CALCU-M=\x I_2: x=(x_1, x_2, , x_n, ), |x_n| 1n

Solution by Steps

step 1

We are given the set

M

defined as

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

M

, we need to analyze the condition

\left|x_{n}\right| \leq \frac{1}{\sqrt{n}}

. This means that for each component

x_n

of the vector

x

, the absolute value of

x_n

must be less than or equal to

\frac{1}{\sqrt{n}}

step 3

The set

M

is a subset of

I_2

, where

I_2

is likely a space of sequences or vectors. Each

x \in M

is a sequence

(x_1, x_2, \ldots)

such that the

n

-th component

x_n

satisfies

\left|x_{n}\right| \leq \frac{1}{\sqrt{n}}

step 4

Therefore,

M

consists of all sequences in

I_2

where each term

x_n

is bounded by

\frac{1}{\sqrt{n}}

Answer

M

is the set of all sequences in

I_2

where each term

x_n

satisfies

\left|x_{n}\right| \leq \frac{1}{\sqrt{n}}

.

Key Concept

Set Definition

Explanation

The set

M

is defined by a condition on the components of the sequences it contains, specifically that each component

x_n

must be less than or equal to

\frac{1}{\sqrt{n}}

in absolute value.

Solution by Steps

step 1

We are given the operator

A: C\left[0, \frac{\pi}{2}\right] \mapsto C\left[0, \frac{\pi}{2}\right]

defined by the formula

A x(t)=\int_{0}^{\frac{\pi}{2}} \sin (t+s) x(s) d s

step 2

To find the norm of the operator

A

, we need to determine

\|A\| = \sup_{\|x\| \leq 1} \|Ax\|

step 3

Consider the function

x(s) = 1

for

s \in \left[0, \frac{\pi}{2}\right]

. Then,

Ax(t) = \int_{0}^{\frac{\pi}{2}} \sin(t+s) ds

step 4

Evaluate the integral:

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+\frac{\pi}{2}) + \cos(t) = -(-\sin(t)) + \cos(t) = \sin(t) + \cos(t)

step 6

The maximum value of

\sin(t) + \cos(t)

on

\left[0, \frac{\pi}{2}\right]

is

\sqrt{2}

, which occurs when

t = \frac{\pi}{4}

step 7

Therefore,

\|Ax\| = \sqrt{2}

when

\|x\| = 1

. Hence, the norm of the operator

A

is

\|A\| = \sqrt{2}

Answer

The norm of the operator

A

is

\sqrt{2}

.

Key Concept

Operator Norm

Explanation

The norm of an operator

A

is the supremum of

\|Ax\|

over all

x

with

\|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 - 1

onto the subspace

V_1

of all polynomials of degree less than or equal to 1 in the space

L^2(0,1)

. The subspace

V_1

is defined as

V_1 = \{ p(t) = a_1 t + a_0 \mid a_0, a_1 \text{ are constants} \}

step 2

The projection of

x(t)

onto

V_1

is given by the polynomial

p(t) = a_1 t + a_0

that minimizes the norm

\| x(t) - p(t) \|_{L^2(0,1)}

step 3

To find

a_0

and

a_1

, we need to solve the following system of equations obtained by setting the inner products

\langle x(t) - p(t), 1 \rangle_{L^2(0,1)} = 0

and

\langle x(t) - p(t), t \rangle_{L^2(0,1)} = 0

step 4

Calculate

\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:

\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:

\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

a_0

:

\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

\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:

\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:

\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

a_0

:

\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

a_1 = -2a_0

into the equation:

1 - 2(-2a_0) = 3a_0 \implies 1 + 4a_0 = 3a_0 \implies 1 = -a_0 \implies a_0 = -1

step 13

Substitute

a_0 = -1

into

a_1 = -2a_0

:

a_1 = -2(-1) = 2

step 14

The projection of

x(t) = 2t - 1

onto

V_1

is

p(t) = a_1 t + a_0 = 2t - 1

Answer

The projection of

x(t) = 2t - 1

onto

V_1

is

p(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.

Solution by Steps

step 1

The problem involves understanding the space

L^{2}(0,1)

. This space consists of all square-integrable functions on the interval

(0,1)

step 2

A function

f

belongs to

L^{2}(0,1)

if \int_{0}^{1} |f(x)|^2 \, dx < \infty

step 3

The notation

(\Delta f)^{2 {0, 1}}

suggests a squared difference or a squared norm in the context of

L^{2}(0,1)

step 4

The expression

\{1, (\Delta f)^2\}

indicates a set containing the number 1 and the squared difference

(\Delta f)^2

Answer

The space

L^{2}(0,1)

consists of all functions whose square is integrable over the interval

(0,1)

.

Key Concept

L^{2}(0,1)

space

Explanation

L^{2}(0,1)

is a function space where the integral of the square of the function over the interval

(0,1)

is finite.

Solution by Steps

step 1

We need to determine if the sequence

x_n = \left(1, \frac{1}{2}, \frac{1}{2^2}, \cdots, \frac{1}{2^{n-1}}, 0, 0, \ldots\right)

converges in the space

I_2

step 2

The space

I_2

consists of all sequences

(a_n)

such that the series

\sum_{n=1}^{\infty} |a_n|^2

converges

step 3

For the given sequence

x_n

, we need to check if

\sum_{k=1}^{\infty} \left(\frac{1}{2^{k-1}}\right)^2

converges

step 4

Simplifying the series, we get

\sum_{k=1}^{\infty} \left(\frac{1}{2^{k-1}}\right)^2 = \sum_{k=1}^{\infty} \frac{1}{4^{k-1}}

step 5

This is a geometric series with the first term

a = 1

and common ratio

r = \frac{1}{4}

step 6

A geometric series

\sum_{k=0}^{\infty} ar^k

converges if |r| < 1. Here, |r| = \frac{1}{4} < 1, so the series converges

step 7

Therefore,

\sum_{k=1}^{\infty} \frac{1}{4^{k-1}}

converges, implying that the sequence

x_n

converges in the space

I_2

Answer

The sequence

x_n

converges in the space

I_2

.

