Show that $L_0 x + L_1 x + L_2 x + L_3 x = 1$ for all values of x.

Question

Here I'm talking about the Lagrange's Interpolation formula for calculating the interpolation with unequal intervals.

I tried too hard but I was unable to achieve the result on the RHS. Please help!

Answer 1

I believe the question is about the coefficients in the Lagrange's interpolation formula:

$$L_f(x)=f(x_0)L_0(x)+f(x_1)L_1(x)+f(x_2)L_2(x)+f(x_3)L_3(x)$$

where $L_f$ is the Lagrange interpolation polynomial (of degree $\le 3$) of the unknown function $f$, which has the properties that $L_f(x_i)=f(x_i)$ for $i=0,1,2,3$. $L_i(x)$ are given as:

$$L_i(x)=\frac{\prod_{j\ne i}(x-x_j)}{\prod_{j\ne i}(x_i-x_j)}$$

If I am right, then... let's proceed with the proof.

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Set $f(x)=1$. You get:

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

and, because $L_f(x_i)=f(x_i)=1$, you have that the left-hand side is a polynomial of degreee $\le 3$, which takes the values $1$ at $x_i, i=0,1,2,3$. This implies that this polynomial is $1$: namely $1$ is another polynomial of degree $\le 3$ with the same property - and if two polynomials of degree $\le 3$ match at four different points, they must be the same polynomial! (Otherwise, their difference $L_f(x)-1$ would have four different zeros.)

Thus $L_f(x)=1$ and so $L_0(x)+L_1(x)+L_2(x)+L_3(x)=1$.