Limit the mixed recurrence

Calculus Level 4

Let a 0 = 1 , b 0 = 2 , a_0 = 1, b_0 = 2, and for n ≥ 0 n\geq 0 a n + 1 = a n + b n 2 , b n + 1 = a n + 1 b n . a_{n+1} = \frac{a_n + b_n}{2},\quad b_{n+1} = \sqrt{a_{n+1}b_n}. If lim ⁡ n → ∞ a n b n = c π d \displaystyle \lim_{n\to\infty } a_n b_n = \dfrac c{\pi^ d} for positive integers c c and d d , find c + d c+d .


The answer is 29.

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Kartik Sharma by Kartik Sharma , 21, India

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2 solutions

Mark Hennings
Mar 21, 2017

Suppose that a 0 = 2 cos ⁡ x a_0 =2\cos x and b 0 = 2 b_0 = 2 for some 0 < x < 1 2 π 0 < x < \tfrac12\pi . It is a simple induction to show that b n = 2 ∏ j = 1 n cos ⁡ ( x 2 j ) a n = b n cos ⁡ ( x 2 n ) n ≥ 1 . b_n \; = \; 2\prod_{j=1}^n\cos\big(\tfrac{x}{2^j}\big) \hspace{1cm} a_n \; = \; b_n \cos\big(\tfrac{x}{2^n}\big) \hspace{2cm} n \ge 1 \;. and a second induction shows that b n = 2 sin ⁡ x 2 n sin ⁡ x 2 n n ≥ 1 b_n \; = \; \frac{2\sin x}{2^n \sin \frac{x}{2^n}} \hspace{2cm} n \ge 1 From this last formula it is clear that β = lim ⁡ n → ∞ b n = 2 sin ⁡ x x \beta \;=\; \lim_{n \to \infty} b_n \; = \; \frac{2\sin x}{x} and therefore that α = lim ⁡ n → ∞ a n = lim ⁡ n → ∞ b n cos ⁡ ( x 2 n ) = β \alpha \; = \; \lim_{n \to \infty} a_n \; = \; \lim_{n \to \infty} b_n \cos\big(\tfrac{x}{2^n}\big) \; = \; \beta In our case, x = 1 3 π x = \tfrac13\pi , and hence α = β = 3 3 π \alpha = \beta = \frac{3\sqrt{3}}{\pi} , making the answer 27 + 2 = 29 27 + 2= \boxed{29} .

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How do you know working out with. Cosine function would help ? I mean a 0 a_{0} could be any function, so how cos x comes to mind ?

Vishal Yadav - 4 years, 2 months ago

You just nailed every bit of it👏👏👏

Divyansh Joshi - 4 years, 1 month ago
Kartik Sharma
Mar 21, 2017

b n + 1 2 b n = a n + 1 \displaystyle \frac{b_{n+1}^2}{b_n} = a_{n+1}

2 b n + 1 2 b n = a n + b n \displaystyle \frac{2b_{n+1}^2}{b_n} = a_n + b_n

2 b n + 1 2 b n = b n 2 b n − 1 + b n \displaystyle \frac{2b_{n+1}^2}{b_n} = \frac{b_n^2}{b_{n-1}} + b_n

Dividing by b n b_n ,

2 b n + 1 2 b n 2 = b n b n − 1 + 1 \displaystyle \frac{2 b_{n+1}^2}{b_n^2} = \frac{b_n}{b_{n-1}} + 1

Substitute t n = b n b n − 1 t_n = \frac{b_n}{b_{n-1}} ,

2 t n + 1 2 − 1 = t n \displaystyle 2t_{n+1}^2 - 1 = t_n

This hints 2 cos ⁡ 2 ( x ) − 1 = cos ⁡ ( 2 x ) 2 \cos^2(x) - 1 = \cos(2x) , so here we go,

t n = cos ⁡ ( x 2 n − 1 ) \displaystyle t_n = \cos\left(\frac{x}{2^{n-1}}\right)

where x = cos ⁡ − 1 ( t 1 ) = cos ⁡ − 1 ( b 1 b 0 ) = cos ⁡ − 1 ( a + b 2 b ) x = \cos^{-1}(t_1) = \cos^{-1}\left(\frac{b_1}{b_0}\right) = \cos^{-1}\left(\sqrt{\frac{a+b}{2b}}\right) .

