February 16th, 2019, 07:19 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010  
February 17th, 2019, 01:39 AM  #12 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  $a$ and $b$ need not be integers, though, just reals  the 'integer endpoints' are $\lceil a \rceil$ and $\lfloor b \rfloor$.
Last edited by cjem; February 17th, 2019 at 01:51 AM. 
February 17th, 2019, 05:24 AM  #13 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 
cjem Thank you very much for your patience. I see your proof now. 
February 17th, 2019, 08:29 AM  #14  
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010  Quote:
If $a$ and $b$ are noninteger reals, there needn't be any integers in [a, b], so that $f$ doesn't have a maximum on the integers.  
February 17th, 2019, 11:32 AM  #15  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  Quote:
Would you mind elaborating? I'm pretty sure I'm correct in working with the endpoints rather than the 'integer endpoints' each time I've done so, but I could be mistaken.  
February 18th, 2019, 07:50 AM  #16 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
If the function has maximum at one of the integer parts and to save time use calculus again : First compare them , $\displaystyle f(\lfloor x \rfloor ) $ R $\displaystyle f(\lceil x \rceil ) \; $ with R I mean relation (smaller or larger ) . $\displaystyle f(\lfloor x \rfloor)  f(1+\lfloor x \rfloor ) $ R $\displaystyle 0$ . Then work with another function : $\displaystyle z(t)=f(t)f(1+t)$ R $\displaystyle 0\; $ for $\displaystyle t\in \mathbb{R}$ . Now find where $\displaystyle \frac{dz}{dt}$ is larger or smaller than $\displaystyle 0$ using calculus . Or $\displaystyle z'( \lfloor x \rfloor ) $ R $\displaystyle 0$ . And all values above must be in the domain . Last edited by idontknow; February 18th, 2019 at 08:19 AM. 
February 18th, 2019, 11:03 AM  #17 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 
I'd like to go back to my original point. Here is the theorem: $\text {If } f(x) \text { is differentiable at every point in } [a, b],$ $\text {and if } c \text { is a maximizing integer of } f \in [a, b],$ $\text {then either } c = \lceil a \rceil, \text { or } c = \lfloor b \rfloor, \text { or}$ $\exists \ y \in (a,\ b) \text { such that } f'(y) = 0 \text { and either } c = \lceil y \rceil \text { or } c = \lfloor y \rfloor.$ Here is a sketch of a proof. If f'(x) is everywhere nonnegative in [a, b], then f(x) is nowhere decreasing in that interval. Therefore either b (if b is an integer) or the integer immediately below b is not less than any other integer in [a, b]. If f'(x) is everywhere nonpositive in [a, b], then f(x) is nowhere increasing in that interval. Therefore either a (if a is an integer) or the integer immediately above a is not less than any other integer in [a, b]. If f(x) has n > 0 points in (a, b) with derivatives of zero, let's call those points y_1 through y_n. So we can divide divide [a, b] into n + 1 closed subintervals such that the first subinterval is [a, y_1], the final subinterval is [y_n, b], and any intervening subinterval is [y_k, y_{k+1}] with 0 < k < n. We now apply the logic of the previous two paragraphs to each subinterval. We can do this because of the intermediate value theorem. The net result is that, in principle, we can always find the maximizing integer if one exists by looking at no more than 2n + 2 integers. That may not be the most efficient way to find a maximizing integer, but it is always a valid way. 

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