Now in an intuitive sense, this is obviously what the answer has to be, of course.
If you stack up a bunch of one-dimensional lines of length 1, 2, 3, ... 10, you end up with roughly half of a two-dimensional square of side 10.
If you stack up a bunch of two-dimensional squares each with sides 1, 2, 3, ..., 10, you end up with roughly a third of a three-dimensional cube of side 10.
So naturally if you stack up a bunch of N-dimensional hypercubes with sides 1, 2, 3, ... infinity, you'll end up with 1/(N+1) of an (N+1)-dimensional hypercube with side infinity.
Now in an intuitive sense, this is obviously what the answer has to be, of course.
If you stack up a bunch of one-dimensional lines of length 1, 2, 3, ... 10, you end up with roughly half of a two-dimensional square of side 10.
If you stack up a bunch of two-dimensional squares each with sides 1, 2, 3, ..., 10, you end up with roughly a third of a three-dimensional cube of side 10.
So naturally if you stack up a bunch of N-dimensional hypercubes with sides 1, 2, 3, ... infinity, you'll end up with exactly 1/(N+1) of an (N+1)-dimensional hypercube with side infinity.
Now in an intuitive sense, INF^(N+1) / (N+1) is obviously what the answer has to be, of course.
If you stack up a bunch of one-dimensional lines of length 1, 2, 3, ... 10, you end up with roughly half of a two-dimensional square of side 10.
If you stack up a bunch of two-dimensional squares each with sides 1, 2, 3, ..., 10, you end up with roughly a third of a three-dimensional cube of side 10.
So naturally if you stack up a bunch of N-dimensional hypercubes with sides 1, 2, 3, ... infinity, you'll end up with exactly 1/(N+1) of an (N+1)-dimensional hypercube with side infinity.
Now in an intuitive sense, INF^(N+1) / (N+1) is obviously what the answer has to be, of course.
If you stack up a bunch of one-dimensional lines of length 1, 2, 3, ... 10, you end up with roughly half of a two-dimensional square of side 10.
If you stack up a bunch of two-dimensional squares each with sides 1, 2, 3, ..., 10, you end up with roughly a third of a three-dimensional cube of side 10.
So naturally if you stack up a bunch of N-dimensional hypercubes with sides 1, 2, 3, ... infinity, you'll end up with exactly 1/(N+1) of an (N+1)-dimensional hypercube with side infinity.
(Parenthetical class chatter)This nerdsnipes multiple researchers hard enough that they miss the next several words, unfortunately for them.
Now in an intuitive sense, INF^(N+1) / (N+1) is obviously what the answer has to be, of course.
If you stack up a bunch of one-dimensional lines of length 1, 2, 3, ... 10, you end up with roughly half of a two-dimensional square of side 10.
If you stack up a bunch of two-dimensional squares each with sides 1, 2, 3, ..., 10, you end up with roughly a third of a three-dimensional cube of side 10.
So naturally if you stack up a bunch of N-dimensional hypercubes with sides 1, 2, 3, ... infinity, you'll end up with exactly 1/(N+1) of an (N+1)-dimensional hypercube with side infinity.
(Parenthetical class chatter)
This nerdsnipes multiple researchers hard enough that they have trouble following the next few sentences, which thankfully aren't vital ones.
Now in an intuitive sense, INF^(N+1) / (N+1) is obviously what the answer has to be, of course.
If you stack up a bunch of one-dimensional lines of length 1, 2, 3, ... 10, you end up with roughly half of a two-dimensional square of side 10.
If you stack up a bunch of two-dimensional squares each with sides 1, 2, 3, ..., 10, you end up with roughly a third of a three-dimensional cube of side 10.
So naturally if you stack up a bunch of N-dimensional hypercubes with sides 1, 2, 3, ... infinity, you'll end up with exactly 1/(N+1) of an (N+1)-dimensional hypercube with side infinity.
(Retroactive parenthetical class chatter)
(This nerdsnipes multiple researchers hard enough that they have trouble following the next few sentences, which thankfully aren't vital ones.)
