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In order to prove the crucial formulas (5.83)?5.85) for the Fourier coefficients an and bn, you must first prove the following: $$\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \cos (m \omega t) d t=\left\{τ/2 if m=n00 if mn\right.(Thisintegralisobviously\(τ\)if\(m=n=0.\))Thereisanidenticalresultwithallcosinesreplacedbysines,andfinally\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \sin (m \omega t) d t=0 \quad \text { for all integers } n \text { and } m$$ where as usual ω=2π/τ. Prove these. [Hint: Use trig identities to replace cos(θ)cos(ϕ) by terms like cos(θ+ϕ) and so on. ]

Short Answer

Expert verified
Use trig identities to show integrals yield specific values or zero based on m and n values.

Step by step solution

01

Understand the Function to Integrate

We need to integrate functions of the form cos(nωt)cos(mωt) over the interval [τ/2,τ/2]. The result depends on whether n is equal to m or not, and whether n and m are both zero or not.
02

Apply Trigonometric Identity

Use the identity cos(A)cos(B)=12[cos(A+B)+cos(AB)] to simplify the integral. Replace A=nωt and B=mωt to get cos(nωt)cos(mωt)=12[cos((n+m)ωt)+cos((nm)ωt)].
03

Integrate the Resulting Expressions

Split the integral into two separate parts, each involving the cosine of a single angle: τ/2τ/2cos(nωt)cos(mωt)dt=12τ/2τ/2cos((n+m)ωt)dt+12τ/2τ/2cos((nm)ωt)dt. Evaluate each of these integrals.
04

Evaluate Special Cases

Consider the three cases: (i) n=m=0, (ii) m=neq0, and (iii) meqn.- For case (i): cos(0)=1, the integral becomes τ/2τ/21dt=τ.- For case (ii): Both terms simplify to τ/2τ/2cos(0)dt=τ2.- For case (iii): Both integrals evaluate to zero due to the periodic nature of cosines with frequencies not equal to zero.
05

Verify Results with Sines

Perform similar steps replacing cosines with sines. Use the identity sin(A)sin(B)=12[cos(AB)cos(A+B)] to simplify, and verify that the integral results in 0 for τ/2τ/2sin(nωt)sin(mωt)dt, again resulting in zero unless both are zero, which does not apply here.
06

Integrate Cosine and Sine Product

Lastly, consider τ/2τ/2cos(nωt)sin(mωt)dt. Since cos and sin are orthogonal over a full period, any such integral is zero. This works for all integer n and m, confirming that the result is zero.

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are the building blocks for simplifying expressions involving trigonometric functions. They provide relationships between functions like sine, cosine, and tangent that allow us to transform complex expressions into simpler ones. In our case, we used the identity:
  • cos(A)cos(B)=12[cos(A+B)+cos(AB)]
This identity lets us express the product of two cosine functions as a sum of two other cosines.
This simplification is particularly useful when integrating products of trigonometric functions. By breaking down the original product into simpler terms, we can more easily compute integrals, which form the core of finding Fourier coefficients.
It's crucial to understand and be able to derive these identities, as they enable us to solve complex mathematical problems more efficiently.
Orthogonality
Orthogonality in the context of trigonometric functions refers to certain pairs of functions being mutually "independent" over specific intervals. This concept is central in Fourier analysis.
Essentially, for functions defined over the same interval, the integral of their product can result in zero, which mathematically states their independence or orthogonality. Key points are:
  • τ/2τ/2cos(nωt)sin(mωt)dt=0 for all integers n and m.
  • τ/2τ/2cos(nωt)cos(mωt)dt=0 if meqn.
This principle of orthogonality is crucial when dealing with series like Fourier series, as it simplifies the integration process by canceling out terms that would typically complicate calculations.
Understanding orthogonality helps clarify why certain terms vanish in integrations, thus simplifying the solution of differential equations and the decomposition of signals into their frequency components.
Integrals
Integrals are fundamental in calculus and are used to find quantities like area, volume, and in this context, Fourier coefficients. The concept involves accumulating quantities over a range, and the integral of a function over a specific interval gives an aggregate measure of that function along that interval. In our problem:
  • τ/2τ/21dt=τ, which finds the 'area' under a flat line across the interval.
  • τ/2τ/2cos((n+m)ωt)dt and τ/2τ/2cos((nm)ωt)dt simplify due to properties of periodic functions, often resulting in zero unless special conditions are met.
Understanding how to evaluate these integrals allows us to compute the Fourier coefficients that represent periodic functions in terms of sine and cosine functions.
This understanding is vital for signal processing, systems analysis, and solving physics problems where periodic functions play a pivotal role.

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Most popular questions from this chapter

Consider an underdamped oscillator (such as a mass on the end of a spring) that is released from rest at position xo at time t=0. (a) Find the position x(t) at later times in the form x(t)=eβt[B1cos(ω1t)+B2sin(ω1t)] That is, find B1 and B2 in terms of xo. (b) Now show that if you let β approach the critical value ωo your solution automatically yields the critical solution. (c) Using appropriate graphing software, plot the solution for 0t20, with xo=1,ωo=1, and β=0,0.02,0.1,0.3, and 1

We know that if the driving frequency ω is varied, the maximum response (A2) of a driven damped oscillator occurs at ωωo (if the natural frequency is ωo and the damping constant β ωo ). Show that A2 is equal to half its maximum value when ωωo±β, so that the full width at half maximum is just 2β. [Hint: Be careful with your approximations. For instance, it's fine to say ω+ωo2ωo, but you certainly mustn't say ωωo0.]

A massless spring has unstretched length lo and force constant k. One end is now attached to the ceiling and a mass m is hung from the other. The equilibrium length of the spring is now l1. (a) Write down the condition that determines l1. Suppose now the spring is stretched a further distance x beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is F=kx. That is, the net force obeys Hooke's law, when x is the distance from the equilibrium position a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal. (b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form U(x)= const +12kx2

Suppose that you have found a particular solution xp(t) of the inhomogeneous equation (5.48) for a driven damped oscillator, so that Dxp=f in the operator notation of (5.49). Suppose also that x(t) is any other solution, so that Dx=f. Prove that the difference xxp must satisfy the corresponding homogeneous equation, D(xxp)=0. This is an alternative proof that any solution x of the inhomogeneous equation can be written as the sum of your particular solution plus a homogeneous solution; that is, x=xp+xh.

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