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Chapter 8: Problem 3

Let \(\eta(t)=e^{-1 / t}\) for \(t>0, \eta(t)=0\) for \(t \leq 0\). a. For \(k \in \mathbb{N}\) and \(t>0, \eta^{(k)}(t)=P_{k}(1 / t) e^{-1 / t}\) where \(P_{k}\) is a polynomial of degree \(2 k\). b. \(\eta^{(k)}(0)\) exists and equals zero for all \(k \in \mathbb{N}\).

Short Answer

Expert verified
\(\eta^{(k)}(t)=P_{k}(1/t)e^{-1/t}\) and \(\eta^{(k)}(0)=0\).

Step by step solution

01

Understand the Function Definition

The given function is described as \( \eta(t) = e^{-1/t} \) for \( t > 0 \) and \( \eta(t) = 0 \) for \( t \leq 0 \). We need to investigate properties of its derivatives, particularly for \( t > 0 \) in part a, and for \( t = 0 \) in part b.
02

Derive \\( \eta^{(k)}(t) \\) for \\( t > 0 \\)

We need to derive \( \eta^{(k)}(t) \) assuming \( t > 0 \). Start with the first derivative: \[ \eta'(t) = \frac{d}{dt} \left( e^{-1/t} \right) = e^{-1/t} \cdot \left( \frac{1}{t^2} \right). \] This is because of the chain rule: \( \frac{d}{dt}(-1/t) = \frac{1}{t^2}. \). The first derivative can be seen as \( P_1(1/t) e^{-1/t} \) with \( P_1(x) = x^2 \), a polynomial of degree 2.
03

Generalize for Higher Derivatives

For the k-th derivative \( \eta^{(k)}(t) = P_k(1/t) e^{-1/t} \), where \( P_k(x) \) is a polynomial of degree \( 2k \). The polynomial arises from repeated application of the product rule and chain rule, leading to more factors of \( \frac{1}{t^2} \) and terms involving powers of \( 1/t \). Each differentiation introduces more factors of \( 1/t \), contributing to the polynomial's degree.
04

Analyze \\( \eta^{(k)}(0)\\)

For \( t \leq 0 \), \( \eta(t) = 0 \), thus for \( t = 0 \), \( \eta^{(k)}(0) \) would intuitively be from \( \lim_{t \to 0^+} \eta^{(k)}(t) \).The exponential term \( e^{-1/t} \) for any \( t > 0 \) tends to 0 very quickly as \( t \to 0^+ \), resulting in all derivatives at \( t = 0 \) equating to 0 because \( e^{-1/t} \) dominates \( (1/t)^n \) for any \( n \).

