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Re: Problems and SolutionsProblem # n+7 Character is who you are when no one is looking. #77 20050812 15:52:10
Re: Problems and SolutionsMinimum? "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #78 20050812 18:37:01
Re: Problems and SolutionsThe 'proper' way to solve this would be to find turning points by differentiating. Why did the vector cross the road? It wanted to be normal. #79 20050812 19:01:53
Re: Problems and Solutionsfor quadratic equations, you can make use of the symmetry of the graphs to find the minimum point, which is halfway in between the two xintercepts #80 20050812 19:12:32
Re: Problems and SolutionsAm I allowed to slightly modify my question? Character is who you are when no one is looking. #81 20050812 19:53:46
Re: Problems and SolutionsIf you are allowed negative values, then they would have to be odd integers or reciprocals of odd integers, otherwise x^x would be imaginary or positive. The lowest value of x^x would be when x= 1/(2n+1), such that 1/(2n+1) is the closest possible to the value of x that gives the lowest value of x^x. This is 1/3, that returns a value of 1.44224927... Why did the vector cross the road? It wanted to be normal. #82 20050812 20:23:33
Re: Problems and SolutionsOops, I made a mistake. Character is who you are when no one is looking. #83 20050812 20:28:30
Re: Problems and SolutionsMathsy, you are right. Last edited by ganesh (20050812 20:29:35) Character is who you are when no one is looking. #85 20050813 21:44:28
Re: Problems and Solutionsd/dx (logx) = 1/x and ∫(1/x)dx = logx Character is who you are when no one is looking. #86 20050813 21:53:01
Re: Problems and SolutionsProblem # n+8 Character is who you are when no one is looking. #87 20050813 22:07:36
Re: Problems and SolutionsI use 10 ones to make these:   Why did the vector cross the road? It wanted to be normal. #88 20050814 09:08:24
Re: Problems and Solutions
Out of interest, you can derive it from the derivative of the exponential function: 2 + 2 = 5, for large values of 2. #89 20050814 09:18:15
Re: Problems and Solutions
How about and . Works in any number base!2 + 2 = 5, for large values of 2. #90 20050815 13:28:53
Re: Problems and SolutionsYou are correct, NIH ! Character is who you are when no one is looking. #91 20050815 16:31:26
Re: Problems and SolutionsWell, at least this one goes UP (rather than those updownleftrightnexttimeaddsubtractmultiplydivide types!) "The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman #92 20050815 17:51:12
Re: Problems and Solutions3 14 39 84 155 Why did the vector cross the road? It wanted to be normal. #93 20050816 09:12:41
Re: Problems and SolutionsFor the one where you have two numbers that add and multiply to give the same thing, I got bored so worked out that if you are given a value of z, then you can work out values of x and y such that x+y=xy=z like so: Why did the vector cross the road? It wanted to be normal. #94 20050816 14:10:51
Re: Problems and Solutions
Mathsy, thats very good! Last edited by ganesh (20050816 14:12:15) Character is who you are when no one is looking. #95 20050816 18:17:01
Re: Problems and Solutions
Why did the vector cross the road? It wanted to be normal. #96 20050817 14:27:32
Re: Problems and SolutionsCorrect, Mathsy! Character is who you are when no one is looking. #97 20050817 18:23:57
Re: Problems and SolutionsThis is all in my head, so there might be a mistake somewhere... Why did the vector cross the road? It wanted to be normal. #98 20050818 14:29:10
Re: Problems and SolutionsProblem # n+11 Character is who you are when no one is looking. #99 20050823 14:25:14
Re: Problems and SolutionsProblem #k Character is who you are when no one is looking. #100 20050823 14:37:18
Re: Problems and Solutionshere it is... just the answer Last edited by kylekatarn (20050823 14:41:39) 