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Topic:
AE911
Replies:
6
Last Post:
Feb 10, 2013 7:18 AM



JT
Posts:
1,448
Registered:
4/7/12


Re: AE911
Posted:
Feb 9, 2013 3:49 AM


On 9 Feb, 09:48, JT <jonas.thornv...@gmail.com> wrote: > On 9 Feb, 09:34, JT <jonas.thornv...@gmail.com> wrote: > > > On 9 Feb, 09:32, JT <jonas.thornv...@gmail.com> wrote: > > > > A mad woman scribbled down AE911 upon the famous paintinghttp://www.aftonbladet.se/nyheter/article16207744.ab > > > > I was thinking it maybe hexadecimal but there is nothing special about > > > 715025 > 7,15025 >71,5025>715,025 >7150,25 >71502,5 or...? > > > Evidently the color code is green in HTML. > > > > Anyone other have any idea what AE911 could be hinting? Well the most > > > probable it is just a doodle of madness. > > > But numbers are intruiging. > > > Let us say the woman was not mad, what would be so revolutionary about > > AE911 it probably meant something to her with the choice of painting, > > she must thought it to be something extrordinary? > > 201/400=0,5025 > > Is it something about math without the need of approxiamtion of > digits?http://www.jiskha.com/display.cgi?id=1350495808 > > Calculus II  Lauren, Wednesday, October 17, 2012 at 3:27pm > please someone help this is due tomorrow and i don't know where to > start with solving this! > Calculus II  Jennifer, Wednesday, October 17, 2012 at 3:34pm > 3^0.5 = 1.732 > 5^0.5 = 2.236 > > on [3,5], > > 3^x  1.732 < 0.01 > 5^x  2.236 < 0.01 > > 3^x < 3^0.5 + 0.01 > x < log(3) (1.742) > 5^x < 5^0.5 + 0.01 > x < log(5) (2.246) > > For the first equation, > x < 0.50521360825 > For the second equation, > x < 0.50275369436 > > x has to be lower than both of these numbers, and greater than 1/2, > and it has to be a rational fraction. > > So start trying numbers that are rational fractions slightly greater > than 1/2 > > 201/400 = 0.5025 > > so P(x) = x^(201/400) is one such polynomial > Calculus II  Lauren, Wednesday, October 17, 2012 at 3:38pm > i dont understand why you raised the 3 and 5 to 1/2 > Calculus II  Steve, Wednesday, October 17, 2012 at 4:26pm > polynomials have integer powers > f(3) = ?3 = 1.732 > f(5) = ?5 = 2.236 > > Since this is calculus II, I assume you know about Taylor Polynomials. > Let's use the polynomial for ?x at x=4 (the midpoint of our interval) > > ?x = 2 + (x4)/4  ... > > at x=3,5, we want p(x)f(x) < .1 > use enough terms to get that accuracy > > if p(x) = 2, > p(3)f(3) = 2?3 = 0.26 > p(5)f(5) = 2?5 = .236 > > if p(x) = 2 + (x4)/4 = 1x/4, > p(3)f(3) = 1.75  ?3 = 0.0179 > p(5)f(5) = 2.25  ?5 = 0.0139 > > Looks like a linear approximation fits the bill. > > Just for grins, what happens if we use a parabola to approximate ?x? > > p(x) = 2 + (x4)/4  (x4)^2/64 > p(3)f(3) = 0.0023 > p(5)f(5) = 0.0017
Who solved that, i want to know the name. Was it 71 ;D



