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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

Just fnishing off some new pages on functions:

Comments, etc welcome. Help make the pages perfect

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi MathsisFun;

Look good from here, thanks.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I learnt increasing functions as x < y ⇒ f(x) ≤ f(y).

ie. A constant function counts as increasing, for example.

I learnt the definition on the page as "strictly increasing".

Probably different people use different definitions though.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi;

I just read monotonic increasing. Looked it up on wikipedia and they said the same thing for x < y ⇒ f(x) ≤ f(y). Preserving the order they call it.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

Good points, thank you.

I have redone Increasing Functions

Please check it out and let me know if it passes.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,490

Hi MathsIsFun;

Glad you went with strictly increasing instead of monotonic. Thanks for the page.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

According to the new definition, a constant function is increasing and decreasing, rather than neither.

Also, I'd say something about how for local extrema, the point should be "in" the interval, rather than on the edge.

For example, with f(x) = x on the interval [1,2], the maximum there would be at x=2. But you wouldn't consider that a local maximum of the function.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

mathsyperson wrote:

According to the new definition, a constant function is increasing and decreasing, rather than neither.

That seems to be a consequence of the definitions, yes. But not strictly increasing or strictly decreasing.

mathsyperson wrote:

Also, I'd say something about how for local extrema, the point should be "in" the interval, rather than on the edge.

For example, with f(x) = x on the interval [1,2], the maximum there would be at x=2. But you wouldn't consider that a local maximum of the function.

In that case there would be NO local extrema?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Well, that problem only happens when you work with a specific interval. So maybe the page should keep the "f(a) ≥ f(x) for all x in the interval", but say it has to be true for all small enough intervals rather than just one.

Or is that complicating things too much?

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

I added "f(a) should be inside the interval, not at one end or the other."

Hopefully that covers it

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

About constant functions

A Constant Function neither increases nor decreases:

You realize that this is wrong? In fact, a Constant Function is both increasing and decreasing (but not strictly) along its whole domain.

Think about it.

*Last edited by JaneFairfax (2009-09-15 13:29:23)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

Missed that one, thanks.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

According to the definition, a function **y=f(x)** is increasing iff

when x1 < x2 then f(x1) ≤ f(x2)

Hence, if **y=f(x)** is non-increasing, then there exist x1 < x2 such that f(x1) > f(x2). But a constant function cannot possibly have f(x1) > f(x2). Therefore a constant function cannot be non-increasing.

By a similar argument a constant function cannot be non-decreasing either. Therefore it must be both increasing and decreasing.

*Last edited by JaneFairfax (2009-09-15 17:23:45)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

I agree.

PS: I really do appreciate the help you guys give.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I think what you meant was that a constant function is neither strictly increasing nor strictly decreasing.

Well, arent constant functions just fascinating? They are both increasing and decreasing, yet neither strictly increasing nor strictly decreasing. Amazing how we take so many things for granted, only to find, on closer inspection, that we have been mistaken all along.

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