Section 2.21

Taylor Series

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Subsection 2.21.2Problems[85/71%]
Problem 2.21.1

Using expansions (3)-(7), prove that:

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(1)

eax=∑n=0∞eax0ann!(x−x0)n,x∈Re^{a x} = \sum_{n = 0}^{\infty } e^{a x_{0}} \frac{a^{n}}{n !}\left(x-x_{0}\right)^{n}, \quad x \in R;

(2)

ax=∑n=0∞ln⁡nan!xn,a>0,x∈Ra^{x} = \sum_{n = 0}^{\infty } \frac{\ln^{n} a}{n !} x^{n}, \quad a > 0, \quad x \in R;

(3)

ch⁡ax=∑n=0∞a2n(2n)!x2n,x∈R\operatorname {ch} a x = \sum_{n = 0}^{\infty } \frac{a^{2 n}}{(2 n) !} x^{2 n}, \quad x \in R;

(4)

sin⁡ax=∑n=0∞(−1)na2n+1(2n+1)!x2n+1,x∈R\sin a x = \sum_{n = 0}^{\infty }(-1)^{n} \frac{a^{2 n+1}}{(2 n+1) !} x^{2 n+1}, \quad x \in R.

Problem 2.21.2

Using expansions (8), (34), (36), (37), prove that:

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(1)

1a+bx=∑n=0∞(−1)nbnan+1xn,∣x∣<∣ab∣,ab≠0\frac{1}{a+b x} = \sum_{n = 0}^{\infty } \frac{(-1)^{n} b^{n}}{a^{n+1}} x^{n}, \quad \left|x\right| < \left|\frac{a}{b}\right|, \quad a b \neq 0;

(2)

1x2+a2=∑n=0∞(−1)na2(n+1)x2n,∣x∣<∣a∣,a≠0\frac{1}{x^{2}+a^{2}} = \sum_{n = 0}^{\infty } \frac{(-1)^{n}}{a^{2(n+1)}} x^{2 n}, \quad \left|x\right| < \left|a\right|, \quad a \neq 0;

(3)

1a2+x2=1a+∑n=1∞(−1)n(2n−1)!!a2n+1(2n)!!x2n,∣x∣<a,a>0\frac{1}{\sqrt{a^{2}+x^{2}}} = \frac{1}{a}+\sum_{n = 1}^{\infty } \frac{(-1)^{n}(2 n-1) ! !}{a^{2 n+1}(2 n) ! !} x^{2 n}, \quad \left|x\right| < a, \quad a > 0;

(4)

a2+x2=a+x22a+∑n=2∞(−1)n−1(2n−3)!!(2n)!!a2n−1x2n,∣x∣<a,a>0\sqrt{a^{2}+x^{2}} = a+\frac{x^{2}}{2 a}+\sum_{n = 2}^{\infty } \frac{(-1)^{n-1}(2 n-3) ! !}{(2 n) ! ! a^{2 n-1}} x^{2 n}, \quad \left|x\right| < a, \quad a > 0.

Problem 2.21.3

Using expansions (11), (12), prove that:

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(1)

ln⁡(a+bx)=ln⁡a+∑n=1∞(−1)n−1(ba)nxnn,∣x∣<a∣b∣,a>0,b≠0\ln (a+b x) = \ln a+\sum_{n = 1}^{\infty }(-1)^{n-1}\left(\frac{b}{a}\right)^{n} \frac{x^{n}}{n}, \quad \left|x\right| < \frac{a}{\left|b\right|}, \quad a > 0, \quad b \neq 0;

(2)

ln⁡a+xa−x=∑n=0∞22n+1(xa)2n+1,∣x∣<a,a>0\ln \frac{a+x}{a-x} = \sum_{n = 0}^{\infty } \frac{2}{2 n+1}\left(\frac{x}{a}\right)^{2 n+1},\left|x\right| < a, a > 0.

Problem 2.21.4

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

e−x2e^{-x^{2}};

(2)

x2(1+x)2\frac{x^{2}}{(1+x)^{2}};

(3)

1(1−x3)2\frac{1}{\left(1-x^{3}\right)^{2}};

(4)

(1−x2)−3/2\left(1-x^{2}\right)^{-3 / 2};

(5)

sin⁡x23\sin \frac{x^{2}}{3};

(6)

ch⁡(x2)\operatorname {ch}(x \sqrt{2});

(7)

x31−2x\frac{x^{3}}{\sqrt{1-2 x}};

(8)

1−x2\sqrt{1-x^{2}}.

Problem 2.21.5

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

e2x+2e−xe^{2 x}+2 e^{-x};

(2)

(1+x)e−x(1+x) e^{-x}

(3)

x2ln⁡(4+x2)x^{2} \ln \left(4+x^{2}\right);

(4)

(1−x)ln⁡(1−x)(1-x) \ln (1-x);

(5)

ln⁡1+x1−x3\ln \sqrt[3]{\frac{1+x}{1-x}}

(6)

(1+x2)arctg⁡x\left(1+x^{2}\right) \operatorname {arctg} x;

(7)

12(sh⁡x+sin⁡x)\frac{1}{2}(\operatorname {sh} x+\sin x);

(8)

arccos⁡x\arccos x.

Problem 2.21.6

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

5x−4x+2\frac{5 x-4}{x+2};

(2)

1x2−2x−3\frac{1}{x^{2}-2 x-3};

(3)

12x2+5x−3\frac{1}{2 x^{2}+5 x-3};

(4)

3x+4x2+x−6\frac{3 x+4}{x^{2}+x-6}

(5)

5−2xx2−5x+6\frac{5-2 x}{x^{2}-5 x+6};

(6)

1(x2+2)2\frac{1}{\left(x^{2}+2\right)^{2}};

(7)

x(x+1)(x2−1)\frac{x}{(x+1)\left(x^{2}-1\right)};

(8)

15−4x2−x4\frac{1}{5-4 x^{2}-x^{4}};

(9)

5−3x2x2+5x−3\frac{5-3 x}{2 x^{2}+5 x-3};

(10)

7−4x6−5x−6x2\frac{7-4 x}{6-5 x-6 x^{2}}.

Problem 2.21.7

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

13x4+10x2+3\frac{1}{3 x^{4}+10 x^{2}+3};

(2)

12x4+5x2+2\frac{1}{2 x^{4}+5 x^{2}+2};

(3)

1(1−x2)(x2+4)\frac{1}{\left(1-x^{2}\right)\left(x^{2}+4\right)};

(4)

2x2+x+3(1−x)2(2+x)\frac{2 x^{2}+x+3}{(1-x)^{2}(2+x)};

(5)

11−x−x2\frac{1}{1-x-x^{2}};

(6)

11+x+x2\frac{1}{1+x+x^{2}}.

Problem 2.21.8

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

1x3+x2+3x+3\frac{1}{x^{3}+x^{2}+3 x+3};

(2)

1x3+x2−4x−4\frac{1}{x^{3}+x^{2}-4 x-4};

(3)

x−2x3+x2+2x+2\frac{x-2}{x^{3}+x^{2}+2 x+2};

(4)

1x4+x3+x2+x+1\frac{1}{x^{4}+x^{3}+x^{2}+x+1}.

Problem 2.21.9

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

ln⁡3−2x2+3x\ln \frac{3-2 x}{2+3 x}

(2)

ln⁡(12−x−x2)\ln \left(12-x-x^{2}\right);

(3)

ln⁡2+x21−2x2\ln \frac{2+x^{2}}{\sqrt{1-2 x^{2}}};

(4)

ln⁡∣x2−2∣x2+1\ln \frac{\left|x^{2}-2\right|}{x^{2}+1};

(5)

ln⁡2+x21−x\ln \frac{2+x^{2}}{1-x};

(6)

(x2+5)ln⁡9−x24−x2\left(x^{2}+5\right) \ln \frac{9-x^{2}}{4-x^{2}}.

