Section 4.6

Expansion into Series

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Problem 1646

By direct division, arranging the polynomials in ascending powers of the variable xx, prove the series expansions:

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(a)

1a−bx=1a+bxa2+b2x2a3+…\dfrac {1}{a-bx} = \dfrac {1}{a}+\dfrac {bx}{a^{2}}+\dfrac {b^{2}x^{2}}{a^{3}}+\ldots;

(b)

1a+bx=1a−bxa2+b2x2a3−…\dfrac {1}{a+bx} = \dfrac {1}{a}-\dfrac {bx}{a^{2}}+\dfrac {b^{2}x^{2}}{a^{3}}-\ldots, (∣x∣<∣ab∣)\left(\left|x\right| < \left|\dfrac {a}{b}\right|\right).

Problem 1647

Using the identity 5−2x6−5x+x2=12−x+13−x\dfrac {5-2x}{6-5x+x^{2}} = \dfrac {1}{2-x}+\dfrac {1}{3-x}, prove the series expansion:

5−2x6−5x+x2=56+1336x+35216x2+…+(12n+13n)xn−1+…,∣x∣<2. \frac{5-2x}{6-5x+x^{2}} = \frac{5}{6}+\frac{13}{36}x+\frac{35}{216}x^{2}+\ldots +\left(\frac{1}{2^{n}}+\frac{1}{3^{n}}\right)x^{n-1}+\ldots , \quad \left|x\right| < 2.
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Problem 1648

Assuming the series expansion

11−x−x2=a0+a1x+a2x2+a3x3+… \frac{1}{1-x-x^{2}} = a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\ldots

to be legitimate, and multiplying both sides of the equality by 1−x−x21-x-x^{2}, prove that a0=1a_{0}=1, a1=1a_{1}=1, a2=2a_{2}=2, a3=3a_{3}=3, a4=5,…a_{4}=5, \ldots and in general that a0,a1,a2,…a_{0}, a_{1}, a_{2}, \ldots are the Fibonacci numbers, defined by the equalities an=an−1+an−2a_{n}=a_{n-1}+a_{n-2} for n≥2n \geq 2, a0=1a_{0}=1, a1=1a_{1}=1.

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Problem 1649

Using the identity

11−x−x2=15(15+12+x+15−12−x), \frac{1}{1-x-x^{2}} = \frac{1}{\sqrt{5}}\left(\frac{1}{\frac{\sqrt{5}+1}{2}+x}+\frac{1}{\frac{\sqrt{5}-1}{2}-x}\right),

find a formula giving the general term ana_{n} of the Fibonacci series.

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Problem 1650

Prove that in the series expansion of the fraction

a+bx+cx2A+Bx+Cx2+Dx3=a0+a1x+a2x2+a3x3+…, \frac{a+bx+cx^{2}}{A+Bx+Cx^{2}+Dx^{3}} = a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\ldots ,

which holds for A≠0A \neq 0 and for sufficiently small values of ∣x∣\left|x\right|, there exists the relation between the coefficients ana_{n}: Aan+Ban−1+Can−2+Dan−3=0Aa_{n}+Ba_{n-1}+Ca_{n-2}+Da_{n-3}=0; n>2n > 2.

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Problem 1651

Expand the polynomial x4−5x3+5x2+x+2x^{4}-5x^{3}+5x^{2}+x+2 in powers of x−2x-2.

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Problem 1652

Expand the polynomial x5+2x4−x2+x+1x^{5}+2x^{4}-x^{2}+x+1 in powers of x+1x+1.

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Problem 1653

Prove the equalities:

sin⁡(a+x)=sin⁡acos⁡+cos⁡asin⁡x, \sin (a+x) = \sin a \cos +\cos a \sin x, cos⁡(a+x)=cos⁡acos⁡x−sin⁡asin⁡x, \cos (a+x) = \cos a \cos x-\sin a \sin x,

by expanding the left-hand sides in powers of xx.

