/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Let \(x_{1}(t)=1\) and \(x_{2}(t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(x_{1}(t)=1\) and \(x_{2}(t)=t^{2}\). Does \(Y=\operatorname{span}\left\\{x_{1}, x_{2}\right\\}\) satisfy the Haar condition if \(Y\) is regarded as a subspace \((a)\) of \(C[0,1],(b)\) of \(C[-1,1]\) ? (To understand what is going on, approximate \(x\) defined by \(x(t)=t^{3}\) in both cases.)

Short Answer

Expert verified
Yes, the span satisfies Haar in both \([0,1]\) and \([-1,1]\).

Step by step solution

01

Understanding Haar Condition

The Haar condition for a set of functions means that they are linearly independent on the interval given. A set of functions like \(\{x_1, x_2\}\) cannot satisfy the Haar condition if there exists a non-zero function in the span that vanishes over a non-zero interval.
02

Identify Function Set

Given functions are \(x_1(t) = 1\) and \(x_2(t) = t^2\). We form \( Y = \text{span}\{x_1, x_2\}\), which includes all linear combinations \( y(t) = c_1 + c_2 t^2 \) for any constants \(c_1\) and \(c_2\).
03

Approximate \(x(t) = t^3\) in \([0,1]\)

To approximate \( x(t) = t^3 \) using \(Y\), find constants \( c_1 \) and \( c_2 \) such that \( c_1 + c_2 t^2 \approx t^3 \). Testing with sample values, notice \( t^3 \otin Y \) as there is no \( c_1 \) and \( c_2 \) satisfying \( t^3 = c_1 + c_2 t^2 \) over \([0,1]\). Thus, \(Y\) does not match \(x\) over \([0,1]\).
04

Approximate \(x(t) = t^3\) in \([-1,1]\)

Similarly, approximate \( t^3 \) using \( Y \) in \([-1,1]\). Since \(Y\) only contains constant and quadratic terms, \(t^3\), having different polynomial order, is inapproximable by \(Y\). No linear combination of \(c_1\) and \(c_2t^2\) equates to \( t^3 \) in \([-1,1]\) as well.
05

Conclusion on Haar Condition

In both \([0,1]\) and \([-1,1]\), no non-trivial function from \(Y\) can approximate \(t^3\), confirming no list of non-zero function vanishes over the interval. Thus, no function of \(t^3\) form negates Haar's requirement, implying Haar condition satisfied on both intervals.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Haar Condition
The Haar Condition is an important concept in functional analysis that ensures a set of functions is linearly independent over a certain interval. To satisfy the Haar Condition, no non-zero function in the span should vanish entirely over any sub-interval. This means if you have a group of functions, such as \(x_1(t) = 1\) and \(x_2(t) = t^2\), the set will satisfy the Haar condition if there isn’t a function that becomes zero over any segment of the interval.

In the context of the exercise, consider the subspaces where these functions reside: \(C[0,1]\) and \(C[-1,1]\). Evaluating the span of \(x_1\) and \(x_2\), represented as \(c_1 + c_2 t^2\), involves checking if the combination could yield a function that equals zero beyond trivial instances (i.e., \(c_1 = c_2 = 0\)). If such a case is impossible, then the functions fulfill the Haar condition).

For practical understanding, approximate another function \(x(t) = t^3\) with these functions to check applicability. To satisfy the Haar Condition on \([0,1]\) or \([-1,1]\), no orientation of \(c_1\) and \(c_2\) works for \(c_1 + c_2 t^2 = t^3\). This forms the basis for proving these functions hold the Haar Condition on these intervals.
Linear Independence
Linear independence signifies that no function in a set can be written as a combination of others. For \(x_1(t) = 1\) and \(x_2(t) = t^2\) to be independent, they should not express others within the same set using scalars other than zero.

Given the problem’s requirements, observe the span \(Y = \operatorname{span}\left\{x_1, x_2\right\}\). This ensures any linear combination like \(c_1 + c_2 t^2\) doesn’t redundantly express one function through the others, provided \(c_1, c_2\) are not both zero at once.

Testing linear independence can practically involve seeing if these two account for another function—in this case, \(t^3\). The attempt to imitate \(t^3\) by simply adjusting \(c_1\) and \(c_2\) confirms their separate existence in the span, thereby affirming their independence. No possible straightforward substitution fully adopts \(t^3\) due to variance in degree, fortifying \(x_1\) and \(x_2\) as linearly independent.
Span of Functions
The span of a set of functions encloses all possible linear combinations obtainable from that set. When you see notations like \(Y = \operatorname{span}\{x_1, x_2\}\), it depicts all functions that can be written as \(a x_1(t) + b x_2(t)\) where \(a\) and \(b\) are scalars.

For the functions in this exercise, their span encompasses constant and quadratic terms \(a + bt^2\). Any attempt to replicate another form, such as \(t^3\), through these functions spotlights the limitations of the span. In both \([0,1]\) and \([-1,1]\), while \(t^3\) is a valid function, it steps outside the established span needing a cubic term.

Understanding function span informs us about what types of functions can be generated or approximated from a set, bridging concepts like function approximation and transformation in analysis while confirming the boundaries of what’s possible within a given framework.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In certain applications, the zeros of the Chebyshev polynomials are of interest. Show that all the zeros of \(T_{n}\) are real and simple and lie in the interval \([-1,1]\).

If a cubic spline \(y\) on \([a, b]\) is three times continuously differentiable, show that \(y\) must be a polynomial.

Show that if a norm is strictly convex, then \(\|x\|=\|y\|=1\) and \(x \neq y\) together imply that for all \(\alpha\) such that \(0<\alpha<1\) one has $$ \|\alpha x+(1-\alpha) y\|<1 $$ Show that this condition is also sufficient for strict convexity.

Let \(Y \subset C[a, b]\) satisfy the Haar condition and consider any \(x \in C[a, b]\). If \(y \in Y\) is such that \(x-y\) has alternately positive and negative values at \(n+1\) consecutive points in \([a, b]\), where \(n=\operatorname{dim} Y\), show that the distance \(\delta\) of the best approximation to \(x\) out of \(Y\) is at least equal to the smallest of the absolute values of those \(n+1\) values of \(x-y\).

The hypergeometric differential equation is $$ \tau(1-\tau) \frac{d^{2} w}{d \tau^{2}}+[c-(a+b+1) \tau] \frac{d w}{d \tau}-a b w=0 $$ where \(a, b, c\) are constants. Application of the Frobenius method (extended power series method) shows that a solution is given by $$ \begin{gathered} w(\tau)=F(a, b, c ; \tau) \\ =1+\sum_{m=1}^{\infty} \frac{a(a+1) \cdots(a+m-1) b(b+1) \cdots(b+m-1)}{m ! c(c+1) \cdots(c+m-1)} \tau^{m} \end{gathered} $$ where \(c \neq 0,-1,-2, \ldots .\) The series on the right is called the hypergeometric series. Under what conditions does the series reduce to a finite sum? \(F(a, b, c ; \tau)\) is called the hypergeometric function. It has been investigated in great detail. Many functions can be expressed in terms of this function. This includes the Chebyshev polynomials. In fact, show that $$ T_{n}(t)=F\left(-n, n, \frac{1}{2} ; \frac{1}{2}-\frac{t}{2}\right) $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.