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Old 2022-01-31, 01:56   #12
a1call
 
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The historic justification in the Wikipedia article states:

0^0 = (a-a)^(n-n) = (a-a)^n/(a-a)^n = 1

(a-a)^n is 0 for all real numbers a and defined values n other than 0. This makes the justification based on divide-by-0 which is undefined. Not sure how this can be considered a valid justification. Corrections/insights are appreciated.

Last fiddled with by a1call on 2022-01-31 at 01:57
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Old 2022-01-31, 14:30   #13
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Quote:
Originally Posted by a1call View Post
The historic justification in the Wikipedia article states:

0^0 = (a-a)^(n-n) = (a-a)^n/(a-a)^n = 1

(a-a)^n is 0 for all real numbers a and defined values n other than 0. This makes the justification based on divide-by-0 which is undefined. Not sure how this can be considered a valid justification. Corrections/insights are appreciated.
00 \lim_{x\rightarrow 0+}\(e^{\frac{1}{x}}\)^{x^{2}}\;=\;1

00 \lim_{x\rightarrow 0+}\(e^{\frac{1}{x}}\)^{x}\;=\;e

00 \lim_{x\rightarrow 0+}\(e^{\frac{1}{x}}\)^{\sqrt{x}}\;=\;\infty

Last fiddled with by Dr Sardonicus on 2022-01-31 at 14:34 Reason: gifnix optsy
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Old 2022-01-31, 16:23   #14
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For almost any real a, a0=1. Why should it be different for a=0?
Limit as a->0 from above of aa is?

Last fiddled with by kriesel on 2022-01-31 at 16:24
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Old 2022-01-31, 16:52   #15
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Quote:
Originally Posted by kriesel View Post
For almost any real a, a0=1. Why should it be different for a=0?
Limit as a->0 from above of aa is?
For almost any real a, 0a=0. Why should it be any different for a=0?

For 00 you can argue that any results is an answer, same for 0/0.

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Old 2022-01-31, 17:10   #16
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I also draw your attention to: https://en.wikipedia.org/wiki/Zero_t...rent_situation

Last fiddled with by paulunderwood on 2022-01-31 at 17:18
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Old 2022-01-31, 19:14   #17
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Quote:
Originally Posted by Dr Sardonicus View Post

Your f(x) is 2 for positive integer x. That's prime, no question.
And large. Let's not forget large. An average of 2 daily lottery jackpots is pretty large I think.

On a bit more serious note, I have a question. If I were to change the formula to:

f(x) = 1ln(-x) + 0x!

Would it be correct to say that the above formula also generates only prime numbers for its whole domain?

(Like vacuous truth?)
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Old 2022-01-31, 20:05   #18
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it is common knowledge that \lim_{n \rightarrow \infty}\frac{sin(n+1)\phi}{sin(n\phi)}=e^{i\phi}

its joke, on 50/50 percent)
If every prime number can be associated with a "prime angle" i.e. tg(x) = 7, What are the unique properties of these "prime angles" in the light of their respected prime counterparts?))
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Old 2022-02-01, 13:02   #19
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Default An Alternative Prime Numbers Formula

Dear Colleague!


What are your views on the following mathematical formula:
C^4 + 3*C^2 + 1
It only contains addition and multiplication, so it's easy to count (exponentiation can be traced back to a series of multiplications).


This formula, when the value of "C" varies from 1 to 10,000, produces 1148 prime numbers.


The first 30 are as follows (C <= 100):

5 29 109 701 2549 4289 10301 21169 84389 161201 281429 812701 1051649 1189189 4106701 5315329 7898909 11326589 14787869 20164589 21395249 24024701 31657501 35170829 37033309 40979201 57312469 65634301 88557509 100030001
 
Old 2022-02-01, 17:30   #20
Batalov
 
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Quote:
Originally Posted by Technical View Post
This formula, when the value of "C" varies from 1 to 10,000, produces 1148 prime numbers.
Then it is surely not a formula for prime numbers.
At best, it is a series that happens to have "some" prime numbers in it.

Not interesting at all. In comparison, formula x2-x+41 produces 40 primes for the first 40 values of x, 1..40.

P.S. Creating clones for the banned user is not allowed. You have already created two: Mikloska and Benkoe. Stop doing that or you will be banned by IP.
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