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f is a polynomial of degree $1007$ ,if $f(k) = 2^k$ for $0\le k \le 1007$ find the value of $f(2015)$

The solution should be $f(2015) = 2^{2014}$ and the polynomial $p(x) = \sum_{k=0}^{1007}{x \choose k}$ but i don't get how we can arrive to that.

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    $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – Hippalectryon Mar 5 '15 at 19:55
  • $\begingroup$ Is $f(k) = 2^k$ true only on natural numbers or also on reals ? $\endgroup$ – servabat Mar 5 '15 at 20:18
  • $\begingroup$ @servabat A polynomial cannot coincide with an exponential on infinitely many values. $\endgroup$ – Slade Mar 6 '15 at 19:16
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The idea is to use the identity $$ {{n}\choose{0}}+ {n \choose 1} + \cdots + {n \choose n} = 2^n $$ to create a polynomial of degree 1007 that has the desired properties. We can see that this is true by the binomial theorem. Observe that $$ (x+y)^n = \sum_{k=0}^{n}{n \choose k} x^{n-k} y^{k}, $$ so letting $x = y =1$ we obtain the desired result.

With this result in mind, we can show the polynomial $$ p(x) = \sum_{k=0}^{1007} {x \choose k} $$ satisfies the desired properties. Furthermore, since we are specifying 1008 points that a 1007 degree polynomial maps to, this must be the unique polynomial of this degree that does this. Finally, calculating $p(2015)$ requires a one last binomial coefficient identity. For any $n$ and $k$, $$ {n \choose k} = {n \choose n-k}. $$ Using this and the previous identity, we see that for any Odd $n = 2n' + 1$, \begin{align*} 2^n &= \sum_{k=0}^{n} {n \choose k} \\ & = \sum_{k=0}^{n'} {n \choose k} + \sum_{k=n'+1}^{n} {n \choose k} \\ & = \sum_{k=0}^{n'} {n \choose k} + \sum_{k=n'+1}^{n} {n \choose n - k}. \end{align*} The latter sum is equal to ${n \choose n'} + {n \choose n' -1 }+ ... + {n \choose 0} $, while the former is equal to ${n \choose 0} + ... + {n \choose n'}$. Therefore, they are equal, so \begin{align*} 2^n &= 2 \sum_{k=0}^{n'} {n \choose k} \\ & \text{and} \\ 2^{n-1} &=\sum_{k=0}^{n'} {n \choose k}. \end{align*} This identity allows us to calculate $p(2015)$. Observe that $2015 = 2 \times 1007 +1$. Then by the above identity,

\begin{align*} p(2015) & = \sum_{k=0}^{1007} {2015 \choose k} \\ & = 2^{2015-1} \\ &= 2^{2014}, \end{align*} as desired.

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  • $\begingroup$ Could you write also the way to calculate $p(2015)$ ? I've never done problem like this. $\endgroup$ – Azazel Mar 6 '15 at 8:25
  • $\begingroup$ Yep, I just added it to the solution now. @Azazel $\endgroup$ – BigMathTimes Mar 6 '15 at 18:35
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Hint $\ $ Write $\,f(x) = c_0 + c_1 x + c_2 x(x-1) + c_3 x(x-1)(x-2) + \cdots$ then successively evaluate at $\,x=0,1,2\ldots$ to deduce $\,c_n = 1/n!\,\ $ (this is essentially Newton interpolation).

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