Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #1 20090914 13:52:41
More About FunctionsJust fnishing off some new pages on functions: "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #2 20090915 06:20:04
Re: More About FunctionsHi MathsisFun; 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. #3 20090915 08:19:14
Re: More About FunctionsI learnt increasing functions as x < y ⇒ f(x) ≤ f(y). Why did the vector cross the road? It wanted to be normal. #4 20090915 08:26:00
Re: More About FunctionsHi; 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. #5 20090915 11:40:24
Re: More About FunctionsGood points, thank you. "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #6 20090915 14:55:24
Re: More About FunctionsHi MathsIsFun; 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. #7 20090915 21:13:30
Re: More About FunctionsAccording to the new definition, a constant function is increasing and decreasing, rather than neither. Why did the vector cross the road? It wanted to be normal. #8 20090915 21:46:17
Re: More About Functions
That seems to be a consequence of the definitions, yes. But not strictly increasing or strictly decreasing.
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 #9 20090916 04:42:03
Re: More About FunctionsWell, 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. Why did the vector cross the road? It wanted to be normal. #10 20090916 11:13:13
Re: More About FunctionsI added "f(a) should be inside the interval, not at one end or the other." "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #11 20090916 11:28:57
Re: More About FunctionsAbout constant functions …
You realize that this is wrong? In fact, a Constant Function is both increasing and decreasing (but not strictly) along its whole domain. Last edited by JaneFairfax (20090916 11:29:23) #12 20090916 11:35:06
Re: More About FunctionsMissed that one, thanks. "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #13 20090916 11:41:31
Re: More About FunctionsAccording to the definition, a function y=f(x) is increasing iff
Hence, if y=f(x) is nonincreasing, 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 nonincreasing. Last edited by JaneFairfax (20090916 15:23:45) #14 20090916 12:07:13
Re: More About FunctionsI agree. "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #15 20090916 17:05:59
Re: More About FunctionsI think what you meant was that a constant function is neither strictly increasing nor strictly decreasing. 