Section 4.7

Series and Operations on Them

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Problem 1689

Find the sum of the series s=1+2x+3x2+4x3+…s = 1+2x+3x^{2}+4x^{3}+\ldots, using the equality:

s=1+x+x2+x3+x4+x5+…++x+x2+x3+x4+x5+…++x2+x3+x4+x5+…++x3+x4+x5+…++x4+x5+…++x5+…+ \begin{aligned} s = {} & 1+x+x^{2}+x^{3}+x^{4}+x^{5}+\ldots + \\ {} & +x+x^{2}+x^{3}+x^{4}+x^{5}+\ldots + \\ {} & +x^{2}+x^{3}+x^{4}+x^{5}+\ldots + \\ {} & +x^{3}+x^{4}+x^{5}+\ldots + \\ {} & +x^{4}+x^{5}+\ldots + \\ {} & +x^{5}+\ldots + \end{aligned}
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Problem 1690

Find the sum of the series s=1+3x+5x2+7x3+…s = 1+3x+5x^{2}+7x^{3}+\ldots, by forming the products xsxs and x2sx^{2}s.

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Problem 1691

Noting that the coefficients of the series

1+4x+9x2+16x3+25x4+… 1+4x+9x^{2}+16x^{3}+25x^{4}+\ldots

are related by an+2−2an+1+an=2a_{n+2}-2a_{n+1}+a_{n}=2, where ana_{n} is the coefficient of xnx^{n}, find the sum of the series by multiplying it by 1−2x+x21-2x+x^{2}.

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Problem 1692

In the series

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

each successive coefficient equals the sum of the three preceding ones. Find the sum of the series.

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Problem 1693

Multiply the two series: s=1+2x+3x2+4x3+…s = 1+2x+3x^{2}+4x^{3}+\ldots and s1=1−2x+3x2−4x3+…s_{1} = 1-2x+3x^{2}-4x^{3}+\ldots.

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Problem 1694

By direct multiplication, prove that

(1+x)sm(x)=sm+1(x), (1+x)s_{m}(x) = s_{m+1}(x),

where sm=1+mx+m(m−1)1⋅2x2+m(m−1)(m−2)1⋅2⋅3x3+…s_{m} = 1+mx+\dfrac {m(m-1)}{1 \cdot 2}x^{2}+\dfrac {m(m-1)(m-2)}{1 \cdot 2 \cdot 3}x^{3}+\ldots.

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Problem 1695

By multiplying series, prove the equality

(1+x+x21⋅2+x31⋅2⋅3+…)×(1+y1+y21⋅2+y31⋅2⋅3+…)=1+(x+y)1+(x+y)21⋅2+(x+y)31⋅2⋅3+… \left(1+x+\frac{x^{2}}{1 \cdot 2}+\frac{x^{3}}{1 \cdot 2 \cdot 3}+\ldots \right) \times \left(1+\frac{y}{1}+\frac{y^{2}}{1 \cdot 2}+\frac{y^{3}}{1 \cdot 2 \cdot 3}+\ldots \right) = 1+\frac{(x+y)}{1}+\frac{(x+y)^{2}}{1 \cdot 2}+\frac{(x+y)^{3}}{1 \cdot 2 \cdot 3}+\ldots
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Problem 1696

Prove the equalities:

ln⁡(1+x)1+x=x−(1+12)x2+(1+12+13)x3−…,∣x∣<1, \frac{\ln (1+x)}{1+x} = x-\left(1+\frac{1}{2}\right)x^{2}+\left(1+\frac{1}{2}+\frac{1}{3}\right)x^{3}-\ldots , \quad \left|x\right| < 1, 12x(1−x2)ln⁡1+x1−x=1+(1+13)x2+(1+13+15)x4+(1+13+15+17)x6+…,∣x∣<1. \frac{1}{2x(1-x^{2})}\ln \frac{1+x}{1-x} = 1+\left(1+\frac{1}{3}\right)x^{2}+\left(1+\frac{1}{3}+\frac{1}{5}\right)x^{4}+\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\right)x^{6}+\ldots , \quad \left|x\right| < 1.
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Problem 1697