Problem 2.21.10

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

ln⁡2x+1x2−4x+4\ln \frac{2 x+1}{x^{2}-4 x+4};

(2)

ln⁡x2−8x+16x+8\ln \frac{x^{2}-8 x+16}{x+8};

(3)

ln⁡10+3x−x24−3x\ln \frac{10+3 x-x^{2}}{4-3 x};

(4)

ln⁡3−x+6x2−2x37\ln \sqrt[7]{3-x+6 x^{2}-2 x^{3}};

(5)

ln⁡2x−33x2+x−10\ln \frac{2 x-3}{3 x^{2}+x-10};

(6)

ln⁡3x−14x2−4x−3\ln \frac{3 x-1}{4 x^{2}-4 x-3}.

Problem 2.21.11

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

cos⁡2x\cos^{2} x;

(2)

sin⁡3xsin⁡5x\sin 3 x \sin 5 x;

(3)

sin⁡xcos⁡2x\sin x \cos^{2} x;

(4)

xsin⁡2xcos⁡3xx \sin 2 x \cos 3 x;

(5)

sin⁡3x\sin^{3} x;

(6)

xcos⁡32xx \cos^{3} 2 x.

Problem 2.21.12

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series.

?
(1)

x2ch⁡2xx^{2} \operatorname {ch}^{2} x;

(2)

xsh⁡2xx \operatorname {sh}^{2} x

(3)

ch⁡3x\operatorname {ch}^{3} x;

(4)

sin⁡5x\sin^{5} x.

Problem 2.21.13

Prove that if a>0a > 0, then

ln⁡(x+a2+x2)=ln⁡a+xa+∑n=1∞(−1)n(2n−1)!!(2n)!!a2n+1(2n+1)x2n+1 \ln \left(x+\sqrt{a^{2}+x^{2}}\right) = \ln a+\frac{x}{a}+\sum _{n = 1}^{\infty } \frac{(-1)^{n}(2 n-1) ! !}{(2 n) ! ! a^{2 n+1}(2 n+1)} x^{2 n+1}

Find the radius of convergence of this series.

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Problem 2.21.14

Expand the function into a Maclaurin series and find the radius of convergence RR of this series.

?
(1)

xln⁡(x+x2+2)x \ln \left(x+\sqrt{x^{2}+2}\right);

(2)

xln⁡(x2+9+x4)x \ln \left(x^{2}+\sqrt{9+x^{4}}\right);

(3)

ln⁡(x3+9+x6)\ln \left(x^{3}+\sqrt{9+x^{6}}\right)

(4)

ln⁡(x3+64+x6)\ln \left(x^{3}+\sqrt{64+x^{6}}\right);

(5)

(x2−1)arcsin⁡2x2\left(x^{2}-1\right) \arcsin 2 x^{2};

(6)

x2arccos⁡2xx^{2} \arccos 2 x

(7)

ln⁡(e1+2x(x2+1+x4))\ln \left(e^{1+2 x}\left(x^{2}+\sqrt{1+x^{4}}\right)\right).

Problem 2.21.15

Expand the function into a Maclaurin series and find the radius of convergence RR of this series.

?
(1)

xarcsin⁡x+1−x2x \arcsin x+\sqrt{1-x^{2}};

(2)

xarccos⁡x−1−x2x \arccos x-\sqrt{1-x^{2}};

(3)

2xarctg⁡x−ln⁡(1+x2)2 x \operatorname {arctg} x-\ln \left(1+x^{2}\right);

(4)

−32+x21−x2+12arcsin⁡x-\frac{3}{2}+\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \arcsin x;

(5)

14ln⁡1+x1−x+12arctg⁡x\frac{1}{4} \ln \frac{1+x}{1-x}+\frac{1}{2} \operatorname {arctg} x;

(6)

ln⁡(π2+x2−x)+arctg⁡x2\ln \left(\pi \sqrt{\frac{2+x}{2-x}}\right)+\operatorname {arctg} \frac{x}{2}.

Problem 2.21.16

Expand the function into a Maclaurin series and find the radius of convergence RR of this series.

?
(1)

xln⁡(x+1+x2)−1+x2x \ln \left(x+\sqrt{1+x^{2}}\right)-\sqrt{1+x^{2}};

(2)

(12+x)e−2x−(12−x)e2x;\left(\frac{1}{2}+x\right) e^{-2 x}-\left(\frac{1}{2}-x\right) e^{2 x} ;

(3)

ln⁡((2+x)2+x)+ln⁡((2−x)2−x)\ln \left((2+x)^{2+x}\right)+\ln \left((2-x)^{2-x}\right);

(4)

xx2+4+4ln⁡(x+x2+4)x \sqrt{x^{2}+4}+4 \ln \left(x+\sqrt{x^{2}+4}\right);

(5)

5+x2x2+1+12ln⁡(x+x2+1)5+\frac{x}{2} \sqrt{x^{2}+1}+\frac{1}{2} \ln \left(x+\sqrt{x^{2}+1}\right).

Problem 2.21.17

Prove that:

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(1)

14(ex+e−x+2cos⁡x)=∑n=0∞x4n(4n)!,−∞<x<+∞\frac{1}{4}\left(e^{x}+e^{-x}+2 \cos x\right) = \sum_{n = 0}^{\infty } \frac{x^{4 n}}{(4 n) !},-\infty < x < +\infty;

(2)

13(ex+2e−x/2cos⁡x32)=∑n=0∞x3n(3n)!,−∞<x<+∞\frac{1}{3}\left(e^{x}+2 e^{-x / 2} \cos \frac{x \sqrt{3}}{2}\right) = \sum_{n = 0}^{\infty } \frac{x^{3 n}}{(3 n) !},-\infty < x < +\infty;

(3)

x2+2x4+12(1−x2)ln⁡(1−x)=∑n=1∞xn+2n(n+2),∣x∣<1\frac{x^{2}+2 x}{4}+\frac{1}{2}\left(1-x^{2}\right) \ln (1-x) = \sum_{n = 1}^{\infty } \frac{x^{n+2}}{n(n+2)},\left|x\right| < 1;

(4)

34x2−x2+12(1−x)2ln⁡(1−x)=∑n=1∞xn+2n(n+1)(n+2),∣x∣<1\frac{3}{4} x^{2}-\frac{x}{2}+\frac{1}{2}(1-x)^{2} \ln (1-x) = \sum_{n = 1}^{\infty } \frac{x^{n+2}}{n(n+1)(n+2)},\left|x\right| < 1.

Problem 2.21.18

Expand the function into a Taylor series in a neighborhood of the point x0x_{0} and find the radius of convergence RR of the resulting series.

?
(1)

1x2−5x+6,x0=1\frac{1}{x^{2}-5 x+6}, x_{0} = 1;

(2)

1(x2−2x+3)2,x0=1\frac{1}{\left(x^{2}-2 x+3\right)^{2}}, x_{0} = 1;

(3)

1(x2−6x+18)2,x0=3\frac{1}{\left(x^{2}-6 x+18\right)^{2}}, x_{0} = 3;

(4)

x3−3x2+4x−1x2−2x+2,x0=1\frac{x^{3}-3 x^{2}+4 x-1}{x^{2}-2 x+2}, x_{0} = 1.

Problem 2.21.19

Expand the function into a Taylor series in a neighborhood of the point x0x_{0} and find the radius of convergence RR of the resulting series.

