Arithmetic-geometric mean

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Prove that the sequences xnx_{n} and yn(n=1,2,…)y_{n} (n=1,2,\ldots ) determined by the following formulas

x1=a,y1=b,xn+1=xnyn,yn+1=xn+yn2, x_{1}=a,y_{1}=b,x_{n+1}=\sqrt{x_{n}y_{n}},y_{n+1}=\frac{x_{n}+y_{n}}{2} ,

have a common limit

μ(a,b)=lim⁡n→∞xn=lim⁡n→∞yn \mu (a,b)=\lim _{n\to \infty }x_{n}=\lim _{n\to \infty }y_{n}

(the arithmetic-geometric mean of the numbers aa and bb).

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Numbers aa and bb are given, with a>b>0a > b > 0. New pairs of numbers are then formed by the formulas: an+1=an+bn2a_{n+1} = \frac{a_{n}+b_{n}}{2}, bn+1=2anbnan+bnb_{n+1} = \frac{2 a_{n} b_{n}}{a_{n}+b_{n}}. Here a0=aa_{0} = a and b0=bb_{0} = b. Prove that anbn=aba_{n} b_{n} = ab, that

an−anbnan+anbn=(a−aba+ab)2n \frac{a_{n}-\sqrt{a_{n} b_{n}}}{a_{n}+\sqrt{a_{n} b_{n}}} = \left(\frac{a-\sqrt{ab}}{a+\sqrt{ab}}\right)^{2^{n}}

and, finally, that lim⁡n→∞an=lim⁡n→∞bn=ab\lim_{n \to \infty } a_{n} = \lim_{n \to \infty } b_{n} = \sqrt{ab}.

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From two given positive numbers aa and bb one can construct a series of new numbers using the equalities: a0=aa_{0} = a, b0=bb_{0} = b, a1=a0+b02a_{1} = \frac{a_{0}+b_{0}}{2}, b1=a0b0b_{1} = \sqrt{a_{0} b_{0}}, and in general, an+1=an+bn2a_{n+1} = \frac{a_{n}+b_{n}}{2}, bn+1=anbnb_{n+1} = \sqrt{a_{n} b_{n}}. Prove that as n→∞n \to \infty the numbers ana_{n} and bnb_{n} tend to a common limit. (Gauss's arithmetic-geometric mean.)

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Two sequences of numbers are formed by the formulas:

an=an−1+bn−12,bn=anbn−1. a_{n} = \frac{a_{n-1}+b_{n-1}}{2}, \qquad b_{n} = \sqrt{a_{n} b_{n-1}}.

Here a0=aa_{0} = a, b0=bb_{0} = b are given positive numbers. Find lim⁡n→∞an\lim_{n \to \infty } a_{n}, which equals lim⁡n→∞bn\lim_{n \to \infty } b_{n}, setting a=bcos⁡φa = b \cos \varphi for a<ba < b and a=bch⁡φa = b \operatorname {ch} \varphi for a>ba > b.

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The sequences {xn}\left\{ x_{n}\right\} and {yn}\left\{ y_{n}\right\} are given recursively:

x1=a>0,y1=b>0xn+1=12(xn+yn),yn+1=12(xn2+yn2),n∈N. \begin{gathered} x_{1} = a > 0, \quad y_{1} = b > 0 \\ x_{n+1} = \frac{1}{2}\left(x_{n}+y_{n}\right), \quad y_{n+1} = \sqrt{\frac{1}{2}\left(x_{n}^{2}+y_{n}^{2}\right)}, \quad n \in \mathbb {N}.\end{gathered}

Prove that:

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(1)

yn⩾xn,n∈N,n⩾2y_{n} \geqslant x_{n}, n \in \mathbb {N}, n \geqslant 2

(2)

the sequence {xn}\left\{ x_{n}\right\} is increasing (n⩾2)(n \geqslant 2);

(3)

the sequence {yn}\left\{ y_{n}\right\} is decreasing (n⩾2)(n \geqslant 2);

(4)

∣yn+1−xn+1∣⩽∣b−1∣/4n\left|y_{n+1}-x_{n+1}\right| \leqslant \left|b-1\right| / 4^{n}.

