Make an assumption for this weird mathematic phenomenon. - How does one number go infinite if it's base number is only 1? (Mathematics Sperging).

[Alex Hogendorp]

Clearly no matter how many times you exponent the number 1 it will remain at 1.

This is what happens when you plot it on a grid using X.

 

[Shidoen]

It looks like a staircase that crashed flat

 

[Alex Hogendorp]

It looks like a staircase that crashed flat

I guess if you exponent the number 1 too much, the answer flatlines.

 

[The Last Stand]

A number multiplied by 1 will result in the same number.

 

[Ughubughughughughughghlug]

The exponent is saying "X times itself for Y times." 1 times 1 is just 1. Why? Well, times is the same as saying add up this many of the same thing. 3*5, for example, could be 3 + 3 + 3 + 3 + 3, or 5 + 5 + 5. Well, if it's 1 and it's once, then that's just 1. Well, say you do the multiplication several times, like 1^5. Well, that's the same as (1 * 1)^4 = 1^3 = (1 * 1)^2 = 1^2 = (1 * 1) = 1, right?

It's worth going through the basic properties of arithmetic, including logarithms and exponents, sometime. I never really appreciated them until halfway through a Math major when I just sat down and did it one day. Before that point I just did it without thinking, just a rule, but when you walk through exactly what it's saying, it pretty much all goes back to just addition (the most basic mathematical operation) in some way.


Edit: The logarithm, BTW, is the "inverse" of exponentiation. Inverse just meaning that if you do something to X to make it Y (f(X) = Y), then doing the inverse to Y will make it back into X. Logarithms "undo" exponents.

 

[The Last Stand]

Here's a better math riddle.

Explain the concept of x^0 = 1. (Why every non zero number raised to the power of zero equals to 1.)

 

[Alex Hogendorp]

Here's a better math riddle.

Explain the concept of x^0 = 1. (Why every non zero number raised to the power of zero equals to 1.)

1 is actually the base in exponents. So anything exponent 0 just gets redirected all the way to 1 since it has nothing to multiply. Think of it like a sequence. 0 is at the very start so it would always equal 1 even if that number is 0. Anyone who knows number theory and algebra would be able to explain it well.
0 is also the only number in the exponent to bring a 0 to 1. All other numbers don't however. Positive integers don't add anything and negative integers bring up as undefined. This is also why negative exponents make fractions.

I double checked and the reason why 1^2^3^4^5 shows up as undefined has to do with the line in the graph failing to go past 1 in the positive x-axis.

 

[The Last Stand]

1 is actually the base in exponents. So anything exponent 0 just gets redirected all the way to 1 since it has nothing to multiply. Think of it like a sequence. 0 is at the very start so it would always equal 1. Anyone who knows number theory and algebra would be able to explain it well.

From this source, I see your reasoning.



It's basically a long winded expression of dividing a number by itself.

 

[Just A Fat Round Bird]

Clearly no matter how many times you exponent the number 1 it will remain at 1.

This is what happens when you plot it on a grid using X.

My guess to what is happening in these three pictures regardless of the mathematics is that this here is specifically a technical issue producing the undefined in case 2. Depending on implementation you may have an expression tree with substitution or similar or something that can hande case three due to the repeated pattern (or perhaps even lazy evaluation, the fuck do I know how or with what it was implemented with). Another potential thing could be that the second produces an bound overstepping value or something, throwing an exception that leads to a default undefined. So it could really be a sideeffect or badly written code for all we know.
On the note of infinite things in computers though. Haskell and other functional languages allow you to do things like define the set of all natural numbers pretty easily and do operations on them as if they are finite. e.g.: you could say nat = [1..], and then odd=2*nat+1 and youd have all odd numbers..(of course would need adjustment to actual syntax) .. and of course you could also just filter nat or use a fold or whatever, there are many ways..
Anyway, the reason why this works is that a stream is defined by build rules or unevaluated expression trees over a building rule and due to lazy evaluation is only evaluated up to the point that is really needed (and usually only if you want to display/print/write the result). Its quite neat so I thought Id share.

 

[SteinserGate]

Sorry for the necro, but it is because Desmos uses "IEEE-754 double-precision floating point numbers", which are 64 bits in size. I found this on some wiki, and it checks out.
I'm a math guy but I'd rather search the internet rather than solve for mystery problem.

53 are used for mantissa (52 for data, 1 for the sign) and the other 11 are used for exponents.
The highest number the significand, mantissa, whatever the fuck you wanna call it, can represent is just under 1, but not 1.
The 11-bit signed integer can only hold 2^11 values, or 2048. If you noticed, it's even. One value has to go to 0, so the rest are left with 2047 values. Odd.
So, the range for positive and negative numbers have to range between -1024 to 1023.

Why?
It uses two's complement method, using the binary digit with the greatest value as the sign (what determines positive/negative) which is 1024.
The biggest binary digit of an 11-bit signed integer would be 1024, represented as 10000000000 in base 2.
A walk through the two's complements see that the absolute binary representation of 1024 is replaced by -1024 in the two's complement method.
Let's say we were solving for -1024 using this method. You would start with its absolute binary representation, 10000000000.
You'd flip all bits, this gets you 01111111111. Then we'd add 1 to the number, which gives us 10000000000.
Obviously, you can't have -1024 = 1024, so 1024 loses out in the end.

You'll find that 2^3^4^5 = 1024. 1024 is not a value that the 11-bit can calculate based on the system in place.
The highest number you can find would be .999(shitload of 9s later) * 1^1023. So since 1^1024 is outside the bounds of the system, it is undefined.
Also, the first and third pictures are the same.