Complex-exponential method for Fourier expansions

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Problem 2965 from Demidovich B.P. "Сборник задач и упражнений по математическому анализу" [Dem97]More problems in Section 5.6 "Fourier Series"

Using the formulas

cos⁡x=12(t+tˉ),sin⁡x=12i(t−tˉ), \cos x=\frac{1}{2}(t+\bar{t}), \quad \sin x=\frac{1}{2i}(t-\bar{t}),

where t=eixt=e^{ix} and tˉ=e−ix\bar{t}=e^{-ix}, expand the following function into a Fourier series: cos⁡2mx\cos^{2m}x (mm a positive integer).

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Problem 2966 from Demidovich B.P. "Сборник задач и упражнений по математическому анализу" [Dem97]More problems in Section 5.6 "Fourier Series"

Using the formulas

cos⁡x=12(t+tˉ),sin⁡x=12i(t−tˉ), \cos x=\frac{1}{2}(t+\bar{t}), \quad \sin x=\frac{1}{2i}(t-\bar{t}),

where t=eixt=e^{ix} and tˉ=e−ix\bar{t}=e^{-ix}, expand the following function into a Fourier series:

qsin⁡x1−2qcos⁡x+q2(∣q∣<1). \frac{q\sin x}{1-2q\cos x+q^{2}} \quad (\left|q\right|<1).
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Problem 2967 from Demidovich B.P. "Сборник задач и упражнений по математическому анализу" [Dem97]More problems in Section 5.6 "Fourier Series"

Using the formulas

cos⁡x=12(t+tˉ),sin⁡x=12i(t−tˉ), \cos x=\frac{1}{2}(t+\bar{t}), \quad \sin x=\frac{1}{2i}(t-\bar{t}),

where t=eixt=e^{ix} and tˉ=e−ix\bar{t}=e^{-ix}, expand the following function into a Fourier series:

1−q21−2qcos⁡x+q2(∣q∣<1). \frac{1-q^{2}}{1-2q\cos x+q^{2}} \quad (\left|q\right|<1).
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Problem 2968 from Demidovich B.P. "Сборник задач и упражнений по математическому анализу" [Dem97]More problems in Section 5.6 "Fourier Series"

Using the formulas

cos⁡x=12(t+tˉ),sin⁡x=12i(t−tˉ), \cos x=\frac{1}{2}(t+\bar{t}), \quad \sin x=\frac{1}{2i}(t-\bar{t}),

where t=eixt=e^{ix} and tˉ=e−ix\bar{t}=e^{-ix}, expand the following function into a Fourier series:

1−qcos⁡x1−2qcos⁡x+q2(∣q∣<1). \frac{1-q\cos x}{1-2q\cos x+q^{2}} \quad (\left|q\right|<1).
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Problem 2969 from Demidovich B.P. "Сборник задач и упражнений по математическому анализу" [Dem97]More problems in Section 5.6 "Fourier Series"

Using the formulas

cos⁡x=12(t+tˉ),sin⁡x=12i(t−tˉ), \cos x=\frac{1}{2}(t+\bar{t}), \quad \sin x=\frac{1}{2i}(t-\bar{t}),

where t=eixt=e^{ix} and tˉ=e−ix\bar{t}=e^{-ix}, expand the following function into a Fourier series:

ln⁡(1−2qcos⁡x+q2)(∣q∣<1). \ln (1-2q\cos x+q^{2}) \quad (\left|q\right|<1).
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Problem 4885 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 11.9 "Fourier Series and Related Questions"

Functions rational in cos⁡x\cos x and sin⁡x\sin x can be expanded in a Fourier series without using the Fourier formulas. To do this we write:

cos⁡x=t2+12t,sin⁡x=t2−12ti, \cos x = \frac{t^{2}+1}{2t}, \qquad \sin x = \frac{t^{2}-1}{2ti},

where t=exit = e^{xi}. The problem then reduces to expanding the function in positive and negative powers of tt. Passing back from tt to xx, we obtain the desired Fourier series expansion.

By the method just described and similar ones, obtain the following expansions:

1−a21−2acos⁡x+a2=1+2∑n=1∞ancos⁡nx;∣a∣<1. \frac{1-a^{2}}{1-2a\cos x+a^{2}} = 1+2\sum _{n=1}^{\infty } a^{n}\cos nx; \quad \left|a\right| < 1 .
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Problem 4886 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 11.9 "Fourier Series and Related Questions"

By the method just described, obtain the following expansion:

1−acos⁡x1−2acos⁡x+a2=∑n=0∞ancos⁡nx;∣a∣<1. \frac{1-a\cos x}{1-2a\cos x+a^{2}} = \sum _{n=0}^{\infty } a^{n}\cos nx; \quad \left|a\right| < 1 .
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Problem 4887 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 11.9 "Fourier Series and Related Questions"

By the method just described, obtain the following expansion:

asin⁡x1−2acos⁡x+a2=∑n=0∞ansin⁡nx;∣a∣<1. \frac{a\sin x}{1-2a\cos x+a^{2}} = \sum _{n=0}^{\infty } a^{n}\sin nx; \quad \left|a\right| < 1 .
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Problem 4888 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 11.9 "Fourier Series and Related Questions"

By the method just described, obtain the following expansion:

ln⁡(1−2acos⁡x+a2)=−2∑n=1∞anncos⁡nx;∣a∣<1. \ln (1-2a\cos x+a^{2}) = -2\sum _{n=1}^{\infty } \frac{a^{n}}{n}\cos nx; \quad \left|a\right| < 1 .
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Problem 4889 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 11.9 "Fourier Series and Related Questions"

By the method just described, obtain the following expansion:

ln⁡∣2sin⁡x2∣−1=cos⁡x+cos⁡2x2+cos⁡3x3+… \ln \left|2\sin \frac{x}{2}\right|^{-1} = \cos x+\frac{\cos 2x}{2}+\frac{\cos 3x}{3}+\ldots
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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.