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February 16th, 2019, 07:19 PM   #11
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Quote:
Originally Posted by cjem View Post
. . . 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 $b$.
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February 17th, 2019, 01:39 AM   #12
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Originally Posted by skipjack View Post
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.
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February 17th, 2019, 05:24 AM   #13
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cjem

Thank you very much for your patience. I see your proof now.
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February 17th, 2019, 08:29 AM   #14
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Quote:
Originally Posted by cjem View Post
$a$ 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.
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February 17th, 2019, 11:32 AM   #15
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Originally Posted by skipjack View Post
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 View Post
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.
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February 18th, 2019, 07:50 AM   #16
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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.
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February 18th, 2019, 11:03 AM   #17
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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
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