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January 13th, 2019, 10:52 AM   #1
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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).
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January 13th, 2019, 11:44 AM   #2
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$\lim \limits_{x\to 0} \dfrac{d^2}{dx^2}\left(\dfrac 1 x \ln(1+x)\right) = \dfrac 2 3 > 0$

It's convex at 0.
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January 13th, 2019, 12:17 PM   #3
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Originally Posted by romsek View Post
$\lim \limits_{x\to 0} \dfrac{d^2}{dx^2}\left(\dfrac 1 x \ln(1+x)\right) = \dfrac 2 3 > 0$

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 12:35 PM.
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January 13th, 2019, 12:49 PM   #4
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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.
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January 13th, 2019, 01:09 PM   #5
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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!
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