For the Lagrange polynomials $l_j(x)$, $l_1(x)+l_2(x)+l_3(x)+l_4(x)=1$

Consider a function $$y=f(x)$$ that takes the values $$f(1)=0, \ f(2)=0.6931, \ f(4)=4.1589, \ f(3)=2.1972,$$ where $$x_1=1, x_2=2, x_3=4, x_4=3.$$ Show that the Lagrange polynomials $$l_1,..,l_4$$ defined by the above data satisfies $$l_1(x)+l_2(x)+l_3(x)+l_4(x)=1 \ \ \forall x. \tag{1}$$

Here we define the Lagrange polynomials, $$l_j(x)$$, by $$l_j(x)=\prod_{{k=1} \\ {k\neq j}}^{n+1}\frac{x-x_k}{x_j-x_k}.$$ I have computed the Lagrange polynomials $$l_j(x)$$ for $$j=1,..,4$$ as follows: \begin{align} l_1(x)&=-\frac{1}{6}(x-2)(x-3)(x-4) \\ l_2(x)&=\frac{1}{2} (x-1)(x-3)(x-4) \\ l_3(x)&=\frac{1}{6}(x-1)(x-2)(x-3) \\ l_4(x)&=-\frac{1}{2} (x-1)(x-2)(x-4). \end{align} Substituting the above equations into $$(1)$$ does appear to satisfy the required equation, but this method is computationally tedious. The answer provided in my textbook uses the fundamental theorem of algebra to explain this result, but I did not understand the explanation.

The sum of the $$l_j(x)$$ is an at most cubic polynomial that has the value $$1$$ at all of $$x=1,2,3,4$$. The difference to $$1$$ is an at most cubic polynomial with $$4$$ roots, of which there is only one.

• Why does the fundamental theorem of algebra come into play? I still don't quite understand.
– M B
May 7 '20 at 12:09
• Because you can not have 4 distinct roots in a non-trivial cubic (or quadratic or linear) polynomial. That is, this uses the lesser version of the fundamental theorem, that you can split off a linear factor for every root and that this factorization is unique. The full version, that every non-trivial polynomial has at least one root, is not used. May 7 '20 at 12:23
• So I understand your first sentence, and it allows me to think of the problem graphically. There exists a unique cubic polynomial that passes through $x_j$ for $j=1,..,4$ at $f(x)=1$. However, i'm unsure of the meaning of your second sentence. Would you be able to expand a little bit more?
– M B
May 7 '20 at 12:44
• Set $p(x)=l_1(x)+l_2(x)+l_3(x)+l_4(x)-1$. Then $p(x_j)=0$ for all $j=1,2,3,4$. On the other hand, $\deg p\le 3$, so that $p$ can have at most 3 roots if it is not the zero polynomial. May 7 '20 at 12:47
• Oh I think I see. We take $p(x)$ to be the zero polynomial, otherwise it violates the fundamental theorem of algebra?
– M B
May 8 '20 at 1:31