Show that the sum of Lagrange Polynomials $\sum_{i=0}^{n} L_{i}(t)=1 \quad \forall t \in R$

Question

I am reviewing a homework problem that is supposed to be really easy but I have trouble wrapping my head around it. For $j=0, \ldots, n \quad t_{j} \neq t_{i}$ if $ i \neq j $ we define the $n$ Lagrange Polynomials as usual:

$$L_{i}(x)=\prod_{j=0 \atop j \neq i}^{n} \frac{x-t_{j}}{t_{i}-t_{j}}$$

The problem asks us to show that $$\sum_{i=0}^{n} L_{i}(t)=1 \quad \forall t \in R$$

Unfortunately, my professor only gave the solution that this is obvious from the definition.

I can see how this is true for any of the nodes $t_j$, however I can't see it for any real number $t$.

I'm trying to show this algebraically (since it's not obvious to me) but I simply can't without getting a huge mess of terms and I feel like I'm missing something essential since this shouldn't take that long.

Could someone provide me with the necessary steps to follow or a full explanation of why this is obvious?

Thank you so much!

Answer 1

So, at the end of the day, the interpolations gives you a polynomial of degree $\# points -1$ that maps some $t_j$ to some $\ell _j$ when you compute $$L(t)=\sum _{k=0}^n\ell _k\cdot L_i(t).$$ Notice that if you take then $\ell _k=1$ you get your sum. Notice further that if you do $L(t)-1=0$ then all the $t_j$ are solutions. But there are $\#points $ of those, so one more than the degree. Can this be possible without $L$ being constant at $1$?

Answer 2

The interpolating polynomial of $f$ on the distinct points $x_0, \cdots, x_n$ is given by $$ p_n(t) = \sum_{i=0}^n f(t_i) L_i(t) $$

In particular, if you take the constant function $f(t)=1$, whose interpolating polynomial is clearly $p_n(t)=1$, it follows that $$ \underbrace{1}_{=p_n(t)} = \sum_{i=0}^n L_i (t). $$

You can obtain similar relations for other particular choices of $f(t)$. For instance, taking $f(t) = t$, you can say that $$ t = \sum_{i=0}^n t_i L_i(t). $$