By direct summation, find SnS_{n} and verify that for ∣x∣<1\left|x\right| < 1

x1+x+2x21+x2+4x41+x4+8x81+x8+…=x1−x, \frac{x}{1+x}+\frac{2x^{2}}{1+x^{2}}+\frac{4x^{4}}{1+x^{4}}+\frac{8x^{8}}{1+x^{8}}+\ldots = \frac{x}{1-x}, x1−x2+x21−x4+x41−x8+x81−x16+…=x1−x. \frac{x}{1-x^{2}}+\frac{x^{2}}{1-x^{4}}+\frac{x^{4}}{1-x^{8}}+\frac{x^{8}}{1-x^{16}}+\ldots = \frac{x}{1-x}.
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Problem 1698

For Morgan's series

x1+x2+x1+x2⋅x21+x4+x1+x2⋅x21+x4⋅x41+x8+… \frac{x}{1+x^{2}}+\frac{x}{1+x^{2}} \cdot \frac{x^{2}}{1+x^{4}}+\frac{x}{1+x^{2}} \cdot \frac{x^{2}}{1+x^{4}} \cdot \frac{x^{4}}{1+x^{8}}+\ldots

one can obtain (for ∣x∣≠1\left|x\right| \neq 1)

Sn=x1−x2n+2−21−x2n+1 S_{n} = x\frac{1-x^{2^{n+2}-2}}{1-x^{2^{n+1}}}

and hence the series always converges, but

S={x,if ∣x∣≤1,1x,if ∣x∣≥1. S = \begin{cases} x, & \text{if } \left|x\right| \leq 1, \\ \dfrac {1}{x}, & \text{if } \left|x\right| \geq 1. \end{cases}

Prove this.

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Problem 1699

Find by direct differentiation the derivative of the series x−x22+x33−x44+…x-\dfrac {x^{2}}{2}+\dfrac {x^{3}}{3}-\dfrac {x^{4}}{4}+\ldots, ∣x∣<1\left|x\right| < 1.

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Problem 1700

Find by direct differentiation the derivative of the series x−x33+x55−x77+…x-\dfrac {x^{3}}{3}+\dfrac {x^{5}}{5}-\dfrac {x^{7}}{7}+\ldots, ∣x∣<1\left|x\right| < 1.

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Problem 1701

Find by direct differentiation the derivative of the series x−x33!+x55!−x77!+…x-\dfrac {x^{3}}{3!}+\dfrac {x^{5}}{5!}-\dfrac {x^{7}}{7!}+\ldots.

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Problem 1702

Find by direct differentiation the derivative of the series 1+x+x21⋅2+x31⋅2⋅3+x41⋅2⋅3⋅4+…1+x+\dfrac {x^{2}}{1 \cdot 2}+\dfrac {x^{3}}{1 \cdot 2 \cdot 3}+\dfrac {x^{4}}{1 \cdot 2 \cdot 3 \cdot 4}+\ldots.

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Problem 1703

Verify that

lim⁡n→∞cos⁡x2⋅cos⁡x4⋅cos⁡x8…cos⁡x2n=sin⁡xx \lim _{n \to \infty } \cos \frac{x}{2} \cdot \cos \frac{x}{4} \cdot \cos \frac{x}{8} \ldots \cos \frac{x}{2^{n}} = \frac{\sin x}{x}

and from this obtain:

12tg⁡x2+14tg⁡x4+18tg⁡x8+…=1x−ctg⁡x \frac{1}{2}\tg \frac{x}{2}+\frac{1}{4}\tg \frac{x}{4}+\frac{1}{8}\tg \frac{x}{8}+\ldots = \frac{1}{x}-\ctg x

and

∑n=1∞122ncos⁡2x2n=1sin⁡2x−1x2. \sum _{n=1}^{\infty } \frac{1}{2^{2n}\cos ^{2}\frac{x}{2^{n}}} = \frac{1}{\sin ^{2}x}-\frac{1}{x^{2}}.
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Problem 1704

