a way to get close to big square roots in your head

First take your number, say 25347 and round that a bit to
2.5*10^4
note that the 10^4 is an even power of 10 so use this formula:
.8+(1/3)*x-(1/100)*x^2
so .8 + (1/3)2.5 = 1.63 and subtract .0625 to get 1.57
for every even power of 10 above 0 move a decimal place to the right, so since 25347 was *10^4 move 2 decimal places over. That gives you 157 which is close to the square root of 25347 (it is actually 159.2)

For odd powers such as 253478
use this formula:
2.4+1.1*x - (1/30)*x^2
so start the same round the number to
2.5 *10^5
2.4 + 1.1*2.5 = 2.4 + 2.75 = 5.15 and then do 1/3 of 2.5 .83 and shift to .083, 5.15-.083 = 5.067
then shift the decimal to the right for every odd power of 10 over 1, 506 the answer is actually around 503.

So with a little practice one can get pretty good at getting close to square roots in one's head.

Why it works:

Here's a graph of the function I listed vs. the square root from 1 to 10

The shape of the graph repeats for 100-1000 with the y axis multiplied by 10

Every even power of 10 such as 4.5*10^4 looks just like 4.5 but with the axis multiplied by a 100. 

The other function works the same way. 

Modulus algorithm

Suppose you want to see if 897354 divides 54
the leading eight of 897354 has 5 digits to the right of it, so looking at the fraction 8/54
.14814814 we can just look at the fractional part when this decimal is shifted to the right 5 places
14814.814 and just the part to the right of the decimal is .814 (rounded to thousandths)
The 9 in 897354 has 4 digits to the right of it so 9/54 is:
.166666 and shifting the decimal 4 digits to the right and dropping the whole number part is .666
so so far we have .814, .666 and doing the same thing for the rest is .63, .555, .925, .074
A fast way to do find these numbers is to calculate 1/54 and then multiply by 8,9,7,3,5,4
We can add these 5 numbers and get: 3.664 the part to the right of the decimal is .664 which is within a couple thousands of the remainder when 897354 dividing 54 = .666 so 897354 does not divide 54 (it has a remainder of 2/3rds.

The nice thing about this algorithm is it takes a process that is usually serial in nature, division you usually do long division from left to right and makes it parallel, if you had 10,000 digits of a number you were dividing you could send to 10,000 computers for example if the 573rd digit was 9,
573, 9, 4714
which would tell it return the 573rd place of the decimal 9/4714:see section B,   and it would return something and then one thread could add up all the returned numbers to find out whether the 10,000 digit number divides 4714.

-B- to be completed

Universal alphabet

Here is a sample of how it looks written...

This says: "My name is Ben."

It looks cryptic but it is really simple, every individual sound is a group of four symbols in a square such as in the above:

This is the B sound in "Ben". The top left symbol in this square is one of x,1,2,3,4,5 and indicates how open the lips of the mouth are, x for closed and 1,2,3,4,5 for progressively more open. If the lips are rounded to form the sound there is a circle around this symbol. As you round and close the lips to form the b sound this is an x with a circle around it.
The bottom left symbol is a letter one of a,b,c,d,e,f,g and indicates where your tongue is in your mouth as the sound is formed. a is top of the mouth against the teeth, b is middle of the roof of the mouth, c is towards the back and roof of the mouth, d is forward near the teeth but halfway between the bottom and top of the mouth, e is back middle of the mouth, and f is bottom and forward against the teeth and g is bottom of the mouth back away from the teeth.
The top right symbol indicates the tone of voice as in music, a is the neutral note and b,c,d,e up to a gain are progressively higher tones of voice.
The bottom right symbol indicates how open the jaw is x for closed and is circled if the throat is closed as in "guh". , 1,2,3,4,5 for more and more open.
A word like "Ben" has a beginning, middle, and end sound hence three groups of four symbols make up the word. If the word is multiple syllables a comma is placed between syllables. Also the stressed syllable of the word has an apostrophe following it.

And that is all there is to it, with this way to describe mechanically the way the lips, tongue, jaw, and voice box are acting every dialect of every language can be written phonetically.

solving for the graceful tree of a graph

The goal of finding a graceful graph is finding numbers to fill in the letters of this graph:

such that the differences between them (to label the edges) are 1,2,3,4,5. Or for larger graphs following that same pattern. 

First one can observe that one equation can be formed from the information as:

But there are five unknowns and that is only one equation, so the way I found to provide the other equations is simply to raise this basic equation to higher powers. So I use:

This provides many solutions but the first one given is:

This solution does work. There might be other ways to find a set of nonlinearly-related equations using the information given that work even better. 

Acid/base battery part 2

\

Above is the battery after running for 24 hours. It held between a quarter and a third of an amp for at least a couple hours, and today 24 hours later it is still reading a few milliamps. These readings were taken by simply placing the electrodes of the voltmeter in each beaker.  The jar in the left held a green liquid aluminum chloride and the one on the right held a whitish aluminum hydroxide. These were prepared by dissolving an excess of aluminum foil in HCl and NaOH and decanting each liquid part into a beaker. Then a tube full of water was carefully inserted between them to form a bridge. You might be able to see in the picture above that eventually an unknown to me orange precipitate forms in the one on the right and a dark precipitate forms in the one on the left. 

