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February 16th, 2019, 07:19 PM   #11
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Quote:
 Originally Posted by cjem . . . or an endpoint (i.e. $a$ or $b$).
You need to specify explicitly that $a$ and $b$ are integers, else the "integer endpoints" of $f$ need not be $a$ and $b$. February 17th, 2019, 01:39 AM   #12
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Quote:
 Originally Posted by skipjack You need to specify explicitly that $a$ and $b$ are integers, else the "integer endpoints" of $f$ need not be $a$ and $b$.
$a$ 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: 551 cjem Thank you very much for your patience. I see your proof now. February 17th, 2019, 08:29 AM   #14
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Quote:
 Originally Posted by cjem $a$ and $b$ need not be integers, though, just reals - the 'integer endpoints' are $\lceil a \rceil$ and $\lfloor b \rfloor$.
For that to work, some of your references to $a$ and $b$ (or to the endpoints) would need to be replaced with references to $\lceil a \rceil$ and $\lfloor b \rfloor$ (or to the "integer endpoints").

If $a$ and $b$ are non-integer 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
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Quote:
 Originally Posted by skipjack If $a$ and $b$ are non-integer reals, there needn't be any integers in [a, b], so that $f$ doesn't have a maximum on the integers.
Good point, I should have included "and if $[a,b]$ contains an integer" in the statement. (Although, depending on your definitions, the statement as written could be vacuously true when $[a,b]$ contains no integers.)

Quote:
 Originally Posted by skipjack For that to work, some of your references to $a$ and $b$ (or to the endpoints) would need to be replaced with references to $\lceil a \rceil$ and $\lfloor b \rfloor$ (or to the "integer endpoints").
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: somewhere Posts: 596 Thanks: 87 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: 551 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 non-negative 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 non-positive 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 sub-intervals such that the first sub-interval is [a, y_1], the final sub-interval is [y_n, b], and any intervening sub-interval is [y_k, y_{k+1}] with 0 < k < n. We now apply the logic of the previous two paragraphs to each sub-interval. 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. Thanks from idontknow Tags confused 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 SarahW Algebra 1 May 14th, 2013 12:50 PM Siedas Algebra 9 March 10th, 2012 01:29 PM Ballser Academic Guidance 60 January 14th, 2012 05:23 PM katfisher1208 Algebra 1 April 20th, 2010 06:59 PM momoftwo New Users 4 August 19th, 2009 06:15 PM

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