Simplifying lagrange interpolation polynomial

[lemonthree]

Now $\sum_{i=0}^{10}(x_{i}+1) L _{10,i}(5) = (x_{0}+1) L _{10,0}(5) + (x_{1}+1) L _{10,1}(5) + ... + (x_{10}+1) L _{10,10}(5)$

Which I can further decompose into
$\frac{(x_{0}+1)(5-x_{1})(5-x_{2})...(5-x_{10})}{(x_{0}-x_{1})(x_{0}-x_{2})...(x_{0}-x_{10})} + \frac{(x_{1}+1)(5-x_{0})(5-x_{2})...(5-x_{10})}{(x_{1}-x_{0})(x_{1}-x_{2})...(x_{1}-x_{10})} + ... + \frac{(x_{10}+1)(5-x_{0})(5-x_{2})...(5-x_{9})}{(x_{10}-x_{0})(x_{10}-x_{2})...(x_{10}-x_{9})}$

since we know the formula for $L _{10,i}(5)$.

I need help simplifying this expression, because I think there should be a trick somewhere to making this question much easier to solve, but I just don't know it yet. I don't think it would be wise to expand this out as there are 11 $ x_{i}$ terms which would mean 11 x 10 = 110 terms in the denominator.

 

[jvicens]

Hello,

I understand your concern about simplifying this expression as it can be quite daunting to deal with a large number of terms. However, there is indeed a trick that can make this expression much easier to solve.

First, let's rewrite the expression in a slightly different form:
$\frac{(x_{0}+1)(x_{1}+1)(x_{2}+1)...(x_{10}+1)(5-x_{0})(5-x_{1})(5-x_{2})...(5-x_{10})}{(x_{0}-x_{1})(x_{0}-x_{2})...(x_{0}-x_{10})(x_{1}-x_{0})(x_{1}-x_{2})...(x_{1}-x_{10})...(x_{10}-x_{0})(x_{10}-x_{2})...(x_{10}-x_{9})}$

Now, let's focus on just one term in the numerator, say $(x_{0}+1)(5-x_{0})$. We can rewrite this as:
$(x_{0}+1)(5-x_{0}) = 5+x_{0}-x_{0}^{2}$.

Notice that the term $x_{0}$ cancels out, leaving us with $5-x_{0}^{2}$.

We can apply this same logic to each term in the numerator, and we will end up with:
$\frac{(5-x_{0}^{2})(5-x_{1}^{2})(5-x_{2}^{2})...(5-x_{10}^{2})}{(x_{0}-x_{1})(x_{0}-x_{2})...(x_{0}-x_{10})(x_{1}-x_{0})(x_{1}-x_{2})...(x_{1}-x_{10})...(x_{10}-x_{0})(x_{10}-x_{2})...(x_{10}-x_{9})}$

Now, we can simplify the expression further by noticing that each term in the denominator has a corresponding term in the numerator with the same exponent. For example, $(x_{0}-x_{1})$ in the denominator has a corresponding term $(5-x_{0}^{2})$ in the numerator. So, we can cancel out these terms, leaving us with:
$\frac{(5