797 decimal places pi is equal to 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914127372458700660631558817488152092096282925409171536436789259036001133053054882046652138414695194151160943305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912983367336244065664308602139494639522473719070217986094370277053921717629317675238467481846766940513200056812714526356082778577134275778960917363717872146844090122495343014654958537105079227968925892354201995611212902196086403441815981362977477130996051870721134999999
So you could round that to 3.14159265358979323846264338327950288419716939937510582097494459230781640628608998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190 7021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853 7105079227968925892354201995611212902196086403441815981362977477130996051870721135 and it would work for most calculations you would need, but it still wouldn’t be exact. If you made a circle the diameter of the known universe, the difference would be way less, 720 orders of magnitude less than the diameter of a proton.
If you need more accuracy than that, you’re on your own.
Nitpick: After the sextuple 9, pi continues …837297804995
Thus, don’t you think that, to [del]797[/del] 767 decimal places, pi should be rendered as 3.14159 … 1135000000 ?
BTW, you needn’t Google to find the first 767 decimals of pi. Using the standard Unix bc calculator
scale = 770
4*a(1)
produces this much precision almost instantaneously.
Using the Chudnovsky algorithm I can calculate 2.5M places in 2 min on this computer, but I don’t have enough ram to do more but here are the first 3000, that are verified.
A big thanks to Thudlow Boink for these links; I’d never heard of that site. it looks to be a gold mine for folks trying to stretch their math wings past the stuff they think they remember from whatever they memorized back in school.
Our intrepid OP would do well to start reading that site beginning with topics he’s sure he understands working out towards the edge of what he thinks he understands then beyond.
I goofed on my adding because I did it by hand and the only reason I was looking at this thread was I waiting for the LOTRO last beta test for update 21 to come on line and while I was typing it my friend, who I was Skyping with said it was up. Yes, on a $3000 computer with at least 3 calculators and Mathcad installed, I was adding by hand.
A small error here but it doesn’t affect the rest of the proof:
f’ = -1/2 * sin(sqrt(x))/sqrt(x)
Or, if we want to use sinc:
f’ = -1/2 * sinc(sqrt(x))*sqrt(x)
And then the second derivative:
f’’ = (sin(sqrt(x))/sqrt(x) + cos(sqrt(x)))/(4x)
I can see how this works; the sin() bits have an extra sqrt(x) factor in them, but this is consistent, so if you factor that out you get a polynomial in terms of x. The cos() bits don’t have the extra sqrt(x) factor, and this is also consistent because you always end up multiplying by the derivative of the inside (having gotten there from the sin()). So it all works out in the end like you say.
You are clearly right–brain fart on my part. I did the math without the sinc (just a sin), then tried to substitute it in after, but obviously screwed it up. Sorry!
That said, there is a sqrt(x) factor hiding in there, which was part of my initial confusion with the approach–I was getting expressions with half-integer powers after playing with the math in your first post. I knew I could just factor out a x^0.5 to get a polynomial but I wondered where the x^0.5 went. Now I see that it disappeared inside the sinc. Although I’m familiar with sinc for image processing purposes, it’s not really part of my mental toolkit for calculus.
From your first post on the subject, I did find that the derivatives came in the form:
P(x)sin(x) + Q(x)cos(x)
…where P(x) is not a polynomial but rather consisted of half-integer powers of x. Ok, I saw from the y=1/4x substitution that the negative powers would be made positive, and that I could further factor out a x^0.5 to then make it a polynomial, but I didn’t really make it further than that.
But I think I see how it all works out now, with the derivatives ping-ponging between sin(sqrt(x)) and cos(sqrt(x)), and the chain rule providing the extra power of 0.5 to “switch” between the half-integer powers and integer (polynomial) powers.
Anyway, I suspect this logic is entirely wrapped up in your mental picture of how sinc(u) works and so the mechanics of it are abstracted away as far as you’re concerned. But your second post to completed the picture for me, so thanks again.
I played this game out to 5,000,000 for the numerator (which is probably too far, actually, for accurate calculations), and saw a bunch of interesting patterns. Multiples of 113 as the numerator are the best representation for pi (and were no better than 113) out a long way (I’ll report further when I have a little more time).
I explicitly want to encourage people in these sorts of investigations, but also feel obligated to point out that the continued fraction for pi has been computed to billions of terms
I actually don’t know the derivative of sinc or such things off the top of my head; I’d sit down and calculate it from scratch each time. For what it’s worth, I personally thought about this barely ever thinking of sinc, or the specific values of f’ and f’‘, at all. The specific value of f’ in terms of sinc I listed only for people who like to see the details, but I myself thought about this by moving to the abstract.
Like I said in post 199, suppose you are given something of the form P(y)f + Q(y)f’, and you wish to take its derivative, knowing only that f’’ is also of this form and that y’ is also a polynomial in y. You can readily go ahead and see that this derivative is also of the same form, increasing the degree of the polynomials at a fixed rate.
It doesn’t matter whether f is cos(sqrt(x)) and y = 1/(4x), or any other things with the same property. This is why I like abstraction; it gets rid of the clutter of unnecessary details, and lets you see only the things that matter. So it becomes clear how to generalize this argument to further situations too (not just the trigonometric situations I happened to use it for here).
So the only important thing is to establish that f’’ is indeed of the form P(y)f + Q(y)f’ for some y whose derivative is also a polynomial in y. All the other details stop mattering once you’ve shown that.
Anyway, whatever way works for you to understand it works; I’m just expanding on how it became easiest for me to understand.
