Expansion of meromorphic functions in simple fractions

[25/4%]
Show
LaTeX
Problem 5.5.10 from Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. "Problems and Solutions in Mathematics" [CJP+11]More problems in Section 5.5 "Series and Normal Families"

Find an explicit formula for a meromorphic function ff whose only singularities are simple poles at −1,−2,−3,…-1,-2,-3,\ldots with residue nn at z=−nz=-n. Prove in detail that your function has all the required properties.

?
Problem 5073 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 12.2 "Direct Summation of Finite Sums and Infinite Series"

Prove the equality φ(x)+φ(x+π2)=4φ(2x)\varphi (x)+\varphi \left(x+\frac{\pi }{2}\right) = 4\varphi (2x), where

φ(x)=∑n=−∞∞1(x−nπ)2−1sin⁡2x. \varphi (x) = \sum _{n=-\infty }^{\infty } \frac{1}{(x-n\pi )^{2}} - \frac{1}{\sin ^{2}x}.
?
Problem 5074 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 12.2 "Direct Summation of Finite Sums and Infinite Series"

Using the result of problem 5073 (the equality φ(x)+φ(x+π2)=4φ(2x)\varphi (x)+\varphi \left(x+\frac{\pi }{2}\right) = 4\varphi (2x), where φ(x)=∑n=−∞∞1(x−nπ)2−1sin⁡2x\varphi (x) = \sum_{n=-\infty }^{\infty } \frac{1}{(x-n\pi )^{2}} - \frac{1}{\sin^{2}x}), show that

1sin⁡2x=∑n=−∞∞1(x−nπ)2. \frac{1}{\sin ^{2}x} = \sum _{n=-\infty }^{\infty } \frac{1}{(x-n\pi )^{2}}.
?
Problem 5075 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 12.2 "Direct Summation of Finite Sums and Infinite Series"

By integration, obtain Euler's equality

ctg⁡x=1x+∑n≠0(1x−nπ+1x+nπ)=1x+∑n=1∞2xx2−n2π2. \operatorname {ctg}x = \frac{1}{x}+\sum _{n \neq 0}\left(\frac{1}{x-n\pi }+\frac{1}{x+n\pi }\right) = \frac{1}{x}+\sum _{n=1}^{\infty } \frac{2x}{x^{2}-n^{2}\pi ^{2}}.
?
Problem 5077 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 12.2 "Direct Summation of Finite Sums and Infinite Series"

Prove Euler's equality giving the expansion into an infinite series of fractions of the function tg⁡x=∑n=0∞2x(2n+12π)2−x2\operatorname {tg}x = \sum_{n=0}^{\infty } \frac{2x}{\left(\frac{2n+1}{2}\pi \right)^{2}-x^{2}}.

?
Problem 5078 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 12.2 "Direct Summation of Finite Sums and Infinite Series"

Prove Euler's equality giving the expansion into an infinite series of fractions of the function 1sin⁡x=1x+∑n=1∞(−1)n⋅2xx2−n2π2\frac{1}{\sin x} = \frac{1}{x}+\sum_{n=1}^{\infty } \frac{(-1)^{n}\cdot 2x}{x^{2}-n^{2}\pi^{2}}.

?
Problem 5079 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 12.2 "Direct Summation of Finite Sums and Infinite Series"

Prove Euler's equality giving the expansion into an infinite series of fractions of the function 1cos⁡x=∑n=1∞(−1)n(2n−1)πx2−(n−12)2π2\frac{1}{\cos x} = \sum_{n=1}^{\infty } \frac{(-1)^{n}(2n-1)\pi }{x^{2}-\left(n-\frac{1}{2}\right)^{2}\pi^{2}}.

?
Problem 5080 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 12.2 "Direct Summation of Finite Sums and Infinite Series"

Prove Euler's equality giving the expansion into an infinite series of fractions of the function 11−e−x−1x−12=∑n=1∞2xx2+4π2n2\frac{1}{1-e^{-x}} - \frac{1}{x} - \frac{1}{2} = \sum_{n=1}^{\infty } \frac{2x}{x^{2}+4\pi^{2}n^{2}}.

