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In order to prove the crucial formulas (5.83)?5.85) for the Fourier coefficients \(a_{n}\) and \(b_{n},\) you must first prove the following: $$\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \cos (m \omega t) d t=\left\\{\begin{array}{ll} \tau / 2 & \text { if } m=n \neq 0 \\ 0 & \text { if } m \neq n \end{array}\right.$$ (This integral is obviously \(\tau\) if \(m=n=0 .\) ) There is an identical result with all cosines replaced by sines, and finally $$\int_{-\tau / 2}^{\tau / 2} \cos (n \omega t) \sin (m \omega t) d t=0 \quad \text { for all integers } n \text { and } m$$ where as usual \(\omega=2 \pi / \tau\). Prove these. [Hint: Use trig identities to replace \(\cos (\theta) \cos (\phi)\) by terms like \(\cos (\theta+\phi) \text { and so on. }]\)

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 \(\cos(n \omega t) \cos(m \omega t)\) over the interval \([-\tau/2, \tau/2]\). 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 \(\cos(A) \cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)]\) to simplify the integral. Replace \( A = n\omega t \) and \( B = m\omega t \) to get \[\cos(n \omega t) \cos(m \omega t) = \frac{1}{2}[\cos((n+m)\omega t) + \cos((n-m)\omega t)].\]
03

Integrate the Resulting Expressions

Split the integral into two separate parts, each involving the cosine of a single angle: \[ \int_{-\tau/2}^{\tau/2} \cos(n \omega t) \cos(m \omega t) \, dt = \frac{1}{2} \int_{-\tau/2}^{\tau/2} \cos((n+m)\omega t) \, dt + \frac{1}{2} \int_{-\tau/2}^{\tau/2} \cos((n-m)\omega t) \, dt. \] Evaluate each of these integrals.
04

Evaluate Special Cases

Consider the three cases: (i) \(n = m = 0\), (ii) \(m = n eq 0\), and (iii) \(m eq n\).- For case (i): \(\cos(0) = 1\), the integral becomes \(\int_{-\tau/2}^{\tau/2} 1 \, dt = \tau.\)- For case (ii): Both terms simplify to \(\int_{-\tau/2}^{\tau/2} \cos(0) \, dt = \frac{\tau}{2}.\)- 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 \( \sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \) to simplify, and verify that the integral results in 0 for \(\int_{-\tau/2}^{\tau/2} \sin(n \omega t) \sin(m \omega t) \, dt \), again resulting in zero unless both are zero, which does not apply here.
06

Integrate Cosine and Sine Product

Lastly, consider \( \int_{-\tau/2}^{\tau/2} \cos(n \omega t) \sin(m \omega t) \, dt \). 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:
  • \(\cos(A) \cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)]\)
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:
  • \(\int_{-\tau/2}^{\tau/2} \cos(n \omega t) \sin(m \omega t) \, dt = 0\) for all integers \(n\) and \(m\).
  • \(\int_{-\tau/2}^{\tau/2} \cos(n \omega t) \cos(m \omega t) \, dt = 0\) if \(m eq n\).
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:
  • \(\int_{-\tau/2}^{\tau/2} 1 \, dt = \tau\), which finds the 'area' under a flat line across the interval.
  • \(\int_{-\tau/2}^{\tau/2} \cos((n+m)\omega t) \, dt\) and \(\int_{-\tau/2}^{\tau/2} \cos((n-m)\omega t) \, dt\) 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

A massless spring has unstretched length \(l_{\mathrm{o}}\) 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 \(l_{1}\). (a) Write down the condition that determines \(l_{1}\). 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 \(F=-k x\). 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 \(+\frac{1}{2} k x^{2}\)

This problem is to refresh your memory about some properties of complex numbers needed at several points in this chapter, but especially in deriving the resonance formula (5.64). (a) Prove that any complex number \(z=x+i y\) (with \(x\) and \(y\) real) can be written as \(z=r e^{i \theta}\) where \(r\) and \(\theta\) are the polar coordinates of \(z\) in the complex plane. (Remember Euler's formula.) (b) Prove that the absolute value of \(z\), defined as \(|z|=r,\) is also given by \(|z|^{2}=z z^{*},\) where \(z^{*}\) denotes the complex conjugate of \(z\) defined as \(z^{*}=x-\)iy. \(\left(\text { c) Prove that } z^{*}=r e^{-i \theta} . \text { (d) Prove that }(z w)^{*}=z^{*} w^{*} \text { and that }(1 / z)^{*}=1 / z^{*}\right.\) (e) Deduce that if \(z=a /(b+i c),\) with \(a, b,\) and \(c\) real, then \(|z|^{2}=a^{2} /\left(b^{2}+c^{2}\right)\).

Consider a simple harmonic oscillator with period \(\tau\). Let \(\langle f\rangle\) denote the average value of any variable \(f(t),\) averaged over one complete cycle: $$\langle f\rangle=\frac{1}{\tau} \int_{0}^{\tau} f(t) d t$$ Prove that \(\langle T\rangle=\langle U\rangle=\frac{1}{2} E\) where \(E\) is the total energy of the oscillator. [Hint: Start by proving the more general, and extremely useful, results that \(\left\langle\sin ^{2}(\omega t-\delta)\right\rangle=\left\langle\cos ^{2}(\omega t-\delta)\right\rangle=\frac{1}{2} .\) Explain why these two results are almost obvious, then prove them by using trig identities to rewrite \(\sin ^{2} \theta\) and \(\left.\cos ^{2} \theta \text { in terms of } \cos (2 \theta) .\right]\)

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscillates is each axle assembly, comprising the axle and its two wheels.) Find the speed of my car at which this oscillation resonates, given the following information: (a) When four \(80-\mathrm{kg}\) men climb into my car, the body sinks by a couple of centimeters. Use this to estimate the spring constant \(k\) of each of the four springs. (b) If an axle assembly (axle plus two wheels) has total mass \(50 \mathrm{kg}\), what is the natural frequency of the assembly oscillating on its two springs? ( \(\mathbf{c}\) ) If the bumps on a road are \(80 \mathrm{cm}\) apart, at about what speed would these oscillations go into resonance?

The potential energy of two atoms in a molecule can sometimes be approximated by the Morse function, $$U(r)=A\left[\left(e^{(R-r) / S}-1\right)^{2}-1\right]$$ where \(r\) is the distance between the two atoms and \(A, R,\) and \(S\) are positive constants with \(S \ll R .\) Sketch this function for \(0 < r < \infty\). Find the equilibrium separation \(r_{\mathrm{o}}\), at which \(U(r)\) is minimum. Now write \(r=r_{\mathrm{o}}+x\) so that \(x\) is the displacement from equilibrium, and show that, for small displacements, \(U\) has the approximate form \(U=\) const \(+\frac{1}{2} k x^{2} .\) That is, Hooke's law applies. What is the force constant \(k ?\)

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