Thread for Non-CA Academic Questions
-
- Posts: 1175
- Joined: June 14th, 2014, 5:03 pm
- Contact:
Re: Thread for Non-CA Academic Questions
Still not sure why you haven't joined the Googology Discord yet btw
Re: Thread for Non-CA Academic Questions
I don't have a discord accountfluffykitty wrote: ↑November 14th, 2019, 6:28 pmStill not sure why you haven't joined the Googology Discord yet btw
not active here but active on discord
Re: Thread for Non-CA Academic Questions
Hold it wait, I just joined. To confirm, I am Moooosey#7155.fluffykitty wrote: ↑November 14th, 2019, 6:28 pmStill not sure why you haven't joined the Googology Discord yet btw
not active here but active on discord
Re: Thread for Non-CA Academic Questions
Are there any online SKIO calculus websites? I've already seen a SKI calculus one.
not active here but active on discord
Re: Thread for Non-CA Academic Questions
what happens if we extend bf with a handful of extra commands:
* = mark/remove marker: add marker to data at pointer
( = check for marker; if none is present, skip to corresponding )
) = check for marker; if one is present, go back to corresponding (
Note: ( and ) are like [ and ] but they do it with markers instead of checking for data
* = mark/remove marker: add marker to data at pointer
( = check for marker; if none is present, skip to corresponding )
) = check for marker; if one is present, go back to corresponding (
Note: ( and ) are like [ and ] but they do it with markers instead of checking for data
not active here but active on discord
Re: Thread for Non-CA Academic Questions
is it possible to run hex13 (https://gitlab.com/apgoucher/hex13) on mac?
already did git clone, which worked
already did git clone, which worked
not active here but active on discord
- gameoflifemaniac
- Posts: 1242
- Joined: January 22nd, 2017, 11:17 am
- Location: There too
Re: Thread for Non-CA Academic Questions
Can someone explain Bowers Exploding Array Function?
I was so socially awkward in the past and it will haunt me for the rest of my life.
Code: Select all
b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$
4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo$24b4o!
Re: Thread for Non-CA Academic Questions
There's an explanation of it here on Googology Wiki which I really liked, though it is a bit long.gameoflifemaniac wrote: ↑December 4th, 2019, 3:01 amCan someone explain Bowers Exploding Array Function?
Re: Thread for Non-CA Academic Questions
I found this, and I'm on mac OS X:
http://www.srm.org.uk/downloads.html
what would I need to put into terminal in order to compile and then run the CBlocksSource stuff?
http://www.srm.org.uk/downloads.html
what would I need to put into terminal in order to compile and then run the CBlocksSource stuff?
not active here but active on discord
Re: Thread for Non-CA Academic Questions
There's been some discussion of this on Discord:
So, the CPP (Crappy packing problem) asks for the least dense way to fill an object with N-ominoes for some N such that you cannot place any more N-ominoes. ECPP (Eventual CPP) asks for the same thing, except that it's possible to place one more N-omino but no more
So far, for the 8x8 grid CPP with tetrominoes and 8x8 grid tetromino-ECPP have solutions (both by Blinkerspawn)
So here's a question: Suppose that E2CPP asks for that, but you can place an N-omino such that it brings you to an ECPP position, and never into any position which eventually leads to ECPP, E3CPP asks for that, but you can place an N-omino such that it brings you to an E2CPP position, and never into any position which eventually leads to E2CPP, etc.
Does anyone have solutions for E2CPP, and E3CPP with tetrominoes on an 8x8 grid?
Code: Select all
#C both these solutions were found by BlinkerSpawn
x = 100, y = 99, rule = LifeHistory
3F.F.F.3F5.3F.2F.2F6.F3.2F.F.F2.2F6.F3.3F.F.F.3F.3F$.F2.3F.2F6.F3.2F.
2F5.F.F2.F2.2F3.F2.F4.F.F2.F2.F.F.2F2.F.F$.F2.F.F.3F5.3F.F2.F6.F.F.2F
2.F.F.2F2.F4.3F.3F2.F2.3F.F.F2$20.F2.3F5.3F.2F2.3F.3F.3F.3F4.3F.3F.2F
$19.F.F.F.F5.F.F.3F2.F2.2F2.F4.F2.F2.2F2.F.F.2F$19.F.F.F.F5.3F.2F2.2F
2.3F.3F2.F2.F2.F3.3F.F.F2$5.10F5.2F.3F2.2F.3F5.3F6.F3.F.F2.F2.3F$5.F
2.3D3.F5.F2.F.F.2F2.2F6.F.F.F4.2F2.3F.F.F2.F$5.F.C.D.3AF4.2F2.3F.F3.
