• @[email protected]
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    1108 days ago

    I’m allways astonished by how many function seemingly have nothing to do with circles and yet somehow a pi managed to snuck itself in

    • @[email protected]
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      598 days ago

      Usually when that happens there’s a way to tie it back to circles, but it’s not always easy to find

      • @[email protected]
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        298 days ago

        You could say you just go round and round hunting for it, but no matter how hard you try you just can’t corner it.

        Well, you could.

      • @xantoxis
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        147 days ago

        there’s a way to tie it back to circles

        Not necessarily circles, but conic sections. When you take a series of a fixed exponent over a variable x, and graph it, that graph is a parabola.

        A parabola is a slice through a cone. Tada, pi appears.

  • Grubberfly 🔮
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    418 days ago

    tbf, the 2nd sum is exactly the first one just multiplied by 1/2. though i get that the progression is natural, even, and odd.

    the last one is definitely odd puzzling, but i cannot intuitively get the first one. how does summing the inverse of triangular number equal 2?

    • Buglefingers
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      118 days ago

      I believe starting with 1/1 which equals 1, you are then adding infinitely (fractions) on top of the 1. So 1, then 1 1/2, ect, so the next full integer to be hit (infinitely down the line) would be 2.

      I don’t do high level math so I hope this explanation is correct or intelligible, this is just how I understand it intuitively

      • @[email protected]
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        8 days ago

        But the first few values are:

        1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28…

        I really don’t see any pattern there showing why it converges to 2 exactly

        Edit:

        After thinking some more, you could write the sum as:

        (Sum from n=1 to infinity of): 2/(n * (n + 1))

        That sum is smaller than the sum of:

        2 * (1/n2) which converges to π2/3

        So I can see why it converges, just not where to.

        • @Bender_on_Fire
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          68 days ago

          I didn’t see the pattern either and had to look it up. Apparently, you can rewrite 1 + 1/(1+2) + 1/(1+2+3)+… as 2(1 - 1/2 + 1/2 - 1/3 +…+1/n - 1/(n + 1)) = 2(1 - 1/(n + 1))

          From there, the limit of 2 is obvious, but I guess you just have to build up intuition with infinite sums to see the reformulation.

        • Buglefingers
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          28 days ago

          So the amount you are adding is getting smaller with each iteration, 1/4 is smaller than 1/2, however you are still adding 1/4 on top of the 1/2, and those two are combined, closer to “1” than either of them independently correct? (1/2 +1/4 =1/3. 1/3>1/2)

          So if the number gets bigger forever than at some point it will eventually hit “1”, since we already started with “1” the next “1” will be “2”

          I hope I’m explaining it well enough, it’s similar to how 3.33(repeating)x3…=10 (though technically for different reasons)

  • @mkwt
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    168 days ago

    1 + 2 + 3 + 4 + 5 + … = -1/12

      • @[email protected]
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        8 days ago

        Not really. The left side comes from trying to evaluate the Reimann zeta function at a point outside its domain, and the right side comes from evaluating the analytic continuation of the zeta function at the same point. The deepest truth it reveals is that applying the definition of a function outside its domain can give you nonsense results.

        It’s one of those ideas like Scrödinger’s cat or pulling yourself up by your bootstraps, where something that was intended to illustrate an absurdity is instead taken as representing some amazing and unintuitive truth.

      • @Tudsamfa
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        8 days ago

        Never understood those circumstances, but I choose to believe that all even numbers summed are -2/12 and all odd numbers therefore 1/12.

  • @Th3D3k0y
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    88 days ago

    Yeah, but is the function for 1 over odd numbers, or 1 over prime numbers …

  • @[email protected]
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    8 days ago

    Tired of waiting 2 seconds for hyperpi to download? Simply do infinity calculations by hand and some basic algebra. Cut out the middle man. Thanks, internet!