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(1)

1x2−12x+40,x0=6\frac{1}{\sqrt{x^{2}-12 x+40}}, x_{0} = 6;

(2)

1x2−10x+29,x0=5\frac{1}{\sqrt{x^{2}-10 x+29}}, x_{0} = 5;

(3)

1x2−6x+18,x0=3\frac{1}{\sqrt{x^{2}-6 x+18}}, x_{0} = 3;

(4)

1x2−4x+8,x0=2\frac{1}{\sqrt{x^{2}-4 x+8}}, x_{0} = 2;

(5)

1x2−6x+363,x0=3\frac{1}{\sqrt[3]{x^{2}-6 x+36}}, x_{0} = 3;

(6)

1x2−8x+243,x0=4\frac{1}{\sqrt[3]{x^{2}-8 x+24}}, x_{0} = 4;

(7)

arcsin⁡1+2x+2x2,x0=−1/2\arcsin \sqrt{1+2 x+2 x^{2}}, x_{0} = -1 / 2;

(8)

(x+1)arctg⁡x−13+x,x0=−1(x+1) \operatorname {arctg} \frac{x-1}{3+x}, x_{0} = -1;

(9)

(x+2)arccos⁡x+2x2+4x+5,x0=−2(x+2) \arccos \frac{x+2}{\sqrt{x^{2}+4 x+5}}, x_{0} = -2.

Problem 2.21.20

Expand the function into a Taylor series in a neighborhood of the point x0x_{0} and find the radius of convergence RR of the resulting series.

?
(1)

(x+1)cos⁡2x,x0=−1(x+1) \cos^{2} x, x_{0} = -1;

(2)

(2x+1)sin⁡xsin⁡(x+1),x0=−1/2(2 x+1) \sin x \sin (x+1), x_{0} = -1 / 2

(3)

sin⁡3x,x0=π/4\sin^{3} x, x_{0} = \pi / 4;

(4)

cos⁡4x,x0=−π/2\cos^{4} x, x_{0} = -\pi / 2;

(5)

cos⁡2x−(x(x−π)+π2/4)cos⁡2x,x0=π/2\cos^{2} x-\left(x(x-\pi )+\pi^{2} / 4\right) \cos 2 x, x_{0} = \pi / 2;

(6)

π+∫π/2xcos⁡2tt−π/2dt,x0=π2\pi +\int_{\pi / 2}^{x} \frac{\cos^{2} t}{t-\pi / 2} d t, x_{0} = \frac{\pi }{2}.

Problem 2.21.21

Expand the function into a Taylor series in a neighborhood of the point x0x_{0} and find the radius of convergence RR of the resulting series.

?
(1)

ln⁡(x2+2x+2),x0=−1\ln \left(x^{2}+2 x+2\right), x_{0} = -1;

(2)

ln⁡(x2−4x+6),x0=2\ln \left(x^{2}-4 x+6\right), x_{0} = 2;

(3)

ln⁡(x2−9x+20),x0=3\ln \left(x^{2}-9 x+20\right), x_{0} = 3;

(4)

ln⁡(x2−10x+30),x0=5\ln \left(x^{2}-10 x+30\right), x_{0} = 5;

(5)

ln⁡x2−2x+32−x,x0=1\ln \frac{x^{2}-2 x+3}{2-x}, x_{0} = 1;

(6)

ln⁡(2+3−x),x0=−1\ln (2+\sqrt{3-x}), x_{0} = -1;

(7)

(x2−10x+26)ln⁡x−28−x,x0=5\left(x^{2}-10 x+26\right) \ln \frac{x-2}{8-x}, x_{0} = 5;

(8)

1+∫2xln⁡5t−7tdt,x0=−21+\int_{2}^{x} \ln \frac{5 t-7}{t} d t, x_{0} = -2;

(9)

2+∫1xln⁡(2t2−4t+11)dt,x0=12+\int_{1}^{x} \ln \left(2 t^{2}-4 t+11\right) d t, x_{0} = 1.

Problem 2.21.22

By multiplying the corresponding series, expand the function into a Maclaurin series and find the radius of convergence of the resulting series:

?
(1)

ln⁡(1+x)1+x\frac{\ln (1+x)}{1+x};

(2)

ex1−x\frac{e^{x}}{1-x};

(3)

arctg⁡x1−x\frac{\operatorname {arctg} x}{1-x};

(4)

ln⁡2(1−x)\ln^{2}(1-x);

(5)

(arctg⁡x)2(\operatorname {arctg} x)^{2}.

Problem 2.21.23

Using differentiation of series (9), prove that:

?
(1)

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

(2)

∑n=0∞(n+1)(n+2)(n+3)3!xn=1(1−x)4,∣x∣<1\sum_{n = 0}^{\infty } \frac{(n+1)(n+2)(n+3)}{3 !} x^{n} = \frac{1}{(1-x)^{4}},\left|x\right| < 1.

Problem 2.21.24

Expand the function f(3)(x)f^{(3)}(x) into a Maclaurin series and find the radius of convergence of the resulting series, if:

?
(1)

f(x)=arctg⁡xf(x) = \operatorname {arctg} x;

(2)

f(x)=arcsin⁡xf(x) = \arcsin x;

(3)

f(x)=18x2arcsin⁡4xf(x) = \frac{1}{8} x^{2} \arcsin 4 x.

Problem 2.21.25

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

arctg⁡1−x1+x\operatorname {arctg} \frac{1-x}{1+x};

(2)

arctg⁡x+1/2x−1/2\operatorname {arctg} \frac{x+1 / 2}{x-1 / 2};

(3)

arctg⁡2x−3x+6\operatorname {arctg} \frac{2 x-3}{x+6};

(4)

arctg⁡2−x1+2x\operatorname {arctg} \frac{2-x}{1+2 x}.

Problem 2.21.26

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

arctg⁡2+x22−x2\operatorname {arctg} \frac{2+x^{2}}{2-x^{2}};

(2)

arctg⁡3−4x26+2x2\operatorname {arctg} \frac{3-4 x^{2}}{6+2 x^{2}};

(3)

arctg⁡2−2x21+4x2\operatorname {arctg} \frac{2-2 x^{2}}{1+4 x^{2}};

(4)

arctg⁡6+x23−2x2\operatorname {arctg} \frac{6+x^{2}}{3-2 x^{2}};

(5)

x2arctg⁡1/3+3x2x2−1x^{2} \operatorname {arctg} \frac{1 / 3+3 x^{2}}{x^{2}-1};

(6)

x3arctg⁡2−x3/21+x3x^{3} \operatorname {arctg} \frac{2-x^{3} / 2}{1+x^{3}}.

Problem 2.21.27

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

arctg⁡1−2x1+2x\operatorname {arctg} \sqrt{\frac{1-2 x}{1+2 x}};

(2)

arctg⁡(x+1+x2)\operatorname {arctg}\left(x+\sqrt{1+x^{2}}\right);

(3)

12arctg⁡2x1−x2\frac{1}{\sqrt{2}} \operatorname {arctg} \frac{\sqrt{2} x}{1-x^{2}};

(4)

x2arctg⁡x1−x2x^{2} \operatorname {arctg} \frac{x}{\sqrt{1-x^{2}}}.

Problem 2.21.28

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

arcctg⁡2x31−x6\operatorname {arcctg} \frac{2 x^{3}}{1-x^{6}};

(2)

arcctg⁡x1−x2\operatorname {arcctg} \frac{x}{\sqrt{1-x^{2}}};

(3)

arcctg⁡2x21−4x4\operatorname {arcctg} \frac{2 x^{2}}{\sqrt{1-4 x^{4}}};

(4)

arcctg⁡x1+1−x2\operatorname {arcctg} \frac{x}{1+\sqrt{1-x^{2}}}.