3F5.F.F.F5.2F.F.F.F.F2.F$5.F2C2.B.A.F$5.FC.3B2.CF4.3F2.2F5.3F.F.F.3F
5.F3.3F2.F3.2F.3F$5.F2.C2.3CF5.F3.F7.F2.3F.2F6.F3.2F2.F.F2.F3.F$5.F3C
A.D2.F4.3F.2F7.F2.F.F.3F5.3F.3F.F.F.2F3.F$5.F.D.2A3DF$5.F3D.A3.F4.2F
2.3F.3F2.2F.3F5.F4.F2.F.F5.3F.3F$5.10F4.F.F.2F2.F.F2.F2.2F6.2F2.F.F.
3F6.F2.F.F$19.2F2.3F.F.F.2F2.3F6.2F.F.F2.F7.F2.3F2$19.3F.3F.F3.F7.3F.
3F5.F3.3F.3F.F.F$19.2F3.F2.F3.F8.F3.F6.2F3.F3.F2.3F$19.F3.3F.3F.3F5.
3F2.F7.2F.3F2.F2.F.F2$19.3F5.3F2.2F.3F.3F.3F.3F2.2F$19.F.F.3F.F.F.2F
3.F2.F.F.F.F.2F3.F$19.F.F5.3F.F3.3F.F.F.3F.3F.2F2$20.2F.F.F.3F.F.F5.
3F.F.F2.F2.3F5.3F.3F$20.F2.F.F.F3.3F6.F2.3F.F.F2.F6.F.F.F.F$19.2F2.3F
.3F.F.F6.F2.F.F.F.F2.F6.F.F.3F2$20.2F.3F.2F2.3F$19.2F2.F.F.2F2.2F$19.
F3.3F.F.F.3F2$19.3F5.3F2.2F.3F.3F.3F.3F2.2F$19.F.F.3F.F.F.2F3.F2.F.F.
F.F.2F3.F$19.F.F5.3F.F3.3F.F.F.3F.3F.2F17.3F$72.F$19.3F2.F2.3F5.2F2.
3F5.2F.F4.F2.3F.3F.2F2.3F$19.F3.F.F.F.F5.3F.2F6.2F.F3.F.F.F3.2F2.F.F$
19.3F.F.F.F.F5.2F2.3F5.F2.3F.F.F.3F.3F.2F2.F6$19.3F.F.F.3F2.2F6.2F.3F
.F3.F.F.3F.3F.3F.3F$20.F2.3F2.F3.F7.F2.F.F.F3.F.F2.F3.F2.F.F.F.F$20.F
2.F.F.3F.2F6.2F2.3F.3F.3F2.F2.3F.3F.F.F2$19.3F2.2F5.3F.2F.3F.3F2.2F2.
F2.F$20.F3.F6.F.F.2F2.F3.F2.2F2.F.F.F$19.3F.2F6.3F.F3.F2.3F.F3.F.F.3F
.F6$20.F2.2F2.2F2.3F.3F.3F.3F.3F2.F2.F3.F3.F.F$19.F.F.F.F.F.F2.F3.F3.
F2.F.F.F.F.F.F.F3.F3.3F.F$19.F.F.2F2.2F2.3F2.F2.3F.3F.F.F.F.F.3F.3F2.
F2.F2$19.F3.3F5.3F.3F.F.F.3F2.2F.3F.3F.F4.F2.3F.3F.2F$19.2F2.2F7.F2.F
.F.F.F.2F3.F3.F3.F2.F.F.F.F2.F2.2F2.F.F$20.2F.3F5.3F.F.F2.F2.3F.2F3.F
2.3F.3F.F.F2.F2.3F.2F2$19.3F.3F.2F.2F6.F.3F.F.F.3F.3F.3F.F.F2.F2.F7.
3F.2F.2F.F$19.2F2.F3.2F.2F5.F2.2F2.F.F.2F2.F.F2.F2.F.F.F.F.F7.F3.2F.
