Many cube roots

Calculus Level 4

Let $x_0, x_1, x_2,\ldots$ be a sequence of real numbers satisfying the recursion,

$x_n = \sqrt[3]{x_{n-1} x_{n-2}x_{n-3}}$

for $n > 2$ .

If $x_0 = 1$ , $x_1 = 1$ and $x_2 =10000$ what is

$\large \lim_{n \to \infty} x_n?$

The answer is 100.

2 solutions

Geoff Pilling
Aug 26, 2016

If $A_n = log(x_n)$ , then $A_0 = 0$ , $A_1 = 0$ , and $A_2 =$ 4, and the recursion relation becomes, $A_n = \frac{A_{n-1}+A_{n-2} + A_{n-3}}{3}$

This series converges to $A_\infty = 2$ .

Therefore, $x_\infty = \boxed{100}$

Thanks to @Rajen Kapur for the technique! :^)

This series converges to $A_\infty = 2$ .

How do you know this?

Pi Han Goh - 4 years, 9 months ago

Rajen Kapur
Aug 26, 2016

Thx $\color {blue}{ @Geoff Pilling }$ for citation. For detailed solution to this recurrence relation, please see my $\color{#3D99F6} {One... Two... Three... Infinity!}$ @Geoff Pilling Link to my solution is here: https://brilliant.org/problems/one-two-three-infinity/

@Rajen Kapur So, can you send me a pointer to "One...Two...Three...Infinity!" ? Somehow I can't find it.

Geoff Pilling - 4 years, 9 months ago

Many cube roots

The answer is 100.

2 solutions

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