\langle \mathbf{y}, \mathbf{z} \rangle}{\langle \mathbf{z}, \mathbf{z} \rangle} \mathbf{z} = \frac{\langle \mathbf{y}, \mathbf{z} \rangle}{\|\mathbf{z}\|_2^2} \mathbf{z} \] where \( \langle ., . \rangle \) is the inner product in \( l^2 \), defined as \( \langle \mathbf{x}, \mathbf{y} \rangle = \sum_{n=1}^\infty x_n y_n \).
We will also need the Taylor series for \( -\ln(1-x) = \sum_{n=1}^\infty \frac{x^n}{n} \) for \( |x| < 1 \), and the sum of a geometric series \( \sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r} \) for \( |r| < 1 \).
Step 3: Detailed Explanation or Calculation:
1. Identify the spanning vector for \( M^\perp \):
The subspace \( M \) is defined by \( \sum_{n=1}^\infty \frac{x_n}{4^n} = 0 \). Let \( \mathbf{z} = \left(\frac{1}{4}, \frac{1}{4^2}, \frac{1}{4^3}, \dots \right) \). The condition for \( \mathbf{x} \in M \) can be written using the inner product: \[ \langle \mathbf{x}, \mathbf{z} \rangle = \sum_{n=1}^\infty x_n \cdot \frac{1}{4^n} = 0 \] This means that \( M \) is the set of all vectors in \( l^2 \) that are orthogonal to \( \mathbf{z} \). By definition, \( M = (\text{span}\{\mathbf{z}\})^\perp \), so \( M^\perp = \text{span}\{\mathbf{z}\} \).
2. Define the vector to be projected:
Let \( \mathbf{y} = \left(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \right) \), so \( y_n = \frac{1}{n} \).
3. Calculate the inner products:
First, compute \( \langle \mathbf{y}, \mathbf{z} \rangle \): \[ \langle \mathbf{y}, \mathbf{z} \rangle = \sum_{n=1}^\infty \frac{1}{n} \cdot \frac{1}{4^n} = \sum_{n=1}^\infty \frac{(1/4)^n}{n} \] Using the Taylor series \( \sum_{n=1}^\infty \frac{x^n}{n} = -\ln(1 - x) \) for \( |x| < 1 \), and plugging in \( x = \frac{1}{4} \): \[ \langle \mathbf{y}, \mathbf{z} \rangle = -\ln\left(1 - \frac{1}{4}\right) = -\ln\left(\frac{3}{4}\right) = \ln\left(\frac{4}{3}\right) \] Now compute \( \langle \mathbf{z}, \mathbf{z} \rangle \): \[ \langle \mathbf{z}, \mathbf{z} \rangle = \sum_{n=1}^\infty \left(\frac{1}{4^n}\right)^2 = \sum_{n=1}^\infty \frac{1}{16^n} \] This is a geometric series with \( a = \frac{1}{16} \) and \( r = \frac{1}{16} \), so: \[ \sum_{n=1}^\infty \frac{1}{16^n} = \frac{1/16}{1 - 1/16} = \frac{1/16}{15/16} = \frac{1}{15} \] 4. Determine the projection and find \( \alpha \):
\[ \text{proj}_{\mathbf{z}}(\mathbf{y}) = \frac{\ln(4/3)}{1/15} \mathbf{z} = 15 \ln\left(\frac{4}{3}\right) \mathbf{z} \] We are told the projection is \( \alpha \ln(4/3) \mathbf{z} \). Equating: \[ \alpha \ln\left(\frac{4}{3}\right) \mathbf{z} = 15 \ln\left(\frac{4}{3}\right) \mathbf{z} \Rightarrow \alpha = 15 \] Step 4: Final Answer:
The value of \( \alpha \) is \( \boxed{15} \).
Step 5: Why This is Correct:
The key was to identify the orthogonal complement \( M^\perp \), express the projection using inner product formulas in Hilbert space \( l^2 \), and correctly evaluate known series.