2F2.F.F$19.3F.3F.F2.F7.F.3F2.F2.3F.F.F2.F2.3F.F.F.3F5.3F.F2.F2.F2.F2$
19.3F.3F5.F3.F.F.3F.3F.F.F5.F.F.3F.F.F3.3F2.F2.3F5.3F.3F.F3.F.F$20.F
2.F.F5.2F2.3F2.F2.F3.3F5.3F.F.F.F.F3.F3.F.F.F.F5.F.F.F.F.F3.3F$19.3F.
F.F6.2F.F.F.3F.3F.F.F6.F2.3F.3F3.3F.F.F.F.F5.3F.F.F.3F2.F2$19.2F.F4.F
2.3F.3F5.F5.3F5.3F2.2F.3F.3F.3F$19.2F.F3.F.F.F3.2F6.F5.F.F.3F.F.F.2F
3.F2.F.F.F.F$19.F2.3F.F.F.3F.3F5.F5.F.F5.3F.F3.3F.F.F.3F6$19.F.F.3F.
2F2.3F.F2.2F6.F7.2F.3F.F3.F.F.3F.3F.3F.3F.F$19.3F.2F2.2F2.2F2.F2.F6.F
.F6.F2.F.F.F3.F.F2.F3.F2.F.F.F.F$19.F.F.3F.F.F.3F3.2F6.F.F5.2F2.3F.3F
.3F2.F2.3F.3F.F.F.F3$18.10F$18.F2.3D3.F$18.F.C.D.3AF$18.F2C2.B.A.F$
18.FC.3B2.CF$18.F2.C2.3CF$18.F3CA.D2.F$18.F3.2A3DF$18.F4.A3.F$18.10F!
#C [[ VIEWONLY ]]
So far, for the 8x8 grid CPP with tetrominoes and 8x8 grid tetromino-ECPP have solutions (both by Blinkerspawn)
So here's a question: Suppose that E2CPP asks for that, but you can place an N-omino such that it brings you to an ECPP position, and never into any position which eventually leads to ECPP, E3CPP asks for that, but you can place an N-omino such that it brings you to an E2CPP position, and never into any position which eventually leads to E2CPP, etc.
Does anyone have solutions for E2CPP, and E3CPP with tetrominoes on an 8x8 grid?
not active here but active on discord
Re: Thread for Non-CA Academic Questions
I'm not sure this is academic, but it's non-CA, at least directly. I'll post it as a puzzle in case someone else wants to think of a solution.
We have a 4x4 grid of values (or you could generalize it, but I am interested in n=4). Assigning variable names, it looks like this:
We can make a domino tiling of this grid. I.e., we use 8 dominos to cover all the grid cell such that each domino covers two adjacent cells and dominoes never overlap (there are apparently 36 ways to do this, but that's not the question).
Suppose we make a domino tiling like this and we swap each adjacent pair. Symbolically, if you did it with all horizontal dominoes, you get this series of swaps: a↔b, c↔d, e↔f, g↔h, i↔j, k↔l, m↔n, o↔p. That's one possible permutation of the values, but each domino tiling defines a different one.
So the first question is: What permutations of these 16 values can you obtain just by applying a series of domino-tiling permutations? Can you obtain all permutations? Just, e.g, even permutations? There is no limit on the number of domino tilings you have to apply in succession but you could come up with a lower bound of log_36 16! (right?) assuming you could get every permutation.
Now I'll reveal that I'm actually only interested in cases where the values are 0 or 1. In this case, a swap of identical values doesn't change anything. But you still have to cover all the grid values at each step.
And further, I'm really only interested in whether you can start with an arbitrary set of 0s and 1s and get a set that is monotonic in rows in columns, i.e. like
And yes, this is very remotely connected to a CGOL thing I'm working on but I wonder if anyone knows the answer. (Not that I don't like puzzles but if someone else likes this one, I am really more interested in the answer than the challenge.)
We have a 4x4 grid of values (or you could generalize it, but I am interested in n=4). Assigning variable names, it looks like this:
Code: Select all
a b c d
e f g h
i j k l
m n o p
Suppose we make a domino tiling like this and we swap each adjacent pair. Symbolically, if you did it with all horizontal dominoes, you get this series of swaps: a↔b, c↔d, e↔f, g↔h, i↔j, k↔l, m↔n, o↔p. That's one possible permutation of the values, but each domino tiling defines a different one.
