Section 7

Taylor Series

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Problem 7.1

Using formulas (3) and (5), prove that:

Formula (3): ez=∑n=0∞znn!,∣z∣<+∞e^{z} = \sum_{n = 0}^{\infty } \frac{z^{n}}{n!}, \quad \left|z\right| < +\infty.

Formula (5): cos⁡z=∑n=0∞(−1)nz2n(2n)!,∣z∣<+∞\cos z = \sum_{n = 0}^{\infty } \frac{(-1)^{n} z^{2 n}}{(2 n)!}, \quad \left|z\right| < +\infty.

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(1)

14(ez+e−z+2cos⁡z)=∑n=0∞z4n(4n)!\frac{1}{4}\left(e^{z}+e^{-z}+2 \cos z\right) = \sum_{n = 0}^{\infty } \frac{z^{4 n}}{(4 n)!};

(2)

13(ez+2e−z/2cos⁡z32)=∑n=0∞z3n(3n)!\frac{1}{3}\left(e^{z}+2 e^{-z / 2} \cos \frac{z \sqrt{3}}{2}\right) = \sum_{n = 0}^{\infty } \frac{z^{3 n}}{(3 n)!}.

Problem 7.2

Using formula (8), prove that:

Formula (8): 11−z=∑n=0∞zn,∣z∣<1\frac{1}{1-z} = \sum_{n = 0}^{\infty } z^{n}, \quad \left|z\right| < 1.

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(1)

2(1+z)3=∑n=0∞(−1)n(n+1)(n+2)zn,∣z∣<1;\frac{2}{(1+z)^{3}} = \sum_{n = 0}^{\infty }(-1)^{n}(n+1)(n+2) z^{n}, \quad \left|z\right| < 1 ;

(2)

z(z+a)(a−z)3=∑n=1∞n2znan+1,∣z∣<∣a∣,a≠0\frac{z(z+a)}{(a-z)^{3}} = \sum_{n = 1}^{\infty } \frac{n^{2} z^{n}}{a^{n+1}},\left|z\right| < \left|a\right|, \quad a \neq 0;

(3)

1z2+a2=∑n=0∞(−1)na−2(n+1)z2n,∣z∣<∣a∣,a≠0\frac{1}{z^{2}+a^{2}} = \sum_{n = 0}^{\infty }(-1)^{n} a^{-2(n+1)} z^{2 n}, \quad \left|z\right| < \left|a\right|, \quad a \neq 0;

(4)

1(1+z2)2=∑n=0∞(n+1)(−1)nz2n,∣z∣<1\frac{1}{\left(1+z^{2}\right)^{2}} = \sum_{n = 0}^{\infty }(n+1)(-1)^{n} z^{2 n}, \quad \left|z\right| < 1;

(5)

z2+4z4+z6(1−z2)4=∑n=1∞n3z2n,∣z∣<1\frac{z^{2}+4 z^{4}+z^{6}}{\left(1-z^{2}\right)^{4}} = \sum_{n = 1}^{\infty } n^{3} z^{2 n}, \quad \left|z\right| < 1;

(6)

(1−z)−m−1=∑n=0∞(n+1)(n+2)…(n+m)m!zn,∣z∣<1,m∈N(1-z)^{-m-1} = \sum_{n = 0}^{\infty } \frac{(n+1)(n+2) \ldots (n+m)}{m!} z^{n}, \quad \left|z\right| < 1, \quad m \in \mathbb {N}.

Problem 7.3

Expand the function in a Taylor series in a neighborhood of the point z=0z = 0:

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(1)

1(1+z)2\frac{1}{(1+z)^{2}};

(2)

1(1−z2)2\frac{1}{\left(1-z^{2}\right)^{2}};

(3)

1(1+z3)2\frac{1}{\left(1+z^{3}\right)^{2}};

(4)

1(1−z6)3\frac{1}{\left(1-z^{6}\right)^{3}};

(5)

1(z+1)(z−2)\frac{1}{(z+1)(z-2)};

(6)

2z−5z2−5z+6\frac{2 z-5}{z^{2}-5 z+6};

(7)

z(z2+1)(z2−4)\frac{z}{\left(z^{2}+1\right)\left(z^{2}-4\right)};

(8)

z3(z2+1)(z−1)\frac{z^{3}}{\left(z^{2}+1\right)(z-1)};

(9)

1(z2−1)2(z2+4)\frac{1}{\left(z^{2}-1\right)^{2}\left(z^{2}+4\right)};

(10)

11+z+z2\frac{1}{1+z+z^{2}};

(11)

2z−14z2−2z+1\frac{2 z-1}{4 z^{2}-2 z+1};

(12)

1(1+z)(1+z2)(1+z4)\frac{1}{(1+z)\left(1+z^{2}\right)\left(1+z^{4}\right)};

(13)

1(1−z4)(1+z+z2+z3)\frac{1}{\left(1-z^{4}\right)\left(1+z+z^{2}+z^{3}\right)}.

