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# System prepended metadata

title: Laplacian Expansion.

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$$\det\begin{pmatrix}
1&x_1&x_1^2&x_1^3&x_1^4\\
1&x_2&x_2^2&x_2^3&x_2^4\\
1&x_3&x_3^2&x_3^3&x_3^4\\
0&1&2x_3&3x_3^2&4x_3^3\\
0&0&2&6x_3&12x_3^2\\
\end{pmatrix}
$$


$$\det\begin{pmatrix}
1&x_1&x_1^2&x_1^3&f(x_1)\\
1&x_2&x_2^2&x_2^3&f(x_2)\\
1&x_3&x_3^2&x_3^3&f(x_3)\\
0&1&2x_3&3x_3^2&f'(x_3)\\
0&0&2&6x_3&f''(x_3)\\
\end{pmatrix}
$$

# Laplacian Expansion.

$$
\det\begin{pmatrix}
a&b&c\\
d&e&f\\
g&h&i\\
\end{pmatrix}=
c\cdot\det\begin{pmatrix}
a&b&1\\
d&e&0\\
g&h&0\\
\end{pmatrix}+f\cdot\det\begin{pmatrix}
a&b&0\\
d&e&1\\
g&h&0\\
\end{pmatrix}+i\cdot\det\begin{pmatrix}
a&b&0\\
d&e&0\\
g&h&1\\
\end{pmatrix}
$$

</br></br></br></br></br></br>

$$
0=\det\begin{pmatrix}
1&x_1&x_1\\
1&x_2&x_2\\
1&x_3&x_3\\
\end{pmatrix}=\det\begin{pmatrix}
1&x_1&x_1\\
1&x_2&0\\
1&x_3&0\\
\end{pmatrix}+\det\begin{pmatrix}
1&x_1&0\\
1&x_2&x_2\\
1&x_3&0\\
\end{pmatrix}+\det\begin{pmatrix}
1&x_1&0\\
1&x_2&0\\
1&x_3&x_3\\
\end{pmatrix}
$$
</br></br></br>
$$
\frac{0}{\det\begin{pmatrix}
1&x_1&x_1^2\\
1&x_2&x_2^2\\
1&x_3&x_3^2\\
\end{pmatrix}}=\frac{\det\begin{pmatrix}
1&x_1&x_1^2\\
1&x_2&x_2^2\\
1&x_3&x_3^2\\
\end{pmatrix}}{\det\begin{pmatrix}
1&x_1&x_1^2\\
1&x_2&x_2^2\\
1&x_3&x_3^2\\
\end{pmatrix}}=\frac{\det\begin{pmatrix}
1&x_1&x_1^2\\
1&x_2&0\\
1&x_3&0\\
\end{pmatrix}}{\det\begin{pmatrix}
1&x_1&(x_1-x_2)(x_1-x_3)\\
1&x_2&0\\
1&x_3&0\\
\end{pmatrix}}+\frac{\det\begin{pmatrix}
1&x_1&0\\
1&x_2&x_2^2\\
1&x_3&0\\
\end{pmatrix}}{\det\begin{pmatrix}
1&x_1&0\\
1&x_2&(x_2-x_1)(x_2-x_3)\\
1&x_3&0\\
\end{pmatrix}}+\frac{\det\begin{pmatrix}
1&x_1&0\\
1&x_2&0\\
1&x_3&x_3\\
\end{pmatrix}}{\det\begin{pmatrix}
1&x_1&x_1^2\\
1&x_2&x_2^2\\
1&x_3&x_3^2\\
\end{pmatrix}}
$$

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



</br></br></br></br>


Let $f(t)=(t-x_1)(t-x_2)(t-x_3)$

**Lebnitz Rule**
$$
f'(t)=(t-x_1)'(t-x_2)(t-x_3)+(t-x_1)(t-x_2)'(t-x_3)+(t-x_1)(t-x_2)(t-x_3)'
$$
$$
=(t-x_2)(t-x_3)+(t-x_1)(t-x_3)+(t-x_1)(t-x_2)
$$

</br></br></br></br></br>

Note that $f'(x_1)=(x_1-x_2)(x_1-x_3)\cdots (x_1-x_n)$


</br></br></br>

**Thm**
$$
\det\begin{pmatrix}
A&B\\0&C\\
\end{pmatrix}=\det(A)\det(C)
$$

</br></br></br></br></br></br>
$$
\det\begin{pmatrix}
1&x_1&x_1^2\\
1&x_2&x_2^2\\
1&x_3&x_3^2\\
\end{pmatrix}=S
$$
$$
\det\begin{pmatrix}
1&x_1&x_1\\
1&x_2&x_2\\
1&x_3&x_3\\
\end{pmatrix}=0
$$
$$
\det\begin{pmatrix}
1&x_1&1\\
1&x_2&1\\
1&x_3&1\\
\end{pmatrix}=0
$$

$$f(t)=at^2+bt+c$$

$$
\det\begin{pmatrix}
1&x_1&f(x_1)\\
1&x_2&f(x_2)\\
1&x_3&f(x_3)\\
\end{pmatrix}=S\cdot a
$$
</br></br></br></br></br></br>