Question

For each natural integer \(n\) we define $$ f_{n}(x)=1+\sum_{k=1}^{n}[\cos k x-\sin k x] . $$ Calculate the value of the integral \(\int_{-\pi}^{\pi}\left|f_{n}(x)\right|^{2} d x\).

Short answer

The integral is \(2\pi(n+1)\).

Step-by-step solution

Understand the Problem

We need to calculate the value of the integral \(\int_{-\pi}^{\pi}\left|f_{n}(x)\right|^{2} dx\). The function \(f_n(x)\) is given as \(f_{n}(x) = 1 + \sum_{k=1}^{n}[\cos kx - \sin kx]\).

Expand the Function

Let's expand \(f_n(x)\) in terms of \(\cos kx\) and \(\sin kx\). The function is \(f_{n}(x) = 1 + \sum_{k=1}^{n}\cos kx - \sum_{k=1}^{n}\sin kx\).

Simplify the Expression

Rewrite the function as \(f_{n}(x) = 1 + \sum_{k=1}^{n}\cos kx - \sum_{k=1}^{n}\sin kx\). Recognizing that this function can have trigonometric identities applied to it to possibly simplify expression further.

Evaluate \(\left|f_n(x)\right|^{2}\)

We need to compute \(\left|f_{n}(x)\right|^2 = \left(1 + \sum_{k=1}^{n} (\cos kx - \sin kx)\right)^2\). Use the formula \((a+b)^2 = a^2 + 2ab + b^2\).

Integrate

Now substitute back into the integral and solve: \[ \int_{-\pi}^{\pi}\left|f_{n}(x)\right|^{2} dx = \int_{-\pi}^{\pi} \left(1 + \sum_{k=1}^{n} \cos kx - \sum_{k=1}^{n} \sin kx\right)^2 dx. \]Apply integration properties and trigonometric integrals, considering orthogonality and the fact that terms like \(\int_{-\pi}^{\pi} \cos kx \ dx = 0\) and \(\int_{-\pi}^{\pi} \sin kx \ dx = 0\) unless \(k=0\).

Calculate Result

Simplify the integral using properties of definite integrals over symmetric intervals and known identities over \([-\pi, \pi]\). Upon solving gives: \[ \int_{-\pi}^{\pi}\left|f_{n}(x)\right|^{2} dx = 2\pi (n+1). \]

Conclusion: Summarize the Solution

The integral of the square of the modulus of \(f_n(x)\) over \([-\pi, \pi]\) results in \(2\pi(n+1)\). This takes into account the orthogonality of trigonometric function terms and symmetry of the interval, yielding the solution.

Key concepts

Trigonometric Functions

Trigonometric functions are the building blocks of many mathematical fields, especially in Fourier analysis. In our function \(f_n(x) = 1 + \sum_{k=1}^{n} (\cos kx - \sin kx)\), we find a precise combination of these functions. Trigonometric functions like \(\cos kx\) and \(\sin kx\) represent wave patterns that repeat at consistent intervals, known as periods. This periodic nature is essential in expressing complex periodic functions in simpler forms.

When dealing with this type of exercise, you expand the function in terms of \(\cos kx\) and \(\sin kx\) as part of the Fourier series. This decomposes any periodic function into the sum of simpler oscillating functions. So here, the sum of cosine and sine functions represents a kind of wave addition.

• Cosine functions are symmetric with respect to the y-axis, meaning they are even functions: \(\cos(-x) = \cos x\).
• Sine functions are asymmetric with respect to the y-axis, meaning they are odd functions: \(\sin(-x) = -\sin x\).

Understanding the properties and symmetry of these functions leads to an accurate evaluation of integrals in problem-solving.

Definite Integrals

A definite integral, such as \(\int_{-\pi}^{\pi} \left|f_n(x)\right|^2 \, dx\), calculates the total accumulation of a function over a specific interval. It can be visualized as the area under the curve of a function on this interval. Integrating \(f_n(x)\) across \([-\pi, \pi]\) gives a measure of how much the function accumulates between these two points. This is essential for understanding the overall behavior of \(f_n(x)\) over one complete period.

In solving this problem, you use properties of symmetry specific to periodic functions over intervals like \([-\pi, \pi]\). These properties simplify the calculations:

• Integrals of \(\cos kx\) and \(\sin kx\) over a full period, unless \(k = 0\), result in zero due to their foundational nature of oscillation.
• The definite integral properties allow you to manipulate and combine integrand components efficiently.

The total integral sum, integrating each component, eventually gives \(2\pi(n+1)\), representing the total contribution of all integrated wave patterns.

Orthogonality of Functions

Orthogonality is a cornerstone concept of Fourier series, especially in terms of trigonometric functions. In mathematics, two functions are orthogonal over an interval if their inner product (integral of their product) equals zero on that interval.

In the context of our exercise, the properties of sine and cosine functions being orthogonal is crucial. For example:

• \(\int_{-\pi}^{\pi} \cos kx \sin mx \, dx = 0\) for \(k eq m\).
• \(\int_{-\pi}^{\pi} \cos kx \cos mx \, dx = 0\) for \(k eq m\) and \(\int_{-\pi}^{\pi} \sin kx \sin mx \, dx = 0\) for \(k eq m\).

This orthogonality, combined with symmetry, allows simplification when evaluating integrals like \(\left|f_n(x)\right|^2\). They eliminate cross-product terms in expansions because they integrate to zero over the given interval, leaving only the necessary components for the solution.