Prove that the sum of Gauss's hypergeometric series

F(α,β,γ,x)=1+αβ1⋅γx+α(α+1)1⋅2⋅β(β+1)γ(γ+1)x2+…,∣x∣<1, F(\alpha , \beta , \gamma , x) = 1+\frac{\alpha \beta }{1 \cdot \gamma }x+\frac{\alpha (\alpha +1)}{1 \cdot 2} \cdot \frac{\beta (\beta +1)}{\gamma (\gamma +1)}x^{2}+\ldots , \quad \left|x\right| < 1,

satisfies the equation

x(x−1)y′′+(−γ+(1+α+β)x)y′+αβy=0, x(x-1)y''+\left(-\gamma +(1+\alpha +\beta )x\right)y'+\alpha \beta y = 0,

where

y=F(α,β,γ,x). y = F(\alpha , \beta , \gamma , x).
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Problem 1705

The Bessel function J0(x)J_{0}(x) is defined by the series

J0(x)=1−x222+x422⋅42−x622⋅42⋅62+… J_{0}(x) = 1-\frac{x^{2}}{2^{2}}+\frac{x^{4}}{2^{2} \cdot 4^{2}}-\frac{x^{6}}{2^{2} \cdot 4^{2} \cdot 6^{2}}+\ldots

Prove that it satisfies the differential equation

xy′′+y′+xy=0. xy''+y'+xy = 0.
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Problem 1706

Prove that

y=1−x32⋅3+x62⋅3⋅5⋅6−x92⋅3⋅5⋅6⋅8⋅9+… y = 1-\frac{x^{3}}{2 \cdot 3}+\frac{x^{6}}{2 \cdot 3 \cdot 5 \cdot 6}-\frac{x^{9}}{2 \cdot 3 \cdot 5 \cdot 6 \cdot 8 \cdot 9}+\ldots

satisfies the differential equation

y′′+xy=0. y''+xy = 0.
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Problem 1707

Having formed the differential equation for the corresponding function, prove the series expansion for ∣x∣<1\left|x\right| < 1:

12(ln⁡(1+x)x)2=12−(1+12)x3+(1+12+13)x24−… \frac{1}{2}\left(\frac{\ln (1+x)}{x}\right)^{2} = \frac{1}{2}-\left(1+\frac{1}{2}\right)\frac{x}{3}+\left(1+\frac{1}{2}+\frac{1}{3}\right)\frac{x^{2}}{4}-\ldots
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Problem 1708

Having formed the differential equation for the corresponding function, prove the series expansion for ∣x∣<1\left|x\right| < 1:

(arctg⁡xx)2=1−(1+13)x22+(1+13+15)x43−… \left(\frac{\arctg x}{x}\right)^{2} = 1-\left(1+\frac{1}{3}\right)\frac{x^{2}}{2}+\left(1+\frac{1}{3}+\frac{1}{5}\right)\frac{x^{4}}{3}-\ldots
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Problem 1709

Having formed the differential equation for the corresponding function, prove the series expansion for ∣x∣<1\left|x\right| < 1:

12arctg⁡x⋅ln⁡(1+x2)=(1+12)x33−(1+12+13+14)x55+… \frac{1}{2}\arctg x \cdot \ln (1+x^{2}) = \left(1+\frac{1}{2}\right)\frac{x^{3}}{3}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\frac{x^{5}}{5}+\ldots
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Problem 1710

Prove the series expansion:

sin⁡(μarcsin⁡x)=μ1x−μ(μ2−1)3!x3+μ(μ2−1)(μ2−32)5!x5−… \sin (\mu \arcsin x) = \frac{\mu }{1}x-\frac{\mu (\mu ^{2}-1)}{3!}x^{3}+\frac{\mu (\mu ^{2}-1)(\mu ^{2}-3^{2})}{5!}x^{5}-\ldots
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Problem 1711