So the first question is: What permutations of these 16 values can you obtain just by applying a series of domino-tiling permutations? Can you obtain all permutations? Just, e.g, even permutations? There is no limit on the number of domino tilings you have to apply in succession but you could come up with a lower bound of log_36 16! (right?) assuming you could get every permutation.
Now I'll reveal that I'm actually only interested in cases where the values are 0 or 1. In this case, a swap of identical values doesn't change anything. But you still have to cover all the grid values at each step.
And further, I'm really only interested in whether you can start with an arbitrary set of 0s and 1s and get a set that is monotonic in rows in columns, i.e. like
Code: Select all
0 0 0 0
0 0 1 1
1 1 1 1
1 1 1 1
Re: Thread for Non-CA Academic Questions
If anyone else cares, I think the answer is yes within parity constraints. The main trick is to use the dominoes to move values around cyclic paths. E.g., consider these two successive domino tilings (a domino is represented by two adjacent cells with the same capital letter).pcallahan wrote: ↑January 30th, 2020, 10:22 pmSuppose we make a domino tiling like this and we swap each adjacent pair. Symbolically, if you did it with all horizontal dominoes, you get this series of swaps: a↔b, c↔d, e↔f, g↔h, i↔j, k↔l, m↔n, o↔p. That's one possible permutation of the values, but each domino tiling defines a different one.
So the first question is: What permutations of these 16 values can you obtain just by applying a series of domino-tiling permutations? Can you obtain all permutations?
Code: Select all
A A E E
D B F F
D B G G
C C H H
Code: Select all
A B E E
A B F F
D C G G
D C H H
It's still a puzzle to create a specific permutation and you are limited by parity unless I'm mistaken. E.g. let the grid be a checkerboard of white and black squares. Any value on a white square is on a black square after an odd number of steps, and back on a white square after an even number of steps.
- gameoflifemaniac
- Posts: 1242
- Joined: January 22nd, 2017, 11:17 am
- Location: There too
Re: Thread for Non-CA Academic Questions
What's the general formula for the area of a regular n-gon with side length 1? I know, not very academic question
I was so socially awkward in the past and it will haunt me for the rest of my life.
Code: Select all
b4o25bo$o29bo$b3o3b3o2bob2o2bob2o2bo3bobo$4bobo3bob2o2bob2o2bobo3bobo$
4bobo3bobo5bo5bo3bobo$o3bobo3bobo5bo6b4o$b3o3b3o2bo5bo9bobo$24b4o!
Re: Thread for Non-CA Academic Questions
You can look that up pretty easily (I checked). If you want to work it out yourself, think of the n-gon as consisting of n isosceles triangles with a base length 1 and angle 360°/n. Divide those into right triangles and use the fact that the short leg (1/2 length) to long leg ratio is tan(180°/n). The area of the isosceles triangle is the product of short and long legs, and you multiply that by n for the total area.gameoflifemaniac wrote: ↑February 11th, 2020, 4:13 pmWhat's the general formula for the area of a regular n-gon with side length 1? I know, not very academic question
Let's see, for a square, n=4, 180/4=45. Each isosceles triangle has area (1/2)(1/2) and that times 4 is 1. For a hexagon, n=6, 180/6=30. The tangent is 1/sqrt(3) I think, so total area is 6*(1/2)(sqrt(3)/2) and so on.
Re: Thread for Non-CA Academic Questions
How incredibly unstable would 1,1,3,3,5,5-hexaazido-2,2,4,4,6,6-hexanitrocyclohexane be?
Would it be more or less stable compared to 1,2,3,4,5,6-hexaazido-1,2,3,4,5,6-hexanitrocyclohexane?
And how would those compare to 1,1,2,2,3,3-hexaazido-4,4,5,5,6,6-hexanitrocyclohexane? (And am I right in thinking that these are more unstable than even, say, 1-diazidocarbamoyl-5-azidotetrazole?)
And are those standard nomenclature?
Would it be more or less stable compared to 1,2,3,4,5,6-hexaazido-1,2,3,4,5,6-hexanitrocyclohexane?
And how would those compare to 1,1,2,2,3,3-hexaazido-4,4,5,5,6,6-hexanitrocyclohexane? (And am I right in thinking that these are more unstable than even, say, 1-diazidocarbamoyl-5-azidotetrazole?)
And are those standard nomenclature?
not active here but active on discord
Re: Thread for Non-CA Academic Questions
A more serious question: how would I name the following group?
