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$M=\left\{x \in I_{2}: \quad x=\left(x_{1}, x_{2}, \ldots, x_{n}, \ldots\right)...
Jun 28, 2024
Solution by Steps
step 1
We are given the set MM defined as M={x∈I2:x=(x1,x2,…,xn,…),∣xn∣≤1n,n=1,2,…}M=\left\{x \in I_{2}: \quad x=\left(x_{1}, x_{2}, \ldots, x_{n}, \ldots\right), \quad\left|x_{n}\right| \leq \frac{1}{\sqrt{n}}, \quad n=1,2, \ldots\right\}
step 2
To understand the set MM, we need to analyze the condition ∣xn∣≤1n\left|x_{n}\right| \leq \frac{1}{\sqrt{n}}. This means that for each component xnx_n of the vector xx, the absolute value of xnx_n must be less than or equal to 1n\frac{1}{\sqrt{n}}
step 3
The set MM is a subset of I2I_2, where I2I_2 is likely a space of sequences or vectors. Each x∈Mx \in M is a sequence (x1,x2,…)(x_1, x_2, \ldots) such that the nn-th component xnx_n satisfies ∣xn∣≤1n\left|x_{n}\right| \leq \frac{1}{\sqrt{n}}
step 4
Therefore, MM consists of all sequences in I2I_2 where each term xnx_n is bounded by 1n\frac{1}{\sqrt{n}}
Answer
MM is the set of all sequences in I2I_2 where each term xnx_n satisfies ∣xn∣≤1n\left|x_{n}\right| \leq \frac{1}{\sqrt{n}}.
Key Concept
Set Definition
Explanation
The set MM is defined by a condition on the components of the sequences it contains, specifically that each component xnx_n must be less than or equal to 1n\frac{1}{\sqrt{n}} in absolute value.
Solution by Steps
step 1
We are given the operator A:C[0,π2]↦C[0,π2]A: C\left[0, \frac{\pi}{2}\right] \mapsto C\left[0, \frac{\pi}{2}\right] defined by the formula Ax(t)=∫0π2sin⁡(t+s)x(s)dsA x(t)=\int_{0}^{\frac{\pi}{2}} \sin (t+s) x(s) d s
step 2
To find the norm of the operator AA, we need to determine ∥A∥=sup⁡∥x∥≤1∥Ax∥\|A\| = \sup_{\|x\| \leq 1} \|Ax\|
step 3
Consider the function x(s)=1x(s) = 1 for s∈[0,π2]s \in \left[0, \frac{\pi}{2}\right]. Then, Ax(t)=∫0π2sin⁡(t+s)dsAx(t) = \int_{0}^{\frac{\pi}{2}} \sin(t+s) ds
step 4
Evaluate the integral: Ax(t)=∫0π2sin⁡(t+s)ds=[−cos⁡(t+s)]0π2=−cos⁡(t+π2)+cos⁡(t)Ax(t) = \int_{0}^{\frac{\pi}{2}} \sin(t+s) ds = \left[-\cos(t+s)\right]_{0}^{\frac{\pi}{2}} = -\cos(t+\frac{\pi}{2}) + \cos(t)
step 5
Simplify the expression: Ax(t)=−cos⁡(t+π2)+cos⁡(t)=−(−sin⁡(t))+cos⁡(t)=sin⁡(t)+cos⁡(t)Ax(t) = -\cos(t+\frac{\pi}{2}) + \cos(t) = -(-\sin(t)) + \cos(t) = \sin(t) + \cos(t)
step 6
The maximum value of sin⁡(t)+cos⁡(t)\sin(t) + \cos(t) on [0,π2]\left[0, \frac{\pi}{2}\right] is 2\sqrt{2}, which occurs when t=π4t = \frac{\pi}{4}
step 7
Therefore, ∥Ax∥=2\|Ax\| = \sqrt{2} when ∥x∥=1\|x\| = 1. Hence, the norm of the operator AA is ∥A∥=2\|A\| = \sqrt{2}
Answer
The norm of the operator AA is 2\sqrt{2}.
Key Concept
Operator Norm
Explanation
The norm of an operator AA is the supremum of ∥Ax∥\|Ax\| over all xx with ∥x∥≤1\|x\| \leq 1. In this case, we evaluated the integral and found the maximum value of the resulting function to determine the norm.
