5
$\begingroup$

Suppose that there is a polynomial $P$ with integer coefficients such that $P(x_i)=y_i$ for $i=1,\ldots,n$. Is it true that the result of Lagrange interpolation through the data $(x_i,y_i)$ is a polynomial with integer coefficients? Suppose there are $2$ cases :

$\bullet$ $x_i$ are integers

$\bullet$ $x_i$ are not all integers

What are the conclusions? Note $\deg P$ is not specified.

$\endgroup$
  • 1
    $\begingroup$ The answer is no if the $x_i$s are not all integers. Just take $P(x) = x^2$, $x_1 = 0, x_2 = \pi$: interpolation will give you the (linear) polynomial $\pi\cdot x$. $\endgroup$ – Marco Golla Jun 5 '14 at 8:14
5
$\begingroup$

The answer is yes if the $x_i$ are integers. Write $$ B(X):=\prod_{i=1}^n (X-x_i), $$ so that $B(X)$ is a monic polynomial in $\mathbf{Z}[X]$. For any $P(X)\in\mathbf{Z}[X]$, we can write $$ P(X) = B(X) Q(X) + R(X) $$ where $Q,R\in\mathbf{Z}[X]$ and $\deg(R)<n$. Here $R(x_i)=P(x_i)=y_i$ for every $i$. Since there is a unique polynomial $f(X)\in\mathbf{Q}[X]$ of degree less than $n$ which satisfies $f(x_i)=y_i$ for $i=1,2,\dots,n$, and both $R(X)$ and the Lagrange interpolation polynomial have the properties required of $f(X)$, it follows that the Lagrange interpolation polynomial equals $R(X)$ and hence has integer coefficients.

The answer is no if the $x_i$ are not integers. For instance put $x_1=y_1=1/2$ (and $n=1$), then $P(X):=X$ satisfies $P(x_1)=y_1$, but the Lagrange interpolation polynomial is $1/2$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Seems to me that it might be wrong even with integers; consider the following conditions for instance: $$P(1)=1$$ $$P(2)=1$$ $$P(3)=2$$ Then the Lagrange interpolation through the data $(1,1), (2,1), (3,2)$ gives: $$P(X)=\frac{1}{2}X^2 - \frac{3}{2}X + 2$$ Even though $P(X)$ is an integer for every integer $X$, its coefficients are not integers.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ But the hypothesis is that there exists a polynomial with integer coefficients going through the data points. There isn't one in your example. $\endgroup$ – Gerry Myerson Nov 8 '15 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.