As in, like, say
1,2,3,4,5-pentaazido-6,7,8,9,10-penta(group)-1,2,3,4,5,6,7,8,9,10-decanitrodecane
I'm gonna guess it's something in parentheses, maybe (trinitromethoxy) ?
EDIT: apparently yes.
Code: Select all
NO2
|
NO2-C-NO2
|
O
|
1,2,3,4,5-pentaazido-6,7,8,9,10-penta(group)-1,2,3,4,5,6,7,8,9,10-decanitrodecane
I'm gonna guess it's something in parentheses, maybe (trinitromethoxy) ?
EDIT: apparently yes.
not active here but active on discord
-
- Posts: 2200
- Joined: August 5th, 2016, 10:27 am
- Location: 拆哪!I repeat, CHINA! (a.k.a. 种花家)
- Contact:
Re: Thread for Non-CA Academic Questions
The base of natural logarithm e has many characteristicss/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
熠熠种花 - Glimmering Garden
Harvest Moon
2-engine p45 gliderless HWSS gun
Small p2070 glider gun
Forgive me if I withhold my enthusiasm.
Harvest Moon
2-engine p45 gliderless HWSS gun
Small p2070 glider gun
Forgive me if I withhold my enthusiasm.
Re: Thread for Non-CA Academic Questions
I have no idea what I’m doing, but I’m just gonna experiment.GUYTU6J wrote: ↑February 24th, 2020, 11:45 pmThe base of natural logarithm e has many characteristicss/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
I’ll start with e2
Notice that each factorial is a factor of the next.
I’ll ignore the first 1/0! because it’s just mess.
That means you can do this:
1/1 + 1/2 + 1/6 = 6/6 + 3/6 + 1/6
If you sum up everything you get this sequence: 1/1, 3/2, 10/6, 41/24
It evaluates to this:
n -> Infinity
Σ[k=0->n-1](n^k)/n!
- toroidalet
- Posts: 1514
- Joined: August 7th, 2016, 1:48 pm
- Location: My computer
- Contact:
Re: Thread for Non-CA Academic Questions
In order to prove that en=em, you would use whichever property is used in the proof that en=e (some more complex relations may use multiple properties, in which case you would want to prove those en's to be equivalent first).
For example, to prove that e3=e4, you would find that d/dx(ln4(x))=1/x, and so by the fundamental theorem of calculus ∫(t=1 to e4)(1/t)dt=ln4(e4)-ln4(1)=1-0=1 (the e3 condition), and so e4 is the same as e3.
For example, to prove that e3=e4, you would find that d/dx(ln4(x))=1/x, and so by the fundamental theorem of calculus ∫(t=1 to e4)(1/t)dt=ln4(e4)-ln4(1)=1-0=1 (the e3 condition), and so e4 is the same as e3.
Any sufficiently advanced software is indistinguishable from malice.
-
- Posts: 2200
- Joined: August 5th, 2016, 10:27 am
- Location: 拆哪!I repeat, CHINA! (a.k.a. 种花家)
- Contact:
Re: Thread for Non-CA Academic Questions
Can we not introduce natural logarithm?
e₅ is the unique number a that a^(∫(t=1 to x)(1/t)dt) = x
e₅ is the unique number a that a^(∫(t=1 to x)(1/t)dt) = x
熠熠种花 - Glimmering Garden
Harvest Moon
2-engine p45 gliderless HWSS gun
Small p2070 glider gun
Forgive me if I withhold my enthusiasm.
Harvest Moon
2-engine p45 gliderless HWSS gun
Small p2070 glider gun
Forgive me if I withhold my enthusiasm.
- BlinkerSpawn
- Posts: 1992
- Joined: November 8th, 2014, 8:48 pm
- Location: Getting a snacker from R-Bee's
Re: Thread for Non-CA Academic Questions
e1 -> e4: This is the equation for an interest rate of 100% applied x times per unit of time. Letting the width of each compounding period become infinitesimally small, the rate of change becomes instantaneous (a derivative) while retaining its rate of 100%/period, and so the derivative is equal to the original function.GUYTU6J wrote: ↑February 24th, 2020, 11:45 pmThe base of natural logarithm e has many characteristics/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
e4 -> e2: Starting with e^0 = 1, repeated application of "equals its own derivative" provides the power series e^x = ∑(n=0 to ∞)(x^n/n!); let x=1.
e4 -> e3: Let y = ln x. Then x = e^y, and implicit differentiation yields 1 = e^y dy/dx, and dy/dx = e^-y = 1/x.
