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If f(x)=Lx for 0<x<2L, and if f(x+2L)=f(x), find a formula for f(x) in the interval L<x<0.

Short Answer

Expert verified
The formula for f(x) in the interval (-L, 0) is f(x) = x - L with x belonging to the interval (-L, 0).

Step by step solution

01

Use the periodic statement

We have the periodic statement f(x+2L)=f(x). Let's substitute the given function f(x)=Lx: f(x+2L)=L(x+2L)=Lx2L=xL
02

Find a suitable value for x

We want to find a formula for f(x) in the interval (L,0). Let's set x=L+y, where y belongs to the interval (0,L). Therefore, x belongs to the interval (L,0), which is the desired interval.
03

Substitute the value of x in the general formula

Now, let's substitute x=L+y in the general formula f(x)=xL: f(L+y)=(L+y)L=yL
04

Final formula

Now we have the formula for f(x) in the interval (L,0): f(x)=xL with x belonging to the interval (L,0).

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Key Concepts

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

Periodicity in Functions
Imagine a heartbeat displayed on a monitor; it goes up and down in a continuous, rhythmic pattern. Just like that heartbeat, periodic functions have a regular pulse; they repeat their values at fixed intervals. In the world of mathematics, this 'heartbeat' is known as the period.

The period of a function is the smallest positive value for which the function's pattern repeats. For example, in the exercise, we are given that the function defined by f(x)=Lx for 0<x<2L repeats itself every 2L units. This means that f(x+2L) must be equal to f(x), defining the function's periodic nature. In essence, no matter how far you travel along the x-axis, if you move a distance of 2L, you will encounter the same function value.

Understanding the periodicity is critical when expanding the function across different intervals. In the step-by-step solution, when asked to find f(x) for L<x<0, recognizing that the function's period is 2L allows us to confidently assert that the function should behave identically in this alternate range, albeit with some necessary transformations.
Transformations of Functions
If the concept of periodic functions is the heartbeat, think of transformations as dance moves that can shift, stretch, or flip that heartbeat in various directions. Transformations of functions are like instructions on how to maneuver a graph around the coordinate plane without changing its shape.

Let's explore some common function dances, shall we?
  • Translation: Think of it as a slide—moving a function up, down, left, or right. In our exercise, the function f(x)=Lx had to be translated to make sense in the interval L,0.
  • Reflection: This is akin to looking in a mirror. The function can be flipped over an axis. If we had f(x)=x, reflecting it over the y-axis would give us f(x)=x.
  • Stretching or Compression: This modifies how 'tall' or 'wide' our function is. If you multiply the function by a number greater than 1, it stretches; if you multiply it by a number between 0 and 1, it compresses.

In the provided exercise, by setting x=L+y, we perform a translation that moves the initially defined function into our target interval. This smart shift is needed to establish the function in the negative domain and helps us appreciate how transformations can unveil new expressions for the same periodic function in different intervals.
Piecewise-Defined Function
Life isn't always a smooth ride, and neither are functions—they can behave differently in different intervals. This is where piecewise-defined functions come into play; they allow a function to have different expressions depending on the part of the domain you're looking at. They are the 'choose-your-own-adventure' stories of the math world.

A piecewise-defined function is like a mathematical chameleon, changing its rule of operation depending on the 'environment' (which, in this case, is the interval of x-values). In the exercise, we see the hint of a piecewise-defined function where f(x)=Lx within the interval 0<x<2L, but the problem asks what happens for L<x<0.

Utilizing the concept of periodicity and transformations, we've deduced a new piece of the function's rule. We end up with a two-part function:
  • For 0<x<2L, we use f(x)=Lx.
  • And we've determined for L<x<0, the function should be defined as f(x)=xL.

This is a simplified representation, but it illustrates the core of what piecewise functions are about. They piece together different behaviors to create a function that can effectively handle a diverse set of conditions.

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Most popular questions from this chapter

Dimensionless variables can be introduced into the wave equation a2uxx=utt in the following manner. Let s=x/L and show that the wave equation becomes a2uss=L2utt Then show that L/a has the dimensions of time, and thus can be used as the unit on the time scale. Finally, let τ=at/L and show the wave equation then reduces to uss=uττ

By writing Laplace's equation in cylindrical coordinates r,θ, and z and then assuming that the solution is axially symmetric (no dependence on θ), we obtain the equation urr+(1/r)ur+uzz=0 Assuming that u(r,z)=R(r)Z(z), show that R and Z satisfy the equations rRn+R+λ2rR=0,Zλ2Z=0 The equation for R is Bessel's equation of order zero with independent variable λ.

Let F(x)=0xf(t)dt. Show that if f is even, then F is odd, and that if f is odd, then F is even.

(a) Sketch the graph of the given function for three periods. (b) Find the Fourier series for the given function. $$ f(x)=\left\{x+1,1x<0,1x,0x<1; \quad f(x+2)=f(x)\right. $$

More Specialized Fourier Scries. Let f be a function originally defined on 0xL. In this section we have shown that it is possible to represent f either by a sine series or by a cosine series by constructing odd or even periodic extensions of f, respectively. Problems 38 through 40 concern some other more specialized Fourier series that converge to the given function f on (0,L).  Let f be extended into (L,2L] inan arbitrary manner. Then extend the resulting function  into (2L,0) as an odd function and elsewhere as a periodic function ofperiod 4L (see  Figure 10.4.6). Showthat this function has a Fourier sine series in terms of the functions  sin(nπx/2L),n=1,2,3,. that is,  f(x)=n=1bnsin(nπx/2L) where bn=1L02Lf(x)sin(nπx/2L)dx  This series converges to the original function on (0,L) (Figure cant copy)

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