Problem 2.21.29

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

arcsin⁡2x1+x2\arcsin \frac{2 x}{1+x^{2}};

(2)

arcsin⁡x1+x2\arcsin \frac{x}{\sqrt{1+x^{2}}};

(3)

1+arcsin⁡⋅x216+x41+\arcsin \cdot \frac{x^{2}}{\sqrt{16+x^{4}}};

(4)

2xarcsin⁡2x31+4x62 x \arcsin \frac{2 x^{3}}{\sqrt{1+4 x^{6}}};

(5)

arcsin⁡(2x1−x2)\arcsin \left(2 x \sqrt{1-x^{2}}\right).

Problem 2.21.30

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

18arccos⁡4x21+16x4\frac{1}{8} \arccos \frac{4 x^{2}}{\sqrt{1+16 x^{4}}};

(2)

xarccos⁡x24+x4x \arccos \frac{x^{2}}{\sqrt{4+x^{4}}};

(3)

arccos⁡x216+x4\arccos \frac{x^{2}}{\sqrt{16+x^{4}}};

(4)

xarccos⁡x9+x2x \arccos \frac{x}{\sqrt{9+x^{2}}};

(5)

x3arccos⁡6x9+x2x^{3} \arccos \frac{6 x}{9+x^{2}};

(6)

xarccos⁡(2x1−x2)x \arccos \left(2 x \sqrt{1-x^{2}}\right).

Problem 2.21.31

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

∫0xe−t2dt\int_{0}^{x} e^{-t^{2}} d t;

(2)

∫0xsin⁡ttdt\int_{0}^{x} \frac{\sin t}{t} d t

(3)

∫0xt2sh⁡tdt\int_{0}^{x} t^{2} \operatorname {sh} t d t;

(4)

∫0x1−ch⁡ttdt\int_{0}^{x} \frac{1-\operatorname {ch} t}{t} d t;

(5)

∫0xarctg⁡ttdt\int_{0}^{x} \frac{\operatorname {arctg} t}{t} d t;

(6)

∫0xtarctg⁡t+102t−5dt\int_{0}^{x} t \operatorname {arctg} \frac{t+10}{2 t-5} d t.

Problem 2.21.32

Expand the given function into a Maclaurin series, using the Maclaurin series for its derivative. Find the radius of convergence of the resulting series.

?
(1)

∫0xln⁡(1+4t)tdt\int_{0}^{x} \frac{\ln (1+4 t)}{t} d t;

(2)

∫0xln⁡1+t1−tdtt\int_{0}^{x} \ln \frac{1+t}{1-t} \frac{d t}{t};

(3)

∫0xdt1−t44\int_{0}^{x} \frac{d t}{\sqrt[4]{1-t^{4}}};

(4)

∫0xtdt1+t4\int_{0}^{x} \frac{t d t}{\sqrt{1+t^{4}}}

(5)

∫0xt2dt1+t2\int_{0}^{x} \frac{t^{2} d t}{\sqrt{1+t^{2}}};

(6)

∫x1ln⁡3+t3−tdt\int_{x}^{1} \ln \frac{3+t}{3-t} d t;

(7)

∫0xt2arctg⁡2t−2t+4dt\int_{0}^{x} t^{2} \operatorname {arctg} \frac{2 t-2}{t+4} d t;

(8)

ln⁡(x+3+x2)+x∫0xe−t2−1tdt\ln \left(x+\sqrt{3+x^{2}}\right)+x \int_{0}^{x} \frac{e^{-t^{2}}-1}{t} d t

(9)

arctg⁡3x+13x−1+x∫0x1−cos⁡ttdt\operatorname {arctg} \frac{3 x+1}{3 x-1}+x \int_{0}^{x} \frac{1-\cos t}{t} d t;

(10)

arccos⁡x21+x4+x∫0xsh⁡t2tdt\arccos \frac{x^{2}}{\sqrt{1+x^{4}}}+x \int_{0}^{x} \frac{\operatorname {sh} t^{2}}{t} d t.

(11)
Problem 2.21.33

Find the first four (nonzero) terms of the Maclaurin series of the given function:

?
(1)

ln⁡(1+ex)\ln \left(1+e^{x}\right);

(2)

(1+x)x(1+x)^{x};

(3)

∫0xtdtln⁡(1+t)\int_{0}^{x} \frac{t d t}{\ln (1+t)};

(4)

esin⁡xe^{\sin x}

(5)

etg⁡xe^{\operatorname {tg} x};

(6)

earcsin⁡xe^{\arcsin x}.

Problem 2.21.34

Show that the function y=∑n=0∞x2n2nn!y = \sum_{n = 0}^{\infty } \frac{x^{2 n}}{2^{n} n !} is a solution of the differential equation y′−xy=0y^{\prime }-x y = 0.

?
Problem 2.21.35

Show that the function y=∑n=0∞xn+1n!y = \sum_{n = 0}^{\infty } \frac{x^{n+1}}{n !} is a solution of the differential equation xy′=(x+1)yx y^{\prime } = (x+1) y.

?
Problem 2.21.36

Show that the function y=∑n=0∞xn(n!)2y = \sum_{n = 0}^{\infty } \frac{x^{n}}{(n !)^{2}} is a solution of the differential equation xy′′+y′−y=0x y^{\prime \prime }+y^{\prime }-y = 0.

?
Problem 2.21.37

Show that the function y=∑n=0∞x2n(2n)!!y = \sum_{n = 0}^{\infty } \frac{x^{2 n}}{(2 n) ! !} is a solution of the differential equation y′′−xy′−y=0y^{\prime \prime }-x y^{\prime }-y = 0.

?
Problem 2.21.38

Show that the function

y=1−x222+x422⋅42⋅62+…+(−1)nx2n((2n)!!)2+… y = 1-\frac{x^{2}}{2^{2}}+\frac{x^{4}}{2^{2} \cdot 4^{2} \cdot 6^{2}}+\ldots +\frac{(-1)^{n} x^{2 n}}{((2 n) ! !)^{2}}+\ldots

satisfies the differential equation xy′′+y′+xy=0x y^{\prime \prime }+y^{\prime }+x y = 0.

?
Problem 2.21.39

Show that the function y=∑n=0∞x4n(4n)!y = \sum_{n = 0}^{\infty } \frac{x^{4 n}}{(4 n) !} satisfies the differential equation y(4)−y=0y^{(4)}-y = 0.

?
Problem 2.21.40

Find the Maclaurin series expansion of the solution of the differential equation satisfying the given initial conditions:

?
(1)

y′−y=0,y(0)=1y^{\prime }-y = 0, y(0) = 1;

(2)

(1+x2)y′−1=0,y(0)=0\left(1+x^{2}\right) y^{\prime }-1 = 0, y(0) = 0;

(3)

y′′+λ2y=0,y(0)=0,y′(0)=λy^{\prime \prime }+\lambda^{2} y = 0, y(0) = 0, y^{\prime }(0) = \lambda

(4)

y′′+xy=0,y(0)=1,y′(0)=0y^{\prime \prime }+x y = 0, y(0) = 1, y^{\prime }(0) = 0;

(5)

(1−x2)y′′−xy′=0,y(0)=0,y′(0)=1\left(1-x^{2}\right) y^{\prime \prime }-x y^{\prime } = 0, y(0) = 0, y^{\prime }(0) = 1;

(6)

(1−x2)y′′−5xy′−4y=0,y(0)=1,y′(0)=0\left(1-x^{2}\right) y^{\prime \prime }-5 x y^{\prime }-4 y = 0, y(0) = 1, y^{\prime }(0) = 0.

Problem 2.21.41

Find the first four (nonzero) terms of the series

y=∑n=0∞anxn y = \sum _{n = 0}^{\infty } a_{n} x^{n}

whose sum satisfies the differential equation y′′−xy′+y=exy^{\prime \prime }-x y^{\prime }+y = e^{x} and the initial conditions y(0)=1,y′(0)=0y(0) = 1, y^{\prime }(0) = 0.

