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’m not sure this reasoning holds. We’re talking about 0, and 0^z with z<0 is division by zero.
I do think it makes sense for it to be 1 in some contexts.
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.