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If \(f(x)=-x\) for \(-L < x < L,\) and if \(f(x+2 L)=f(x),\) find a formula for \(f(x)\) in the interval \(L < x < 2 L\) in the interval \(-3 L < x < -2 L\)

Short Answer

Expert verified
Question: Write down the formulas for the function \(f(x) = -x\) in the intervals \(L < x < 2L\) and \(-3L < x < -2L\), given the condition \(f(x+2L) = f(x)\). Answer: In the interval \(L < x < 2L\), the function \(f(x) = -x + 2L\). In the interval \(-3L < x < -2L\), the function \(f(x) = -x - 6L\).

Step by step solution

01

Understand the given condition

In the given exercise, the function \(f(x)=-x\) and we are given a condition \(f(x+2L)=f(x)\). This condition tells us that the function repeats itself every \(2L\) units. In other words, the function has a period of \(2L\).
02

Find formula in interval \(L < x < 2L\)

We know that the function repeats itself every \(2L\) units. Therefore, in the interval \(L < x < 2L\), we would have \(f(x)=f(x-2L)\). Given the initial function \(f(x)=-x\), we can substitute \(x-2L\) instead of \(x\) in the function to find the formula for this interval: $$f(x)=-x$$ $$f(x-2L)=-(x-2L)$$ Now, simplifying the expression: $$f(x-2L)=-x+2L$$ So in the interval \(L < x < 2L\), the function \(f(x)=-x+2L\).
03

Find formula in interval \(-3L < x < -2L\)

Similarly, for the interval \(-3L < x < -2L\), we can find the formula using the given condition \(f(x+2L)=f(x)\). In this case, we would have \(f(x)=f(x+6L)\). Substituting \(x+6L\) instead of \(x\) in the initial function \(f(x)=-x\): $$f(x)=-x$$ $$f(x+6L)=-(x+6L)$$ Now, simplifying the expression: $$f(x+6L)=-x-6L$$ Therefore, in the interval \(-3L < x < -2L\), the function \(f(x)=-x-6L\).
04

Final formulas

We have found the formulas for the function \(f(x)\) in the specified intervals: 1. In the interval \(L < x < 2L\), the function \(f(x) = -x + 2L\). 2. In the interval \(-3L < x < -2L\), the function \(f(x) = -x - 6L\).

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. They describe the rate of change of a quantity as a function of another quantity, often time. In the context of periodic functions, differential equations can be used to model waves, oscillations, and other phenomena that repeat at regular intervals.

For example, the simple harmonic motion of a spring-mass system can be described by the second-order linear differential equation \( m\frac{d^2x}{dt^2} + kx = 0 \), where \(x(t)\) is the position of the mass at time \(t\), \(m\) is the mass, and \(k\) is the spring constant. The general solutions to such equations often contain sine and cosine functions, which are inherently periodic, reflecting the periodic nature of the system being modeled.
Boundary Value Problem
Boundary value problems (BVPs) are a type of differential equation where the solution is determined based on known values, or boundary conditions, at the edges of the domain. In the exercise, although the problem does not directly involve differential equations, it presents a condition that is akin to a boundary value problem in the sense that the behavior of the function beyond its initial interval is constrained by given conditions.

Specifically, the periodic nature of the function acts as a boundary condition, leading to the solution in different intervals. When solving BVPs, the known conditions at the boundaries guide us to find a precise solution that fulfills the requirements throughout the entire domain, which, in our case, is the extension of the periodic function outside the initial interval.
Function Transformation
Function transformation involves changing a function's graph in various ways, including shifting, reflecting, stretching, or compressing. In the exercise, the fundamental concept of transforming the function \(f(x) = -x\) is applied through horizontal shifts to satisfy the periodic condition \(f(x + 2L) = f(x)\).

This transformation can be visualized on a graph where the original line will be shifted horizontally by \(2L\) units to align with the periodic condition. In practice, this means if you were to draw the graph of \(f(x)\), every \(2L\) units along the horizontal axis, the graph would repeat its shape.

Understanding Periodic Shifts

When \(f(x)\) is given to be periodic, we can translate it across the horizontal axis to find what the function looks like on different intervals. For instance, shifting \(f(x)\) to the right by \(2L\) units gives us the function for \(L < x < 2L\). The same principle can be applied in both directions to explore the periodicity of \(f(x)\) as shown in the step-by-step solution presented previously.

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