b n b 0 = t n t n − 1 t n − 2 ⋯ t 1 = ∏ k = 1 n cos ⁡ ( x 2 k − 1 ) \displaystyle \frac{b_n}{b_0} = t_n t_{n-1} t_{n-2} \cdots t_1 = \prod_{k=1}^n{\cos\left(\frac{x}{2^{k-1}}\right)}

The product can be found by multiplying and dividing by sin ⁡ ( x 2 n − 1 ) \sin\left(\frac{x}{2^{n-1}}\right) .

b n = b sin ⁡ ( 2 x ) 2 n sin ⁡ ( x 2 n − 1 ) \displaystyle b_n = b \frac{\sin\left(2x\right)}{2^n\sin\left(\frac{x}{2^{n-1}}\right)}

β = b sin ⁡ ( 2 x ) 2 x lim ⁡ n → ∞ 1 sin ⁡ ( x 2 n − 1 ) x 2 n − 1 \displaystyle \beta = \frac{b\sin(2x)}{2x} \lim_{n\rightarrow\infty} \frac{1}{\frac{\sin\left(\frac{x}{2^{n-1}}\right)}{\frac{x}{2^{n-1}}}}

β = b sin ⁡ ( x ) cos ⁡ ( x ) x = b a + b 2 b a − b 2 b cos ⁡ − 1 ( a + b 2 b ) \displaystyle \beta = \frac{b\sin(x)\cos(x)}{x} = \frac{b \sqrt{\frac{a+b}{2b}} \sqrt{\frac{a-b}{2b}}}{\cos^{-1}\left(\sqrt{\frac{a+b}{2b}}\right)}

Now using 2 cos ⁡ − 1 ( x ) = cos ⁡ − 1 ( 2 x 2 − 1 ) 2\cos^{-1}(x) = \cos^{-1}(2x^2 -1) ,

β = b 2 − a 2 cos ⁡ − 1 ( a b ) \displaystyle \beta = \frac{\sqrt{b^2 - a^2}}{\cos^{-1}\left(\frac{a}{b}\right)}

It can be shown easily that α = β \alpha = \beta .

6 Helpful 0 Interesting 0 Brilliant 0 Confused

It has some good history. Shwaub -Shoenberg mean. Shwaub proved it by some rather geometrical rigor. Shoenberg's proof was a very different kind. He proved that b n 2 − a n 2 cos ⁡ − 1 ( a n b n ) \frac{\sqrt{b_n^2 - a_n^2}}{\cos^{-1}\left(\frac{a_n}{b_n}\right)} is constant.

In fact a similar kind also prompted 14-year old Gauss, it was -

a n = a n + b n 2 , b n = a n b n a_{n} = \frac{a_n + b_n}{2}, b_{n} = \sqrt{a_{n}b_n}

a 0 = a , b 0 = b a_0 = a, b_0 = b .

Can anyone give me a hint how to do it?

Kartik Sharma - 4 years, 2 months ago

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@Mark Hennings A t n t_n can be found just like I did above but no further success.

