Approximate formulas from Taylor expansions and their errors

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For small ∣x∣\left|x\right|, derive simple approximate formulas for the following expressions:

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(a)

1R2−1(R+x)2 (R>0)\dfrac {1}{R^{2}}-\dfrac {1}{(R+x)^{2}} \ (R>0);

(b)

1+x1−x3−1−x1+x3\sqrt[3]{\dfrac {1+x}{1-x}}-\sqrt[3]{\dfrac {1-x}{1+x}};

(c)

Ax[1−(1+x100)−n]\dfrac {A}{x}\left[1-\left(1+\dfrac {x}{100}\right)^{-n}\right];

(d)

ln⁡2ln⁡(1+x100)\dfrac {\ln 2}{\ln \left(1+\dfrac {x}{100}\right)}.

For small absolute value of xx, derive an approximate formula of the form

x=αsin⁡x+βtg⁡x x=\alpha \sin x+\beta \tg x

accurate to the x5x^{5} term.

Apply this formula to the approximate computation of the arc length for a small angle.

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Estimate the relative error of the following Chebyshev rule: a circular arc is approximately equal to the sum of the two equal legs of an isosceles triangle standing on the chord subtended by the arc, whose height is 43\sqrt{\dfrac {4}{3}} times the sagitta of the circular segment.

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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}.

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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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}.

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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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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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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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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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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If aa, bb, cc are the sides of a spherical triangle, and AA, BB, CC the opposite angles, then

cos⁡c=cos⁡acos⁡b+sin⁡asin⁡bcos⁡C. \cos c = \cos a \cos b + \sin a \sin b \cos C.

What does this formula become if aa, bb and cc are small?

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A body falls in a medium whose resistance is proportional to the first power of the speed. The distance traveled in tt seconds is expressed by the formula

s=mgta−m2ga2(1−e−atm), s = \frac{mgt}{a}-\frac{m^{2}g}{a^{2}}\left(1-e^{-\frac{at}{m}}\right),

where mm is the mass, aa the coefficient of friction. Find approximate formulas for ss when atm\dfrac {at}{m} is large, and also for the case when atm\dfrac {at}{m} is very small.

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The problem of "the globe in a string bag." The Earth's globe was girdled with an inextensible thread. Then the thread was lengthened by 1 cm and pulled taut by tugging at the knot. Compute the height to which it rose, with a relative error of no more than one ten-thousandth of a percent (take the radius of the globe to be 6400 km6400\ \text{km}).

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[МГЛ+04] Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. Избранные задачи по вещественному анализу. 2-е изд., Невский Диалект; БХВ-Петербург, 2004.