- #1

geoffrey159

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## Homework Statement

Let ##x_1,...,x_n## be distinct real numbers, and ## P = \prod_{i=1}^n(X-x_i)##.

If for ##i=1...n ##, ##L_i = \frac{\prod_{j \neq i}^n(X-x_j)}{\prod_{j\neq i}(x_i-x_j)}##, show that for any polynomial A (single variable and real coefs), the rest of the euclidian division of A by P is ##\sum_{i=1}^n A(x_i)L_i ##

## Homework Equations

## Q = \sum_{i=1}^n A(x_i)L_i ##

## The Attempt at a Solution

Hello, could you tell me if my proof is correct please? It seems correct to me but I have a doubt.

Let ##(B,R)## the quotient and the rest of the euclidian division of A by P. We have that ##\text{deg}(R)<\text{deg}(P) = n ##, and ##A = BP+R##.

Since ## Q(x_i) = A(x_i) = R(x_i)##, then for all ##i = 1...n ##, ##(R-Q)(x_i) = 0 ##, so ##X-x_i## divides ##R-Q##

But for ##i\neq j##, ##X-x_i## and ##X-x_j ## are relatively prime and divide ##R-Q##, so ## P | R-Q##.

However, ##\text{deg}(R-Q) \le \max(\text{deg}(R),\text{deg(Q)}) \le n-1 < n =\text{deg}(P) ##, so ## R-Q = 0 ##

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