The trick was finding an acid and a base that conduct electricity well, I found it hard to find information on the internet, but aluminum chloride has about 750 ohms of resistance and aluminum hydroxide has 14,000. So an obvious way to improve the battery would be to find a better base for conducting electricity than aluminum hydroxide. 

Redistribution network

I was thinking to study a network like so: suppose you have a graph, this one is simple but they don't have to be:

Now consider this matrix, the letters on the left are just labeling the rows: 

Now imagine what each cell in the matrix represents is how much of each different variable each node has in its inventory. Here I've started every node off with 1 unit of it's own variable, but they could start in any way.

Then they go through this cycle, every node divides what it has in it's inventory among the nodes it is connected to. For example A in the graph diagram is connected to B,E, and F. and it has in it's inventory A, so it will give (1/3)A to B, (1/3)A to E, and (1/3)A to F. At the same time every node distributes what is in it's inventory to all the other nodes. So after one iteration you have:

So F has 5 connections to A, D, E, G, and H. It gets a share from each of those nodes that depends on the number of nodes those nodes are connected to. Like row F and column E is 1/4 because E is connected to 4 nodes of which F is one, and E had previously just had one E in it's row.

After another iteration (rounded to maximum of 3 decimal places):
[0.261, 0.083, 0.111, 0.113, 0.133, 0.05 , 0.1     , 0.05  , 0.0   ,  0.0    ],
[0.083, 0.305, 0.0    , 0.145, 0.083, 0.116, 0.0    , 0.0    , 0.083, 0.0    ],
[0.111, 0.0    , 0.277, 0.062, 0.145, 0.05  , 0.0    , 0.062, 0.062, 0.125],
[0.15  , 0.194, 0.083, 0.258, 0.05  , 0.05  , 0.1    , 0.113, 0.083, 0.125],
[0.177, 0.11  , 0.194, 0.05  , 0.279, 0.116, 0.1    , 0.05  , 0.063, 0.0    ],
[0.083, 0.194, 0.083, 0.063, 0.145, 0.316, 0.125, 0.125, 0.125, 0.125],
[0.066, 0.0    , 0.0    , 0.05  , 0.05  , 0.05  , 0.225, 0.05  , 0.063, 0.125],
[0.066, 0.0    , 0.083, 0.113, 0.05  , 0.1    , 0.1    , 0.363, 0.125, 0.125],
[0.0    , 0.111, 0.083, 0.083, 0.063, 0.1    , 0.125, 0.125, 0.333, 0.125],
[0.0    , 0.0    , 0.083, 0.063, 0.0    , 0.05  , 0.125, 0.063, 0.063, 0.25  ]

Here row A and column A is .261 because A gets 1/3 of every part of B's inventory but B had 1/3 A so A gets 1/9 from B. And similarly 1/12 and 1/15 from E and F, added together that is .261.

After 100 iterations:

[[0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088],
 [0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088],
[0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088, 0.088],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
[0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147, 0.147]
[0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
[0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118, 0.118],
 [0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058],
[0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058, 0.058],

It ends up all of these numbers are approached as limits! 30 Iterations is correct to about 3 decimal places and they get closer and closer.

**EDIT**
I found a way to find these numbers exactly. Imagine that the process iterated until it reached the limits. That would mean in another iteration, the amount going into each node would match the amount going out. That would mean that the new amount in A for example would be 1/3 B, 1/4 E, 1/5 F. So we can write out the equations:

Here there would be 10 equations one for each node, but you remove one or otherwise the solutions would be 0 for all variables. In the removed equations place you put A+B+C+D+E+F+G+H+J+K=1. That's the other constraint on the system. Then the solutions are:

And it appears that the bottom number of these fractions is twice the number of edges in the graph and the top number is the number of edges coming from the vertex.

Concerning isometries

I noticed if you have two graphs labeled any way, like:

You can set up a set of equations like so for each of them, for the graph on the left it is:
a) a+m=b+c+f
b) b+m=a+c+d+e
c) c=m+a+b+d+e+f
d) d=m+b+c+e
e) e=m+b+c+d+f
f)  f=m+a+c+e
Where each lettered equation is a letter for a vertex +m = the vertices that lettered vertex is connected to summed.
The set of equations for the figure on the right in the above image is:
a) a+m=b+e+f
b) b+m=a+c+d+e+f
c) c+m=b+d+e
d) d+m=b+c+f
e) e+m=a+b+c+f
f) f+m=a+b+d+e

Now solving each system respectively gives:

It's not too hard to see that these are the same solutions but with some of the letters having exchanged roles.

The idea is the graph is uniquely identified by the system of equations up to renaming vertices, and the solution for each variable is unique for the system of equations up to renaming variables, so isometrical graphs have the same set of solutions up to renaming variables.

** I did find that it can't always distinguish between a graph and the same graph where a vertex is connected to itself but it seems to work for simple graphs