however, 0^0 isn’t 0x0, that would be 0^2. 0^1 is 0x1, anything^0 is… well, it’s 1. afaik there isn’t am equivalent mathematical expression to n^0, it’s multiplying a number by itself -1x, or something equally mind melting.
actually, I thought of a (maybe) helpful way to visualise this.
x^-n is equivalent to 1÷(x^n), so 10^-1 is one tenth, 10^-2 is one hundredth, so on. the number, x, appears in the equation n times.
you can view positive exponents as the inverse, (x^n)÷1. likewise, the number appears n times.
so what happens for x^0? well, zero is neither positive nor negative. and to maintain consistency, x must appear in the equation zero times. so what you’re left with is 1÷1, regardless of what number you input as x.
I’d imagine you want something defined recursively like multiplication
( 0x = 0 )
( xy = x(y-1)+ x ) ( y > 0 ).
So it needs to be
( x^0 = c ) (c is some constant)
( x^y = xx^{y-1} ) (( y > 0 ) (to see why, replace multiplication with exponentiation and addition with multiplication).
So what could ( c ) be? Well, the recursive exponentiation definition we want refers to ( x^0 ) in ( x^1 ). ( x^1 ) must be ( x ) by the thing we wish to capture in the formalism (multiplication repeated a single time). So the proposed formalism has ( x = x^1 = xx^0 = xc ). So ( cx = x ) hence ( c = 1 ), the multiplicative identity. Anything else would leave exponentiation to a zeroth power undefined, require a special case for a zeroth power and make the base definition that of ( x^1 ), or violate the intuition that exponentiation is repeated multiplication.
On an unrelated note, it’d be nice if Lemmy had Mathjax. I just wrote all this on mobile with that assumption, and I’m not rewriting now that I know better.
it makes graphs look nicer.
however, 0^0 isn’t 0x0, that would be 0^2. 0^1 is 0x1, anything^0 is… well, it’s 1. afaik there isn’t am equivalent mathematical expression to n^0, it’s multiplying a number by itself -1x, or something equally mind melting.
actually, I thought of a (maybe) helpful way to visualise this.
x^-n is equivalent to 1÷(x^n), so 10^-1 is one tenth, 10^-2 is one hundredth, so on. the number, x, appears in the equation n times.
you can view positive exponents as the inverse, (x^n)÷1. likewise, the number appears n times.
so what happens for x^0? well, zero is neither positive nor negative. and to maintain consistency, x must appear in the equation zero times. so what you’re left with is 1÷1, regardless of what number you input as x.
I’d imagine you want something defined recursively like multiplication
So it needs to be
On an unrelated note, it’d be nice if Lemmy had Mathjax. I just wrote all this on mobile with that assumption, and I’m not rewriting now that I know better.
Indeed, my mind is melted.