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Acording to Mathmetica which I consider the last word on maths 0^0 is inditerminate

Ok. They don't change the result much.

Quote from: alancalverd on 18/03/2022 16:04:31Quote from: hamdani yusuf on 18/03/2022 10:46:211 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.Oh yes it does! (1 * Δ)^{2} = Δ^{2} ≠ 1(1+Δ) ^{2} = 1 + 2Δ + Δ^{2} ≠ 1.Ok. They don't change the result much.

Quote from: hamdani yusuf on 18/03/2022 10:46:211 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.Oh yes it does! (1 * Δ)^{2} = Δ^{2} ≠ 1(1+Δ) ^{2} = 1 + 2Δ + Δ^{2} ≠ 1.

1 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.

Squeeze the trigger slowly and you'll appreciate the enormous difference between "not much" and "unstable".

It isn't. Only you suggested it was.

You mentioned that 0 x 0 was very stable but that depends on how you measure stability.

0 * 0 is extremely stable. The operands can be perturbed heavily without changing the result (e.g. by multiplying them with some constants). Adding one of the operands with a finite constant doesn't change the result. Adding both operands with a finite constant does change the result.

Anyway, using this notion of proportional difference, 0 x 0 and also 0 + 0 isn't stable, it's terrifyingly unstable. Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded. No other choice of input numbers is that bad.

A simple example of a trigger instability is the equation y = 1/x. As x passes through zero its value switches between +∞ and -∞.

There are seven indeterminate forms involving 0, 1, and infinity:https://mathworld.wolfram.com/Indeterminate.htmlIf complex infinity is allowed as well, then six additional indeterminate forms result:

Anyway, using this notion of proportional difference, 0 x 0 and also 0 + 0 isn't stable, it's terrifyingly unstable. Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded. No other choice of input numbers is that bad.Can you show how?

0 + 0 = 0 = the expected answer for 0 + 0.Perturb the input values slightly and consider δ + ε where δ, ε are arbitrarily small but positive.Then δ + ε ≠ 0, instead δ + ε = something small but positive. The percentage change from the expected answer, 0, is then undefined: (something positive) / 0 x 100% - the division by 0 is a problem.Best Wishes.

Perturbation by multiplication already implies ratio. Percentage change simply converts perturbation from addition to multiplication.

0^0 is highly debated in the mathematical community. Why is 0^0 undefined? Does 0^0=0? Does 0^0=1? In this video I'll address the 0^0 meaning and give you a solid 0^0 proof. SPOILER ALERT: 0^0=1 proof can be found just using the definition of how we define exponents.

Timeline---0:00 Opening Quote from Donald Knuth0:18 The Integer Exponentiation Context2:36 The Discrete Math Argument5:57 Examples from Polynomials7:55 The Abstract Algebra Argument14:00 Indeterminate Form and The History of 0^0: Team Euler vs. Team Cauchy17:00 Handling of 0^0 in Programming and Softwares18:12 Ending words