I Help with rewriting a compound inequality

AI Thread Summary
The discussion revolves around rewriting a compound inequality by breaking it down into two separate inequalities for easier manipulation. Participants explore the concept of integer-convex functions, particularly focusing on the expression g(k+1) + g(k-1) - 2g(k) and its implications for convexity. The importance of maintaining the sign when manipulating inequalities is emphasized, as it affects the validity of the operations. Clarifications are made regarding the relevance of optimal points in the context of the discussion. Overall, the thread highlights the process of simplifying complex mathematical problems through systematic approaches.
Andrea94
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Help with rewriting optimality conditions for integer-convex functions.
See attached screenshot.
Stumped on this, I'll take anything at this point (hints, solution, etc).

help.png
 
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Big problems are to be split into smaller problems. You have two inequalities. Write the first one down and manipulate ! Then idem number two.

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And: what does integer-convex mean for e.g. $$g(k+1) + g(k-1) - 2g(k) \quad \textsf{?} $$

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BvU said:
Big problems are to be split into smaller problems. You have two inequalities. Write the first one down and manipulate ! Then idem number two.

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Great, thanks! Didn't even think about the fact that I could do each inequality separately.
 
BvU said:
And: what does integer-convex mean for e.g. $$g(k+1) + g(k-1) - 2g(k) \quad \textsf{?} $$

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What do you mean? The expression you've written is the same thing as $$\Delta g(k) - \Delta g(k-1)$$ but I'm not sure how that is relevant.
 
Andrea94 said:
What do you mean? The expression you've written is the same thing as $$\Delta g(k) - \Delta g(k-1)$$ but I'm not sure how that is relevant.
Right, but ##\Delta g(k) - \Delta g(k-1)## can be ##\ge 0## or ##\le 0## for an integer-convex function ... whereas ... :wink:

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BvU said:
Right, but ##\Delta g(k) - \Delta g(k-1)## can be ##\ge 0## or ##\le 0## for an integer-convex function ... whereas ... :wink:

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Still not sure what I'm supposed to see here 😅
 
The problem with inequalities is that you can only multiply (or divide) left and right with something positive. As it happens, for integer-convex functions the sign of ##g(k+1) + g(k-1) - 2g(k)## is ...

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BvU said:
The problem with inequalities is that you can only multiply (or divide) left and right with something positive. As it happens, for integer-convex functions the sign of ##g(k+1) + g(k-1) - 2g(k)## is ...

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Hm, the only thing I can think of is that if ##k## is optimal and nonzero, then ##g(k+1) + g(k-1) - 2g(k)## is always positive since for optimal ##k## we have ##\Delta g(k) > 0## and ##\Delta g(k-1) < 0##. Is this what you mean?
 
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Not sure where your 'optimal' comes from (it seems to live in your context, but not in the context of this thread ?).

But: yes, for a convex function the second derivative is always positive and so is this ##g(k+1) + g(k-1) - 2g(k)##.

I figured it might help in manipulating the inequalities ...

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  • #11
BvU said:
Not sure where your 'optimal' comes from (it seems to live in your context, but not in the context of this thread ?).

But: yes, for a convex function the second derivative is always positive and so is this ##g(k+1) + g(k-1) - 2g(k)##.

I figured it might help in manipulating the inequalities ...

##\ ##
Ohh I see, thanks a lot for the help!
 
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