Sequence iteration and inequality

Algebra Level 4

For positive real number $c$ and integer $p>1$ , sequence $\{a_{n}\}$ is such that $a_1>c^{\frac{1}{p}}$ and $a_{n+1}=\dfrac{p-1}{p}a_{n}+\dfrac{c}{p}a_{n}^{1-p}$ for all positive integers $n$ .

Is it always true that $a_{n}>a_{n+1}>c^{\frac{1}{p}}$ ?

Show your religious proof.

Bonus: Can you generalize it to real number $p>1$ ?

Yes No

1 solution

Mark Hennings
Sep 26, 2019

If $a_n > c^{\frac1p}$ then $\frac{a_n^p - c}{a_n - c^{\frac1p}} \; = \; p\theta^{p-1} \; < \; pa_n^{p-1}$ for some $c < \theta < a_n$ , and hence $a_n^p - c < pa_n^{p-1}(a_n - c^{\frac1p})$ , which implies that $a_{n+1} > c^{\frac1p}$ . Moreover $a_n - a_{n+1} \; = \; \tfrac{1}{p}a_n^{1-p}(a_n^p - c) > 0$ and hence $a_n > a_{n+1}$ .

The result now follows by induction.

Unfortunately, the problem only shows that $a_1>c^{\frac{1}{p}}$ , not $a_n>c^{\frac{1}{p}}$ . And how do you find $\theta$ ?

Alice Smith - 1 year, 8 months ago

I have shown that $a_n > c^{\frac1p}$ implies that $a_{n+1} > c^{\frac1p}$ . Since $a_1 > c^{\frac1p}$ , the fact that $a_n > c^{\frac1p}$ for all $n \ge 1$ follows by induction.

The existence of $\theta$ follows from the Mean Value Theorem, which tells us that if $f$ is continuous on $[a,b]$ and differentiable in $(a,b)$ , then there exists $\theta \in (a,b)$ such that $\frac{f(b) - f(a)}{b-a} \; = \; f'(\theta)$ I am using $f(x) = x^p$ on the interval $[c^{\frac1p},a_n]$ .

Mark Hennings - 1 year, 8 months ago

Oh, that should help, thanks!

Alice Smith - 1 year, 8 months ago

Sequence iteration and inequality

1 solution

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