Is That Special?

Calculus Level 5

Define f 0 ( x ) = 1 − x 2 f_0(x)=\sqrt{1-x^2} .

Now, for n > 0 n>0 recursively define f n ( x ) = 1 2 ( f n − 1 ( x ) + f n − 1 ( x − 1 2 n − 1 ) ) f_n(x)=\frac{1}{2}\left(f_{n-1}(x)+f_{n-1}\left(x-\frac{1}{2^{n-1}}\right)\right) .

For n > 0 n>0 , the domain of f n ( x ) f_n(x) is − 1 + ∑ k = 0 n − 1 1 2 k ≤ x ≤ 1 -1+\displaystyle \sum_{k=0}^{n-1} \frac{1}{2^k}\leq x \leq 1

If lim ⁡ n → ∞ f n ( 1 ) = π a b \displaystyle \lim_{n\to \infty} f_n(1)=\frac{\pi^a}{b} for positive integers a , b a,b , find a + b a+b .

Helpful hints:

  • f 2 ( x ) = 1 − x 2 + 1 − ( x − 1 ) 2 2 + 1 − ( x − . 5 ) 2 + 1 − ( x − 1.5 ) 2 2 2 f_2(x)=\dfrac{\dfrac{\sqrt{1-x^2}+\sqrt{1-(x-1)^2}}{2}+\dfrac{\sqrt{1-(x-.5)^2}+\sqrt{1-(x-1.5)^2}}{2}}{2}


The answer is 5.

Trevor Arashiro by Trevor Arashiro , 22, USA

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1 solution

Trevor Arashiro
Jan 29, 2016

The biggest hint is given in which f 2 ( x ) f_2(x) is shown to be the average of ( \large ( the average of ( \large ( the functions f ( x ) f(x) and f ( x − 1 ) f(x-1) ) \large) and the average of ( \large ( f ( x − . 5 ) f(x-.5) and f ( x − 1.5 ) f(x-1.5) ) \large ) ) \large ) . This continuous averaging will eventually be the expected value of a point on the upper half of the circle y = 1 − x 2 y=\sqrt{1-x^2} which is π / 4 \pi/4

Purple, red, green, and blue represent f 0 , 1 , 2 , 3 ( x ) f_{0,1,2,3}(x) respectively.

Using AVT (essentially this is geometry and not really calculus since the area of any 2D figure is base*avg height).

Avg height= A r e a b a s e = π / 2 2 = π 4 \frac{Area}{base}=\frac{\pi/2}{2}=\boxed{\frac{\pi}{4}}

If you try to evaluate this algebraically as a sum, it comes out to

lim ⁡ n → ∞ ∑ k = 1 2 n + 1 1 − ( 1 − k − 1 2 n ) 2 2 n + 1 \lim \limits_{n\to \infty} \dfrac{\displaystyle \sum_{k=1}^{2^{n+1}}\sqrt{1-\left(1-\dfrac{k-1}{2^n}\right)^2}}{2^{n+1}} .

I have no idea how to solve this by hand and wolfram can only handle this up to n = 12 n=12

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The limit lim ⁡ n → ∞ ∑ k = 1 2 n + 1 1 − ( 1 − k − 1 2 n ) 2 2 n + 1 \lim \limits_{n\to \infty} \dfrac{\displaystyle \sum_{k=1}^{2^{n+1}}\sqrt{1-\left(1-\dfrac{k-1}{2^n}\right)^2}}{2^{n+1}} is a Riemann sum for ∫ 0 2 1 − ( 1 − x ) 2 2 d x = π 4 . \int_0^2 \frac{\sqrt{1 - (1 - x)^2}}{2} \ dx = \frac{\pi}{4}.

Jon Haussmann - 5 years, 4 months ago

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Oh, wow, never thought of it this way. I don't have too much experience working with Riemann Sums but I think I see how this works. Thanks for the insight!

Trevor Arashiro - 5 years, 4 months ago

Why does the average converge to pi/4?

Agnishom Chattopadhyay - 5 years, 4 months ago

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