Prove the series expansion:

cos⁡(μarcsin⁡x)=1−μ21⋅2x2+μ2(μ2−22)4!x4−… \cos (\mu \arcsin x) = 1-\frac{\mu ^{2}}{1 \cdot 2}x^{2}+\frac{\mu ^{2}(\mu ^{2}-2^{2})}{4!}x^{4}-\ldots
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Problem 1712

Prove that cos⁡(μarccos⁡x)=Asin⁡μπ2+Bcos⁡μπ2\cos (\mu \arccos x) = A\sin \dfrac {\mu \pi }{2}+B\cos \dfrac {\mu \pi }{2}, where AA and BB are the series of the two preceding problems (1710 and 1711).

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Problem 1713

Prove the series expansion:

emarcsin⁡x=1+mx+∑ν=1∞m(m2+12)(m2+32)…(m2+(2ν−1)2)(2ν+1)!x2ν+1+∑ν=0∞m2(m2+22)(m2+42)…(m2+4ν2)(2ν+2)!x2ν+2. e^{m\arcsin x} = 1+mx+\sum _{\nu =1}^{\infty } \frac{m(m^{2}+1^{2})(m^{2}+3^{2})\ldots (m^{2}+(2\nu -1)^{2})}{(2\nu +1)!}x^{2\nu +1}+\sum _{\nu =0}^{\infty } \frac{m^{2}(m^{2}+2^{2})(m^{2}+4^{2})\ldots (m^{2}+4\nu ^{2})}{(2\nu +2)!}x^{2\nu +2}.
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Problem 1714

From problem 1713 (emarcsin⁡x=1+mx+∑ν=1∞m(m2+12)(m2+32)…(m2+(2ν−1)2)(2ν+1)!x2ν+1+∑ν=0∞m2(m2+22)(m2+42)…(m2+4ν2)(2ν+2)!x2ν+2e^{m\arcsin x} = 1+mx+\sum_{\nu =1}^{\infty } \frac{m(m^{2}+1^{2})(m^{2}+3^{2})\ldots (m^{2}+(2\nu -1)^{2})}{(2\nu +1)!}x^{2\nu +1}+\sum_{\nu =0}^{\infty } \frac{m^{2}(m^{2}+2^{2})(m^{2}+4^{2})\ldots (m^{2}+4\nu^{2})}{(2\nu +2)!}x^{2\nu +2}) obtain

arcsin⁡x=2∑n=0∞(2n)!(n!)2⋅(2n+1)(x2)2n+1; \arcsin x = 2\sum _{n=0}^{\infty } \frac{(2n)!}{(n!)^{2} \cdot (2n+1)}\left(\frac{x}{2}\right)^{2n+1}; (arcsin⁡x)2=12∑n=1∞((n−1)!)2(2n)!(2x)(2n). (\arcsin x)^{2} = \frac{1}{2}\sum _{n=1}^{\infty } \frac{((n-1)!)^{2}}{(2n)!}(2x)^{(2n)}.
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Problem 1715

By expanding eexe^{e^{x}} in two ways, verify that

131+232!+333!+434!+…=5e, \frac{1^{3}}{1}+\frac{2^{3}}{2!}+\frac{3^{3}}{3!}+\frac{4^{3}}{4!}+\ldots = 5e, 141+242!+343!+444!+…=15e. \frac{1^{4}}{1}+\frac{2^{4}}{2!}+\frac{3^{4}}{3!}+\frac{4^{4}}{4!}+\ldots = 15e.
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Problem 1716

