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        riddles >> hard >> 98th Digit
        
(Message started by: william wu on Oct 28^th, 2002, 2:01am)

Title: 98th Digit
Post by william wu on Oct 28^th, 2002, 2:01am

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.

Title: Re: 98th Digit
Post by James Fingas on Oct 28^th, 2002, 7:14am

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 98^th digit of (sqrt(2)+1)⁵⁰⁰ is the same as the 98th digit of sqrt(2)*sum_i=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.

Title: Re: 98th Digit
Post by James Fingas on Oct 28^th, 2002, 7:33am

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
a_i = 2*a_i-1+a_i-2

This is sort of like phi, the golden section, in this respect. Large powers of phi also approach integers.

Title: Re: 98th Digit
Post by towr on Oct 28^th, 2002, 10:01am

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..)

Title: Re: 98th Digit
Post by Icarus on Oct 28^th, 2002, 10:10am

on 10/28/02 at 07:33:07, James Fingas wrote:

I'm postulating that it approximates the series:

a₁ = 2
a₂ = 2
a_i = 2*a_i-1+a_i-2

This is sort of like phi, the golden section, in this respect. Large powers of phi also approach integers.

The sequence it approximates is:

a₁ = 2
a₂ = 6
a_i = 2*a_i-1+a_i-2

The closed form expression for this sequence is

a_n = (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!

Title: Re: 98th Digit
Post by towr on Oct 28^th, 2002, 10:34am

on 10/28/02 at 10:10:48, Icarus wrote:

The closed form expression for this sequence is

a_n = (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!

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..

Title: Re: 98th Digit
Post by Eric Yeh on Oct 28^th, 2002, 12:01pm

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

Title: Re: 98th Digit
Post by towr on Oct 28^th, 2002, 1:09pm

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..

Title: Re: 98th Digit
Post by Eric Yeh on Oct 28^th, 2002, 1:18pm

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)?

Title: Re: 98th Digit
Post by towr on Oct 28^th, 2002, 1:40pm

wait.. what am I saying again?

euhms.. (it's late here)..

ke.. it's 9.. :p
(must go sleep.....)

(really it's 9)

Title: Re: 98th Digit
Post by Icarus on Oct 28^th, 2002, 3:34pm

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 a_n to show that a_n = (1+sqrt(2))ⁿ + (1-sqrt(2))ⁿ must always be an integer.

Title: Re: 98th Digit
Post by towr on Oct 28^th, 2002, 11:18pm

is it just me, or is f(x, y) := (1 + sqrt(y))^x + (1 - sqrt(y))^x integer for every x,y >= 0 ?

Title: Re: 98th Digit
Post by Eric Yeh on Oct 29^th, 2002, 5:31am

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

Title: Re: 98th Digit
Post by Icarus on Oct 29^th, 2002, 3:34pm

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!