Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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You are not logged in. #1 2009-09-14 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 2009-09-15 06:20:04
Re: More About FunctionsHi MathsisFun; In mathematics, you don't understand things. You just get used to them. 90% of mathematicians do not understand 90% of currently published mathematics. I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on. #3 2009-09-15 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 2009-09-15 08:26:00
Re: More About FunctionsHi; In mathematics, you don't understand things. You just get used to them. 90% of mathematicians do not understand 90% of currently published mathematics. I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on. #5 2009-09-15 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 2009-09-15 14:55:24
Re: More About FunctionsHi MathsIsFun; In mathematics, you don't understand things. You just get used to them. 90% of mathematicians do not understand 90% of currently published mathematics. I am willing to wager that over 75% of the new words that appeared were nothing more than spelling errors that caught on. #7 2009-09-15 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 2009-09-15 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 2009-09-16 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 2009-09-16 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 2009-09-16 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 (2009-09-16 11:29:23) #12 2009-09-16 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 2009-09-16 11:41:31
Re: More About FunctionsAccording to the definition, a function y=f(x) is increasing iff
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. Last edited by JaneFairfax (2009-09-16 15:23:45) #14 2009-09-16 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 2009-09-16 17:05:59
Re: More About FunctionsI think what you meant was that a constant function is neither strictly increasing nor strictly decreasing. |