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Problem 1654

In the expansion for ln⁡(1+x)\ln (1+x), replace xx by −x-x; starting from the series for ln⁡(1+x)\ln (1+x) and ln⁡(1−x)\ln (1-x), obtain Gregory's series:

ln⁡1+x1−x=2(x+x33+x55+…),∣x∣<1. \ln \frac{1+x}{1-x} = 2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\ldots \right), \quad \left|x\right| < 1.
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Problem 1655

Obtain the equality

ln⁡(N+1)−ln⁡N=2(12N+1+13(2N+1)3+15(2N+1)5+…),N>0. \ln (N+1)-\ln N = 2\left(\frac{1}{2N+1}+\frac{1}{3(2N+1)^{3}}+\frac{1}{5(2N+1)^{5}}+\ldots \right), \quad N > 0.
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Problem 1656

From the series for exe^{x} obtain the expansion for e−xe^{-x}, as well as the expansions:

ch⁡x≡ex+e−x2=1+x22!+x44!+…, \ch x \equiv \frac{e^{x}+e^{-x}}{2} = 1+\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\ldots , sh⁡x≡ex−e−x2=x+x33!+x55!+…, \sh x \equiv \frac{e^{x}-e^{-x}}{2} = x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\ldots ,

valid for every xx.

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Problem 1657

Using Euler's formula eix=cos⁡x+isin⁡xe^{ix} = \cos x+i\sin x, find the sum 1+cos⁡x+cos⁡2x+…+cos⁡(n−1)x1+\cos x+\cos 2x+\ldots +\cos (n-1)x.

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Problem 1658

Using Euler's formula eix=cos⁡x+isin⁡xe^{ix} = \cos x+i\sin x, find the sum cos⁡x+cos⁡3x+cos⁡5x+…+cos⁡(2n−1)x\cos x+\cos 3x+\cos 5x+\ldots +\cos (2n-1)x.

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Problem 1659

Using Euler's formula eix=cos⁡x+isin⁡xe^{ix} = \cos x+i\sin x, find the sum sin⁡x+sin⁡2x+sin⁡3x+…+sin⁡nx\sin x+\sin 2x+\sin 3x+\ldots +\sin nx.

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Problem 1660

Using Euler's formula eix=cos⁡x+isin⁡xe^{ix} = \cos x+i\sin x, find the sum sin⁡x−sin⁡2x+…−sin⁡2nx\sin x-\sin 2x+\ldots -\sin 2nx.

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Problem 1661

Using Euler's formula eix=cos⁡x+isin⁡xe^{ix} = \cos x+i\sin x, find the sum sin⁡2x−sin⁡23x+sin⁡25x−…+(−1)n−1sin⁡2(2n−1)x\sin^{2}x-\sin^{2}3x+\sin^{2}5x-\ldots +(-1)^{n-1}\sin^{2}(2n-1)x.

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Problem 1662

Using Euler's formula eix=cos⁡x+isin⁡xe^{ix} = \cos x+i\sin x, find the sum xsin⁡a+x2sin⁡2a+…+xn−1sin⁡(n−1)xx\sin a+x^{2}\sin 2a+\ldots +x^{n-1}\sin (n-1)x.

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Problem 1663

Prove the equality

∑n=0∞x4n(4n)!=14(ex+e−x+2cos⁡x). \sum _{n=0}^{\infty } \frac{x^{4n}}{(4n)!} = \frac{1}{4}(e^{x}+e^{-x}+2\cos x).
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Problem 1664

Prove the equality

∑n=0∞x3n(3n)!=13(ex+2e−x/2cos⁡x32). \sum _{n=0}^{\infty } \frac{x^{3n}}{(3n)!} = \frac{1}{3}\left(e^{x}+2e^{-x/2}\cos \frac{x\sqrt{3}}{2}\right).
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Problem 1665

Prove the equality

excos⁡x=∑n=0∞2n/2cos⁡πn4n!xn. e^{x}\cos x = \sum _{n=0}^{\infty } \frac{2^{n/2}\cos \frac{\pi n}{4}}{n!}x^{n}.

Here, as elsewhere, it is assumed that 0!=10! = 1.