?
Problem 2.21.42

Find the first four (nonzero) terms of the series

y=∑n=0∞an(x−1)n y = \sum _{n = 0}^{\infty } a_{n}(x-1)^{n}

whose sum satisfies the differential equation y′′=(y′)2+xyy^{\prime \prime } = \left(y^{\prime }\right)^{2}+x y and the initial conditions y(1)=1,y′(1)=0y(1) = 1, y^{\prime }(1) = 0.

?
Problem 2.21.43

Find the first five (nonzero) terms of the series

y=∑n=0∞anxn y = \sum _{n = 0}^{\infty } a_{n} x^{n}

whose sum satisfies the differential equation y′=x3+y2y^{\prime } = x^{3}+y^{2} and the initial condition y(0)=1y(0) = 1.

?
Problem 2.21.44

Using the fact that the function y=arcsin⁡x1−x2y = \frac{\arcsin x}{\sqrt{1-x^{2}}} satisfies the differential equation (1−x2)y′−xy=1\left(1-x^{2}\right) y^{\prime }-x y = 1 and the condition y(0)=0y(0) = 0, prove that

arcsin⁡x1−x2=∑n=0∞(2n)!!(2n+1)!!x2n+1,∣x∣<1 \frac{\arcsin x}{\sqrt{1-x^{2}}} = \sum _{n = 0}^{\infty } \frac{(2 n) ! !}{(2 n+1) ! !} x^{2 n+1}, \quad \left|x\right| < 1
?
Problem 2.21.45

Expand the function into a Maclaurin series and find the radius of convergence RR of the resulting series:

?
(1)

1(1+x)(1+x2)(1+x4)(1+x8)\frac{1}{(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)};

(2)

xsin⁡α1−2xcos⁡α+x2\frac{x \sin \alpha }{1-2 x \cos \alpha +x^{2}}

(3)

xcos⁡α−x21−2xcos⁡α+x2\frac{x \cos \alpha -x^{2}}{1-2 x \cos \alpha +x^{2}};

(4)

ln⁡(1−2xcos⁡α+x2)\ln \left(1-2 x \cos \alpha +x^{2}\right);

(5)

arccos⁡(1−2x2)\arccos \left(1-2 x^{2}\right);

(6)

(arcsin⁡xx)2\left(\frac{\arcsin x}{x}\right)^{2}.

Problem 2.21.46

Let f(x)=arth⁡xf(x) = \operatorname {arth} x be the function inverse to the function th x=sh⁡x/ch⁡xx = \operatorname {sh} x / \operatorname {ch} x. Expand f(x)f(x) into a Maclaurin series and find the radius of convergence of the series.

?
Problem 2.21.47

Prove that:

?
(1)

(arctg⁡xx)2=1+∑n=1∞(−1)n(1+13+…+12n+1)x2nn+1,∣x∣<1\left(\frac{\operatorname {arctg} x}{x}\right)^{2} = 1+\sum_{n = 1}^{\infty }(-1)^{n}\left(1+\frac{1}{3}+\ldots +\frac{1}{2 n+1}\right) \frac{x^{2 n}}{n+1}, \quad \left|x\right| < 1;

(2)

(ln⁡(1+x)x)2=1+2∑n=1∞(−1)n(1+12+…+1n+1)xnn+2\left(\frac{\ln (1+x)}{x}\right)^{2} = 1+2 \sum_{n = 1}^{\infty }(-1)^{n}\left(1+\frac{1}{2}+\ldots +\frac{1}{n+1}\right) \frac{x^{n}}{n+2}, ∣x∣<1;\left|x\right| < 1 ;

(3)

arctg⁡x⋅ln⁡(1+x2)=2∑n=1∞(−1)n−1(1+12+13+…+12n)x2n+12n+1\operatorname {arctg} x \cdot \ln \left(1+x^{2}\right) = 2 \sum_{n = 1}^{\infty }(-1)^{n-1}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{2 n}\right) \frac{x^{2 n+1}}{2 n+1}, ∣x∣<1;\left|x\right| < 1 ;

(4)

12x(1−x2)ln⁡1+x1−x=1+∑n=1∞(1+13+…+12n+1)x2n,∣x∣<1\frac{1}{2 x\left(1-x^{2}\right) \ln \frac{1+x}{1-x}} = 1+\sum_{n = 1}^{\infty }\left(1+\frac{1}{3}+\ldots +\frac{1}{2 n+1}\right) x^{2 n}, \quad \left|x\right| < 1;

(5)

16ln⁡(1−x3)−12ln⁡(1−x)−13arctg⁡x3x+2=∑n=0∞x3n+23n+2\frac{1}{6} \ln \left(1-x^{3}\right)-\frac{1}{2} \ln (1-x)-\frac{1}{\sqrt{3}} \operatorname {arctg} \frac{x \sqrt{3}}{x+2} = \sum_{n = 0}^{\infty } \frac{x^{3 n+2}}{3 n+2},

∣x∣<1;\left|x\right| < 1 ;

(6)

16ln⁡(x2+x+1)−13ln⁡(1−x)+13arctg⁡x3x+2=∑n=0∞x3n+13n+1\frac{1}{6} \ln \left(x^{2}+x+1\right)-\frac{1}{3} \ln (1-x)+\frac{1}{\sqrt{3}} \operatorname {arctg} \frac{x \sqrt{3}}{x+2} = \sum_{n = 0}^{\infty } \frac{x^{3 n+1}}{3 n+1},

∣x∣<1\left|x\right| < 1.

Problem 2.21.48

Prove that:

?
(1)

(arcsin⁡x)3=∑n=1∞12⋅(2n)!(2n+1)(n!)2(1+132+…+1(2n−1)2)(x2)2n+1(\arcsin x)^{3} = \sum_{n = 1}^{\infty } \frac{12 \cdot (2 n) !}{(2 n+1)(n !)^{2}}\left(1+\frac{1}{3^{2}}+\ldots +\frac{1}{(2 n-1)^{2}}\right)\left(\frac{x}{2}\right)^{2 n+1}, ∣x∣<1;\left|x\right| < 1 ;

(2)

ln⁡(x+1+x2)1+x2=∑n=0∞(−1)n22n(n!)2(2n+1)!x2n+1,∣x∣<1\frac{\ln \left(x+\sqrt{1+x^{2}}\right)}{\sqrt{1+x^{2}}} = \sum_{n = 0}^{\infty }(-1)^{n} \frac{2^{2 n}(n !)^{2}}{(2 n+1) !} x^{2 n+1},\left|x\right| < 1;

(3)

(ln⁡(x+1+x2))2=∑n=0∞22n−1((n−1)!)2(2n)!(−1)n−1x2n,∣x∣<1\left(\ln \left(x+\sqrt{1+x^{2}}\right)\right)^{2} = \sum_{n = 0}^{\infty } \frac{2^{2 n-1}((n-1) !)^{2}}{(2 n) !}(-1)^{n-1} x^{2 n},\left|x\right| < 1;

(4)

1+x2ln⁡(x+1+x2)=x−∑n=1∞(−1)n22n(n!)22n(2n+1)!x2n+1\sqrt{1+x^{2}} \ln \left(x+\sqrt{1+x^{2}}\right) = x-\sum_{n = 1}^{\infty }(-1)^{n} \frac{2^{2 n}(n !)^{2}}{2 n(2 n+1) !} x^{2 n+1}, ∣x∣<1\left|x\right| < 1.