Problem 7.4

Expand the function

f(z)=2z2+2z−7z2+z−2 f(z) = \frac{2 z^{2}+2 z-7}{z^{2}+z-2}

in a Taylor series in a neighborhood of the point z=−1z = -1.

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Problem 7.5

Expand the function

f(z)=z2+2z(z+1)3 f(z) = \frac{z^{2}+2 z}{(z+1)^{3}}

in a Taylor series in a neighborhood of the point z=1z = 1.

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Problem 7.6

Expand the functions in a Taylor series in a neighborhood of the point z=0z = 0:

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(1)

sin⁡2z\sin^{2} z;

(2)

cos⁡3z\cos^{3} z;

(3)

sin⁡4z+cos⁡4z\sin^{4} z+\cos^{4} z;

(4)

cos⁡2z+ch⁡2z\cos^{2} z+\operatorname {ch}^{2} z;

(5)

e2sin⁡ze^{2} \sin z;

(6)

ch⁡z⋅cos⁡z\operatorname {ch} z \cdot \cos z.

Problem 7.7

Expand the function

f(z)=(z2−4z+5)e4z−z2 f(z) = \left(z^{2}-4 z+5\right) e^{4 z-z^{2}}

in a Taylor series in a neighborhood of the point z=2z = 2.

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Problem 7.8

Expand the function

f(z)=(z23+2z−1)cos⁡(2z+6) f(z) = \left(\frac{z^{2}}{3}+2 z-1\right) \cos (2 z+6)

in a Taylor series in a neighborhood of the point z=−3z = -3.

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Problem 7.9

Find the first three nonzero terms of the Taylor series expansion, in a neighborhood of the point z=0z = 0, of the function:

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(1)

tg⁡z\operatorname {tg} z;

(2)

z(1−z2)sin⁡z\frac{z}{\left(1-z^{2}\right) \sin z};

(3)

ez+cos⁡ze^{z+\cos z}.

Problem 7.10

Using the method of undetermined coefficients, show that the coefficients AnA_{n} of the power series

11−z−z2=∑n=0∞Anzn \frac{1}{1-z-z^{2}} = \sum _{n = 0}^{\infty } A_{n} z^{n}

are determined by the conditions A0=1,A1=1,An+1=An+An−1,n∈NA_{0} = 1, A_{1} = 1, A_{n+1} = A_{n}+A_{n-1}, n \in \mathbb {N}. The numbers AnA_{n} are called the Fibonacci numbers. Prove that

An=15[(5+12)n+1−(1−52)n+1],n∈N. A_{n} = \frac{1}{\sqrt{5}}\left[\left(\frac{\sqrt{5}+1}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right], \quad n \in \mathbb {N}.
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Problem 7.11

Determine the order mm of the zero z=az = a of the function f(z)f(z), if:

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(1)

f(z)=(cos⁡3z−cos⁡5z)2(1−cos⁡2z)3,a=0f(z) = (\cos 3 z-\cos 5 z)^{2}(1-\cos 2 z)^{3}, \quad a = 0;

(2)

f(z)=(z2−π2)3sin⁡3z,a=πf(z) = \left(z^{2}-\pi^{2}\right)^{3} \sin^{3} z, \quad a = \pi;

(3)

f(z)=(z2+π2)2(e2z−1)4,a=πif(z) = \left(z^{2}+\pi^{2}\right)^{2}\left(e^{2 z}-1\right)^{4}, \quad a = \pi i;

(4)

f(z)=(z4+2z3−2z−1)2(eiπz+1)3,a=−1f(z) = \left(z^{4}+2 z^{3}-2 z-1\right)^{2}\left(e^{i \pi z}+1\right)^{3}, a = -1.

