A small group of collaborators are looking for geometries that exist outside the matrix of spacetime.

The idea is that space and time may not be foundational. They might emerge from some deeper mathematical structure.

Could the Buddhist law of "dependent origination" could be a geometry of this kind?

https://zachperlman.substack.com/p/the-shape-of-dependent-origination

#physics #buddhism #math #philosophy

Hi!
Your hourly hectoc is here:

726825

The goal is to combine the 6 numbers to a total of 100. You can use the mathematical operations + - * / ^ and the parenthesis ( ). Numbers can be combined, but you have to use all 6 of them and are not allowed to change the order. See https://hectoc.seism0saurus.de for an example. Please use CW "solution".

Have fun! Your hourly hectoc bot
by @seism0saurus

#math #mentalcalculator #hectoc

☆ Π MTH ~CLVii~✧⁰¹⁵⁷ Π∞ßτ З³²² ⁰⁰כ

Studying the Koch curve can be unsettling. The curve starts with 2 points

•----•

After applying the algorithm, the 2 points become 5 points

•
/ \
▪︎--▪︎ ▪︎--▪︎

When we apply the algorithm again, the 5 points become 17 points, because each segment generates 3 points
\[4×3+5=17\]
And the 17 points generate 16 segments, where for each one we have 3 points, so that
\[16×3+17=65\]
Finally, we see the recursive law
\[p=(u-1)×3+u\]
Which results in
\[p=4u-3\]
Generating the OEIS A052539
\[2\space5\space17\space65\space257 \dotsc\]
However, the recursive formulation
\[a_{n+1}=4a_{n}-3;n\ge0\]
is unnecessary because the direct formulation is evident
\[a_n=4^n+1;n\ge0\]
Note that this expression can be deduced using first-order linear recurrence; so that the sequence
\[a_0\space a_1\space a_2\space\dotsc \space a_n\space \dotsc \]
Where the terms are obtained by the recurrence relation
\[a_n=a_{n-1}×k+r; n\ge1;k\neq1\]
So it is a fact that
\[\begin{align}
a_0&=a_0\\
a_1&=a_0×k+r\\
a_2&=a_1×k+r\\
&=(a_0×k+r)×k+r\\
&=a_0×k^2+rk+r\\
&=a_0k^2+r(k+1)\\
&\vdots\\
a_n&=a_0k^n + r(k^{n-1}+k^{n-2}+\dotsc+1)
\end{align}\]
Using the formula for a finite geometric sequence
\[S_n=\alpha_1\frac{q^n-1}{q-1}\]
To sum the terms
\[\overbrace{1\space k\space k²\space\dotsc\space k^{n-1}}^n\]
we have
\[a_n=a_0k^n+r\frac{k^n-1}{k-1}\]
Thus, for the sequence
\[2\space5\space17\space65\space257 \dotsc\]
We have
\[\left\{\begin{align}
a_0 &=2\\
k&=4\\
r&=-3
\end{align}\right.\]
Hence
\[a_n=2×4^n+(-3)\frac{4^n-1}{4-1}\]
It follows that
\[a_n=4^n+1;n\ge0\]
\[■\]

αΩ

~ ∇ ~

#math #art

[ BRAZIL ⋅ З³²² LND Lunedì χViii Maggio MMχχVi ⁰⁰כ ]

Greetings fellow humans,
Here is a new hectoc challenge for you:

191733

The goal is to combine the 6 numbers to a total of 100. You can use the mathematical operations + - * / ^ and the parenthesis ( ). Numbers can be combined, but you have to use all 6 of them and are not allowed to change the order. See https://hectoc.seism0saurus.de for an example. Please use CW "solution".

Have fun! Your hourly hectoc bot
by @seism0saurus

#math #mentalcalculator #hectoc