Problem 2.21.49

Using the expansion xex−1=∑n=0∞Bnn!xn\frac{x}{e^{x}-1} = \sum_{n = 0}^{\infty } \frac{B_{n}}{n !} x^{n}, where B1=1,B2k+1=0,k∈NB_{1} = 1, B_{2 k+1} = 0, k \in N (example 5), prove that:

?
(1)

xctg⁡x=1+∑n=1∞(−1)n22nB2n(2n)!x2n,∣x∣<πx \operatorname {ctg} x = 1+\sum_{n = 1}^{\infty }(-1)^{n} \frac{2^{2 n} B_{2 n}}{(2 n) !} x^{2 n},\left|x\right| < \pi;

(2)

ln⁡sin⁡xx=∑n=1∞(−1)n22n−1B2nn(2n)!x2n,∣x∣<π\ln \frac{\sin x}{x} = \sum_{n = 1}^{\infty }(-1)^{n} \frac{2^{2 n-1} B_{2 n}}{n(2 n) !} x^{2 n},\left|x\right| < \pi;

(3)

tg⁡x=∑n=1∞(−1)n22n(1−22n)(2n)!B2nx2n−1,∣x∣<π\operatorname {tg} x = \sum_{n = 1}^{\infty }(-1)^{n} \frac{2^{2 n}\left(1-2^{2 n}\right)}{(2 n) !} B_{2 n} x^{2 n-1},\left|x\right| < \pi;

(4)

xsin⁡x=1+∑n=1∞(−1)n2−22n(2n)!B2nx2n,∣x∣<π\frac{x}{\sin x} = 1+\sum_{n = 1}^{\infty }(-1)^{n} \frac{2-2^{2 n}}{(2 n) !} B_{2 n} x^{2 n},\left|x\right| < \pi;

(5)

ln⁡cos⁡x=∑n=1∞(−1)n22n−1(22n−1)n(2n)!B2nx2n,∣x∣<π2\ln \cos x = \sum_{n = 1}^{\infty }(-1)^{n} \frac{2^{2 n-1}\left(2^{2 n}-1\right)}{n(2 n) !} B_{2 n} x^{2 n},\left|x\right| < \frac{\pi }{2}.

Problem 2.21.50

Let the Maclaurin series expansion of the function 1/cos⁡x1 / \cos x be written in the form

1cos⁡x=∑n=1∞(−1)nE2n(2n)!x2n. \frac{1}{\cos x} = \sum _{n = 1}^{\infty }(-1)^{n} \frac{E_{2 n}}{(2 n) !} x^{2 n}.

Prove that

E0=1,E0+C2n2E2+C2n4E4+…+C2n2nE2n=0,n∈N E_{0} = 1, \quad 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 N
?
Problem 2.21.51

Prove that

ln⁡tg⁡(π2+x2)=∑n=0∞(−1)nE2n(2n+1)(2n)!x2n+1,∣x∣<π2 \ln \operatorname {tg}\left(\frac{\pi }{2}+\frac{x}{2}\right) = \sum _{n = 0}^{\infty }(-1)^{n} \frac{E_{2 n}}{(2 n+1)(2 n) !} x^{2 n+1}, \quad \left|x\right| < \frac{\pi }{2}

where the coefficients E2nE_{2 n} are defined in problem 50.

?
Problem 2.21.52

Using the definition of the functions of a complex variable cos⁡z\cos z and sin⁡z\sin z (formulas (17) and (18)), prove that:

?
(1)

sin⁡z⋅cos⁡z=12sin⁡2z\sin z \cdot \cos z = \frac{1}{2} \sin 2 z;

(2)

sin⁡2z+cos⁡2z=1\sin^{2} z+\cos^{2} z = 1.

Problem 2.21.53

Using the definition of the functions of a complex variable sh⁡z\operatorname {sh} z and ch⁡z\operatorname {ch} z (formulas (15) and (16)), prove that:

?
(1)

ch⁡2z−sh⁡2z=1\operatorname {ch}^{2} z-\operatorname {sh}^{2} z = 1;

(2)

ch⁡2z=ch⁡2z+sh⁡2z\operatorname {ch} 2 z = \operatorname {ch}^{2} z+\operatorname {sh}^{2} z.

Problem 2.21.54

Expand the function into a series of powers of zz:

?
(1)

ezsin⁡ze^{z} \sin z

(2)

ch⁡zcos⁡z\operatorname {ch} z \cos z

(3)

ezctg⁡αcos⁡z,sin⁡x≠0e^{z \operatorname {ctg} \alpha } \cos z, \quad \sin x \neq 0;

(4)

ezcos⁡αcos⁡(zsin⁡α)e^{z \cos \alpha } \cos (z \sin \alpha ).

Problem 2.21.55

Using term-by-term differentiation, compute the sum of the series:

?
(1)

x+x33+x55+…+x2n+12n+1+…x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots +\frac{x^{2 n+1}}{2 n+1}+\ldots

(2)

x2+x42+x63+…+x2nn+…x^{2}+\frac{x^{4}}{2}+\frac{x^{6}}{3}+\ldots +\frac{x^{2 n}}{n}+\ldots

Problem 2.21.56

Using integration, compute the sum of the series:

?
(1)

∑n=1∞n2xn−1\sum_{n = 1}^{\infty } n^{2} x^{n-1};

(2)

∑n=0∞(2n+1)x2nn!\sum_{n = 0}^{\infty } \frac{(2 n+1) x^{2 n}}{n !}.

Problem 2.21.57

Compute the sum of the series.

?
(1)

∑n=0∞ln⁡nxn!\sum_{n = 0}^{\infty } \frac{\ln^{n} x}{n !};

(2)

∑n=0∞(−1)nln⁡nx2nn!\sum_{n = 0}^{\infty } \frac{(-1)^{n} \ln^{n} x}{2^{n} n !}

(3)

∑n=1∞xnn\sum_{n = 1}^{\infty } \frac{x^{n}}{n};

(4)

∑n=1∞nxn+1\sum_{n = 1}^{\infty } n x^{n+1};

(5)

∑n=0∞(2n+1)xn\sum_{n = 0}^{\infty }(2 n+1) x^{n};

(6)

∑n=1∞(−1)nx2n2n+1\sum_{n = 1}^{\infty }(-1)^{n} \frac{x^{2 n}}{2 n+1}.

Problem 2.21.58

Compute the sum of the series.

?
(1)

∑n=1∞n(n+2)xn\sum_{n = 1}^{\infty } n(n+2) x^{n};

(2)

∑n=1∞(−1)n−1n(n+1)xn\sum_{n = 1}^{\infty }(-1)^{n-1} n(n+1) x^{n};

(3)

∑n=0∞(3n+1)x3nn!\sum_{n = 0}^{\infty } \frac{(3 n+1) x^{3 n}}{n !};

(4)

∑n=0∞(−1)n(2n2+1)(2n)!x2n\sum_{n = 0}^{\infty } \frac{(-1)^{n}\left(2 n^{2}+1\right)}{(2 n) !} x^{2 n}

(5)

∑n=0∞n2+12nn!xn\sum_{n = 0}^{\infty } \frac{n^{2}+1}{2^{n} n !} x^{n}

(6)

∑n=0∞(−1)nn3(n+1)!xn\sum_{n = 0}^{\infty } \frac{(-1)^{n} n^{3}}{(n+1) !} x^{n}.

Problem 2.21.59

Compute the sum of the series.