Problem 7.12

Find the Taylor series expansion, in a neighborhood of the point z=0z = 0, of the function satisfying the conditions indicated below:

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(1)

f′(z)=f(z),f(0)=1f^{\prime }(z) = f(z), \quad f(0) = 1;

(2)

(1+z2)f′(z)=1,f(0)=0\left(1+z^{2}\right) f^{\prime }(z) = 1, f(0) = 0;

(3)

f′′(z)+λ2f(z)=0,f(0)=0,f′(0)=λf^{\prime \prime }(z)+\lambda^{2} f(z) = 0, f(0) = 0, f^{\prime }(0) = \lambda;

(4)

(1−z2)f′′(z)−zf′(z)=0,f(0)=0,f′(0)=1\left(1-z^{2}\right) f^{\prime \prime }(z)-z f^{\prime }(z) = 0, \quad f(0) = 0, \quad f^{\prime }(0) = 1;

(5)

zf′′(z)+f′(z)+zf(z)=0,f(0)=0,f′(0)=1z f^{\prime \prime }(z)+f^{\prime }(z)+z f(z) = 0, \quad f(0) = 0, \quad f^{\prime }(0) = 1;

(6)

(1−z2)f′′(z)−5zf′(z)−4f(z)=0,f(0)=0,f′(0)=1\left(1-z^{2}\right) f^{\prime \prime }(z)-5 z f^{\prime }(z)-4 f(z) = 0, \quad f(0) = 0, \quad f^{\prime }(0) = 1.

Problem 7.13

Let the function f(z)f(z) be regular in some neighborhood of the point z=0z = 0 and satisfy the conditions

f(0)=0,f(z)=z+f(z2) f(0) = 0, \quad f(z) = z+f\left(z^{2}\right)

Prove that f(z)=∑n=0∞z2nf(z) = \sum_{n = 0}^{\infty } z^{2^{n}}.

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Problem 7.14

Let the function f(z)f(z) be regular in some neighborhood of the point z=0z = 0 and satisfy the conditions

f(0)=1,f′(z)=(1+qz)f(q2z), f(0) = 1, \quad f^{\prime }(z) = (1+q z) f\left(q^{2} z\right),

where qq is a given real number such that ∣q∣⩽1\left|q\right| \leqslant 1. Prove that

f(z)=∑n=0∞znn!qn(n+1)2 f(z) = \sum _{n = 0}^{\infty } \frac{z^{n}}{n!} q^{\frac{n(n+1)}{2}}
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Problem 7.15

Using formula (12) (Example 4), prove that:

Formula (12): zez−1=∑n=0∞Bnn!zn,∣z∣<2π\frac{z}{e^{z}-1} = \sum_{n = 0}^{\infty } \frac{B_{n}}{n!} z^{n}, \quad \left|z\right| < 2 \pi.

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(1)

tg⁡z=∑n=1∞(−1)n22n(1−22n)(2n)!B2nz2n−1,∣z∣<π2\operatorname {tg} z = \sum_{n = 1}^{\infty }(-1)^{n} \frac{2^{2 n}\left(1-2^{2 n}\right)}{(2 n)!} B_{2 n} z^{2 n-1}, \quad \left|z\right| < \frac{\pi }{2};

(2)

zsin⁡z=1+∑n=1∞(−1)n2−22n(2n)!B2nz2n,∣z∣<π\frac{z}{\sin z} = 1+\sum_{n = 1}^{\infty }(-1)^{n} \frac{2-2^{2 n}}{(2 n)!} B_{2 n} z^{2 n},\left|z\right| < \pi.

Problem 7.16

Let the function f(z)f(z) be regular in a simply connected domain DD containing the point z0z_{0}. Prove that for any point z∈Dz \in D the formula holds

f(z)=f(z0)+∑k=1nf(k)(z0)k!(z−z0)k+1n!∫z0z(z−t)nf(n+1)(t)dt f(z) = f\left(z_{0}\right)+\sum _{k = 1}^{n} \frac{f^{(k)}\left(z_{0}\right)}{k!}\left(z-z_{0}\right)^{k}+\frac{1}{n!} \int _{z_{0}}^{z}(z-t)^{n} f^{(n+1)}(t) d t
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Problem 7.17

Prove that

1cos⁡z=∑n=0∞(−1)nE2n(2n)!z2n,∣z∣<π2, \frac{1}{\cos z} = \sum _{n = 0}^{\infty }(-1)^{n} \frac{E_{2 n}}{(2 n)!} z^{2 n}, \quad \left|z\right| < \frac{\pi }{2},

where the coefficients E2nE_{2 n} (the Euler numbers) are determined by the conditions

E0=1,E2k−1=0(k∈N),E0+C2n2E2+C2n4E4+…+C2n2nE2n=0,n∈N. \begin{gathered} E_{0} = 1, \quad E_{2 k-1} = 0 \quad (k \in \mathbb {N}), \\ E_{0}+C_{2 n}^{2} E_{2}+C_{2 n}^{4} E_{4}+\ldots +C_{2 n}^{2 n} E_{2 n} = 0, \quad n \in \mathbb {N}. \end{gathered}
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