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Chapter 10: Problem 3

Apply Hake's Theorem to \(g(x):=(1-x)^{-1 / 2}\) for \(x \in[0,1)\) and \(g(1):=0\).

Short Answer

Expert verified
Hake's Theorem can be applied to the function \(g(x) = (1 - x)^{-1 / 2}\) as it satisfies all the conditions imposed by the theorem. Thus, the integral of \(g(x)\) on the interval [0,1] exists and is finite.

Step by step solution

01

Understand the function

Analyze the given function \(g(x) = (1 - x)^{-1 / 2}\). This function represents the reciprocal of the square root of the quantity (1 - x). It's defined for \(x \in [0,1)\) and \(g(1) = 0\)
02

Identify the conditions

To apply Hake's Theorem, two major conditions need to be met. First, the function should be defined almost everywhere. Secondly, it should tend to a finite limit as \(x\) approaches the end point. We can see that the function is defined almost everywhere except at \(x = 1\), but \(g(1)\) is given as 0, which makes the function defined everywhere.
03

Apply Hake's Theorem

Since the given conditions of Hake's Theorem are satisfied, we can say that the given function \(g(x)\) is finite almost everywhere and tends to a finite limit at \(x = 1\). Thus Hake's Theorem applies, and the integral \( \int_{0}^{1} g(x) dx \) exists and its value is finite.

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Most popular questions from this chapter

Let \(k:[0,1] \rightarrow \mathbb{R}\) be defined by \(k(x):=0\) if \(x \in[0,1]\) is 0 or is irrational, and \(k(m / n):=n\) if \(m, n \in \mathbb{N}\) have no common integer factors other than \(1 .\) Show that \(k \in \mathcal{R}^{*}[0,1]\) with integral equal to 0 . Also show that \(k\) is not continuous at any point, and not bounded on any subinterval \([c, d]\) with \(c

Let \(f \in \mathcal{R}^{*}[a, b]\), let \(g\) be monotone on \([a, b]\) and suppose that \(f \geq 0\). Then there exists \(\xi \in[a, b]\) such that \(\int_{a}^{b} f g=g(a) \int_{a}^{\xi} f+g(b) \int_{\xi}^{b} f .\) (This is a form of the Second Mean Value Theorem for integrals.)

If \(f_{k}(x):=k\) for \(x \in[1 / k, 2 / k]\) and \(f_{k}(x):=0\) elsewhere on \([0,2]\), show that \(f_{k}(x) \rightarrow 0\) but that \(\int_{0}^{2} f_{k}=1\)

Let \(f(x):=\cos x\) for \(x \in[0, \infty)\). Show that \(f \notin \mathcal{R}^{*}[0, \infty)\).

Let \(L_{1}(x):=x \ln |x|-x\) for \(x \neq 0\) and \(L_{1}(0):=0\), and let \(l_{1}(x):=\ln |x|\) if \(x \neq 0\) and \(l_{1}(0):=\) 0. If \([a, b]\) is any interval, show that \(l_{1} \in \mathcal{R}^{*}[a, b]\) and that \(\int_{a}^{b} \ln |x| d x=L_{1}(b)-L_{1}(a)\)
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