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To determine if each function can be represented by a Fourier series, we need to check if the function is periodic and well-defined (no discontinuities or singularities) on the given interval [tex]\(-\pi < x < \pi\)[/tex]. A Fourier series representation is only valid for periodic functions, and the series should converge to the function values everywhere except possibly at points where the function has jump discontinuities.
1. [tex]\(f(x) = x^3\)[/tex]:
- The function [tex]\(f(x) = x^3\)[/tex] is not periodic with period [tex]\(2\pi\)[/tex]. Therefore, it cannot be represented by a Fourier series on the interval [tex]\(-\pi < x < \pi\)[/tex].
2. [tex]\(f(x) = 4x - 5\)[/tex]:
- The function [tex]\(f(x) = 4x - 5\)[/tex] is not periodic with period [tex]\(2\pi\)[/tex]. Therefore, it cannot be represented by a Fourier series on the interval [tex]\(-\pi < x < \pi\)[/tex].
3. [tex]\(f(x) = \frac{2}{x}\)[/tex]:
- The function [tex]\(f(x) = \frac{2}{x}\)[/tex] has a singularity at [tex]\(x = 0\)[/tex]. Therefore, it cannot be represented by a Fourier series because it is not well-defined at [tex]\(x = 0\)[/tex] on the interval [tex]\(-\pi < x < \pi\)[/tex].
4. [tex]\(f(x) = \frac{1}{x - 5}\)[/tex]:
- The function [tex]\(f(x) = \frac{1}{x - 5}\)[/tex] has a singularity at [tex]\(x = 5\)[/tex]. Since [tex]\(x = 5\)[/tex] lies outside the interval [tex]\(-\pi < x < \pi\)[/tex], we might initially consider the function to be well-defined on the interval. However, it is not periodic with period [tex]\(2\pi\)[/tex]. Therefore, it cannot be represented by a Fourier series.
5. [tex]\(f(x) = \tan x\)[/tex]:
- The function [tex]\(f(x) = \tan x\)[/tex] has singularities at [tex]\(x = \frac{\pi}{2}\)[/tex] and [tex]\(x = -\frac{\pi}{2}\)[/tex] within the interval [tex]\(-\pi < x < \pi\)[/tex]. Therefore, it cannot be represented by a Fourier series because it is not well-defined at these points within the interval.
6. [tex]\(f(x) = y\)[/tex] where [tex]\(x^2 + y^2 = 9\)[/tex]:
- This represents a circle of radius 3 centered at the origin. If we let [tex]\(x = 3\cos(t)\)[/tex] and [tex]\(y = 3\sin(t)\)[/tex] where [tex]\(t\)[/tex] is the parameter, we can parameterize [tex]\(y\)[/tex] as a function of [tex]\(x\)[/tex]. The appropriate parameterization on the interval [tex]\(-\pi < x < \pi\)[/tex] is:
[tex]\[ y = \pm\sqrt{9 - x^2} \][/tex]
- This function is periodic since it repeats itself as [tex]\(x\)[/tex] takes values over the interval, and [tex]\(\sqrt{9 - x^2}\)[/tex] is well-defined and continuous on the interval [tex]\(-\pi < x < \pi\)[/tex]. Thus, the upper (or lower) half of the circle could be represented by a Fourier series. Therefore, [tex]\(y = \pm\sqrt{9 - x^2}\)[/tex] can be represented by a Fourier series.
Hence the Fourier series representation is valid for:
- [tex]\(f(x) = y\)[/tex] where [tex]\(x^2 + y^2 = 9\)[/tex] (with the understanding that we consider it as [tex]\(y = \pm\sqrt{9 - x^2}\)[/tex]).
1. [tex]\(f(x) = x^3\)[/tex]:
- The function [tex]\(f(x) = x^3\)[/tex] is not periodic with period [tex]\(2\pi\)[/tex]. Therefore, it cannot be represented by a Fourier series on the interval [tex]\(-\pi < x < \pi\)[/tex].
2. [tex]\(f(x) = 4x - 5\)[/tex]:
- The function [tex]\(f(x) = 4x - 5\)[/tex] is not periodic with period [tex]\(2\pi\)[/tex]. Therefore, it cannot be represented by a Fourier series on the interval [tex]\(-\pi < x < \pi\)[/tex].
3. [tex]\(f(x) = \frac{2}{x}\)[/tex]:
- The function [tex]\(f(x) = \frac{2}{x}\)[/tex] has a singularity at [tex]\(x = 0\)[/tex]. Therefore, it cannot be represented by a Fourier series because it is not well-defined at [tex]\(x = 0\)[/tex] on the interval [tex]\(-\pi < x < \pi\)[/tex].
4. [tex]\(f(x) = \frac{1}{x - 5}\)[/tex]:
- The function [tex]\(f(x) = \frac{1}{x - 5}\)[/tex] has a singularity at [tex]\(x = 5\)[/tex]. Since [tex]\(x = 5\)[/tex] lies outside the interval [tex]\(-\pi < x < \pi\)[/tex], we might initially consider the function to be well-defined on the interval. However, it is not periodic with period [tex]\(2\pi\)[/tex]. Therefore, it cannot be represented by a Fourier series.
5. [tex]\(f(x) = \tan x\)[/tex]:
- The function [tex]\(f(x) = \tan x\)[/tex] has singularities at [tex]\(x = \frac{\pi}{2}\)[/tex] and [tex]\(x = -\frac{\pi}{2}\)[/tex] within the interval [tex]\(-\pi < x < \pi\)[/tex]. Therefore, it cannot be represented by a Fourier series because it is not well-defined at these points within the interval.
6. [tex]\(f(x) = y\)[/tex] where [tex]\(x^2 + y^2 = 9\)[/tex]:
- This represents a circle of radius 3 centered at the origin. If we let [tex]\(x = 3\cos(t)\)[/tex] and [tex]\(y = 3\sin(t)\)[/tex] where [tex]\(t\)[/tex] is the parameter, we can parameterize [tex]\(y\)[/tex] as a function of [tex]\(x\)[/tex]. The appropriate parameterization on the interval [tex]\(-\pi < x < \pi\)[/tex] is:
[tex]\[ y = \pm\sqrt{9 - x^2} \][/tex]
- This function is periodic since it repeats itself as [tex]\(x\)[/tex] takes values over the interval, and [tex]\(\sqrt{9 - x^2}\)[/tex] is well-defined and continuous on the interval [tex]\(-\pi < x < \pi\)[/tex]. Thus, the upper (or lower) half of the circle could be represented by a Fourier series. Therefore, [tex]\(y = \pm\sqrt{9 - x^2}\)[/tex] can be represented by a Fourier series.
Hence the Fourier series representation is valid for:
- [tex]\(f(x) = y\)[/tex] where [tex]\(x^2 + y^2 = 9\)[/tex] (with the understanding that we consider it as [tex]\(y = \pm\sqrt{9 - x^2}\)[/tex]).
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