The Octopus Polynomial

Consider $f(x)-\frac{1}{x}$ . Actually, since this is not a polynomial, consider $xf(x)-1$ , the previous expression multiplied by x. This new polynomial is of degree 9 (an octic times x plus a constant), and we know its roots, since $f(x)=\frac{1}{x}$ implies $xf(x)-1=x*\frac{1}{x}-1=1-1=0$ . Since we know (all) nine of its roots are $\frac{1}{2^i}$ for $0 \le i \le 8$ , we know that $xf(x)-1=a(x-1)(x-2)...(x-2^8)$ . Since $f(x)$ is itself a polynomial, the RHS must have constant term -1 so $xf(x)$ can have constant term 0. Therefore $a(-1)(-2)...(-2^8)=-1$ , and dividing out -1 gives $a(-2)(-4)...(-2^8)=1$ . Since there are an even number of terms, all the negatives on the LHS cancel out and leave us with $a*2^{1+2+...+8}=a*2^{36}=1$ . Dividing by $2^{36}$ gives us $a=2^{-36}$ . We're almost there! When multiplying by x, the constant term ( $f(0)$ ) becomes the coefficient of x, and subtracting 1 doesn't change that. So we simply need to find the x term of $xf(x)-1$ , which is by Vieta's the sum of the numbers $2^{36-i}$ when $2^i$ is a root, multiplied by a. The sum mentioned is $2^{28}+2^{29}+...+2^{36}=2^{28}(1+2+4+...+2^8)=2^{28}*(2^9-1)$ , so multiplying by $2^{-36}$ , or dividing by $2^{36}$ , gives us $f(0)=\frac{2^{28}*(2^9-1)}{2^{36}}=\frac{2^9-1}{2^8}$ . Since $2^9-1$ is odd, it is relatively prime to $2^8$ . Answer: $2^9+2^8-1=512+256-1=768-1=767$ .

Let's denote $g(x) = x f(x) - 1$ , then we have $\text{deg}(g) = 9$ and for all $0 \leq i \leq 8$ , $g(2^i) = 2^i \times f(2^i) - 1 = 0$

From Bezout theorem, we have $x f(x) - 1 = g(x) = \alpha (x - 1)(x - 2) \cdots (x - 2^8)$

Setting $x = 0$ , we obtain $\alpha$ . Note that we don't even have to determine $\alpha$ , since it is only an intercept factor that normalizes out $-1$ from the lefthand side. Now we have $f(x) = \frac{1}{x} (\alpha (x - 1)(x - 2)\cdots (x - 2^8) + 1)$

So the value of $f(0)$ equals the coefficient of $x^1$ of the polynomial above, which can be computed via Vieta's theorem, to be $\frac{511}{256}$ .

This gives the answer of $767$ to this problem

Consider the polynomial $g(x)=xf(x)-1$ . Note that degree of $g$ is $9$ and $g(2^i)=0$ for $i=0,1,...,8$ . So $g(x)$ has nine roots, so for some constant $A$ , $g(x)=A \prod \limits_{i=0}^{8} (2^i-x)$ . Because $g(0)=0\times f(0) - 1,$ we get $-1 = g(0) =A \prod \limits_{i=0}^{8} (2^i)$ , so $A=-1/\prod \limits_{i=0}^{8} (2^i)$ . Thus, $g(x)=\frac{-1}{\prod \limits_{i=0}^{8} (2^i)} \prod \limits_{i=0}^{8} (2^i-x)$

Now consider $\frac {g(x) + 1} { x} = \frac { \left( \frac{-1}{\prod \limits_{i=0}^{8} (2^i)} \prod \limits_{i=0}^{8} (2^i-x)\right) + 1 } {x}$ . The constant term in the numerator is $0$ , hence we have a polynomial, which will be $f(x)$ . As such, we can calculate $f(0)$ by looking at the linear term in the numerator.

We see that the linear term of the numerator is $\sum \limits_{i=0}^8 \frac{1}{2^i}=\frac{2^8+2^7+...+1}{2^8}=\frac{2^9-1}{2^8}=\frac{511}{256}$ . So the answer is $511+256=767.$

f(0) = (2^9 - 1)/2^8 = 511/256

so a + b = 767