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Key Concepts

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

Derivatives
Derivatives, in real analysis, are fundamental tools for measuring how a function changes. When you have a function like \( \eta(t) = e^{-1/t} \) for \( t > 0 \), finding the derivative involves applying rules to determine how the function behaves as \( t \) changes slightly. To compute the first derivative of \( \eta(t) \), you apply the chain rule, which helps us understand how composite functions are differentiated.
  • The derivative \( \eta'(t) = \frac{d}{dt}(e^{-1/t}) \) involves differentiating \(-1/t \), which results in \( e^{-1/t} \cdot (1/t^2) \).
  • Each subsequent derivative will introduce more factors like \( 1/t^2 \), increasing the complexity and degree of the resulting expression.
By continuing this differentiation process multiple times, you're assessing how each derivative modifies the rate of change even further.
Polynomials
Polynomials are mathematical expressions involving sums of powers of variables. In this exercise, polynomials appear in the form of \( P_k(1/t) \), with \( P_k(x) \) being a polynomial of degree \( 2k \). This relates to the derivatives of \( \eta(t) \), where each derivative introduces more powers of \( 1/t \).
  • As you compute higher derivatives, the expression for each \( k \)-th derivative embeds these polynomials naturally.
  • This is due to repeated application of the chain and product rules, increasing the degree by 2 with each differentiation.
Polynomials are crucial in simplifying and understanding complex expressions, such as those arising in advanced calculus and real analysis.
Chain Rule
The chain rule is an essential concept in calculus used to differentiate composite functions. It allows us to understand how changes in one function affect another when they are composed.
  • For \( \eta(t) = e^{-1/t} \), the chain rule helps differentiate by addressing the outer function \( e^{u} \) and the inner function \( u = -1/t \).
  • This lets us compute derivatives like \( \frac{d}{dt} e^{-1/t} = e^{-1/t} \cdot \frac{d}{dt}(-1/t) \).
Applying the chain rule repeatedly is key to breaking down complex expressions into manageable parts, making it easier to find derivatives in real analysis.
Limits
In real analysis, limits are used to describe the behavior of functions as inputs approach a value. They become especially relevant when evaluating functions at points where they might be otherwise undefined, such as \( t = 0 \) in this example.
  • The exercise involves computing \( \eta^{(k)}(0) \) using \( \lim_{t \to 0^+} \eta^{(k)}(t) \).
  • As \( t \to 0^+ \), the term \( e^{-1/t} \) tends immediately to 0 faster than any polynomial expression related to \( 1/t \).
Understanding limits helps determine the values of functions at points where direct substitution is not possible.
Mathematical Functions
Mathematical functions, like \( \eta(t) \), serve as rules or relationships that assign each input a specific output. These functions are the backbone of real analysis, describing continuous transitions and behaviors across their domains.
  • In \( \eta(t) \) for \( t > 0 \), the function is defined as \( e^{-1/t} \), which decays very rapidly as \( t \) approaches zero.
  • For \( t \leq 0 \), this function is set to 0, showcasing a piecewise definition that ensures continuity and controlled behavior over different intervals.
By studying these functions through derivatives, limits, and more, we gain deeper insight into their properties and applications in various mathematical fields.

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Most popular questions from this chapter

Let \(\operatorname{sinc} x=(\sin \pi x) / \pi x(\operatorname{sinc} 0=1)\). a. If \(a>0, \widehat{\chi}_{|-a, a|}(x)=\chi_{[a, a]}^{V}(x)=2 a \operatorname{sinc} 2 a x\). b. Let \(\mathcal{H}_{a}=\left\\{f \in L^{2}: \hat{f}(\xi)=0\right.\) (a.c.) for \(\left.|\xi|>a\right\\}\). Then \(\mathcal{T}\) is a Hilbert space and \(\\{\sqrt{2 a}\) sinc \((2 a x-k): k \in Z\\}\) is an orthonormal basis for \(\mathcal{H}\). c. (The Sampling Theorem) If \(f \in \mathcal{T}_{a}\), then \(f \in C_{0}\) (after modification on a null set), and \(f(x)=\sum_{-\infty}^{\infty} f(k / 2 a) \operatorname{sinc}(2 a x-k)\), where the series converges both uniformly and in \(L^{2}\). (In the terminology of signal analysis, a signal of bandwidth \(2 a\) is completely determined by sampling its values at a sequence of points \(\\{k / 2 a\\}\) whose spacing is the reciprocal of the bandwidth.)

Given \(a>0\), let \(f(x)=e^{-2 \pi x} x^{a-1}\) for \(x>0\) and \(f(x)=0\) for \(x \leq 0\). a. \(f \in L^{1}\), and \(f \in L^{2}\) if \(a>\frac{1}{2}\). b. \(\hat{f}(\xi)=\Gamma(a)[(2 \pi)(1+i \xi)]^{-a}\). (Here we are using the branch of \(z^{a}\) in the right half plane that is positive when \(z\) is positive. Cauchy's theorem may be used to justify the complex substitution \(y=(1+i \xi) x\) in the integral defining \(\widehat{f_{0}}\) ) c. If \(a, b>\frac{1}{2}\) then $$ \int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} d x=\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} $$

Let \(\phi(x)=e^{-|x| / 2}\) on \(\mathbb{R}\). Use the Fourier transform to derive the solution \(u=f * \phi\) of the differential equation \(u-u^{\prime \prime}=f\), and then check directly that it works. What hypotheses are needed on \(f\) ?

