Finite Taylor series of Lagrange Basis functions

Question

Lagrange basis functions are defined as follows $$L_{j}(x) = \prod_{i\neq j} \frac{x-x_{i}}{x_{j}-x_{i}} $$

then $$\ln\Big(L_{j}(x)\Big) = \ln\Big(\prod_{i\neq j} \frac{x-x_{i}}{x_{j}-x_{i}} \Big) = \sum_{i \neq j } \ln\Big( \frac{x-x_{i}}{x_{j}-x_{i}} \Big) $$

if we derivate we have:

$$ \frac{L'_{j}(x)}{L_{j}(x)} =\sum_{i \neq j} \frac{\frac{1}{x_{j}-x_{i}}}{\frac{x-x_{i}}{x_{j}-x_{i}}} = \sum_{i \neq j} \frac{1}{x-x_{i}} $$ then

$$ L_{j}^{(1)}(x) = L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big) $$

if we use the product rule for derivatives we have that:

$$ L''_{j}(x) = L'_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)+L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)' \\ = L'_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)+L_{j}(x) \Big( \sum_{i \neq j} \frac{-1}{(x-x_{i})^2} \Big) $$ we know $ L'_{j}(x) $ so $$ \\ = L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)\Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)-L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{(x-x_{i})^2} \Big)\\ = L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)^2 -L_{j}(x) \Big( \sum_{i \neq j} \frac{1}{(x-x_{i})^2} \Big) $$ finally $$ L_{j}^{(2)}(x) = L_{j}(x) \Big\{\Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)^2 - \sum_{i \neq j} \frac{1}{(x-x_{i})^2} \Big\} $$

if we continue this way we get:

$$ L_{j}^{(3)}(x) = L_{j}(x)\Big\{ \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big)^3 -3 \Big( \sum_{i \neq j} \frac{1}{x-x_{i}} \Big) \Big( \sum_{i \neq j} \frac{1}{(x-x_{i})^2} \Big) +2\Big( \sum_{i \neq j} \frac{1}{(x-x_{i})^3} \Big) \Big\} $$

I want to compute the Taylor Series of $ L_{j}(x) $ and I need to find $ L_{j}^{(n)}(x) $ . Do you know an expression for $ L_{j}^{(n)}(x) $ or something that could help ? Thank you.

Answer 1

Since $L_j$ is a polynomial, in order to compute the Taylor series, you don't need to compute the derivatives: Taylor series of a polynomial is the polynomial itself because the derivatives are eventually zero.

If $L_{j}$ is a polynomial of degree $n-1$ then, by Vieta's formulas, the coefficient of $x^k$ for $k=0,1,\dots,n-1$ is $$a_k:=[x^k]L_{j}(x) = [x^k]\prod_{i\neq j} \frac{x-x_{i}}{x_{j}-x_{i}}=\frac{(-1)^{n-1-k} \sum_{|I|=n-1-k}\prod_{i\in I}x_{i}}{\prod_{i\neq j} (x_{j}-x_{i})}$$ where the sum is taken over all subsets $I$ of $\{1,2,\dots,j-1,j+1,\dots,n\}$ of cardinality $n-1-k$. Hence the Taylor expansion of $L_j$ at $0$ is $$L_j(x)=\sum_{k=0}^{n-1}a_kx^k.$$ If you need the Taylor expansion at $x_0$ then consider $$\begin{align}L_j(x)&=\sum_{k=0}^{n-1}a_k((x-x_0+x_0)^k=\sum_{k=0}^{n-1}a_k\sum_{i=0}^{k}\binom{k}{i}x_0^{k-i}(x-x_0)^i\\ &=\sum_{i=0}^{n-1}b_i(x-x_0)^i \end{align}$$ where $\displaystyle b_i=\sum_{k=i}^{n-1}a_k\binom{k}{i}x_0^{k-i}.$

For example, if $x_1=1$, $x_2=2$ and $x_3=3$, then the Taylor expansion of $L_2$ at $-2$ is $$L_2(x)=\frac{(x-1)(x-3)}{(2-1)(2-3)}= -15+8(x+2)-(x+2)^2$$ where $a_0=-3$, $a_1=4$, $a_2=-1$, and $$\begin{align} b_2&=\sum_{k=2}^{3-1}a_k\binom{k}{2}(-2)^{k-2}=a_2=-1,\\ b_1&=\sum_{k=1}^{3-1}a_k\binom{k}{1}(-2)^{k-1}=a_1-4a_2=4-4\cdot (-1)=8,\\ b_0&=\sum_{k=0}^{3-1}a_k\binom{k}{0}(-2)^{k-0}=a_0-2a_1+4a_2=-3-2\cdot 4+4\cdot (-1)=-15. \end{align}$$

Answer 2

Taylor's Theorem says that if $f$ has derivatives $f^{(1)},f^{(2)},...,f^{(n)},f^{(n+1)}$ defined on an interval $[a,x]$ then

$$f(x) - T_{a}^{n}\{f\}(x) = \frac{f^{(n+1)}(\lambda)}{(n+1)!}(x-a)^{(n+1)}$$

for some $\lambda \in (a,x)$ and where

$$T_{a}^{n}\{f\}(x) = \sum_{k=0}^{n} { \frac{f^{(k)}(a)(x-a)^k}{k!} } $$

if we take $f(x) = a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\ \ $ a polynomial of degree $n$... note that $f^{(n+1)}(x) = 0$. For example $h(x)=x^2$ has derivatives $h^{(1)}(x)=2x$, $h^{(2)}(x)=2$ and finally $h^{(3)}(x) =0$

Then

$$f(x) - T_{a}^{n}\{f\}(x) = \frac{f^{(n+1)}(\lambda)}{(n+1)!}(x-a)^{(n+1)} = \frac{0}{(n+1)!}(x-a)^{(n+1)} = 0$$ and this implies that

$$ T_{a}^{n}\{f\}(x) = f(x) = a_{n}x^n+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} $$

Thus the Taylor Series of a polynomial is the polynomial. This is just a particular case.