Noting that sec⁡x\sec x is an even function, we may conclude that the expansion of sec⁡x\sec x has the form

sec⁡x=∑n=0∞En(2n)!x2n, \sec x = \sum _{n=0}^{\infty } \frac{E_{n}}{(2n)!}x^{2n},

where EnE_{n} are the corresponding coefficients (the Euler numbers). Multiplying term by term by the equality

cos⁡x=∑m=0∞(−1)m(2m)!x2m, \cos x = \sum _{m=0}^{\infty } \frac{(-1)^{m}}{(2m)!}x^{2m},

obtain a recurrence formula for the Euler numbers:

En(2n)!−En−12!(2n−2)!+En−24!(2n−4)!−…=0. \frac{E_{n}}{(2n)!}-\frac{E_{n-1}}{2!(2n-2)!}+\frac{E_{n-2}}{4!(2n-4)!}-\ldots = 0.

Since E0=1E_{0}=1, it follows from this formula that

E1=1,E2=5,E3=61,E4=1385,… E_{1}=1, \quad E_{2}=5, \quad E_{3}=61, \quad E_{4}=1385, \quad \ldots

Verify this.

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Problem 1717

It is known that 1−x2ctg⁡x2=∑Bn(2n)!x2n1-\dfrac {x}{2}\ctg \dfrac {x}{2} = \sum \dfrac {B_{n}}{(2n)!}x^{2n}, where BnB_{n} are coefficients (called the Bernoulli numbers). Prove that for the Bernoulli numbers there exists the recurrence formula:

Bn2(2n)!−Bn−1233!(2n−2)!+Bn−225⋅5!(2n−4)!−…+(−1)n−1B122n−1(2n−1)!2!+(−1)n2n22n+1(2n+1)!=0. \frac{B_{n}}{2(2n)!}-\frac{B_{n-1}}{2^{3}3!(2n-2)!}+\frac{B_{n-2}}{2^{5} \cdot 5!(2n-4)!}-\ldots +\frac{(-1)^{n-1}B_{1}}{2^{2n-1}(2n-1)!2!}+\frac{(-1)^{n}2n}{2^{2n+1}(2n+1)!} = 0.

From this formula it follows that

B1=16,B2=130,B3=142,B4=130,… B_{1}=\frac{1}{6}, \quad B_{2}=\frac{1}{30}, \quad B_{3}=\frac{1}{42}, \quad B_{4}=\frac{1}{30}, \ldots

Verify this.

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Problem 1718

Write the first five terms of the expansion in powers of xx for the function ln⁡(1+ex)\ln (1+e^{x}).

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Problem 1719

Write the first five terms of the expansion in powers of xx for the function esin⁡xe^{\sin x}.

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Problem 1720

Write the first five terms of the expansion in powers of xx for the function (1+x)x(1+x)^{x}.

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Problem 1721

Write the first five terms of the expansion in powers of xx for the function 1ex−1−1x\dfrac {1}{e^{x}-1}-\dfrac {1}{x}.

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Problem 1722

Prove that for small ba2\dfrac {b}{a^{2}} the approximate equality a2+b≈a+b2a\sqrt{a^{2}+b} \approx a+\dfrac {b}{2a} has an error approximately equal to a8(ba2)2\dfrac {a}{8}\left(\dfrac {b}{a^{2}}\right)^{2}, and use this equality to compute the values of the radicals:

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(a)

2\sqrt{2};

(b)

24\sqrt{24};

(c)

84\sqrt{84};

(d)

235\sqrt{235};

(e)

240\sqrt{240}.

Problem 1723

Prove the approximate formula

an+xn≈a+xnan−1 \sqrt[n]{a^{n}+x} \approx a+\frac{x}{na^{n-1}}

and use it to compute the roots: 2455\sqrt[5]{245}, 1297\sqrt[7]{129}, 5159\sqrt[9]{515}, 102710\sqrt[10]{1027}.

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Problem 1724

There is the following approximate formula for extracting roots:

Nn=an+bn≈a+2ab2nN−(n+1)b, \sqrt[n]{N} = \sqrt[n]{a^{n}+b} \approx a+\frac{2ab}{2nN-(n+1)b},

where ana^{n} is an exact power close to NN. Prove that the error of this formula is approximately equal to n2−112n3(ban)3\dfrac {n^{2}-1}{12n^{3}}\left(\dfrac {b}{a^{n}}\right)^{3}.

