Laplacian Expansion.

$$\det\begin{pmatrix} 1&x_1&x_1^2&x_1^3&x_1^4\\ 1&x_2&x_2^2&x_2^3&x_2^4\\ 1&x_3&x_3^2&x_3^3&x_3^4\\ 0&1&2x_3&3x_3^2&4x_3^3\\ 0&0&2&6x_3&12x_3^2\\ \end{pmatrix} $$ $$\det\begin{pmatrix} 1&x_1&x_1^2&x_1^3&f(x_1)\\ 1&x_2&x_2^2&x_2^3&f(x_2)\\ 1&x_3&x_3^2&x_3^3&f(x_3)\\ 0&1&2x_3&3x_3^2&f'(x_3)\\ 0&0&2&6x_3&f''(x_3)\\ \end{pmatrix} $$ # Laplacian Expansion. $$ \det\begin{pmatrix} a&b&c\\ d&e&f\\ g&h&i\\ \end{pmatrix}= c\cdot\det\begin{pmatrix} a&b&1\\ d&e&0\\ g&h&0\\ \end{pmatrix}+f\cdot\det\begin{pmatrix} a&b&0\\ d&e&1\\ g&h&0\\ \end{pmatrix}+i\cdot\det\begin{pmatrix} a&b&0\\ d&e&0\\ g&h&1\\ \end{pmatrix} $$ $$ 0=\det\begin{pmatrix} 1&x_1&x_1\\ 1&x_2&x_2\\ 1&x_3&x_3\\ \end{pmatrix}=\det\begin{pmatrix} 1&x_1&x_1\\ 1&x_2&0\\ 1&x_3&0\\ \end{pmatrix}+\det\begin{pmatrix} 1&x_1&0\\ 1&x_2&x_2\\ 1&x_3&0\\ \end{pmatrix}+\det\begin{pmatrix} 1&x_1&0\\ 1&x_2&0\\ 1&x_3&x_3\\ \end{pmatrix} $$ $$ \frac{0}{\det\begin{pmatrix} 1&x_1&x_1^2\\ 1&x_2&x_2^2\\ 1&x_3&x_3^2\\ \end{pmatrix}}=\frac{\det\begin{pmatrix} 1&x_1&x_1^2\\ 1&x_2&x_2^2\\ 1&x_3&x_3^2\\ \end{pmatrix}}{\det\begin{pmatrix} 1&x_1&x_1^2\\ 1&x_2&x_2^2\\ 1&x_3&x_3^2\\ \end{pmatrix}}=\frac{\det\begin{pmatrix} 1&x_1&x_1^2\\ 1&x_2&0\\ 1&x_3&0\\ \end{pmatrix}}{\det\begin{pmatrix} 1&x_1&(x_1-x_2)(x_1-x_3)\\ 1&x_2&0\\ 1&x_3&0\\ \end{pmatrix}}+\frac{\det\begin{pmatrix} 1&x_1&0\\ 1&x_2&x_2^2\\ 1&x_3&0\\ \end{pmatrix}}{\det\begin{pmatrix} 1&x_1&0\\ 1&x_2&(x_2-x_1)(x_2-x_3)\\ 1&x_3&0\\ \end{pmatrix}}+\frac{\det\begin{pmatrix} 1&x_1&0\\ 1&x_2&0\\ 1&x_3&x_3\\ \end{pmatrix}}{\det\begin{pmatrix} 1&x_1&x_1^2\\ 1&x_2&x_2^2\\ 1&x_3&x_3^2\\ \end{pmatrix}} $$ THerefore $$ \frac{\det\begin{pmatrix} 1&x_1&x_1^n\\ 1&x_2&x_2^n\\ 1&x_3&x_3^n\\ \end{pmatrix}}{\det\begin{pmatrix} 1&x_1&x_1^2\\ 1&x_2&x_2^2\\ 1&x_3&x_3^2\\ \end{pmatrix}}=\frac{x_1^n}{(x_1-x_2)(x_1-x_3)}+\frac{x_2^n}{(x_2-x_1)(x_2-x_3)}+\frac{x_3^n}{(x_3-x_1)(x_3-x_2)} $$ **Laplacian Balancing** Let $f(t)=(t-x_1)(t-x_2)\cdots(t-x_n)$. Then we have $$ \frac{x_1^k}{f'(x_1)}+\frac{x_2^k}{f'(x_2)}+\cdots+\frac{x_n^k}{f'(x_n)}=\begin{cases} 0;&k=0,1,2,...,n-2\\\\ 1&k=n-1 \end{cases}$$ Hint: $$ f'(t)=(t-x_2)\cdots(t-x_n)+(t-x_1)\cdots(t-x_n)+ $$ **Laplacian Interpolation Polynomials** $$f(t)=(t-x_1)(t-x_2)(t-x_3)(t-y)$$ **Laplacian Balancing** when $k=0,1,2$ $$ 0=\frac{x_1^k}{(x_1-x_2)(x_1-x_3)(x_1-y)}+\frac{x_2^k}{(x_2-x_1)(x_2-x_3)(x_2-y)}+\frac{x_3^k}{(x_3-x_1)(x_3-x_2)(x_3-y)}+\frac{y^k}{(y-x_1)(y-x_2)(y-x_3)} $$ Then we have $$ -\frac{y^k}{(y-x_1)(y-x_2)(y-x_3)}=\frac{x_1^k}{(x_1-x_2)(x_1-x_3)(x_1-y)}+\frac{x_2^k}{(x_2-x_1)(x_2-x_3)(x_2-y)}+\frac{x_3^k}{(x_3-x_1)(x_3-x_2)(x_3-y)} $$ $$ -y^k=x_1^k\cdot \frac{(y-x_2)(y-x_3)}{(x_1-x_2)(x_1-x_3)}\cdot(-1) + x_2^k\cdot \frac{(y-x_1)(y-x_3)}{(x_2-x_1)(x_2-x_3)}\cdot(-1) + x_3^k\cdot \frac{(y-x_1)(y-x_2)}{(x_3-x_1)(x_3-x_2)}\cdot(-1) $$ $$ y^k=x_1^k\cdot \frac{(y-x_2)(y-x_3)}{(x_1-x_2)(x_1-x_3)} + x_2^k\cdot \frac{(y-x_1)(y-x_3)}{(x_2-x_1)(x_2-x_3)} + x_3^k\cdot \frac{(y-x_1)(y-x_2)}{(x_3-x_1)(x_3-x_2)} $$ Then for any polynomial of degree 0,1,2, we have $$ f(y)=f(x_1)\cdot \frac{(y-x_2)(y-x_3)}{(x_1-x_2)(x_1-x_3)} + f(x_2)\cdot \frac{(y-x_1)(y-x_3)}{(x_2-x_1)(x_2-x_3)} + f(x_3)\cdot \frac{(y-x_1)(y-x_2)}{(x_3-x_1)(x_3-x_2)} $$ # Application Ex. Find a polynomial of degree 2 such that $f(1)=3$, $f(2)=6$, $f(3)=7$ **Solution:** $$ f(y)=3\cdot\underbrace{\frac{(y-2)(y-3)}{(1-2)(1-3)}}_{f_1(y)} +6\cdot \underbrace{\frac{(y-1)(y-3)}{(2-1)(2-3)}}_{f_2(y)}+7\cdot \underbrace{\frac{(y-1)(y-2)}{(3-1)(3-2)}}_{f_3(y)} $$ $$ f(y)=3f_1(y)+6f_2(y)+7f_3(y) $$ $y=$|$f$|$f_1$|$f_2$|$f_3$ -|-|-|-|- 1|3|1|0|0 2|6|0|1|0 3|7|0|0|1 Let $f(t)=(t-x_1)(t-x_2)(t-x_3)$ **Lebnitz Rule** $$ f'(t)=(t-x_1)'(t-x_2)(t-x_3)+(t-x_1)(t-x_2)'(t-x_3)+(t-x_1)(t-x_2)(t-x_3)' $$ $$ =(t-x_2)(t-x_3)+(t-x_1)(t-x_3)+(t-x_1)(t-x_2) $$ Note that $f'(x_1)=(x_1-x_2)(x_1-x_3)\cdots (x_1-x_n)$ **Thm** $$ \det\begin{pmatrix} A&B\\0&C\\ \end{pmatrix}=\det(A)\det(C) $$ $$ \det\begin{pmatrix} 1&x_1&x_1^2\\ 1&x_2&x_2^2\\ 1&x_3&x_3^2\\ \end{pmatrix}=S $$ $$ \det\begin{pmatrix} 1&x_1&x_1\\ 1&x_2&x_2\\ 1&x_3&x_3\\ \end{pmatrix}=0 $$ $$ \det\begin{pmatrix} 1&x_1&1\\ 1&x_2&1\\ 1&x_3&1\\ \end{pmatrix}=0 $$ $$f(t)=at^2+bt+c$$ $$ \det\begin{pmatrix} 1&x_1&f(x_1)\\ 1&x_2&f(x_2)\\ 1&x_3&f(x_3)\\ \end{pmatrix}=S\cdot a $$