0^0
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11-24-2023, 05:43 PM
(This post was last modified: 11-25-2023 06:13 PM by Albert Chan.)
Post: #1
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0^0
(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: 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 |
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11-24-2023, 05:56 PM
Post: #2
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RE: 0^0
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. 41CV†,42S,48GX,49G,DM42,DM41X,17BII,15CE,DM15L,12C,16CE |
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11-24-2023, 08:14 PM
Post: #3
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RE: 0^0
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 |
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11-24-2023, 09:41 PM
Post: #4
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RE: 0^0
(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 |
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11-25-2023, 12:47 PM
(This post was last modified: 11-25-2023 05:16 PM by Albert Chan.)
Post: #5
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RE: 0^0
(11-24-2023 09:41 PM)Albert Chan Wrote: Limit depends on relative rate of base and exponent shrink to 0 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. |
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