Let's say you have a computer program that counts natural

>>8338590
Actually, -1/12 means you got the result in the past

>>8338590
Sorry op, I actually ran the program and got it all printing a fraction of a second before I started it.

>>8338590
clearly this is incorrect, because there is no computer that can count up to infinity, so there is no computer that can execute such program.

but could imagine a settings where gods solve problem that require infinite time to calculate with this trick of making the computer wait k seconds each operation, it would be pretty rad actually

Since numbers are stored in memory spaces of n bytes, your program would eventually only print the number 111111... (n times) after 11111... seconds

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>>8338626
>>8338634
You [math]\textbf{can}[/math] make a computer to count to infinity, and you only need a constant amount of memory for that, too. Just make it print some symbol after another, and interpret it as a natural number in unary.

>>8338684
Ok but then if would print - 1/12, because the series of printed values 1+1+1+1..... is less than the series of second spent waiting 1+2+3+...., but the series 1+1+1+1 is greater than - 1/12, therefore

-1/12<=1+1+1+....<=1+2+3+4+......=-1/12

So 1+1+1+1....=-1/12

It does not matter how many times you call your print function, or what is the argument, your program will always have - 1/12 as output

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>>8338744
>because the series of printed values 1+1+1+1..... is less than the series of second spent waiting 1+2+3+....
You misunderstood. If you have [math]k[/math] symbols on the screen, you wait for [math]k[/math] seconds. Then you put another symbol after the old ones, so that you have [math]k + 1[/math] symbols, and you wait for [math]k + 1[/math] seconds.

>It does not matter how many times you call your print function, or what is the argument, your program will always have - 1/12 as output
Nope. You can't have [math]-\frac{1}{12}[/math] symbols on the screen. Nor is the program even designed to erase existing symbols -- Only to add new ones.

But I just realized that the program still wouldn't work, because the next wait time would still need to be stored in memory.

>>8338595
This. Mathematicians are 1/12 of a second faster than OP starting the program.

Vedä vittus

i was about to run the program but the numbers appeared on my screen before i compiled it

>>8339080
Takas ylikselle

>>8338590

You're conflating normal summation with Ramanujan summation of divergent series. They share some properties, but not the ones you're thinking of. Your mistake is like taking the rules for addition and just applying them to the numerators and denominators of fractions so that you get 3/5 as the answer for 1/2 + 2/3 instead of 7/6.

That is because in this case '=' does not mean equals to but means associated with.

>>8338590
[eqn]\zeta(-1) = -\frac{1}{12} \text{ is } not \text{ the same as } \sum_{k\in\mathbb{N}} k.[/eqn] God I hate you all sometimes.

>>8339763
dude what does that big backwards 3 thing have to do with anything?

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Hey, guys, I have a question for number theorists here:

How many numbers of form 2^x * 3^y * 5^z are between 1 and 2?

This question comes easily from music, actually the answer is the number of keys of the octave in a mathematically perfect piano. I hope that the answer is finite. But unfortunately I have no idea how to find it. Any ideas?

>>8339846
Oops, I think it's infinite even for 2^x * 3^y.
Every 3^y is between some 2^x and 2^{x+1}
Divide it by 2^{x+1} and you get that 3^y/2{x+1} is between 1/2=2^x/2^{x+1} and 1=2^{x+1}/2^{x+1}.
Invert fractions and you get that 2^{x+1}/3^y is between 1 and 2.

Sad to see

>>8338590

you actually just proved it yourself by knowing the answer was infinity before even starting the program.

>>8339763
Whats the difference serious question

>>8339807
>>8339943
Are you legit serious?

If so we will share the word so we can lay this meme to rest.

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>>8339943
The summation is only an equivalent definition for s such that Re(s) > 1. The actual definition of the zeta function is pic related

>>8339954
im serious

>>8339962
oh, so why does the -1/12 meme even exist? wtf even numberphile made a vid on it and didnt talk anything about the bounds for the rieman zetta func

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>>8339972

>why does the -1/12 meme even exist?

I don't understand your question. Assigning a summation value to a divergent series can be written like 1 + 2 + 3 ... = -1/12. But it's actually different from how you'd add numbers in mundane situations like on a calculator for an accounting problem. It looks shocking and counter-intuitive to people who don't understand what summation value assignment in the context of divergent series means, which is why it gets mentioned a lot by people who don't study that sort of mathematics.what else is there to explain about it?

>>8339972
It's just a really pedantic way of using the analytic continuation as if it were the function itself. It shouldn't be an equals sign; it would be more correct to have a symbol meaning "can be represented as" or "is extended to."

>>8339991

>it would be more correct to have a symbol meaning "can be represented as" or "is extended to."

Where do you draw the line though? All mathematical operations require different rules / approaches depending on which set of numbers you're operating on.

>>8340002
Equality should be unambiguous, and the relevant group should be stated explicitly. You could use the actual symbol with context, but it has to be made clear. There are multiple valid continuations for things like the factorial or zeta-style summation. You can't just say 1 + 2 + 3 + ... = -1/12 without any context because there are plenty of situations where it doesn't make sense.

>>8340019
>Equality should be unambiguous

tumblr

>>8340061
kek

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SHUT UP AND POST MORE ASIAN BITCHES