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Problem 1725

Using the formula of problem 1724 (Nn=an+bn≈a+2ab2nN−(n+1)b\sqrt[n]{N} = \sqrt[n]{a^{n}+b} \approx a+\dfrac {2ab}{2nN-(n+1)b}), compute the following roots:

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(a)

303\sqrt[3]{30};

(b)

703\sqrt[3]{70};

(c)

5003\sqrt[3]{500};

(d)

2505\sqrt[5]{250};

(e)

60\sqrt{60};

(f)

84\sqrt{84}.

Problem 1726

Using the binomial theorem and the equality

23=141283=541+31253, \sqrt[3]{2} = \frac{1}{4}\sqrt[3]{128} = \frac{5}{4}\sqrt[3]{1+\frac{3}{125}},

compute the value of 23\sqrt[3]{2} to 10 digits.

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Problem 1727

Using the binomial theorem and the identity

2=1550=751+149 \sqrt{2} = \frac{1}{5}\sqrt{50} = \frac{7}{5}\sqrt{1+\frac{1}{49}}

compute 2\sqrt{2} to 10 digits.

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Problem 1728

Using the series for ln⁡(1+x)\ln (1+x) at x=−110,−125x = -\dfrac {1}{10}, -\dfrac {1}{25} and 180\dfrac {1}{80}, as well as the identities

ln⁡910=2ln⁡3−ln⁡2−ln⁡5,ln⁡2425=3ln⁡2+ln⁡3−2ln⁡5, \ln \frac{9}{10} = 2\ln 3-\ln 2-\ln 5, \qquad \ln \frac{24}{25} = 3\ln 2+\ln 3-2\ln 5, ln⁡8180=4ln⁡3−4ln⁡2−ln⁡5 \ln \frac{81}{80} = 4\ln 3-4\ln 2-\ln 5

compute the natural logarithms of the numbers 2,3,52, 3, 5 and 1010 to 10 decimal places.

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Problem 1729

Prove that the quantities

Δln⁡n=ln⁡(n+1)−ln⁡n, \Delta \ln n = \ln (n+1)-\ln n, Δ2ln⁡n=ln⁡(n+2)−2ln⁡(n+1)+ln⁡n, \Delta ^{2} \ln n = \ln (n+2)-2\ln (n+1)+\ln n, Δ3ln⁡n=ln⁡(n+3)−3ln⁡(n+2)+3ln⁡(n+1)−ln⁡n \Delta ^{3} \ln n = \ln (n+3)-3\ln (n+2)+3\ln (n+1)-\ln n

have, respectively, the following series expansions:

Δln⁡n=2(12n+1+13(2n+1)3+15(2n+1)5+…), \Delta \ln n = 2\left(\frac{1}{2n+1}+\frac{1}{3(2n+1)^{3}}+\frac{1}{5(2n+1)^{5}}+\ldots \right), Δ2ln⁡n=−(1(n+1)2+12(n+1)4+13(n+1)6+…), \Delta ^{2} \ln n = -\left(\frac{1}{(n+1)^{2}}+\frac{1}{2(n+1)^{4}}+\frac{1}{3(n+1)^{6}}+\ldots \right), Δ3ln⁡n=2(8(2n+3)3+48(2n+3)5+312(2n+3)7+…). \Delta ^{3} \ln n = 2\left(\frac{8}{(2n+3)^{3}}+\frac{48}{(2n+3)^{5}}+\frac{312}{(2n+3)^{7}}+\ldots \right).
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Problem 1730

Prove that the tabular difference of ordinary five-digit logarithms of numbers, i.e. lg⁡(n+1)−lg⁡n\lg (n+1)-\lg n, for 1000<n<100001000 < n < 10000 is approximately expressed by the formula Δ=105⋅Mn=43429n\Delta = 10^{5} \cdot \dfrac {M}{n} = \dfrac {43429}{n}; M=0,43429M = 0{,}43429 is the modulus of common logarithms, Δ\Delta being expressed in hundred-thousandths.

