Let \(p(x)\) be a polynomial of degree 6 such that \(p(2^n) = \frac{1}{2^n}\) for \(n=0, 1, 2, ..., 6\) Find \(p(0)\)
+129899 This is DEFINITELY tedious to solve by hand !!!
I'll do the set-up, but I'm going to rely on this website to do the "heavy-lifting"
https://matrix.reshish.com/gauss-jordanElimination.php
We have this system
a + b + c + d + e + f + g = 1
64 a + 32b + 16c + 8d + 4e + 2f + g = 1/2
4096a + 1024b + 256c +64d + 16e + 4f + g =1/4
262144a + 32768b + 4096c + 512d + 64e + 8f + g = 1/8
16777216a + 1048576b + 65536c + 4096d + 256e + 16f + g = 1/16
1073741824a + 33554432b + 1048576c + 32768d+1024e + 32f + g = 1/32
68719476736a + 1073741824b + 16777216c + 262144d+ 64f + g = 1/64
We don't need to worry about anything but the answer for "g" since this will = p(0)
And "g" = 127 / 64 = p(0)
WHEW !!!!
+119 I'm pretty sure there's a better way to do this. I've seen this question before. I have just forgotten how to approach it. Maybe let q(n)=p(2^n)-1/(2^n) for 0 <= n <=6 and then q(n)=n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)? Something like that? I could be wrong about the better way, though. It's just that 127/64 is a pretty nice number.
