Problem 17

Question

Using the fact that $\phi_{n}$ and $\phi_{m}$ are orthogonal for $n \neq m$ we have $$\begin{aligned} \left\|\phi_{m}(x)+\phi_{n}(x)\right\|^{2} &=\int_{a}^{b}\left[\phi_{m}(x)+\phi_{n}(x)\right]^{2} d x=\int_{a}^{b}\left[\phi_{m}^{2}(x)+2 \phi_{m}(x) \phi_{n}(x)+\phi_{n}^{2}(x)\right] d x \\ &=\int_{a}^{b} \phi_{m}^{2}(x) d x+2 \int_{a}^{b} \phi_{m}(x) \phi_{n}(x) d x+\int_{a}^{b} \phi_{n}^{2}(x) d x \\ &=\left\|\phi_{m}(x)\right\|^{2}+\left\|\phi_{n}(x)\right\|^{2}. \end{aligned}$$

Step-by-Step Solution

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Answer

The square of the norm sum equals the sum of the norm squares due to orthogonality.

1Step 1: Expand the Square

To solve the problem, we start by expanding the expression inside the integral. We have $ \|\phi_m(x) + \phi_n(x)\|^2 = \int_a^b [\phi_m(x) + \phi_n(x)]^2 \, dx $. Expanding the square gives us $ \phi_m^2(x) + 2\phi_m(x)\phi_n(x) + \phi_n^2(x) $.

2Step 2: Integrate Each Term Separately

After expanding, we integrate each term separately over the interval from $ a $ to $ b $: $ \int_a^b \phi_m^2(x) \, dx + 2 \int_a^b \phi_m(x)\phi_n(x) \, dx + \int_a^b \phi_n^2(x) \, dx $.

3Step 3: Utilize Orthogonality Condition

Given that $ \phi_n $ and $ \phi_m $ are orthogonal when $ n eq m $, the cross-term $ \int_a^b \phi_m(x)\phi_n(x) \, dx $ is zero. So, $ 2 \int_a^b \phi_m(x)\phi_n(x) \, dx = 0 $.

4Step 4: Express the Result

The expression simplifies to $ \int_a^b \phi_m^2(x) \, dx + \int_a^b \phi_n^2(x) \, dx $, which is equivalent to $ \|\phi_m(x)\|^2 + \|\phi_n(x)\|^2 $. This is the result established by the orthogonality condition.

Key Concepts

Integral CalculusOrthogonality ConditionFunction Expansion

Integral Calculus

Integral calculus is a branch of calculus focusing on the concept of integration, which is essentially the reverse process of differentiation. In simple terms, integration is about finding the total accumulative effect of a function over a certain interval or area.

Definite Integrals: These involve the evaluation of the integral over specific bounds, such as from point $a$ to point $b$. The result is a numerical value, representing the net area under the curve of the function.
Indefinite Integrals: In contrast, these don't have specific boundaries and are used to derive antiderivatives of functions, usually expressed with a constant of integration $C$.

Understanding integration is crucial for evaluating expressions involving functions over intervals, as seen in the exercise, where we calculate the integral of different terms separately.
When you integrate a squared function, like $\phi_{m}^{2}(x)$, you're essentially summing up the infinite small parts of the curve represented by this function over a set interval. This facilitates the evaluation of more complex expressions by breaking them down into manageable parts.

Orthogonality Condition

The orthogonality condition is an essential concept when discussing orthogonal functions. In mathematics, two functions are considered orthogonal over a given interval if their inner product is zero. This implies that they are "perpendicular" in the function space.
The concept stems from vector spaces where two vectors are orthogonal if their dot product is zero. Similarly, for functions $\phi_{m}(x)$ and $\phi_{n}(x)$, they are orthogonal if:\[\int_{a}^{b} \phi_{m}(x)\phi_{n}(x) \, dx = 0\]when $n eq m$.

This property simplifies many calculations in mathematical problems, as seen in the exercise where the cross term vanishes due to orthogonality. If you have a term like $2\int_{a}^{b} \phi_{m}(x)\phi_{n}(x) \, dx$, orthogonality allows it to become zero, reducing complexity.
Orthogonality is crucial in fields such as signal processing and quantum mechanics where distinct functions often represent different states or signals, and their orthogonal nature ensures they do not interfere with one another.

Function Expansion

Function expansion involves expressing complex functions as a sum or series of simpler, known functions. This process is useful when analyzing the behavior or solving equations involving complex functions.
When dealing with orthogonal functions, they can be particularly effective in making a complex problem more manageable by expanding it in terms of a series.

Expanding a Function: The exercise illustrates a perfect example of this by using expressions such as $(\phi_{m}(x) + \phi_{n}(x))^2$. By expanding this, we break it into more straightforward, individual terms that are easier to integrate and interpret.
Utility in Calculus: Function expansion can simplify the process of integration, essential in tackling problems across various mathematics and engineering fields.
Basis Functions and Series: Orthogonal functions often form a basis set where numerous functions within that space can be expressed as a linear combination of these basis functions, similar to how vectors in a plane can be expressed using unit vectors.

By expanding functions and leveraging the properties of orthogonal functions, we can efficiently solve complex problems in a structured manner.

Problem 17

Problem 18

Other exercises in this chapter

Problem 16

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Problem 17

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Problem 18

Since $f(x)$ is an odd function, we expand in a sine series: $$b_{n}=\frac{2}{\pi} \int_{0}^{\pi} x^{3} \sin n x d x=\frac{2}{\pi}\left(-\left.\frac{x^{3}}{n}

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Problem 19

The function in Problem 7 is continuous at $x=\pi / 2$ so \\[ \frac{3 \pi}{2}=f\left(\frac{\pi}{2}\right)=\pi+\sum_{n=1}^{\infty} \frac{2}{n}(-1)^{n+1} \sin \

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