I'm studying Lagrange's formula for polynomial interpolation and I cannot seem understand why for $n=3$

$$L_0(x)+L_1(x)+L_2(x)+L_3(x) = 1$$

for all real x.

In my textbook it says as a hint that expanding $P_n(x)$ is unnecessary and to use Lagrange's formula:


with some ideal {$y_0,y_1,y_2,y_3$}.

I have verified it numerically that for ($x_0,y_0$), ($x_1,y_1$), ($x_2,y_2$), ($x_3,y_3$) where $y_{0..3}=1$ and $x_{0..3}$ are distinct that this does indeed equal $1$.

Am I right in choosing $y_{0..3}=1$? Are there any suggestions to point me in the right direction?


Note: $$L_i(x)=\frac{(x-x_0)...(x-x_{i-1)})(x-x_{i+1)})...(x-x_n)}{(x_i-x_0)...(x_i-x_{i-1})(x_i-x_{i+1})...(x_i-x_n)}, i=0..n$$

  • $\begingroup$ This will be difficult to answer for those of us who do not know what you mean by $L_0(x)$, $L_1(x)$, and so on. $\endgroup$ Feb 6 '13 at 6:02
  • $\begingroup$ Thanks, I fixed the question. $\endgroup$
    – Dacto
    Feb 6 '13 at 6:08

The polynomial on the left of your first display, call it $Q(x)$, has degree at most $3$. It's easy to check that $Q(x_0)=Q(x_1)=Q(x_2)=Q(x_3)=1$ (for one thing, $L_i(x_j)$ is $1$ if $i=j$, zero otherwise). So $Q(x)-1$ is a polynomial of degree at most $3$ with at least $4$ roots. Do you understand why the only way that can happen is if $Q(x)-1$ is identically zero?

It follows that $Q(x)$ is identically $1$, as was to be proved.


Consider the uniqueness of the interpolating polynomial.

That is, there is one and only one polynomial with degree $\leq n$ that interpolates the given points.

If you're allowed to treat that as a known result, then it gets you most of the way there.


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