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Problem 1666

Prove the equality

ch⁡x⋅cos⁡x=∑n=0∞(−1)n22nx4n(4n)!. \ch x \cdot \cos x = \sum _{n=0}^{\infty } (-1)^{n}\frac{2^{2n}x^{4n}}{(4n)!}.
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Problem 1667

Prove the equality

exctg⁡acos⁡x=∑n=0∞cos⁡nasin⁡na⋅xnn!. e^{x\ctg a}\cos x = \sum _{n=0}^{\infty } \frac{\cos na}{\sin ^{n}a} \cdot \frac{x^{n}}{n!}.
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Problem 1668

Prove the equality

(1+x2)m/2sin⁡(marctg⁡x)=∑ν=0∞(−1)νm(m−1)…(m−2ν)(2ν+1)!x2ν+1,∣x∣<1. (1+x^{2})^{m/2}\sin (m\arctg x) = \sum _{\nu =0}^{\infty } (-1)^{\nu }\frac{m(m-1)\ldots (m-2\nu )}{(2\nu +1)!}x^{2\nu +1}, \quad \left|x\right| < 1.
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Problem 1669

Prove the equality

(1+x2)−m/2cos⁡(marctg⁡x)=∑ν=0∞(−1)νm(m+1)…(m+2ν−1)(2ν)!x2ν,∣x∣<1. (1+x^{2})^{-m/2}\cos (m\arctg x) = \sum _{\nu =0}^{\infty } (-1)^{\nu }\frac{m(m+1)\ldots (m+2\nu -1)}{(2\nu )!}x^{2\nu }, \quad \left|x\right| < 1.
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Problem 1670

Applying Lagrange's formula to φ(x)−φ(π)\varphi (x)-\varphi (\pi ), where

φ(x)=x2+sin⁡x1+sin⁡2x2+…+sin⁡nxn+cos⁡(n+12)x(2n+1)sin⁡x2, \varphi (x) = \frac{x}{2}+\frac{\sin x}{1}+\frac{\sin 2x}{2}+\ldots +\frac{\sin nx}{n}+\frac{\cos \left(n+\frac{1}{2}\right)x}{(2n+1)\sin \frac{x}{2}},

prove that for 0<x<2π0 < x < 2\pi the equality

sin⁡x1+sin⁡2x2+…+sin⁡nxn+…=π−x2. \frac{\sin x}{1}+\frac{\sin 2x}{2}+\ldots +\frac{\sin nx}{n}+\ldots = \frac{\pi -x}{2}.

holds.

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Problem 1671

Applying Lagrange's formula to φ(x)−φ(π2)\varphi (x)-\varphi \left(\dfrac {\pi }{2}\right), where

φ(x)=sin⁡x1+sin⁡3x3+…+sin⁡(2n−1)x2n−1+cos⁡2nx4nsin⁡x, \varphi (x) = \frac{\sin x}{1}+\frac{\sin 3x}{3}+\ldots +\frac{\sin (2n-1)x}{2n-1}+\frac{\cos 2nx}{4n\sin x},

prove that for 0<x<π0 < x < \pi the equality

sin⁡x1+sin⁡3x3+sin⁡5x5+…+sin⁡(2n−1)x2n−1+…=π4. \frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\ldots +\frac{\sin (2n-1)x}{2n-1}+\ldots = \frac{\pi }{4}.

holds.

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Problem 1672

Using the binomial theorem, obtain the expansion

11−x2=1+12x2+1⋅32⋅4x4+1⋅3⋅52⋅4⋅6x6+…,∣x∣<1. \frac{1}{\sqrt{1-x^{2}}} = 1+\frac{1}{2}x^{2}+\frac{1 \cdot 3}{2 \cdot 4}x^{4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}x^{6}+\ldots , \quad \left|x\right| < 1.
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Problem 1673

Prove that for ∣x∣<1\left|x\right| < 1 the equalities hold:

1(1−x)2=1+2x+3x2+4x3+…=∑n=0∞(n+1)xn, \frac{1}{(1-x)^{2}} = 1+2x+3x^{2}+4x^{3}+\ldots = \sum _{n=0}^{\infty } (n+1)x^{n}, 1(1−x)3=1+3x+6x2+10x3+…=∑n=0∞(n+1)(n+2)2xn. \frac{1}{(1-x)^{3}} = 1+3x+6x^{2}+10x^{3}+\ldots = \sum _{n=0}^{\infty } \frac{(n+1)(n+2)}{2}x^{n}.
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Problem 1674

Prove the equality:

1a2+x2=1a−12x2a3+1⋅32⋅4x4a5−1⋅3⋅52⋅4⋅6x6a7+…,∣x∣<a. \frac{1}{\sqrt{a^{2}+x^{2}}} = \frac{1}{a}-\frac{1}{2}\frac{x^{2}}{a^{3}}+\frac{1 \cdot 3}{2 \cdot 4}\frac{x^{4}}{a^{5}}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\frac{x^{6}}{a^{7}}+\ldots , \quad \left|x\right| < a.
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Problem 1675

Prove the equality:

(a+x)n=an+nxan−1+n(n−1)1⋅2x2an−2+…,∣x∣<a. (a+x)^{n} = a^{n}+nxa^{n-1}+\frac{n(n-1)}{1 \cdot 2}x^{2}a^{n-2}+\ldots , \quad \left|x\right| < a.
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Problem 1676

Expand the function 1+x\sqrt{1+x} in powers of xx.