?
(1)

1+∑n=1∞(2n−1)!!(2n)!!xn1+\sum_{n = 1}^{\infty } \frac{(2 n-1) ! !}{(2 n) ! !} x^{n};

(2)

13(x2)+1⋅43⋅6(x2)2+1⋅4⋅73⋅6⋅9(x2)3+…;\frac{1}{3}\left(\frac{x}{2}\right)+\frac{1 \cdot 4}{3 \cdot 6}\left(\frac{x}{2}\right)^{2}+\frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9}\left(\frac{x}{2}\right)^{3}+\ldots ;

(3)

∑n=1∞xnn(n+1)\sum_{n = 1}^{\infty } \frac{x^{n}}{n(n+1)};

(4)

∑n=1∞(−1)n−1x2nn(2n−1)\sum_{n = 1}^{\infty } \frac{(-1)^{n-1} x^{2 n}}{n(2 n-1)}

(5)

∑n=0∞(2x)4n+14n+1\sum_{n = 0}^{\infty } \frac{(2 x)^{4 n+1}}{4 n+1}

(6)

∑n=1∞n3xn\sum_{n = 1}^{\infty } n^{3} x^{n}

Problem 2.21.60

Compute the sum of the series.

?
(1)

∑n=1∞n4xn\sum_{n = 1}^{\infty } n^{4} x^{n}

(2)

∑n=1∞xnn(n+1)(n+2)\sum_{n = 1}^{\infty } \frac{x^{n}}{n(n+1)(n+2)};

(3)

∑n=0∞(−1)nx3n+13n+1\sum_{n = 0}^{\infty }(-1)^{n} \frac{x^{3 n+1}}{3 n+1}

(4)

∑n=2∞(−1)nxn−1(n−1)(n+2)\sum_{n = 2}^{\infty } \frac{(-1)^{n} x^{n-1}}{(n-1)(n+2)}

(5)

∑n=0∞(−1)nxn(2n)!\sum_{n = 0}^{\infty }(-1)^{n} \frac{x^{n}}{(2 n) !};

(6)

∑n=0∞n2x2(2n+1)!\sum_{n = 0}^{\infty } \frac{n^{2} x^{2}}{(2 n+1) !}.

Problem 2.21.61

Compute the sum of the series.

?
(1)

∑n=1∞((n−1)!)2(2n)!(2x)2n\sum_{n = 1}^{\infty } \frac{((n-1) !)^{2}}{(2 n) !}(2 x)^{2 n};

(2)

∑n=1∞sin⁡nxn!\sum_{n = 1}^{\infty } \frac{\sin n x}{n !}.

Problem 2.21.62

The coefficients of the series

1+x+2x2+4x3+7x4+13x5+24x6+… 1+x+2 x^{2}+4 x^{3}+7 x^{4}+13 x^{5}+24 x^{6}+\ldots

have the property that each coefficient (starting from the fourth) equals the sum of the three preceding ones. Find the sum of this series.

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Problem 2.21.63

Compute the sum of the numerical series.

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(1)

∑n=1∞n2n!\sum_{n = 1}^{\infty } \frac{n^{2}}{n !};

(2)

∑n=0∞2n(n+1)n!\sum_{n = 0}^{\infty } \frac{2^{n}(n+1)}{n !};

(3)

∑n=0∞(−1)nn(2n+1)!\sum_{n = 0}^{\infty } \frac{(-1)^{n} n}{(2 n+1) !};

(4)

∑n=1∞n23n\sum_{n = 1}^{\infty } \frac{n^{2}}{3^{n}}

(5)

1+12⋅13+1⋅32⋅4⋅15+1⋅3⋅52⋅4⋅6⋅17+…1+\frac{1}{2} \cdot \frac{1}{3}+\frac{1 \cdot 3}{2 \cdot 4} \cdot \frac{1}{5}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \frac{1}{7}+\ldots;

(6)

1−12+1⋅32⋅4−1⋅3⋅52⋅4⋅6+…1-\frac{1}{2}+\frac{1 \cdot 3}{2 \cdot 4}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}+\ldots

Problem 2.21.64

Compute the sum of the numerical series.

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(1)

∑n=1∞1n2(n+1)2\sum_{n = 1}^{\infty } \frac{1}{n^{2}(n+1)^{2}};

(2)

∑n=1∞1n2(n+1)2(n+2)2\sum_{n = 1}^{\infty } \frac{1}{n^{2}(n+1)^{2}(n+2)^{2}};

(3)

11⋅2−12⋅3+13⋅4−14⋅5+…+(−1)n+1n(n+1)+…\frac{1}{1 \cdot 2}-\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}-\frac{1}{4 \cdot 5}+\ldots +\frac{(-1)^{n+1}}{n(n+1)}+\ldots;

(4)

11⋅2⋅3+13⋅4⋅5+15⋅5⋅7+…\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{3 \cdot 4 \cdot 5}+\frac{1}{5 \cdot 5 \cdot 7}+\ldots;

(5)

∑n=1∞1n(2n+1)\sum_{n = 1}^{\infty } \frac{1}{n(2 n+1)};

(6)

∑n=2∞(−1)nn2+n−2\sum_{n = 2}^{\infty } \frac{(-1)^{n}}{n^{2}+n-2}.

Problem 2.21.65

Compute the sum of the numerical series.

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(1)

∑n=0∞1(n+1)(2n+1)\sum_{n = 0}^{\infty } \frac{1}{(n+1)(2 n+1)};

(2)

∑n=0∞(−1)n(n+1)(2n+1)\sum_{n = 0}^{\infty } \frac{(-1)^{n}}{(n+1)(2 n+1)};

(3)

∑n=1∞1n(2n+3)\sum_{n = 1}^{\infty } \frac{1}{n(2 n+3)};

(4)

∑n=0∞(−1)n3n+1\sum_{n = 0}^{\infty } \frac{(-1)^{n}}{3 n+1}

(5)

∑n=1∞1n(3n+1)\sum_{n = 1}^{\infty } \frac{1}{n(3 n+1)}

(6)

∑n=1∞(−1)nn(3n+1)\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{n(3 n+1)}.

Problem 2.21.66

Let an>0a_{n} > 0, n∈Nn \in N, and let the series ∑n=1∞1an\sum_{n = 1}^{\infty } \frac{1}{a_{n}} diverge. Find the sum a1a2+x+a1a2+x⋅a2a3+x+a1a2+x⋅a2a3+x⋅a3a4+x+…\frac{a_{1}}{a_{2}+x}+\frac{a_{1}}{a_{2}+x} \cdot \frac{a_{2}}{a_{3}+x}+\frac{a_{1}}{a_{2}+x} \cdot \frac{a_{2}}{a_{3}+x} \cdot \frac{a_{3}}{a_{4}+x}+\ldots assuming that x>0x > 0.

…+a1a2…an(a2+x)(a3+x)…(an+1+x)+… \ldots +\frac{a_{1} a_{2} \ldots a_{n}}{\left(a_{2}+x\right)\left(a_{3}+x\right) \ldots \left(a_{n+1}+x\right)}+\ldots
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Problem 2.21.67

Find the sum of the series

x1−x2+x21−x4+x41−x8+…+x2n1−x2n+1+… \frac{x}{1-x^{2}}+\frac{x^{2}}{1-x^{4}}+\frac{x^{4}}{1-x^{8}}+\ldots +\frac{x^{2^{n}}}{1-x^{2^{n+1}}}+\ldots
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Problem 2.21.68

Find the sum of the series ∑n=1∞xn+1(1−xn)(1−xn+1)\sum_{n = 1}^{\infty } \frac{x^{n+1}}{\left(1-x^{n}\right)\left(1-x^{n+1}\right)} for ∣x∣<1\left|x\right| < 1 and for ∣x∣>1\left|x\right| > 1.

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Problem 2.21.69

Find the sum of the series

1!x+1+2!(x+1)(x+2)+…+n!(x+1)(x+2)…(x+n)+… \frac{1 !}{x+1}+\frac{2 !}{(x+1)(x+2)}+\ldots +\frac{n !}{(x+1)(x+2) \ldots (x+n)}+\ldots

where x≠−k(k∈N)x \neq -k \quad (k \in N).

