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 June 26th, 2019, 11:37 PM #11 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2218 You needn't use "$\mathbb N^{*}$", as $\mathbb N$ already means the integers greater than zero. The original wording was "ascending function", not "strictly ascending function". A constant function is ascending, but not strictly ascending, so it's not excluded by the original wording. Hence $f(n) = 4\ \forall n \in \mathbb N^{*}$ satisfies all the originally stated conditions. June 27th, 2019, 06:44 AM   #12
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Quote:
 Originally Posted by skipjack You needn't use "$\mathbb N^{*}$", as $\mathbb N$ already means the integers greater than zero. The original wording was "ascending function", not "strictly ascending function". A constant function is ascending, but not strictly ascending, so it's not excluded by the original wording. Hence $f(n) = 4\ \forall n \in \mathbb N^{*}$ satisfies all the originally stated conditions.
Hello,

I do not understand! I think the function $\displaystyle f(n)=C$ where $\displaystyle C$ is a constant is neither ascending and neither descending.
Why can not we consider that the function $\displaystyle f(n)=C$, where $\displaystyle C$ is a constant, is descending function?
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Really, if we consider $\displaystyle f(n)=C$ where $\displaystyle C$ is a constant, then from the imposed condition it follows that $\displaystyle C=4$, but first to prove that the function $\displaystyle f(n)=C$, where $\displaystyle C$ is a constant, is an ascending function and is not an descending function. Thank you very much!

All the best,

Integrator

Last edited by skipjack; June 28th, 2019 at 02:20 AM. June 27th, 2019, 03:09 PM #13 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2218 The function $\displaystyle f(n)=C$, where $\displaystyle C$ is a constant, is both an ascending function and a descending function. The usual terms are "increasing" and "decreasing", rather than "ascending" and "descending". June 28th, 2019, 07:22 AM   #14
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Quote:
 Originally Posted by skipjack The function $\displaystyle f(n)=C$, where $\displaystyle C$ is a constant, is both an ascending function and a descending function. The usual terms are "increasing" and "decreasing", rather than "ascending" and "descending".
Hello,

I do not understand how the constant function can be increasing and decreasing at the same time. How can we fully solve the proposed problem? Thank you very much!

All the best,

Integrator

Last edited by skipjack; June 28th, 2019 at 07:51 AM. June 28th, 2019, 08:00 AM #15 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2218 This article explains how "increasing" and "decreasing" are used. It also explains how adding the word "strictly" changes the meaning. June 28th, 2019, 08:52 PM   #16
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Quote:
 Originally Posted by skipjack This article explains how "increasing" and "decreasing" are used. It also explains how adding the word "strictly" changes the meaning.
Hello,

OK! In conclusion, how do we find all the functions that meet the condition in the problem? Let I understand that $\displaystyle f(n)=4$ and $\displaystyle f(n)=n$ are the only functions that meet the condition in the problem? If so, then how do we prove this? Thank you very much!

All the best,

Integrator

Last edited by skipjack; June 28th, 2019 at 11:00 PM. Tags functions 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 Nij Calculus 2 November 25th, 2015 06:20 AM kfarnan Calculus 5 September 10th, 2011 03:51 AM UltraMath Algebra 3 September 4th, 2011 01:43 PM panky Calculus 0 December 31st, 1969 04:00 PM

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