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Let $a_1,\ldots,a_n$ real numbers and $x_1,\ldots,x_n$ be distinct real numbers. Show that there is a polynomial of degree at most $n-1$ for which $f(x_i) = a_i$ for all $i$.

My idea was to prove the statement for $a_1 = 1$ and $a_2 = \cdots = a_n = 0$. We then have $f(x_1) = 1$ and $f(x_m) = 0$ for $1 < n \leq m$. Then I have to construct such a function and see how to construct a function for any sequence of $a_i's$ and $x_i's$.

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    $\begingroup$ Lagrange interpolation. $\endgroup$ Commented Mar 4, 2016 at 21:55
  • $\begingroup$ Can you show it the way the question wants me to? $\endgroup$ Commented Mar 4, 2016 at 22:00
  • $\begingroup$ The Lagrange interpolation uses precisely that idea. Just specialize it to $a_i=1$ and the rest $0$. $\endgroup$ Commented Mar 4, 2016 at 22:01
  • $\begingroup$ @AndréNicolas Can you explain how it uses my idea because I don't see it. $\endgroup$ Commented Mar 4, 2016 at 22:02
  • $\begingroup$ But how do we prove it for all sequences? $\endgroup$ Commented Mar 4, 2016 at 22:03

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For your specific approach, let $$f_1(x)=\frac{(x-x_2)(x-x_3)\cdots(x-x_n)}{(x_1-x_2)(x_1-x_3)\cdots (x_1-x_n)}.$$ Then $f_1(x)=1$ when $x=x_1$ and $f_1(x)=0$ for all the other $x_i$.

Construct $f_i(x)$ similarly. Then our interpolating polynomial is a suitable linear combination of the $f_i(x)$.

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  • $\begingroup$ The degree is EXACTLY $n-1$, not at most so why does the question say that? $\endgroup$ Commented Mar 4, 2016 at 22:18
  • $\begingroup$ Certainly the $f_i(x)$ have degree exactly $n-1$. But when you take the linear combination $\sum a_if_i(x)$, there may be cancellation that results in a lower degree polynomial. Spectacularly so if all the $a_i$ are $0$! $\endgroup$ Commented Mar 4, 2016 at 22:21
  • $\begingroup$ What if I have another sequence of real numbers $b_1,b_2,\ldots,b_n$. Show that there is a polynomial of degree at most $2n-1$ for which $f(x_i) = a_i$ and $f'(x_i) = b_i$. $\endgroup$ Commented Mar 4, 2016 at 22:23
  • $\begingroup$ Post it and you will get answers. $\endgroup$ Commented Mar 4, 2016 at 22:25
  • $\begingroup$ As a new question? $\endgroup$ Commented Mar 4, 2016 at 22:28
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$f(X)= \sum_i a_i{{(X-x_1)...\overline{(X-x_i)}..(X-x_n)}\over{(x_i-x_1)..(x_i-x_n)}}$

where $\overline{(X-x_i)}$ means the expression $(X-x_i)$ does not appears in the product.

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