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Chapter 11: Problem 17

Represent \(f(x)\) as an integral (13) $$f(x)=\left\\{\begin{array}{ccc} \pi-x & \text { if } & 0 < x < \pi \\ 0 & \text { if } & x > \pi \end{array}\right.$$

Short Answer

Expert verified
Express \(f(x)\) using the characteristic function in the integral: \(f(x) = \int_{0}^{x} (-1 \cdot \chi_{(0,\pi)}(t)) dt\).

Step by step solution

01

Understand the Function

Given the function \(f(x)\), it behaves as \(\pi - x\) for \(0 < x < \pi\) and as \(0\) for \(x > \pi\). We want to express this in integral form, which means finding a scenario where \(f(x)\) is represented by an integral expression that matches this piecewise definition.
02

Set Up the Integral Representation

Consider the function \(h(x) = \pi - x\) and note that we need an integral that provides this value for \(0 < x < \pi\) and 0 elsewhere. This requires using a characteristic function \(\chi_{(0,\pi)}(x)\), which is 1 for \(0 < x < \pi\) and 0 otherwise, such that \(f(x) = \int_{0}^{x} ( ext{{some function in }} t) dt\) fits the definition.
03

Apply the Characteristic Function

One way to express \(f(x)\) is to use an integral form that involves the characteristic function: \[ f(x) = \int_{0}^{x} ( ext{{h(t) \(\cdot\) \(\chi_{(0,\pi)}(t)\)}}) dt \]. This would ensure that the integral evaluates to \(\pi - x\) for \(0 < x < \pi\) and 0 for \(x > \pi\).
04

Integrate to Match the Piecewise Function

Choose \(h(t) = -1\) for the active interval, making the integral result \(\int_{0}^{x} (-1 \cdot \chi_{(0,\pi)}(t)) dt = -(x-0) = -x\). Thus, \(f(x) = \pi - x\) within the specified interval. This is equivalent to adding \(\pi\) to \(-x\) which matches the desired behavior for the integral representation of \(f(x)\).

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Key Concepts

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

Piecewise Function
A piecewise function is a mathematical expression that has different expressions or rules for different parts of its domain. In simpler terms, it's a function defined by multiple sub-functions, each applying to a certain interval or condition. This type of function is particularly useful for modeling situations where a rule changes at a certain point.

For example, in the exercise you have, the function is divided into two distinct behaviors:
  • For the interval \(0 < x < \pi\), the function is \(f(x) = \pi - x\), which means you subtract \(x\) from \(\pi\) within this range.
  • For \(x > \pi\), the function simply takes the value \(0\).
These divisions help make mathematical models more accurate by allowing them to change behavior at specified intervals.
Characteristic Function
In integral calculus, a characteristic function is a functional tool that helps to define a piecewise function within an integral. Basically, it's a way to switch the integral 'on' or 'off' over certain intervals.

For the problem at hand:
  • The characteristic function \(\chi_{(0,\pi)}(x)\) is 1 when \(x\) is in the interval \(0 < x < \pi\) and 0 elsewhere.
  • This allows the integral to only "activate" and return meaningful results when \(x\) is within the desired range.
This function is crucial for the integral representation of a piecewise function because it helps in accurately capturing the behavior only in specified sub-intervals.
Integral Calculus
Integral calculus, a branch of mathematics, deals with finding the total accumulation of quantities, such as areas under a curve. It uses integrals to solve problems related to accumulation and area.

In the given exercise, the task was to express a piecewise function \(f(x)\) as an integral. Here's how this works:
  • The integral \((\int_{0}^{x} h(t) \cdot \chi_{(0,\pi)}(t) \, dt))\) computes the accumulation of a function \(h(t)\) modified by a characteristic function.
  • This integral must satisfy the behavior of \(f(x) = \pi - x\) for \(0 < x < \pi\) and zero beyond this interval.
  • By incorporating \(\chi_{(0,\pi)}(t)\), the integral respects the specified interval boundaries, only evaluating where the piecewise definition of \(f(x)\) requires it.
Integral calculus allows these complex behaviors to be generalized and solved using a uniform approach.

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Most popular questions from this chapter

Find the Fourier transform of \(f(x)\) (without using Table \(\mathbb{U}\) in Sec. 11.10 ). Show the details. $$f(x)=\left\\{\begin{aligned}-1 & \text { if }-1

Is the given function even or odd? Find its Fourier series. Sketch or graph the function and some partial sums. (Show the details of your work.) $$f(x)=\left\\{\begin{array}{ll} 2 & \text { if }-2 < x < 0 \\ 0 & \text { if } 0 < x < 2 \end{array}\right.$$

Find (a) the Fourier cosine series, (b) the Fourier sine series. Sketch \(f(x)\) and its two periodic extensions. (Show the details of your work.) $$f(x)=\pi-x \quad(0 < x < \pi)$$

Find (a) the Fourier cosine series, (b) the Fourier sine series. Sketch \(f(x)\) and its two periodic extensions. (Show the details of your work.) $$f(x)=\left\\{\begin{array}{ll} 1 & (0 < x < 1) \\ 2 & (1 < x < 2) \end{array}\right.$$

Find a general solution of the ODE \(y^{\prime \prime}+a^{2} y=r(t)\) with \(r(f)\) as given. (Show the details of your work.) $$r(t)=\frac{\pi}{4}|\sin t| \text { if }-\pi
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