gistfile1.txt

Ok, this isn't quite rigorous, but I bet it's right anyway =P
Imagine F is discontinuous at some point, and jumps by like, 1. Well, if you
put x1 and x2 really close together around the point, we'll violate the
constraint, right? So let's make F nice and smooth instead. But let's say
somewhere F has a derivative of 1. Well, if we put x1 and x2, say, 0.01 apart,
then (x2-x1)^2 = 0.0001 but F is gonna change by 0.01, so that's out.

But however small we make the derivative, we can always squeeze the two points
closer together so that the function changes too fast.
Clearly it's just never allowed to change at all. F(x) = C.

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This is basically

a = sin x
b = cos x

a^7 + a^-3 = b^7 + b^-3

Now, that's a slightly nasty equation, but it has one really trivial solution,
if a=b than it clearly holds.

sin x = cos x gives us x = pi/4 + n*pi

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We want to know how many digits 125^100 has. The first thing we can do to make
our lives easier is note that 125 = 5^3, so we're really looking at 5^300. Now,
asking how many digits something has is a lot like asking how many powers of 10
it is. Now we have a really big power of 5, but we want to talk about it in
terms of powers of 10. What's missing? Powers of 2. What do we know about
powers of 2? Well, computer scientists know that 2^10 is almost equal to 1000,
or 10^3. This comes up in how we address memory in computers. People talk about
kilobytes, which literally means thousands of bytes, but technically a kilobyte
is defined as 1024 or 2^10 bytes.

So.

2^10 = 1024

2^10 = 10^3 * 1.024

Let's throw some powers of 5 into that

2^10 * 5^10 = 5^10 * 10^3 * 1.024

10^10 = 5^10 * 10^3 * 1.024

10^7 = 5^10 * 1.024

5^10 = 10^7 / 1.024

Nice! But we don't want 5^10, we want 5^300, so raise both sides...

5^300 = 10^210 / 1.024^30

So we're really close, the question is what is 1.024^30?

Well, 1.024 is really close to 1, so this looks like a good time for binomial
expansion.

0.024 is really close to 0.025 which is 1/40, so let's work with that instead.

(1 + 1/40)^30 = 1 + 30/40 + 30*29/40*40*2 + 30*29*28/40*40*40*6

~= 1 + 3/4 + 9/32 + 9/128 = 269/128

ok, I'm just gonna apply some intuition at this point and say that this series
is not going to get up to 10. And everywhere we've rounded, we've rounded up,
so let's say that 1.024^30 is greater than 1 and less than 10.

So basically, 125^100 = 5^300 is a little less than 10^210, but still greater
than 10^209, so it has 210 digits.