Therefore, ln x is the antiderivative of 1/x, and the needed upper bound for the integral is the unique t such that ln t - ln 1 = 1. But 1 = e^0, so ln t = 1 and t = e.
- Entity Valkyrie 2
- Posts: 1756
- Joined: February 26th, 2019, 7:13 pm
- Contact:
Re: Thread for Non-CA Academic Questions
I'm currently composing a violin concerto
Bx222 IS MY WORST ENEMY.
Please click here for my own pages.
My recent rules:
StateInvestigator 3.0
B3-kq4ej5i6ckn7e/S2-i34q6a7
B3-kq4ej5y6c/S2-i34q5e
Move the Box
Please click here for my own pages.
My recent rules:
StateInvestigator 3.0
B3-kq4ej5i6ckn7e/S2-i34q6a7
B3-kq4ej5y6c/S2-i34q5e
Move the Box
-
- Posts: 2200
- Joined: August 5th, 2016, 10:27 am
- Location: 拆哪!I repeat, CHINA! (a.k.a. 种花家)
- Contact:
Re: Thread for Non-CA Academic Questions
I've figured out a formal proof for e₁→e₁=e₄. Define a function f(x) = lim(t→∞)(1+x/t)^t, then f(1) = e₁ andBlinkerSpawn wrote: ↑February 28th, 2020, 11:57 pme1 -> e4: This is the equation for an interest rate of 100% applied x times per unit of time. Letting the width of each compounding period become infinitesimally small, the rate of change becomes instantaneous (a derivative) while retaining its rate of 100%/period, and so the derivative is equal to the original function.GUYTU6J wrote: ↑February 24th, 2020, 11:45 pmThe base of natural logarithm e has many characteristics/defnitions, e.g.
e₁=lim(x→∞)(1+1/x)^x
e₂=∑(n=0 to ∞)(1/n!)
e₃ is the unique solution x for ∫(t=1 to x)(1/t)dt = 1
e₄ is the unique number a such that f(x)=a^x is equal to its own derivative and has unit tangent slope at (0,f(0))
How to prove their equivalency, i.e. e₁=e₂=e₃=e₄?
...
f(x) = lim(t→∞)(1+x/t)^t = lim(t→∞)(1+1/(t/x))^((t/x)·x) = (lim(t/x→∞)(1+1/(t/x))^(t/x))^x = (f(1))^x = e₁^x
f(x) is equal to its own derivative because
(lim(t→∞)(1+(x/t))^t)' = lim(t→∞)(t·(1+(x/t))^(t-1)·(1/t)) = lim(t→∞)(1+(x/t))^(t-1) = lim(t→∞)(1+(x/t))^t
f(0) = 1, so the tangent slope at (0,f(0)) is indeed 1.
Therefore e₁=e₄. The remaining part about e₄→e₄=e₂ and e₄=e₃ is what you've said, and e₅ is a corollary from e₄=e₃.
---
What if we start from e₂ or e₃ and prove they are equal to e₁? (Their uniqueness is easy to see, btw)
熠熠种花 - Glimmering Garden
Harvest Moon
2-engine p45 gliderless HWSS gun
Small p2070 glider gun
Forgive me if I withhold my enthusiasm.
Harvest Moon
2-engine p45 gliderless HWSS gun
Small p2070 glider gun
Forgive me if I withhold my enthusiasm.
Re: Thread for Non-CA Academic Questions
Can someone explain the symbols \models, \prec, and V_k found in pages like these?
https://web.archive.org/web/20190727152 ... fo/Worldly
https://web.archive.org/web/20190727125 ... Reflecting
They're blocking my understanding of worldlies and stuff
https://web.archive.org/web/20190727152 ... fo/Worldly
https://web.archive.org/web/20190727125 ... Reflecting
They're blocking my understanding of worldlies and stuff
not active here but active on discord
- BlinkerSpawn
- Posts: 1992
- Joined: November 8th, 2014, 8:48 pm
- Location: Getting a snacker from R-Bee's
Re: Thread for Non-CA Academic Questions
It looks like a "model" is essentially a set large enough that logical axioms and formulae can be interpreted and manipulated as subsets of said set.
Someone better-trained in set theory than myself will have to refine that.
Someone better-trained in set theory than myself will have to refine that.