Hacker Newsnew | past | comments | ask | show | jobs | submitlogin

I think it's important to know that this really isn't true. Maths at all times is a subjective language. Maths notation is imprecise "intentionally confused", or just ad-hoc defined all the time.

When you see something like.

    f(x) = summation(x^n, n=0, 10)
We conveniently ignore that this polynomial is defined at x=0 despite 0^0 not making any sense by ad-hoc defining 0^0 = 1 in this context.


> We conveniently ignore that this polynomial is defined at x=0 despite 0^0 not making any sense by ad-hoc defining 0^0 = 0 in this context.

What do you think the polynomial is? I ask, because in all situations similar to this that I've encountered it made sense to define 0^0 as 1, not as 0. If you genuinely have a case where 0^0 = 0 makes consistent sense then I'd be interested in understanding it.

So, what do you think the summation actually is when expanded?


Sorry you're right, it should be 0^0 = 1.

An example where 0^0 = 0 occurs when dealing with areas. The measure of the real line in the plane is zero but it's also a rectangle with sides (0, inf) and we define the area of a rectangle to be l*w.

You usually see this written as inf × 0 = 0 but you sometimes you see the interpretation as 1/0 × 0 = 0^0 = 0. And you know this is a an ad-hoc definition because you're not allowed to algebraically manipulate it at all.


It makes perfect sense, everyone knows:

0^0 = 0

1^0 = 1

0^1 = 1

1^1 = 0


That's just lazy, and I haven't ever read a good text use summation where any of the terms would be outright undefined if evaluated.


You've never seen a textbook present the series form of e^x?

    exp(x) = summation(x^n/n!, n=0, inf)
exp(0) = 1 is perfectly valid but the identity above only holds for x=0 only if we define 0^0 = 1. And we do!

Fair, infinite series are pretty esoteric. How about derivatives. The power rule:

    d/dx x^n = n x^(n-1)
This identity doesn't hold for n = 1 and x = 0 unless 0^0 = 1.

Eh maybe not, programmers don't use calculus that often. But surely statistics!

    (1 + x)^n = summation((n choose k)x^k, k=0, n)
Take x = 0 and n = 0

    (1 + x)^n = (1 + 0)^0 = 1 = 0^0 = x^0 = x^0 + ... + x^n
0^0 is undefined in general but locally we sometimes need to define it.


Ok, fair, I think I have seen those. Personally I would prefer noting the special case even when using the fairly standard degenerate case of 0^0=1, but I agree that a lot of people don't.


actually 0^0 is usually chosen equal to 1... which probably still supports your argument.


Corrected! Thank you!


What isn't really true?




Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: