Math Forum Discussions

JT

Posts: 436
Registered: 4/7/12

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

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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 + (x-4)/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 + (x-4)/4 = 1-x/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 + (x-4)/4 - (x-4)^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