?
Problem 5512 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

ctg⁡(z−a)+ctg⁡a=∑n=−∞∞(1z−a−nπ+1a+nπ). \ctg (z-a)+\ctg a = \sum _{n=-\infty }^{\infty } \left(\frac{1}{z-a-n\pi }+\frac{1}{a+n\pi }\right).
?
Problem 5514 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

1cos⁡z=π∑n=1∞(−1)n(2n−1)z2−(2n−1)2π24. \frac{1}{\cos z} = \pi \sum _{n=1}^{\infty } \frac{(-1)^{n}(2n-1)}{z^{2}-(2n-1)^{2}\frac{\pi ^{2}}{4}}.
?
Problem 5515 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

sin⁡zcos⁡2z=∑n=−∞∞(−1)n(z−(2n+1)π2)2. \frac{\sin z}{\cos ^{2} z} = \sum _{n=-\infty }^{\infty } \frac{(-1)^{n}}{\left(z-(2n+1)\frac{\pi }{2}\right)^{2}}.
?
Problem 5516 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

π2cos⁡2πz=∑n=−∞∞1(z+12−n)2. \frac{\pi ^{2}}{\cos ^{2} \pi z} = \sum _{n=-\infty }^{\infty } \frac{1}{\left(z+\frac{1}{2}-n\right)^{2}}.
?
Problem 5517 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

eazez−1=1z+∑n=1∞2zcos⁡2πna−4πnsin⁡2πnaz2+4n2π2;0<a<1. \frac{e^{az}}{e^{z}-1} = \frac{1}{z}+\sum _{n=1}^{\infty } \frac{2z\cos 2\pi na-4\pi n\sin 2\pi na}{z^{2}+4n^{2}\pi ^{2}}; \quad 0 < a < 1.
?
Problem 5518 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

sin⁡azsin⁡πz=2π∑n=1∞(−1)nn⋅sin⁡naz2−n2;−π<a<π. \frac{\sin az}{\sin \pi z} = \frac{2}{\pi }\sum _{n=1}^{\infty } \frac{(-1)^{n}n\cdot \sin na}{z^{2}-n^{2}}; \quad -\pi < a < \pi .
?
Problem 5519 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

ch⁡azsh⁡πz=1πz+1π∑n=1∞(−1)n2zcos⁡naz2+n2;−π⩽a⩽π. \frac{\ch az}{\sh \pi z} = \frac{1}{\pi z}+\frac{1}{\pi }\sum _{n=1}^{\infty } \frac{(-1)^{n}2z\cos na}{z^{2}+n^{2}}; \quad -\pi \leqslant a \leqslant \pi .
?
Problem 5520 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

1ch⁡2z−cos⁡2z=14z2+4z2∑n=1∞(−1)nnsh⁡πn⋅14z4+n4π4. \frac{1}{\ch 2z-\cos 2z} = \frac{1}{4z^{2}}+4z^{2}\sum _{n=1}^{\infty } \frac{(-1)^{n}n}{\sh \pi n}\cdot \frac{1}{4z^{4}+n^{4}\pi ^{4}}.
?
Problem 5521 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

zsin⁡zsin⁡z−zcos⁡z=3z+∑n=1∞2zz2−λn2;tg⁡λn=λn>0. \frac{z\sin z}{\sin z-z\cos z} = \frac{3}{z}+\sum _{n=1}^{\infty } \frac{2z}{z^{2}-\lambda _{n}^{2}}; \quad \tg \lambda _{n} = \lambda _{n} > 0.
?
Problem 5541 from Гюнтер Н.М., Кузьмин Р.О. "Сборник задач по высшей математике" [ГюКу03]More problems in Section 14.5 "Expansion of Functions into Partial Fractions"

Prove the validity of the expansion:

Γ(a)Γ(x)Γ(a+x)=1x+∑n=1∞(−1)n(a−1)(a−2)…(a−n)1⋅2⋅3⋯n⋅1x+n;a>0. \frac{\Gamma (a)\Gamma (x)}{\Gamma (a+x)} = \frac{1}{x}+\sum _{n=1}^{\infty } \frac{(-1)^{n}(a-1)(a-2)\ldots (a-n)}{1\cdot 2\cdot 3\cdots n}\cdot \frac{1}{x+n}; \quad a > 0.
?
Problem 22.1 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 22 "Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products"

Prove the formulas:

?
(1)

1ez−ea=e−a[−12+1z−a+∑n=1∞2(z−a)(z−a)2+4n2π2]\frac{1}{e^{z}-e^{a}} = e^{-a}\left[-\frac{1}{2}+\frac{1}{z-a}+\sum_{n = 1}^{\infty } \frac{2(z-a)}{(z-a)^{2}+4 n^{2} \pi^{2}}\right];

(2)

πsin⁡πz=∑n=−∞∞(−1)nz−n\frac{\pi }{\sin \pi z} = \sum_{n = -\infty }^{\infty } \frac{(-1)^{n}}{z-n};

(3)

1ch⁡2z−cos⁡2z=14z2+∑n=1∞n(−1)nsh⁡πn2πz24z4+n4π4\frac{1}{\operatorname {ch} 2 z-\cos 2 z} = \frac{1}{4 z^{2}}+\sum_{n = 1}^{\infty } \frac{n(-1)^{n}}{\operatorname {sh} \pi n} \frac{2 \pi z^{2}}{4 z^{4}+n^{4} \pi^{4}}.

