0^0

[Albert Chan]

(11-21-2023 08:30 PM)Johnh Wrote:  On the question of what 0^0 equals, obviously it's a conundrum, but I'm sure proper mathematicians are clear on what it should be. But here's my take, and the answer is 1:

If you say that 0^0 should be the limit of x^x as x approaches zero, then you can easily see the result heading towards 1 ie:

0.1^0.1 = 0.79433
0.01^0.01 = 0.95499
0.001^0.001 = 0.99312
0.0001^0.0001 = 0.99908
0.00001^0.00001 = 0.99988 etc

And barring a typo above, then I think all calculators will agree if considered that way.

It may be OT to post to original thread, HP 48GX Collector’s Edition, so I start a new one.

You can make 0^0 be anything you want!
There are many ways to make this happen, below setup is probably simplest.

ln(x) / ln(x) = 1
ln(x) * (ln(a)/ln(x)) = ln(a)

Exponentiate both side: x ^ (ln(a)/ln(x)) = a

Note that this is an identity! No need to take limit to check.

(x → 0+) --> (ln(x) → -∞) --> (1/ln(x) → 0-)
If ln(a) is finite, we have both base and exponent approaches 0 (at different rate) --> 0^0 = a

[Werner]

What I have always learned:
- 0^0 is undefined. period. So, that is what the 49 and up return in exact mode: '?'.
- however - because the limit of f(x)^g(x) of x->0 is 1. for a whole class of functions f and g, AND for historical compatibility (and the binomial theorem etc), 0.^0. (approx mode) is 1.
So, you have it both ways ;-)

Werner
PS. Remark that limits are not values..
sin(x)/x is undefined for x=0, but its limit for x->0. is 1. This is a similar problem, really.

[klesl]

interesting videos
https://www.youtube.com/watch?v=X65LEl7GFOw
https://www.youtube.com/watch?v=r0_mi8ngNnM
I agree with Werner, 0^0 is undefined with limit equals to 1 as x approaches to 0

[Albert Chan]

(11-24-2023 08:14 PM)klesl Wrote:  I agree with Werner, 0^0 is undefined with limit equals to 1 as x approaches to 0

0^0 is undefined because there is *no* 1 value for limit, we can make it to anything.
What Werner is saying is many f^g cases (f→0, g→0), the limit is 1
For convenience, we may define 0^0=1, but this is not the truth.

Limit depends on relative rate of base and exponent shrink to 0
For example, we have identity: ε ^ (1/ln(ε)) = e

Make exponent shrink to zero a tiny bit slower, we get 0^0=0
Cas> limit(ε ^ (1/(-ln(ε))^0.99999), ε, 0, 1)      → 0

Or course, if exponent shrink slower still, we too get 0^0=0
Cas> limit(ε ^ (1/ln(-ln(ε))), ε, 0, 1)                  → 0

Let x=1/ε, we get exactly the exponent of blackpenredpen video
Cas> limit((1/x) ^ (1/ln(ln(x))), x, ∞)          → 0

Let's confirm without Cas. Let limit = a, y = ln(ln(x)), (y→∞) too
ln(a) = -e^y / y --> L'Hopital's Rule --> -e^y / 1 = -∞
a = exp(-∞) = 0      QED





We are now ready to prove blackpenredpen video limit, as x → ∞

(√(x+1) - √x) = 1/(√(x+1) + √x) ≈ (1/2) * (1/√x)      // 1 is nothing, compare against ∞

  (1/2) ^ (1/ln(ln(x))) ≈ (1/2) ^ 0 = 1
(1/√x) ^ (1/ln(ln(x))) = √((1/x) ^ (1/ln(ln(x)))) ≈ √(0) = 0

limit((√(x+1) - √x) ^ (1/ln(ln(x))), x=∞) = 1 * 0 = 0

[Albert Chan]

(11-24-2023 09:41 PM)Albert Chan Wrote:  Limit depends on relative rate of base and exponent shrink to 0
For example, we have identity: ε ^ (1/ln(ε)) = e

If absolute of exponent shrink slower, we get 0^0=0
Cas> limit(ε ^ (1/(-ln(ε))^0.99999), ε, 0, 1)      → 0

If absolute of exponent shrink faster, we get 0^0=1
Cas> limit(ε ^ (1/(-ln(ε))^1.00001), ε, 0, 1)      → 1

Cas> assume(p > 0)
Cas> limit(ε ^ (ε^p), ε, 0, 1)      → 1

ε^0, ε^√ε, ε^(ε*(polynomial of ε)) ... all have 0^0=1
This explained why define 0^0=1 is so useful, even though it is not the truth.