Question

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\ne 0\\ 0& \text{ if }m\ne n\end{array}$$\right.$$(Thisintegralisobviously\text{(}\tau \text{)}if\text{(}m=n=0.\text{)})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 $\omega =2\pi /\tau$. Prove these. [Hint: Use trig identities to replace $\cos (\theta )\cos (\varphi )$ by terms like $\cos (\theta +\varphi )\text{ and so on. }]$

Short answer

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

Step-by-step solution

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.

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 $$\cos (n\omega t)\cos (m\omega t)=\frac{1}{2}[\cos ((n+m)\omega t)+\cos ((n-m)\omega t)].$$

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.

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 ${\int }_{-\tau /2}^{\tau /2}1dt=\tau .$- 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.

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.

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.

Key concepts

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 $meqn$.

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:

• ${\int }_{-\tau /2}^{\tau /2}1dt=\tau$, 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.