Problem 22.2 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 22 "Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products"

Prove the formulas:

?
(1)

π2cos⁡2πz=∑n=−∞∞1(z+12−n)2\frac{\pi^{2}}{\cos^{2} \pi z} = \sum_{n = -\infty }^{\infty } \frac{1}{\left(z+\frac{1}{2}-n\right)^{2}};

(2)

1ch⁡z=∑n=1∞(−1)n+1πz2+(n−12)2π2\frac{1}{\operatorname {ch} z} = \sum_{n = 1}^{\infty } \frac{(-1)^{n+1} \pi }{z^{2}+\left(n-\frac{1}{2}\right)^{2} \pi^{2}};

(3)

ez−e−zez+e−z=∑n=1∞2zz2+(n−12)2π2\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}} = \sum_{n = 1}^{\infty } \frac{2 z}{z^{2}+\left(n-\frac{1}{2}\right)^{2} \pi^{2}};

(4)

eazez−1=1z+∑n=1∞2z⋅cos⁡2πna−4πnsin⁡2πnaz2+4n2π2,0<a<1\frac{e^{a z}}{e^{z}-1} = \frac{1}{z}+\sum_{n = 1}^{\infty } \frac{2 z \cdot \cos 2 \pi n a-4 \pi n \sin 2 \pi n a}{z^{2}+4 n^{2} \pi^{2}}, 0 < a < 1.

Problem 22.3 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 22 "Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products"

Let F(z)F(z) be an entire function satisfying the inequality

∣F(z+iy)∣⩽Mea∣y∣,−π<a<π, \left|F(z+i y)\right| \leqslant M e^{a\left|y\right|}, \quad -\pi < a < \pi ,

for all real xx and yy. Prove that

πF(z)sin⁡πz=∑n=−∞∞(−1)nF(n)z−n and πF(z)cos⁡πz=∑n=−∞∞(−1)n+1F(n+12)z−n−12. \begin{aligned} \frac{\pi F(z)}{\sin \pi z} & = \sum _{n = -\infty }^{\infty }(-1)^{n} \frac{F(n)}{z-n} \\ \text{ and } \frac{\pi F(z)}{\cos \pi z} & = \sum _{n = -\infty }^{\infty }(-1)^{n+1} \frac{F\left(n+\frac{1}{2}\right)}{z-n-\frac{1}{2}}. \end{aligned}
?
Problem 22.4 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 22 "Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products"

Prove that

?
(1)

sin⁡azsin⁡πz=2π∑n=1∞(−1)nnsin⁡naz2−n2,−π<a<π;\frac{\sin a z}{\sin \pi z} = \frac{2}{\pi } \sum_{n = 1}^{\infty } \frac{(-1)^{n} n \sin n a}{z^{2}-n^{2}},-\pi < a < \pi ;

(2)

ch⁡azsh⁡πz=1πz+1π∑n=1∞(−1)n2zcos⁡naz2+n2,−π<a<π;\frac{\operatorname {ch} a z}{\operatorname {sh} \pi z} = \frac{1}{\pi z}+\frac{1}{\pi } \sum_{n = 1}^{\infty }(-1)^{n} \frac{2 z \cos n a}{z^{2}+n^{2}},-\pi < a < \pi ;

(3)

πsin⁡πz=∑n=−∞∞ei(πn+αn−αz)z−n,−π<α<π\frac{\pi }{\sin \pi z} = \sum_{n = -\infty }^{\infty } \frac{e^{i(\pi n+\alpha n-\alpha z)}}{z-n},-\pi < \alpha < \pi;

(4)

∑n=−∞∞(−1)nsin⁡α(z−n)z−n=0,−π<α<π\sum_{n = -\infty }^{\infty }(-1)^{n} \frac{\sin \alpha (z-n)}{z-n} = 0,-\pi < \alpha < \pi;

(5)

π2sin⁡2πz=πα+∑n=−∞∞cos⁡α(z−n)(z−n)2,0⩽α<π\frac{\pi^{2}}{\sin^{2} \pi z} = \pi \alpha +\sum_{n = -\infty }^{\infty } \frac{\cos \alpha (z-n)}{(z-n)^{2}}, 0 \leqslant \alpha < \pi.

