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Topic: 98th Digit (Read 1038 times) |
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william wu
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Not sure what section this puzzle should go into, since I don't know how to attack it:
Find the 98-th digit to the right of the decimal point in the
decimal expansion of (sqrt(2)+1)^(500).
My friend Yosen suggested multiplying that expression by 10^98 and taking the floor. We're not sure if this works though, since computers usually don't have decimal accuracy 98 digits deep.
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James Fingas
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Re: 98th Digit
« Reply #1 on: Oct 28th, 2002, 7:14am » |
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I think you'll have to use the binomial theorem and split the expression apart, but I haven't found the answer yet. Here's what I have:
The 98th digit of (sqrt(2)+1)500 is the same as the 98th digit of sqrt(2)*sumi=0..249nchoosek(500,2i+1)*2i. That sum is one very large integer. Very very large! Maybe you could use a large-integer package on this or something...
I haven't made any progress simplifying it, however.
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James Fingas
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Re: 98th Digit
« Reply #2 on: Oct 28th, 2002, 7:33am » |
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VERY COOL!
I'ts zero, but I'm not sure why. Try the following on your calculator:
1 =
Ans * (sqrt(2)+1) =
=
=
=
Now watch what happens to the decimal parts! Yowsa!
I'm postulating that it approximates the series:
a1 = 2
a2 = 2
ai = 2*ai-1+ai-2
This is sort of like phi, the golden section, in this respect. Large powers of phi also approach integers.
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| « Last Edit: Oct 28th, 2002, 7:38am by James Fingas » |
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towr
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Re: 98th Digit
« Reply #3 on: Oct 28th, 2002, 10:01am » |
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after a bit of searching I found ((sqrt(2) + 1)^x - (1 - sqrt(2))^x)/(2 * sqrt(2)) = b(n)
where also b(n) = 2*b(n-1) + b(n-2), with b(0)=0 and b(1) = 1
analogous with fibonnaci and the golden number phi
soooo
((sqrt(2) + 1)^n - (1 - sqrt(2))^n)/(2 * sqrt(2)) = b(n)
=>
2* sqrt(2)* b(n) = ((sqrt(2) + 1)^n - (1 - sqrt(2))^n)
=>
2* sqrt(2)* b(n) + (1 - sqrt(2))^n = (sqrt(2) + 1)^n
(1 - sqrt(2))^500 is 0 far past the first 98 decimals (about 4.094089478·10^(-192) )
so we need the decimal part of 2*sqrt(2)* b(n)
Now, there's another integer series a(n) = 2*a(n-1) +a(n-1) where a(0)=a(1)=1 , for which it holds that a(n)/b(n) = sqrt(2) to a certain approximation (in the limit n-> infinity they're identical)..
As a result 2*sqrt(2) *b(n) = 2*a(n) (which is an integer). And that is why the 98th decimal is 0 (likewise the other first 190 or so)
thanks to
http://mathworld.wolfram.com/PythagorassConstant.html
(equations 4 and 5 (mistake in 5 btw))
(finally after an hour waiting my connection is working again..)
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| « Last Edit: Oct 28th, 2002, 10:29am by towr » |
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Icarus
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Re: 98th Digit
« Reply #4 on: Oct 28th, 2002, 10:10am » |
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on Oct 28th, 2002, 7:33am, James Fingas wrote:
I'm postulating that it approximates the series:
a1 = 2
a2 = 2
ai = 2*ai-1+ai-2
This is sort of like phi, the golden section, in this respect. Large powers of phi also approach integers. |
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The sequence it approximates is:
a1 = 2
a2 = 6
ai = 2*ai-1+ai-2
The closed form expression for this sequence is
an = (1+sqrt(2))n + (1-sqrt(2))n
By the recursion, it is obvious all the values are integers, but the second term is less than .5 in absolute value, so when it is raised to the power 500, the first ~190 decimal places are all 0. Therefore the first term also needs to have about the same number of zeros after the decimal point for the sum to be an integer. so not only is the 98th decimal zero, so are the next 90 decimals!
