Puzzle Time - after the decimal point

[Klaus]

Hm, using the binomial equation (a+b)(a-b)=a^2-b^2, I can transform sqrt3 + sqrt2 to 1/(sqrt3-sqrt2). Right direction?

[jon-nyc]

That’s not where I was headed.

I can give a bigger hint.

Let sqrt(3) be a, sqrt(2) be b.

We’re looking for (a+b)^n for n=10^9

Here’s the big hint:

:::

What happens if you do the binomial expansion of (a+b)^n then add (a-b)^n to it?

:::

[Klaus]

Too big of a hint (sqrt3 -sqrt2)^10^9 is too small to influence the first digit after the dot.

[jon-nyc]

It is a huge hint

[Klaus]

I'd say the digit must be 9. Or 0 if I screwed up.

If you apply the binomial theorem to the term you suggest and do some simplification, you get a whole number minus (sqrt3 - sqrt2)^n, which gets very small.

[jon-nyc]

Right. We know that by adding an infinitesimal positive number to the number we seek, we get an integer.

Therefore the first many many digits after the decimal must be 9.

Note this works for any even n sufficiently large.

[jon-nyc]

You can already see the pattern with quite small n.

[Klaus]

@jon-nyc said in Puzzle Time - after the decimal point:

You can already see the pattern with quite small n.

Yes, but I was confused for a while because that pattern switches when you try larger numbers.

I was at first thinking that this was a rounding error and then tried again with a "big float" library, but still, with n >=50 the first digit seems to be 0.

However, I assume that it's just a particularly numerically unstable formula.

[jon-nyc]

@Klaus

Has to be because (sqrt(3)-sqrt(2))^n stays tiny and positive. Same thing happened on my excel, too.

[Klaus]

Effect of increasing the precision to 100 bits, and then 500 bits:

[jon-nyc]

Cool.

[jon-nyc]

The ‘official’ answer came out today.

SOLUTION: If you try entering "(sqrt(2) + sqrt(3))^1,000,000,000" into Google or your calculator, you're likely to find that you get only the dozen or so most significant figures; that is, you don't get a nearly accurate enough answer to see what happens after the decimal point.

But you can try smaller powers and see what happens. For example, the decimal expansion of the 10th power of the square root of 2 plus the square root of 3 begins 95049.9999895. A bit of experimentation shows that each even power of the square root of 2 plus the square root of 3 seems to be just a hair below some integer. So the solution to the puzzle seems to be 9. But why?

If we merely square the number "square root of 2 plus square root of 3," we get about 9.9. If we play with 10 minus this number, we discover that it's equal to the square of the square root of 2 minus the square root of 3. Aha!

Yes, (sqrt(3) + sqrt(2))^2n + (sqrt(3) - sqrt(2))^2n is always an integer, because when you expand it, the odd-power terms cancel and the even ones are integers. But (sqrt(3) - sqrt(2))^2n is very small, about 10^-n, so the first roughly n digits of (sqrt(3) + sqrt(2))^2n past the decimal point are all 9's.

[The idea for this puzzle comes from the 29th Annual Virginia Tech Mathematics Contest, 2007