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August 5 edit

Are there theorems of number theory that have so far needed Calculus for being proven? edit

HOTmag (talk) 14:44, 5 August 2016 (UTC)[reply]

Well it isn't exactly easy to even formulate the Prime number theorem without using the logarithm function. Category:Theorems in analytic number theory gives a few more like that. Dmcq (talk) 15:13, 5 August 2016 (UTC)[reply]

Well, I suspect Prime number theorem is not purely number-theoretical, because its very formulation involves terms (like the natural logarithm as you've already pointed out) which are not recognized by Peano System. Anyways, thank you for two theorems, which are purely number-theoretical: Chen's theorem, and Friedlander–Iwaniec theorem. Please notice that all of the other theorems are not purely number-theoretical. HOTmag (talk) 15:23, 5 August 2016 (UTC)[reply]

See also Waring's problem#The number G(k), in which some logs appear in the results even though they are not part of the problem statement. Loraof (talk) 15:54, 5 August 2016 (UTC)[reply]

The whole subject of analytic number theory is the application of mathematical analysis to number theory. My impression is that it's the bulk of number theory these days. Probably most recent results in number theory are answers to your question.

One very famous example is Fermat's last theorem, for which the only known proofs use deep methods from analysis. --Trovatore (talk) 19:48, 6 August 2016 (UTC)[reply]

Oh, thank you for reminding me this important theorem. Now I wonder if it's provable in Peano system, because if it's not, then it can be used as a much simpler example (than Goodstein's Theorem) for theorems - which are unprovable in Peano system - even though they are formulated in Peano language. HOTmag (talk) 22:26, 6 August 2016 (UTC)[reply]

I think that's totally open as of now. Unless I haven't heard about it (which is totally possible; I'm not all that looped in these days), no one has reported either a proof in Peano arithmetic or a reason there shouldn't be one. A very plausible state of affairs is that the analytic proofs can be converted by sufficiently careful technical attention into a PA proof, but that it would be so tedious that no one would ever actually do it. I don't know whether anyone has made a convincing argument that it ought (or ought not) to be possible. --Trovatore (talk) 22:35, 6 August 2016 (UTC)[reply]

Thx. Now I also wonder, if this theorem can really be formulated in Peano language. How would you formulate the term ∃(a,b,c)(a^b=c) in Peano language? HOTmag (talk) 22:45, 6 August 2016 (UTC)[reply]

It can be done. It requires some technical tricks. You need to code up the sequence a, a^2, ..., a^b. I think you can do this using the Chinese remainder theorem. --Trovatore (talk) 22:49, 6 August 2016 (UTC)[reply]

The information you've supplied is really a bit sketchy, but I'm sure you've made an effort to give me all you've had, so I thank you anyway. HOTmag (talk) 23:24, 6 August 2016 (UTC)[reply]

Well, no, I could have thought about it more and supplied more details, but that would have taken more time. If you're interested in that sort of detail, it's better to work it out for yourself anyway. --Trovatore (talk) 00:26, 7 August 2016 (UTC)[reply]

As for the Prime Number Theorem, it turned out to be "purely number-theoretical". See our article, the section on Elementary Proofs and the refs. It's provable in a somewhat weaker system than Peano & later verified by automatic methods even. The Erdos-Selberg "elementary" (in a vaguer mathematical sense meaning no complex analysis) proof was a key step.John Z (talk) 02:14, 13 August 2016 (UTC)[reply]

calculating the average odds? edit

Hello, this is not homework, just a problem that has been bothering me. Say the chance of a hurricane (in the next year) is 10%, the chance of an earthquake is 20% and the chance of a blizzard is 30%. I want to know the odds that only one of these things is going to happen.

I know how to do that in a very "sisyphean" way: [(1-0.1)*(1-0.2)*0.3]+[(1-0.1)*0.2*(1-0.3)]+[0.1*(1-0.2)*(1-0.3)]=0.398 so that's 39.8%.

The problem is this way is not practical if you have 20 different factors instead of 3.

Do you know a faster way to solve this sort of problem? maybe some equation that takes the sum of the odds, or the average odds, or the product of the odds, or the number of different factors?

Carl — Preceding unsigned comment added by 85.64.207.154 (talk) 23:53, 5 August 2016 (UTC)[reply]

Start by computing the chance that none of them happen: P = (1-0.1) × (1-0.2) × (1-0.3).

Then it's easier to sum the chance of each one happening without any of the others:

P × 0.1/(1-0.1) + P × 0.2/(1-0.2) + P × 0.3/(1-0.3) = P × (0.1/0.9 + 0.2/0.8 + 0.3/0.7)

PrimeHunter (talk) 00:14, 6 August 2016 (UTC)[reply]

Thank you, it works great.

Do you happen to know where I can find an explanation as to why or how it works? 85.64.207.154 (talk) 00:31, 6 August 2016 (UTC)[reply]

It's just a rewriting of your own calculation (I swapped the order). P × 0.1/(1-0.1) is your 0.1*(1-0.2)*(1-0.3), while P × 0.2/(1-0.2) is your (1-0.1)*0.2*(1-0.3), and P × 0.3/(1-0.3) is your (1-0.1)*(1-0.2)*0.3. PrimeHunter (talk) 01:07, 6 August 2016 (UTC)[reply]

To clarify even further: Suppose you have 5 events, and let