Math Forum Discussions

JT

Posts: 570
Registered: 4/7/12

Re: AE911
Posted: Feb 9, 2013 4:03 AM

Plain Text

On 9 Feb, 09:49, JT <jonas.thornv...@gmail.com> wrote:
> On 9 Feb, 09:48, JT <jonas.thornv...@gmail.com> wrote:
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> > On 9 Feb, 09:34, JT <jonas.thornv...@gmail.com> wrote:
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> > > On 9 Feb, 09:32, JT <jonas.thornv...@gmail.com> wrote:
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> > > > A mad woman scribbled down AE911 upon the famous paintinghttp://www.aftonbladet.se/nyheter/article16207744.ab
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> > > > 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.
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> > > > 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.
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> > > 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?
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> > 201/400=0,5025
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> > Is it something about math without the need of approxiamtion of
> > digits?http://www.jiskha.com/display.cgi?id=1350495808
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> > 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
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> > on [3,5],
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> > 3^x - 1.732 < 0.01
> > 5^x - 2.236 < 0.01
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> > 3^x < 3^0.5 + 0.01
> > x < log(3) (1.742)
> > 5^x < 5^0.5 + 0.01
> > x < log(5) (2.246)
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> > For the first equation,
> > x < 0.50521360825
> > For the second equation,
> > x < 0.50275369436
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> > x has to be lower than both of these numbers, and greater than 1/2,
> > and it has to be a rational fraction.
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> > So start trying numbers that are rational fractions slightly greater
> > than 1/2
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> > 201/400 = 0.5025
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> > 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
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> > 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)
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> > ?x = 2 + (x-4)/4 - ...
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> > at x=3,5, we want |p(x)-f(x)| < .1
> > use enough terms to get that accuracy
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> > if p(x) = 2,
> > p(3)-f(3) = 2-?3 = 0.26
> > p(5)-f(5) = 2-?5 = -.236
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> > 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
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> > Looks like a linear approximation fits the bill.
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> > Just for grins, what happens if we use a parabola to approximate ?x?
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> > p(x) = 2 + (x-4)/4 - (x-4)^2/64
> > p(3)-f(3) = 0.0023
> > p(5)-f(5) = -0.0017
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> Who solved that, i want to know the name. Was it 71 ;D

Does x^(201/400) has something todo with factorisation of
primeproducts?