Key Concept

Convergence in

I_2

space

Explanation

A sequence converges in the

I_2

space if the series of the squares of its terms converges.

Solution by Steps

step 1

We need to determine the relative compactness of the set

A

in the space

l_{2}

. The set

A

is defined as A=\left\{x \in l_{2}: x=\left(x_{1}, x_{2}, \ldots, x_{n}, \ldots\right), \left|x_{n}\right|<n \text{ for } n=1,2, \ldots, 10, \left|x_{n}\right| \leq \frac{1}{n} \text{ for } n>10\right\}

step 2

To check for relative compactness in

l_{2}

, we need to verify if the set

A

is totally bounded. A set is totally bounded if for every \epsilon > 0, there exists a finite number of balls of radius

\epsilon

that cover the set

step 3

Consider the elements

x \in A

. For

n \leq 10

, \left|x_{n}\right| < n. For n > 10,

\left|x_{n}\right| \leq \frac{1}{n}

step 4

We need to show that the sequence

\{x_n\}

is bounded in

l_{2}

. For

n \leq 10

, \left|x_{n}\right| < n implies that \sum_{n=1}^{10} \left|x_{n}\right|^2 < \sum_{n=1}^{10} n^2

step 5

For n > 10,

\left|x_{n}\right| \leq \frac{1}{n}

implies that

\sum_{n=11}^{\infty} \left|x_{n}\right|^2 \leq \sum_{n=11}^{\infty} \left(\frac{1}{n}\right)^2

step 6

The series

\sum_{n=11}^{\infty} \left(\frac{1}{n}\right)^2

converges because it is a p-series with p=2 > 1

step 7

Therefore,

\sum_{n=1}^{\infty} \left|x_{n}\right|^2

is bounded, which implies that the set

A

is bounded in

l_{2}

step 8

Since

A

is bounded and the space

l_{2}

is a Hilbert space, the set

A

is relatively compact if it is totally bounded

step 9

To show that

A

is totally bounded, we need to cover

A

with a finite number of balls of radius

\epsilon

. Given the boundedness of

A

, we can find such a finite cover

step 10

Hence, the set

A

is relatively compact in the space

l_{2}

Answer

The set

A

is relatively compact in the space

l_{2}

.

Key Concept

Relative Compactness

Explanation

Relative compactness in

l_{2}

involves showing that the set is totally bounded, which means it can be covered by a finite number of balls of any given radius.

Solution by Steps

step 1

We start with the given integral expression for the operator

A

:

A x(1) = \int_{1}^{2} \sqrt{t+s} x(s) \, ds

step 2

To find the norm of the operator

A

, we need to evaluate the integral and then determine the supremum of the resulting expression

step 3

First, we evaluate the integral

\int_{1}^{2} \sqrt{t+s} x(s) \, ds

step 4

Assuming

x(s)

is a function from reals to reals, we can write the norm of the operator

A

as:

\|A\| = \left\| \int_{1}^{2} \sqrt{t+s} x(s) \, ds \right\|

step 5

The norm of the operator

A

is given by the absolute value of the integral:

\|A\| = \left| \int_{1}^{2} \sqrt{t+s} x(s) \, ds \right|

step 6

To find the supremum, we consider the maximum value of the integral over all possible functions

x(s)

in the given space

step 7

Therefore, the norm of the operator

A

is:

\|A\| = \sup_{x(s)} \left| \int_{1}^{2} \sqrt{t+s} x(s) \, ds \right|

Answer

\|A\| = \sup_{x(s)} \left| \int_{1}^{2} \sqrt{t+s} x(s) \, ds \right|

Key Concept

Norm of an Operator

Explanation

The norm of an operator is determined by evaluating the integral and finding the supremum of the resulting expression over all possible functions in the given space.

Generated Graph

Solution by Steps

step 1

We start by integrating the given function

(a_1 t + a_0)^2

with respect to

t

from

t=0

to

t=1

step 2

The integral is

\int_0^1 (a_1 t + a_0)^2 \, dt

step 3

Expanding the integrand, we get

(a_1 t + a_0)^2 = a_1^2 t^2 + 2 a_1 a_0 t + a_0^2

step 4

Now, integrate each term separately:

\int_0^1 a_1^2 t^2 \, dt + \int_0^1 2 a_1 a_0 t \, dt + \int_0^1 a_0^2 \, dt.

step 5

The integrals are:

\int_0^1 a_1^2 t^2 \, dt = \frac{a_1^2}{3}, \quad \int_0^1 2 a_1 a_0 t \, dt = a_1 a_0, \quad \int_0^1 a_0^2 \, dt = a_0^2.

step 6

Adding these results, we get:

\int_0^1 (a_1 t + a_0)^2 \, dt = \frac{a_1^2}{3} + a_1 a_0 + a_0^2.

Answer

\int_0^1 (a_1 t + a_0)^2 \, dt = \frac{a_1^2}{3} + a_1 a_0 + a_0^2

Key Concept

Integration of a polynomial function

Explanation

The integral of a polynomial function can be found by integrating each term separately and then summing the results.

AskSia

Plus