Kartik Sharma - 4 years, 2 months ago

If a > b > 0 a > b > 0 , define a 0 = a a_0 = a , b 0 = b b_0 = b and, inductively, a n + 1 = 1 2 ( a n + b n ) , b n + 1 = a n b n a_{n+1} \; = \; \tfrac12(a_n + b_n) \hspace{0.5cm},\hspace{0.5cm} b_{n+1} \; = \; \sqrt{a_nb_n} If 0 < b n < a n 0 < b_n < a_n then it follows that 0 < b n < b n + 1 < a n + 1 < a n 0 < b_n < b_{n+1} < a_{n+1} < a_n and moreover 0 < a n + 1 − b n + 1 < 1 2 ( a n − b n ) 0 < a_{n+1} - b_{n+1} < \tfrac12(a_n - b_n) Thus the sequence ( a n ) (a_n) is decreasing and bounded below, and the sequence ( b n ) (b_n) is increasing and bounded above. Thus both sequences converge. Since 0 < a n − b n < 2 − n ( a − b ) 0 < a_n - b_n < 2^{-n}(a-b) , both sequences converge to the same limit M ( a , b ) M(a,b) , called the arithmetic-geometric mean of a a and b b .

Now define I ( a , b ) = ∫ 0 ∞ d x ( x 2 + a 2 ) ( x 2 + b 2 ) I(a,b) \; = \; \int_0^\infty \frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}} for any 0 < b < a 0 < b < a . Note that π 2 a = ∫ 0 ∞ d x x 2 + a 2 ≤ I ( a , b ) ≤ ∫ 0 ∞ d x x 2 + b 2 = π 2 b \frac{\pi}{2a} \; = \; \int_0^\infty \frac{dx}{x^2 + a^2} \; \le \; I(a,b) \; \le \; \int_0^\infty \frac{dx}{x^2 + b^2} \; = \; \frac{\pi}{2b} This integral can be expressed in term of the complete elliptic integral of the first kind.

If 0 < b < a 0 < b < a and we define a ^ = 1 2 ( a + b ) \hat{a} = \tfrac12(a+b) , b ^ = a b \hat{b} = \sqrt{ab} , then the substitution 2 x = t − a b t 2x = t - \tfrac{ab}{t} gives I ( a ^ , b ^ ) = ∫ 0 ∞ d x ( x 2 + a ^ 2 ) ( x 2 + b ^ 2 ) = 1 2 ∫ − ∞ ∞ d x ( x 2 + a ^ 2 ) ( x 2 + b ^ 2 ) = ∫ 0 ∞ d t ( t 2 + a 2 ) ( t 2 + b 2 ) = I ( a , b ) \begin{aligned} I(\hat{a},\hat{b}) & = \int_0^\infty \frac{dx}{\sqrt{(x^2+\hat{a}^2)(x^2+\hat{b}^2)}} \; = \; \tfrac12\int_{-\infty}^\infty \frac{dx}{\sqrt{(x^2+\hat{a}^2)(x^2+\hat{b}^2)}} \\ & = \int_0^\infty \frac{dt}{\sqrt{(t^2+a^2)(t^2+b^2)}} \; = \; I(a,b) \end{aligned}

Putting this all together, we have π 2 a n ≤ I ( a n , b n ) = I ( a , b ) ≤ π 2 b n n ≥ 1 \frac{\pi}{2 a_n} \; \le \; I(a_n,b_n) \; = \; I(a,b) \; \le \; \frac{\pi}{2b_n} \hspace{1cm} n \ge 1 and so, letting n → ∞ n \to \infty , π 2 M ( a , b ) ≤ I ( a , b ) ≤ π 2 M ( a , b ) \frac{\pi}{2M(a,b)} \; \le \; I(a,b) \; \le \; \frac{\pi}{2M(a,b)} so that M ( a , b ) = π 2 I ( a , b ) M(a,b) \; = \; \frac{\pi}{2I(a,b)}

Mark Hennings - 4 years, 2 months ago

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Well, I appreciate what you did but does this come out of experience -

"Now define I ( a , b ) = ⋯ I(a,b) = \cdots " ?

I find it difficult to follow.

Kartik Sharma - 4 years, 2 months ago

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@Kartik Sharma – Experience, or knowing Gauss' result. The tricky part is chasing the change of variable through to show that I I does not change after one stage of AM/GM calculation. The formula for I I is readily available (Wikipedia, Mathworld), the proof is less easy to find.

Mark Hennings - 4 years, 2 months ago

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