User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 October 8th, 2019, 04:39 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Inequality #2 without calculus Given $\displaystyle x^y < y^x$, where $\displaystyle x>y \geq \lambda >0$, find the range of $\displaystyle \lambda$. Without calculus! Last edited by skipjack; October 9th, 2019 at 01:37 AM. October 8th, 2019, 05:33 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 Here is the solution. $\displaystyle x^y < y^x \:$ ; $\displaystyle x^{1/x} f(1+t)\geq f(\lambda ) \: \Rightarrow \: t>(1+\frac{1}{t})^{t}\geq \lambda$. $\displaystyle t$-diverges , $\displaystyle (1+1/t)^{t}$-converges and is increasing. $\displaystyle \lambda \leq \sup\{ (1+1/t)^{t} \}=\lim_{t\rightarrow \infty }(1+1/t )^{t} =e <\lim_{t\rightarrow \infty} t =\infty$. $\displaystyle \lambda \in [e,\infty )$. Last edited by skipjack; October 9th, 2019 at 01:40 AM. October 8th, 2019, 04:24 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 670 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics How is your solution "without calculus"? The function $x^y$ is undefined without calculus. Then you claim that $f(t) > f(1 + t)$ which is probably impossible to prove without calculus. Then you talk about convergence, and take a limit. What does "without calculus" mean to you? Thanks from idontknow Tags calculus, inequality Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Algebra 6 October 9th, 2019 06:49 AM StillAlive Calculus 5 September 2nd, 2016 11:45 PM Happy Math Books 0 December 13th, 2014 03:20 AM KO1337 Academic Guidance 2 May 22nd, 2014 04:41 AM tdod Calculus 3 December 14th, 2011 11:24 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      