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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 \(p(0)\)

 Feb 21, 2021
 #1
avatar+118069 
+1

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  !!!!

 

cool cool cool

 Feb 21, 2021
 #2
avatar+118 
+1

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.

 Feb 21, 2021

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