Prove that if

x1=a>0,y1=b>0,xn+1=xnyn,yn+1=0,5(xn+yn),n∈N, \begin{gathered} x_{1} = a > 0, \quad y_{1} = b > 0, \\ x_{n+1} = \sqrt{x_{n} y_{n}}, \quad y_{n+1} = 0,5\left(x_{n}+y_{n}\right), \quad n \in \mathbb {N},\end{gathered}

then:

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(1)

the sequence {xn}\left\{ x_{n}\right\} is increasing, while the sequence {yn}\left\{ y_{n}\right\} is decreasing;

(2)

both sequences {xn}\left\{ x_{n}\right\} and {yn}\left\{ y_{n}\right\} are bounded;

(3)

∣yn+1−xn+1∣⩽∣b−a∣/2n\left|y_{n+1}-x_{n+1}\right| \leqslant \left|b-a\right| / 2^{n}.

The sequences {xn}\left\{ x_{n}\right\} and {yn}\left\{ y_{n}\right\} satisfy the conditions:

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(1)

x1=a>0,y1=b>0,xn+1=12(xn+yn),yn+1=xnyn,n∈Nx_{1} = a > 0, y_{1} = b > 0, x_{n+1} = \frac{1}{2}\left(x_{n}+y_{n}\right), y_{n+1} = \sqrt{x_{n} y_{n}}, n \in \mathbb {N};

(2)

x1=a>0,y1=b>0,xn+1=12(xn+yn),yn+1=2xnynxn+ynx_{1} = a > 0, y_{1} = b > 0, x_{n+1} = \frac{1}{2}\left(x_{n}+y_{n}\right), y_{n+1} = \frac{2 x_{n} y_{n}}{x_{n}+y_{n}}, n∈Nn \in \mathbb {N}.

Prove that the sequences {xn}\left\{ x_{n}\right\} and {yn}\left\{ y_{n}\right\} converge and lim⁡n→∞xn=lim⁡n→∞yn\lim_{n \rightarrow \infty } x_{n} = \lim_{n \rightarrow \infty } y_{n}. In case 2), find this limit.

Prove Gauss's formula

∫0π/2dφa2cos⁡2φ+b2sin⁡2φ=∫0π/2dψa12cos⁡2ψ+b12sin⁡2ψ,0<b<a,a1=(a+b)/2,b1=ab. \begin{gathered} \int _{0}^{\pi / 2} \frac{d \varphi }{\sqrt{a^{2} \cos ^{2} \varphi +b^{2} \sin ^{2} \varphi }} = \int _{0}^{\pi / 2} \frac{d \psi }{\sqrt{a_{1}^{2} \cos ^{2} \psi +b_{1}^{2} \sin ^{2} \psi }}, \\ 0 < b < a, \quad a_{1} = (a+b) / 2, \quad b_{1} = \sqrt{a b}.\end{gathered}
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Let a>b>0,a1=(a+b)/2,b1=ab,an=(an−1+bn−1)/2,bn=an−1bn−1,n=2,3,…a > b > 0, a_{1} = (a+b) / 2, b_{1} = \sqrt{a b}, a_{n} = \left(a_{n-1}+b_{n-1}\right) / 2, b_{n} = \sqrt{a_{n-1} b_{n-1}}, n = 2,3, \ldots Then there exists a finite limit (see problem 244,1) in [1,§8][1, \S 8])

lim⁡n→∞an=lim⁡n→∞bn=c(a,b). \lim _{n \rightarrow \infty } a_{n} = \lim _{n \rightarrow \infty } b_{n} = c(a, b).

Prove that ∫0π/2dφa2cos⁡2φ+b2sin⁡2φ=π2c(a,b)\int_{0}^{\pi / 2} \frac{d \varphi }{\sqrt{a^{2} \cos^{2} \varphi +b^{2} \sin^{2} \varphi }} = \frac{\pi }{2 c(a, b)}.

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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.
[ККЧ+03a] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Предел. Непрерывность. Дифференцируемость. 2-е изд., ФИЗМАТЛИТ, 2003.
[ККЧ+03b] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.