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towr
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Re: 98th Digit
« Reply #5 on: Oct 28th, 2002, 10:34am » |
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on Oct 28th, 2002, 10:10am, Icarus wrote:
The closed form expression for this sequence is
an = (1+sqrt(2))n + (1-sqrt(2))n
By the recursion, it is obvious all the values are integers, but the second term is less than .5 in absolute value, so when it is raised to the power 500, the first ~190 decimal places are all 0. Therefore the first term also needs to have about the same number of zeros after the decimal point for the sum to be an integer. so not only is the 98th decimal zero, so are the next 90 decimals!
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Actually, they'd have to be 9's, since (1-sqrt(2))n > 0 for even n..
But since it's an approximation (as is mine) it might be either 0 or 9..
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| « Last Edit: Oct 28th, 2002, 10:35am by towr » |
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Eric Yeh
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Re: 98th Digit
« Reply #6 on: Oct 28th, 2002, 12:01pm » |
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Towr,
I agree that it has to be 9 rather than 0 bc 500 is even. But it cannot be 0, bc Icarus' formula is not an approximation -- its exact.
Best,
Eric
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towr
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Re: 98th Digit
« Reply #7 on: Oct 28th, 2002, 1:09pm » |
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I don't agree it's exact..
Sure the closed formula is exact for the series a(n) = 2(a(n-1) +a(n-1), but that series is just an approximation..
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| « Last Edit: Oct 28th, 2002, 1:32pm by towr » |
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Eric Yeh
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Re: 98th Digit
« Reply #8 on: Oct 28th, 2002, 1:18pm » |
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I'm sorry, I don't understand what you're saying.
Are you saying that after defining An = (1+sqrt(2))^n + (1-sqrt(2))^n that you do not have A1 = 2, A2 = 6, Ai = 2*A(i-1)+A(i-2)?
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towr
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Re: 98th Digit
« Reply #9 on: Oct 28th, 2002, 1:40pm » |
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wait.. what am I saying again?
euhms.. (it's late here)..
ke.. it's 9.. :p
(must go sleep.....)
(really it's 9)
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Icarus
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Re: 98th Digit
« Reply #10 on: Oct 28th, 2002, 3:34pm » |
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Yeah... that's what happens when you try to rush a reply during lunch. All the digits are 9s, not 0. This eventually occured to me, but I had to wait until after work to fix it. Towr beat me to the punch too. Slipped in his first post while I was preparing mine.
As for Towr's confusion, I assume that in his late-night groggy state, he misunderstood that I was only using the sequence an
to show that an = (1+sqrt(2))n + (1-sqrt(2))n must always be an integer.
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"Pi goes on and on and on ...
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I wonder: Which is larger
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towr
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Re: 98th Digit
« Reply #11 on: Oct 28th, 2002, 11:18pm » |
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is it just me, or is f(x, y) := (1 + sqrt(y))^x + (1 - sqrt(y))^x integer for every x,y >= 0 ?
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Eric Yeh
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Re: 98th Digit
« Reply #12 on: Oct 29th, 2002, 5:31am » |
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Well, surely you want x,y to be integers, but otherwise yes. All the even powers are integers, and all the odd powers cancel. You can even change your 1 to any integer z.
Alternatievly, the recursion is also always integral. (As are the first two powers.)
Best,
Eric
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| « Last Edit: Oct 29th, 2002, 5:33am by Eric Yeh » |
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Icarus
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Re: 98th Digit
« Reply #13 on: Oct 29th, 2002, 3:34pm » |
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True, Towr, it is fairly easy to see that (1+sqrt(2))n + (1-sqrt(2))n is always an integer. I guess I got my blinders on with the recursion formula bit and didn't consider easier ways to see the same thing. In my defense, I did type it up in 15 minutes on my lunch break. Not a lot time to think it out!
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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