Problem 22.5 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 22 "Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products"

Prove the formulas:

?
(1)

e(α+β)z−1(eαz−1)(eβz−1)=(1α+1β)1z+∑k=1∞(2αzα2z2+4k2π2+2βzβ2z2+4k2π2)\frac{e^{(\alpha +\beta ) z}-1}{\left(e^{\alpha z}-1\right)\left(e^{\beta z}-1\right)} = \left(\frac{1}{\alpha }+\frac{1}{\beta }\right) \frac{1}{z}+\sum_{k = 1}^{\infty }\left(\frac{2 \alpha z}{\alpha^{2} z^{2}+4 k^{2} \pi^{2}}+\frac{2 \beta z}{\beta^{2} z^{2}+4 k^{2} \pi^{2}}\right), αβ≠0,α≠β;\alpha \beta \neq 0, \alpha \neq \beta ;

(2)

ctg⁡(z−a)+ctg⁡a=z∑n=−∞∞1(a−πn)(z−a+πn),sin⁡a≠0\operatorname {ctg}(z-a)+\operatorname {ctg} a = z \sum_{n = -\infty }^{\infty } \frac{1}{(a-\pi n)(z-a+\pi n)}, \sin a \neq 0;

(3)

sin⁡acos⁡z−cos⁡a=∑n=−∞∞2aa2−(z+2πn)2,sin⁡a≠0\frac{\sin a}{\cos z-\cos a} = \sum_{n = -\infty }^{\infty } \frac{2 a}{a^{2}-(z+2 \pi n)^{2}}, \sin a \neq 0;

(4)

cos⁡asin⁡a−sin⁡z=∑n=−∞∞π−2a(z−a+2πn)(z+a+(2n−1)π),cos⁡a≠0\frac{\cos a}{\sin a-\sin z} = \sum_{n = -\infty }^{\infty } \frac{\pi -2 a}{(z-a+2 \pi n)(z+a+(2 n-1) \pi )}, \cos a \neq 0.

Problem 22.6 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 22 "Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products"

Let f(z)f(z) be a meromorphic function with poles a1,a2,…a_{1}, a_{2}, \ldots and zeros b1,b2,…b_{1}, b_{2}, \ldots (each zero is written as many times as its multiplicity, and each pole as many times as its order), where the point z=0z = 0 is neither a zero nor a pole of the function f(z)f(z). Suppose that

f′(z)f(z)=∑n=1∞1z−bn−∑n=1∞1z−an, \frac{f^{\prime }(z)}{f(z)} = \sum _{n = 1}^{\infty } \frac{1}{z-b_{n}}-\sum _{n = 1}^{\infty } \frac{1}{z-a_{n}},

and both series converge uniformly in every bounded part of the plane. Prove that

f(z)=∏n=1∞(1−zbn)∏n=1∞(1−zan)f(0) f(z) = \frac{\prod _{n = 1}^{\infty }\left(1-\frac{z}{b_{n}}\right)}{\prod _{n = 1}^{\infty }\left(1-\frac{z}{a_{n}}\right)} f(0)

and both products also converge uniformly in every bounded part of the plane.

?
Problem 22.8 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 22 "Expansion of Meromorphic Functions into Series of Simplest Fractions and into Infinite Products"

Prove the equalities:

∑n=1∞1n2+a2=π2a(cth⁡aπ−1aπ)∑n=1∞1(n2+a2)2=14a4(π2a2sh⁡2aπ+aπcth⁡aπ−2) \begin{aligned} \sum _{n = 1}^{\infty } \frac{1}{n^{2}+a^{2}} & = \frac{\pi }{2 a}\left(\operatorname {cth} a \pi -\frac{1}{a \pi }\right) \\ \sum _{n = 1}^{\infty } \frac{1}{\left(n^{2}+a^{2}\right)^{2}} & = \frac{1}{4 a^{4}}\left(\frac{\pi ^{2} a^{2}}{\operatorname {sh}^{2} a \pi }+a \pi \operatorname {cth} a \pi -2\right) \end{aligned}

assuming that none of the denominators in these equalities vanishes.

?

References

[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.