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Problem 1677

Expand the function ln⁡a+xa−x\ln \dfrac {a+x}{a-x} in powers of xx.

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Problem 1678

Expand the function ln⁡(1−x+x2)\ln (1-x+x^{2}) in powers of xx.

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Problem 1679

Expand the function ln⁡(2−3x+x2)\ln (2-3x+x^{2}) in powers of xx.

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Problem 1680

Expand the function (1+x)2xln⁡(1+x)\dfrac {(1+x)^{2}}{x}\ln (1+x) in powers of xx.

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Problem 1681

Prove the equality

ln⁡(1−2xcos⁡φ+x2)=−2∑n=1∞cos⁡nφnxn,∣x∣<1. \ln (1-2x\cos \varphi +x^{2}) = -2\sum _{n=1}^{\infty } \frac{\cos n\varphi }{n}x^{n}, \quad \left|x\right| < 1.
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Problem 1682

Verify that

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 1683

Verify that

asin⁡x1−2acos⁡x+a2=∑n=1∞ansin⁡nx,∣a∣<1. \frac{a\sin x}{1-2a\cos x+a^{2}} = \sum _{n=1}^{\infty } a^{n}\sin nx, \quad \left|a\right| < 1.
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Problem 1684

By a suitable change of variable, obtain the series expansion:

ln⁡x+1x−1=2(1x+13x3+15x5+…),∣x∣>1. \ln \frac{x+1}{x-1} = 2\left(\frac{1}{x}+\frac{1}{3x^{3}}+\frac{1}{5x^{5}}+\ldots \right), \quad \left|x\right| > 1.
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Problem 1685

By a suitable change of variable, obtain the series expansion:

ln⁡x=2(x−1x+1+13(x−1x+1)3+15(x−1x+1)5+…),x>0. \ln x = 2\left(\frac{x-1}{x+1}+\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{3}+\frac{1}{5}\left(\frac{x-1}{x+1}\right)^{5}+\ldots \right), \quad x > 0.
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Problem 1686

By a suitable change of variable, obtain the series expansion:

x1+x=xx+1+12(xx+1)2+1⋅32⋅4(xx+1)3+…,x>−12. \frac{x}{\sqrt{1+x}} = \frac{x}{x+1}+\frac{1}{2}\left(\frac{x}{x+1}\right)^{2}+\frac{1 \cdot 3}{2 \cdot 4}\left(\frac{x}{x+1}\right)^{3}+\ldots , \quad x > -\frac{1}{2}.
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Problem 1687

Obtain the series expansion:

ln⁡(x+1+x2)=∑n=0∞(−1)n1⋅3⋅…⋅(2n−1)2⋅4⋅…⋅2nx2n+12n+1,∣x∣<1. \ln \left(x+\sqrt{1+x^{2}}\right) = \sum _{n=0}^{\infty } (-1)^{n}\frac{1 \cdot 3 \cdot \ldots \cdot (2n-1)}{2 \cdot 4 \cdot \ldots \cdot 2n}\frac{x^{2n+1}}{2n+1}, \quad \left|x\right| < 1.
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Problem 1688

Obtain the series expansion:

ln⁡2(1−1−x)x=12⋅x2+1⋅32⋅4x24+1⋅3⋅52⋅4⋅6x36+1⋅3⋅5⋅72⋅4⋅6⋅8x48+…,∣x∣<1. \ln \frac{2\left(1-\sqrt{1-x}\right)}{x} = \frac{1}{2} \cdot \frac{x}{2}+\frac{1 \cdot 3}{2 \cdot 4}\frac{x^{2}}{4}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\frac{x^{3}}{6}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8}\frac{x^{4}}{8}+\ldots , \quad \left|x\right| < 1.
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