/sci/ - Science & Math

>>9012061
Gilliland (1981) found evidence for a decrease in diameter of 0.01% per century:
http://adsabs.harvard.edu/full/1981ApJ...248.1144G
...which was disputed by Sveshnikov (2002) to be observational error:
http://adsabs.harvard.edu/full/2002AstL...28..115S

P{both correct} = 0.9*0.3 = 0.27 = p
P{both incorrect} = 0.1*0.7 = 0.07 = q
P{world ends} = p/(p+q) = 0.27/ 0.34 = 0.79

Do the commercial DNA tests
(ancestryDNA, 23andMe, etcetera)
detect Neanderthal components?

>>8784624
>cancel out
no

https://en.wikipedia.org/wiki/Coand%C4%83_effect

the first ratio can be written
[math] \displaystyle \frac{a^a+b^b}{a^b+b^a}= \lambda e^{x_1}+(1- \lambda)e^{x_2}[/math]
where
[math] \displaystyle \lambda= \frac{b^a}{a^b+b^a}[/math]
and
[math] \displaystyle x_1= \left( \frac{a}{b} \right)^a=e^{a( \ln a- \ln b)}[/math]
and
[math] \displaystyle x_2= \left( \frac{b}{a} \right)^b=e^{b( \ln b- \ln a)}[/math]
then the result from convexity of [math]e^x[/math] is a bit simpler.

the first ratio can be written
\displaystyle \frac{a^a+b^b}{a^b+b^a}= \lambda e^{x_1}+(1- \lambda)e^{x_2}
where
\displaystyle \lambda= \frac{b^a}{a^b+b^a}
and
\displaystyle x_1= \left( \frac{a}{b} \right)^a=e^{a( \ln a- \ln b)}
and
\displaystyle x_2= \left( \frac{b}{a} \right)^b=e^{b( \ln b- \ln a)}
then the result from convexity of [math]e^x[/math] is a bit simpler.

>>8784688
>convexity and monotonicity
...of [math]e^x[/math] and [math] \ln x[/math], respectively

it's between 15 and 16 because
[math]3 \sqrt{26}= \sqrt{234}[/math] and
[math]15= \sqrt{225}< \sqrt{234}< \sqrt{256}=16[/math]
EZ-PZ

>>8614811
>what is the superior engineering major again
>and again
>and again and again
>and again and again and again
giveitup fgt pls

>>8616170
>>8616172
>>8616174
L0Lno fgt pls

>>8615999
>It's a fucking mystery.

>>8615861
>how do you think euler avoided conscription?
...maybe bcoz he was blind, you retard.

>>8614697
all terms positive so the series approaches
the limit from below, giving an opportunity to
estimate the truncation error

I combined pairs of terms to get a faster-converging expression:
[math] \displaystyle \pi =3+ \frac{6}{3 \times 15}+ \frac{6}{15 \times 63}+ \frac{6}{35 \times 143}+ \cdots[/math]

>>8613389
Thank you for showing this, I've been having
a Hell of a time tonight and just couldn't see it.

← Nilakantha expression
[math] \displaystyle \pi =3+ \frac{4}{2 \times 3 \times 4}- \frac{4}{4 \times 5 \times 6}+ \frac{4}{6 \times 7 \times 8}- \cdots[/math]

second try at the Nilakantha expression:
[math] \displaystyle \pi = 3-4 \sum_{n \geq 1} \frac{(-1)^n}{2n(2n+1)(2n+2)}[/math]

this is the similar Nilakantha expression
[math] \displaystyle \pi =3-4 \sum_{n \geq 1} \frac{(-1)^n}{2n(2n+1)(2n+2)}[/math]

>>8613345
my mistake, the second factor in the denom
should be (n + 1)

I found pic related written in the back of an old reference book, with no explanation. It looks similar to a Nilakantha expression, can /sci/ confirm?
[math] \displaystyle \pi =3- \frac{1}{1 \times 2 \times 3}+ \frac{1}{2 \times 3 \times 5}- \frac{1}{3 \times 4 \times 7}+ \cdots[/math]

here's the solution I found with some modular algebra:
[math]208341=(47)(83)A+(17)(83)B+(17)(47)C[/math]
remainders on division by 17:
[math]6 \equiv 8A(mod~17)[/math] so A = 5,22,39...
remainders on division by 47:
[math]37 \equiv B(mod~47)[/math] so B = 37,84,131...
remainders on division by 83:
[math]11 \equiv 52C(mod~83)[/math] so C = 5,88,171...
picking B = 37 (for example) gives the choices
(A, C) = (5, 171), (22, 88), (39, 5)

here's the solution I found with some modular algebra:
[math]208341=(47)(83)A+(17)(83)B+(17)(47)C[/math]
remainders on division by 17:
[math]6 \equiv 8 \times \textup{A}(mod~17)~so~ \textup{A}=5,22,39, \cdots[/math]
remainders on division by 47:
[math]37 \equiv 1 \times \textup{B}(mod~47)~so~ \textup{B}=37,84,131, \cdots[/math]
remainders on division by 83:
[math]11 \equiv 52 \times \textup{C}(mod~83)~so~ \textup{C}=5,88,171, \cdots[/math]
picking B=37 (for example) gives the choices
(A, C) = (5, 171), (22, 88), (39, 5)

>>8580762
>within one-half part per trillion
>it's only ~39 ppb by my calculation
We were both wrong, it's 39 per trillion.