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Problem 2.21.70

Using the expansion of the integrand into a series, compute the integral

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(1)

∫01ln⁡11−xdx\int_{0}^{1} \ln \frac{1}{1-x} d x

(2)

∫01ln⁡x⋅ln⁡(1−x)dx\int_{0}^{1} \ln x \cdot \ln (1-x) d x

(3)

∫0+∞xe2πx−1dx\int_{0}^{+\infty } \frac{x}{e^{2 \pi x}-1} d x

(4)

∫0+∞xdxex+1\int_{0}^{+\infty } \frac{x d x}{e^{x}+1}

Problem 2.21.71

Using the corresponding expansions, compute to within 10−410^{-4}:

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(1)

cos⁡1∘\cos 1^{\circ };

(2)

cos⁡10∘\cos 10^{\circ };

(3)

sin⁡10∘\sin 10^{\circ };

(4)

sin⁡1∘\sin 1^{\circ };

(5)

1303\sqrt[3]{130};

(6)

ln⁡1,2\ln 1,2.

Problem 2.21.72

Compute to within 10−310^{-3}:

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(1)

tg⁡9∘\operatorname {tg} 9^{\circ };

(2)

5003\sqrt[3]{500};

(3)

arctg⁡(π/10)\operatorname {arctg}(\pi / 10);

(4)

683\sqrt[3]{68}

(5)

ln⁡5\ln 5;

(6)

arcsin⁡(1/3)\arcsin (1 / 3).

Problem 2.21.73

Compute to within 10−510^{-5}:

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(1)

sin⁡18∘\sin 18^{\circ };

(2)

ee.

Problem 2.21.74

Using the equality π/6=arcsin⁡(1/2)\pi / 6 = \arcsin (1 / 2), compute the number π\pi by means of the Maclaurin series for the function arcsin⁡x\arcsin x to within 10−410^{-4}.

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Problem 2.21.75

Using the equality

π4=4arctg⁡15−arctg⁡1239 \frac{\pi }{4} = 4 \operatorname {arctg} \frac{1}{5}-\operatorname {arctg} \frac{1}{239}

compute the number π\pi to within 10−910^{-9}.

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Problem 2.21.76

Using the Maclaurin series for f(x)f(x), compute the integral ∫01f(x)dx\int_{0}^{1} f(x) d x to within 10−310^{-3}, if:

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(1)

f(x)=sin⁡xxf(x) = \frac{\sin x}{x}

(2)

f(x)=x3cos⁡xf(x) = \sqrt[3]{x} \cos x

(3)

f(x)=cos⁡x2f(x) = \cos x^{2}

(4)

f(x)=sh⁡xxf(x) = \frac{\operatorname {sh} x}{x}

(5)

f(x)=sin⁡x2f(x) = \sin x^{2}

(6)

f(x)=11+x4f(x) = \frac{1}{\sqrt{1+x^{4}}}.

Problem 2.21.77

Compute to within 10−310^{-3}:

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(1)

∫12exxdx\int_{1}^{2} \frac{e^{x}}{x} d x

(2)

∫24e1xdx\int_{2}^{4} e^{\frac{1}{x}} d x

(3)

∫01/2arctg⁡xxdx\int_{0}^{1 / 2} \frac{\operatorname {arctg} x}{x} d x;

(4)

∫01/3dx1−x23\int_{0}^{1 / 3} \frac{d x}{\sqrt[3]{1-x^{2}}};

(5)

∫510ln⁡(1+x2)xdx\int_{5}^{10} \frac{\ln \left(1+x^{2}\right)}{x} d x;

(6)

∫01/2arcsin⁡xxdx\int_{0}^{1 / 2} \frac{\arcsin x}{x} d x;

(7)

∫2+∞dx1+x3dx\int_{2}^{+\infty } \frac{d x}{1+x^{3}} d x;

(8)

∫01xxdx\int_{0}^{1} x^{x} d x.

Problem 2.21.78

Prove that if there exist numbers δ>0,λ>0\delta > 0, \lambda > 0 such that for all x∈(x0−δ;x0+δ)x \in \left(x_{0}-\delta ; x_{0}+\delta \right) and for all n=0,1,2,…n = 0,1,2, \ldots the inequality ∣f(n)(x)∣⩽λn\left|f^{(n)}(x)\right| \leqslant \lambda^{n} holds, then at every point xx of the interval (x0−δ;x0+δ)(x_{0}-\delta ; x_{0}+\delta ) the function f(x)f(x) is represented by the Taylor series (1) converging to it.

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Problem 2.21.79

Let there exist a constant M>0M > 0 such that for all n∈Nn \in N the inequality ∣f(n)(x0)∣⩽M\left|f^{(n)}\left(x_{0}\right)\right| \leqslant M holds. Does it follow that:

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(1)

the Taylor series (1) converges on some interval (x0−δ;x0+δ)\left(x_{0}-\delta ; x_{0}+\delta \right);

(2)

the function f(x)f(x) is represented in some neighborhood of the point x0x_{0} by a Taylor series converging to it?

Problem 2.21.80

Prove that for the function

f(x)={e−1/x2, if x≠00, if x=0 f(x) = \begin{cases} e^{-1 / x^{2}}, & \text{ if } & x \neq 0 \\ 0, & \text{ if } & x = 0 \end{cases}

one can form the Maclaurin series, but the sum of this series does not coincide with f(x)f(x) for x≠0x \neq 0.

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Problem 2.21.81

Let Sn(x)=∑k=0n(−1)kx2k+1(2k+1)!S_{n}(x) = \sum_{k = 0}^{n}(-1)^{k} \frac{x^{2 k+1}}{(2 k+1) !}. Is it true that Sn(x)⇉sin⁡xS_{n}(x) \rightrightarrows \sin x, x∈Rx \in R?

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Problem 2.21.82

Prove that if the coefficients of the power series ∑n=0∞anxn\sum_{n = 0}^{\infty } a_{n} x^{n} satisfy the inequalities ∣an∣<Mn!(n=1,2…)\left|a_{n}\right| < \frac{M}{n !} \quad (n = 1,2 \ldots ), where M>0M > 0 is a constant, then:

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(1)

the sum f(x)f(x) of this series is infinitely differentiable at any point a∈Ra \in R;

(2)

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n = 0}^{\infty } \frac{f^{(n)}(a)}{n !}(x-a)^{n} for any x∈Rx \in R.

Problem 2.21.83

Let the function ff be infinitely differentiable on the segment [−1;1][-1 ; 1] and f(n)(x)⩾0f^{(n)}(x) \geqslant 0 for n=0,1,2,…n = 0,1,2, \ldots and for all x∈[−1;1]x \in [-1 ; 1]. Prove that on the interval (−1,1)(-1,1) the function f(x)f(x) is represented by a power series converging to it

f(x)=∑n=0∞anxn f(x) = \sum _{n = 0}^{\infty } a_{n} x^{n}
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Problem 2.21.84

Prove that if the function ff is infinitely differentiable on RR and satisfies the conditions:

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(1)

there exists a constant M>0M > 0 such that for all x∈Rx \in R and for all n∈Nn \in N the inequality ∣f(n)(x)∣⩽M\left|f^{(n)}(x)\right| \leqslant M holds;

(2)

f(1/n)=0,n∈Nf(1 / n) = 0, n \in N, then f(x)≡0f(x) \equiv 0.

Problem 2.21.85

Prove that the Maclaurin series for the function

f(x)=∑n=0∞e−ncos⁡n2x f(x) = \sum _{n = 0}^{\infty } e^{-n} \cos n^{2} x

converges only at x=0x = 0.

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