Math Forum Discussions

JT

Posts: 570
Registered: 4/7/12

Re: AE911
Posted: Feb 9, 2013 3:48 AM

Plain Text

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