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 January 13th, 2019, 09:52 AM #1 Senior Member   Joined: Oct 2016 From: Arizona Posts: 209 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. Is this convex? I'm trying to use Jensen's to prove an inequality, but my solution depends on $$\frac{1}{x} \ln(1+x)$$ being convex when $x>0$. I'm not completely sure if this is true. The second derivative is inconclusive (at least it seems like that). January 13th, 2019, 10:44 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 $\lim \limits_{x\to 0} \dfrac{d^2}{dx^2}\left(\dfrac 1 x \ln(1+x)\right) = \dfrac 2 3 > 0$ It's convex at 0. January 13th, 2019, 11:17 AM   #3
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 Originally Posted by romsek $\lim \limits_{x\to 0} \dfrac{d^2}{dx^2}\left(\dfrac 1 x \ln(1+x)\right) = \dfrac 2 3 > 0$ It's convex at 0.
That doesn’t make sense to me. Convex at 0? What I mean is isn't it supposed to be convex on an interval?

Sorry, but I was hoping that it would be convex when $x>0$, so on $(0,\infty)$. By the way, this is my first proof using Jensen's so I'm still learning.

Last edited by ProofOfALifetime; January 13th, 2019 at 11:35 AM. January 13th, 2019, 11:49 AM   #4
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 Originally Posted by ProofOfALifetime That doesn’t make sense to me. Convex at 0? What I mean is isn't it supposed to be convex on an interval? Sorry, but I was hoping that it would be convex when $x>0$, so on $(0,\infty)$. By the way, this is my first proof using Jensen's so I'm still learning.
Ok, my bad, This refers to the function being convex in an infinitesimal interval $(0,\delta)$, but at any rate the 2nd derivative of your function is positive in the interval $(0,\infty)$ so your function is convex for all non-negative reals. January 13th, 2019, 12:09 PM   #5
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 Originally Posted by romsek Ok, my bad, This refers to the function being convex in an infinitesimal interval $(0,\delta)$, but at any rate the 2nd derivative of your function is positive in the interval $(0,\infty)$ so your function is convex for all non-negative reals.
Thank you thank you thank you! This is all I needed to complete the proof I was doing! I appreciate it!  Tags convex 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 Brazen Calculus 0 January 28th, 2013 12:38 PM galc127 Algebra 0 August 31st, 2012 10:23 PM Vasily Applied Math 1 June 30th, 2012 02:57 PM magyaddello Algebra 0 November 26th, 2010 08:42 AM frederico Real Analysis 0 April 6th, 2009 11:31 AM

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