22\. Since \(F\) commutes with rotations, the Fourier transform of a radial function is radial; that is, if \(F \in L^{1}\left(\mathbb{R}^{n}\right)\) and \(F(x)=f(|x|)\), then \(\widehat{F}(\xi)=g(|\xi|)\), where \(f\) and \(g\) are related as follows. a. Let \(J(\xi)=\int_{S} e^{i x \xi} d \sigma(x)\) where \(\sigma\) is surface measure on the unit sphere \(S\) in \(\mathbb{R}^{n}\) (Theorem 2.49). Then \(J\) is radial - say, \(J(\xi)=j(|\xi|)\) - and \(g(\rho)=\int_{0}^{\infty} j(2 \pi r \rho) f(r) r^{n-1} d r .\) b. \(J\) satisfies \(\sum_{1}^{n} \partial_{k}^{2} J+J=0\). c. \(j\) satisfies \(\rho j^{\prime \prime}(\rho)+(n-1) j^{\prime}(\rho)+\rho j(\rho)=0\). (This equation is a variant of Bessel's equation. The function \(j\) is completely determined by the fact that it is a solution of this equation, is smooth at \(\rho=0\), and satisfies \(j(0)=\sigma(S)=\) \(2 \pi^{n / 2} / \Gamma(n / 2)\). In fact, \(j(\rho)=(2 \pi)^{n / 2} \rho^{(2-n) / 2} J_{(n-2) / 2}(\rho)\) where \(J_{\alpha}\) is the Bessel function of the first kind of order \(\alpha .)\) d. If \(n=3, j(\rho)=4 \pi \rho^{-1} \sin \rho\). (Set \(f(\rho)=\rho j(\rho)\) and use (c) to show that \(f^{\prime \prime}+f=0\). Alternatively, use spherical coordinates to compute the integral defining \(J(0,0, \rho)\) directly.) (use induction on \(k\) ), and in particular, $$ h_{k}(x)=\frac{(-1)^{k}}{\left[\pi^{1 / 2} 2^{k} k !\right]^{1 / 2}} e^{x^{2} / 2}\left(\frac{d}{d x}\right)^{k} e^{-x^{2}} . $$ f. Let \(H_{k}(x)=e^{x^{2} / 2} h_{k}(x)\). Then \(H_{k}\) is a polynomial of degree \(k\), called the \(k\) th normalized Hermite polynomial. The linear span of \(H_{0}, \ldots, H_{m}\) is the set of all polynomials of degree \(\leq m\). (The kth Hermite polynomial as usually defined is \(\left[\pi^{1 / 2} 2^{k} k !\right]^{1 / 2} H_{k}\).) g. \(\left\\{h_{k}\right\\}_{0}^{\infty}\) is an orthonormal basis for \(L^{2}(\mathbb{R})\). (Suppose \(f \perp h_{k}\) for all \(k\), and let \(g(x)=f(x) e^{-x^{2} / 2}\). Show that \(\widehat{g}=0\) by expanding \(e^{-2 \pi i \xi \cdot x}\) in its Maclaurin series and using (f).) h. Define \(A: L^{2} \rightarrow L^{2}\) by \(A f(x)=(2 \pi)^{1 / 4} f(x \sqrt{2 \pi})\), and define \(\tilde{f}=\) \(A^{-1} \mathcal{F} A f\) for \(f \in L^{2}\). Then \(A\) is unitary and \(\widetilde{f}(\xi)=(2 \pi)^{-1 / 2} \int f(x) e^{-i \xi x} d x\). Moreover, \(\widetilde{T f}=-i T(\widetilde{f})\) for \(f \in \mathcal{S}\), and \(\widetilde{h}_{0}=h_{0}\); hence \(\widetilde{h}_{k}=(-i)^{k} h_{k}\). Therefore, if \(\phi_{k}=A h_{k} .\left\\{\phi_{k}\right\\}_{0}^{\infty}\) is an orthonormal basis for \(L^{2}\) consisting of eigenfunctions for \(\mathcal{F}\); namely, \(\widehat{\phi}_{k}=(-i)^{k} \phi_{k}\).

If \(f\) is locally integrable on \(\mathbb{R}^{n}\) and \(g \in C^{k}\) has compact support, then \(f * g \in C^{k}\).
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