Solution by Steps
step 1
We need to find the projection of the function x(t)=2t−1x(t) = 2t - 1 onto the subspace V1V_1 of all polynomials of degree less than or equal to 1 in the space L2(0,1)L^2(0,1). The subspace V1V_1 is defined as V1={p(t)=a1t+a0∣a0,a1 are constants}V_1 = \{ p(t) = a_1 t + a_0 \mid a_0, a_1 \text{ are constants} \}
step 2
The projection of x(t)x(t) onto V1V_1 is given by the polynomial p(t)=a1t+a0p(t) = a_1 t + a_0 that minimizes the norm ∥x(t)−p(t)∥L2(0,1)\| x(t) - p(t) \|_{L^2(0,1)}
step 3
To find a0a_0 and a1a_1, we need to solve the following system of equations obtained by setting the inner products ⟨x(t)−p(t),1⟩L2(0,1)=0\langle x(t) - p(t), 1 \rangle_{L^2(0,1)} = 0 and ⟨x(t)−p(t),t⟩L2(0,1)=0\langle x(t) - p(t), t \rangle_{L^2(0,1)} = 0
step 4
Calculate ⟨x(t)−p(t),1⟩L2(0,1)=∫01(2t−1−a1t−a0) dt=0\langle x(t) - p(t), 1 \rangle_{L^2(0,1)} = \int_0^1 (2t - 1 - a_1 t - a_0) \, dt = 0
step 5
Simplify the integral: ∫01(2t−1−a1t−a0) dt=∫01(2t−a1t−1−a0) dt=∫01((2−a1)t−1−a0) dt\int_0^1 (2t - 1 - a_1 t - a_0) \, dt = \int_0^1 (2t - a_1 t - 1 - a_0) \, dt = \int_0^1 ((2 - a_1)t - 1 - a_0) \, dt
step 6
Evaluate the integral: ∫01((2−a1)t−1−a0) dt=[(2−a1)t22−(1+a0)t]01=2−a12−(1+a0)=0\int_0^1 ((2 - a_1)t - 1 - a_0) \, dt = \left[ \frac{(2 - a_1)t^2}{2} - (1 + a_0)t \right]_0^1 = \frac{2 - a_1}{2} - (1 + a_0) = 0
step 7
Solve for a0a_0: 2−a12−1−a0=0  ⟹  2−a12=1+a0  ⟹  2−a1=2+2a0  ⟹  −a1=2a0  ⟹  a1=−2a0\frac{2 - a_1}{2} - 1 - a_0 = 0 \implies \frac{2 - a_1}{2} = 1 + a_0 \implies 2 - a_1 = 2 + 2a_0 \implies -a_1 = 2a_0 \implies a_1 = -2a_0
step 8
Calculate ⟨x(t)−p(t),t⟩L2(0,1)=∫01t(2t−1−a1t−a0) dt=0\langle x(t) - p(t), t \rangle_{L^2(0,1)} = \int_0^1 t(2t - 1 - a_1 t - a_0) \, dt = 0
step 9
Simplify the integral: ∫01t(2t−1−a1t−a0) dt=∫01(2t2−t−a1t2−a0t) dt=∫01((2−a1)t2−t−a0t) dt\int_0^1 t(2t - 1 - a_1 t - a_0) \, dt = \int_0^1 (2t^2 - t - a_1 t^2 - a_0 t) \, dt = \int_0^1 ((2 - a_1)t^2 - t - a_0 t) \, dt
step 10
Evaluate the integral: ∫01((2−a1)t2−t−a0t) dt=[(2−a1)t33−t22−a0t22]01=2−a13−12−a02=0\int_0^1 ((2 - a_1)t^2 - t - a_0 t) \, dt = \left[ \frac{(2 - a_1)t^3}{3} - \frac{t^2}{2} - \frac{a_0 t^2}{2} \right]_0^1 = \frac{2 - a_1}{3} - \frac{1}{2} - \frac{a_0}{2} = 0
step 11
Solve for a0a_0: 2−a13−12−a02=0  ⟹  2−a13=12+a02  ⟹  2−a1=32+3a02  ⟹  4−2a1=3+3a0  ⟹  1−2a1=3a0\frac{2 - a_1}{3} - \frac{1}{2} - \frac{a_0}{2} = 0 \implies \frac{2 - a_1}{3} = \frac{1}{2} + \frac{a_0}{2} \implies 2 - a_1 = \frac{3}{2} + \frac{3a_0}{2} \implies 4 - 2a_1 = 3 + 3a_0 \implies 1 - 2a_1 = 3a_0
step 12
Substitute a1=−2a0a_1 = -2a_0 into the equation: 1−2(−2a0)=3a0  ⟹  1+4a0=3a0  ⟹  1=−a0  ⟹  a0=−11 - 2(-2a_0) = 3a_0 \implies 1 + 4a_0 = 3a_0 \implies 1 = -a_0 \implies a_0 = -1
step 13
Substitute a0=−1a_0 = -1 into a1=−2a0a_1 = -2a_0: a1=−2(−1)=2a_1 = -2(-1) = 2
step 14
The projection of x(t)=2t−1x(t) = 2t - 1 onto V1V_1 is p(t)=a1t+a0=2t−1p(t) = a_1 t + a_0 = 2t - 1
Answer
The projection of x(t)=2t−1x(t) = 2t - 1 onto V1V_1 is p(t)=2t−1p(t) = 2t - 1.
Key Concept
Projection of a function onto a subspace
Explanation
The projection of a function onto a subspace minimizes the distance between the function and the subspace, and is found by solving for the coefficients that satisfy the orthogonality conditions.
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