Because i keep tugging on people’s sleeves
By tha5 I mean I I keep saying Hello?
you mean how you kept posting 3 posts in a row
Well that is a trick I know
Because
HahahaH
well its nice that your back
Thank you
Do you know how to do isometric
i dont know much geometry
Okay do you know how to use pen and really complex stuff
ive learned a little of the pen
and i can figure out how to do stuff with basic blocks
but im not able to do super complex stuff
Okay can you build me a isometric cube and make it duplicate it self left forever and make it them look good
i think i can do that
ill send a link
here's the link:
i may or may not have just used a picture of a isometric cube
It drives me crazy that in all those videos they call 3x+1 an "equation." Equations have equal signs! f(x)=3x+1 is an equation.
In this problem, we're not looking for one number, but rather a sequence of numbers, generated by calling this function:
$$$f(x) = \lbrace{x/2, {\rm if}\ x\ {\rm even}\over 3x+1, {\rm if}\ x\ {\rm odd}}$$$
repeatedly, using the value of $$f(x)$$ from one iteration as the value of $$x$$ in the next iteration. The claim is that for any starting value, eventually you get to a repeating 4, 2, 1:
This claim does not hold for other functions. For example, 3x+3 gets to a different repeating subsequence:
3x+5 doesn't seem to get to a repeating subsequence in the first 40 iterations:
So I maximized the screen to fit as many values in the picture as I could:
At the bottom of the picture, steps 47–49 have the values 38, 19, 62, and so do steps 39–41. If you run this yourself, you'll see that this is indeed an eight-step cycle.
And I can't get "result pic" to make a full-width display, but here's a runnable 4x+3:
The values get bigger and bigger until step 25, then get smaller and smaller until steps 74–76, with values 44, 22, 11, so you might think it would then repeat, multiplying the values by 11, but that's not what happens. Instead, it grows to step 101, then back down to steps 151–153, with values 12, 6, 3. Aha, that's the loop for 3x+3, so it'll loop this time too, right? No, it starts growing again, until step 179, whose value is 18,014,398,509,481,984. Then it shrinks until 231–233, whose values are 4, 2, 1. So now it has to loop forever, right? Nope, it starts growing again, until step 260, whose value is 36,028,797,018,963,970. But, what's this? The value at step 261 is 18,014,398,509,481,984, same as step 179. Is this a cycle? Yes, it seems to be. The value at step 261+(261−179)=343 is indeed our friend 18...984. The length of the cycle is 82. Step 343+82=425 has the same value. The first 178 values aren't repeated, so step 179 is the beginning of this function's cyclic behavior.
you mean x/2 or 0,5x ? or am i just dumb
Yeah, I'm the dumb one. I'll fix it.
Hi
3x+1 can be solved like this. I hate myself for writing this ALL…
The 3x+1 problem asks the following: Suppose we start with a positive integer, and if it is odd then multiply it by 3 and add 1, and if it is even, divide it by 2. Then repeat this process as long as you can. Do you eventually reach the integer 1, no matter what you started with?
For instance, starting with 5, it is odd, so we apply 3x+1. We get 16, which is even, so we divide by 2. We get 8, and then 4, and then 2, and then 1. So yes, in this case, we eventually end up at 1.
That’s it! It’s an addictive and fun problem but it is frustratingly difficult to solve completely.
So far, it is known that all numbers less than 20⋅258, or about 5.8 billion billion, eventually reach 1 when this process is applied.
But what about the general case?
Let’s define the Collatz function T on the positive integers by:
T(x)={3x+12x2x is oddx is even.
Then the conjecture states that the sequence x,T(x),T(T(x)),T(T(T(x))),… has 1as a term. Notice that T(1)=2 and T(2)=1, so 1 is a cyclic point of T.
We can draw a graph on the positive integers in which we connect x to T(x) with an arrow for all x, and color it red if x is odd and black if x is even.