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Problem 1731

Prove that the tabular difference of logarithms of sines given one minute apart, expressed in hundred-thousandths, can be given by the approximate formula

Δlg⁡sin⁡x≈12,6ctg⁡x. \Delta \lg \sin x \approx 12{,}6\ctg x.
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Problem 1732

Prove that formulas of the same kind for sines and for logarithms of tangents have the form:

Δsin⁡x≈29,1cos⁡x,Δlg⁡tg⁡x≈25,2sin⁡2x. \Delta \sin x \approx 29{,}1\cos x, \qquad \Delta \lg \tg x \approx \frac{25{,}2}{\sin 2x}.
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Problem 1733

The approximate equalities sin⁡x≈x\sin x \approx x, tg⁡x≈x\tg x \approx x have high accuracy for small angles (in some cases the more accurate formulas sin⁡x≈x−x36\sin x \approx x-\dfrac {x^{3}}{6}, tg⁡x≈x+x33\tg x \approx x+\dfrac {x^{3}}{3} are needed). At what visual angle is the diameter of a factory chimney of radius 2 m seen from a distance of 10 km?

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Problem 1734

The disk of the Sun is seen at an angle of 30′30'. How many times greater is the distance to the Sun than the diameter of the Sun?

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Problem 1735

A railway embankment has a grade of 0,0120{,}012. What angle does it make with the horizontal?

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Problem 1736

In a day the Earth traverses approximately an arc of 1∘1^{\circ } of its orbit, whose radius is 150⋅106150 \cdot 10^{6} km. Determine by how much the Earth's path deviates from a straight line in 1 second (27 km) and in 1 minute?

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Problem 1737

An arc of 1∘1^{\circ } of the Earth's meridian is about 112 km. By how much is it longer than its chord?

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Problem 1738

Prove the approximate formula: l=13hl = \sqrt{13h}, where hh is the height of the observer above the horizon in meters, and ll is the distance to the horizon in kilometers.

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Problem 1739

In measuring the distance between two points, the reading on a stretched tape showed 12 m. Knowing that the sag of the tape was 20 cm, and taking the shape of the tape to be an arc of a circle, determine the distance between the points with a correction for the sag.

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Problem 1740

Using series, compute sin⁡1∘\sin 1^{\circ } to within 10−510^{-5}.

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Problem 1741

Using series, compute cos⁡1∘\cos 1^{\circ } to within 10−510^{-5}.

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Problem 1742

Using series, compute sin⁡10∘\sin 10^{\circ } to within 10−410^{-4}.

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Problem 1743

Compute π\pi to within 10−1010^{-10} using the identity

π4=4arctg⁡15−arctg⁡1239. \frac{\pi }{4} = 4\arctg \frac{1}{5}-\arctg \frac{1}{239}.
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Problem 1744

Shanks computed π\pi to 707 digits using the identity of problem 1743 (π4=4arctg⁡15−arctg⁡1239\dfrac {\pi }{4} = 4\arctg \dfrac {1}{5}-\arctg \dfrac {1}{239}). How many terms in the expansions of arctg⁡15\arctg \dfrac {1}{5} and arctg⁡1239\arctg \dfrac {1}{239} did he have to retain?

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Problem 1745

One of Huygens's rules for computing the circumference of a circle is equivalent to the equality

2π=Pn+2pn3r, 2\pi = \frac{P_{n}+2p_{n}}{3r},

where PnP_{n} and pnp_{n} are the perimeters of the circumscribed and inscribed regular polygons, and rr is the radius of the circle. Prove that the error of this equality is approximately equal to 30n4\dfrac {30}{n^{4}}.

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