Hermitte Interpolation

The Goal: A More “Snug” Fit

With Lagrange Interpolation , we found a polynomial $P_n(x)$ that matched $n+1$ function values $f(x_i)$.

With Hermite, we want a polynomial $H(x)$ that matches both the function value $f(x_i)$ and the derivative value $f^{\prime }(x_i)$ at all $n+1$ points.

• Given: We have $n+1$ nodes ($x_0,\dots ,x_n$).

• Conditions: For each node $x_i$, we have two conditions:

  1. $H(x_i)=f(x_i)$

  2. $H^{\prime }(x_i)=f^{\prime }(x_i)$

• Total Conditions: This gives us $2\times (n+1)=2n+2$ total conditions.

• Resulting Polynomial: To satisfy $2n+2$ conditions, we need a polynomial with $2n+2$ coefficients, which is a polynomial of degree (at most) $2n+1$.

---

The Construction: Building $H(x)$

Just like Lagrange, we’ll build $H(x)$ as a weighted sum. But this time, we have weights for both $f(x_i)$ and $f^{\prime }(x_i)$:

$H(x)={\sum }_{i=0}^n\left[{\alpha }_i(x)f(x_i)+{\beta }_i(x)f^{\prime }(x_i)\right]$

Our job is to find the basis polynomials ${\alpha }_i(x)$ and ${\beta }_i(x)$.

To make this work, we need to enforce our $2n+2$ conditions. This leads to 4 requirements for our basis polynomials at any node $x_j$:

1. $H(x_j)=f(x_j)$:

   • We need the $f(x_j)$ term to be 1: ${\alpha }_j(x_j)=1$.

   • All other $f(x_i)$ terms must be 0: ${\alpha }_i(x_j)=0$ (for $i\ne j$).

   • All $f^{\prime }(x_i)$ terms must be 0: ${\beta }_i(x_j)=0$ (for all $i$).

2. $H^{\prime }(x_j)=f^{\prime }(x_j)$:

   • We need the $f^{\prime }(x_j)$ term to be 1: ${\beta }_j^{\prime }(x_j)=1$.

   • All other $f^{\prime }(x_i)$ terms must be 0: ${\beta }_i^{\prime }(x_j)=0$ (for $i\ne j$).

   • All $f(x_i)$ terms must be 0: ${\alpha }_i^{\prime }(x_j)=0$ (for all $i$).

---

The “Aha!” Moment: Using $L_i(x{)}^2$

Your notes show the key insight. We need to find ${\alpha }_i(x)$ and ${\beta }_i(x)$ that satisfy these conditions. Let’s look at the Lagrange polynomial $L_i(x)$.

• We know $L_i(x_j)=0$ for $i\ne j$.

• If we use $(L_i(x){)}^2$, it’s still zero at $x_j$ (for $i\ne j$).

• Critically, the derivative of $(L_i(x){)}^2$ is $2L_i(x)L_i^{\prime }(x)$. At $x_j$ (for $i\ne j$), this derivative is $2L_i(x_j)L_i^{\prime }(x_j)=2\cdot 0\cdot L_i^{\prime }(x_j)=0$.

This $(L_i(x){)}^2$ term almost gives us what we need! It satisfies most of the “zero” conditions. We just need to multiply it by a simple linear factor $(ax+b)$ to fix the remaining conditions at $x=x_i$.

---

Deriving the Basis Polynomials (as in your notes)

Your notes perfectly follow this logic to find the 4 unknown coefficients ($a_i,b_i,c_i,d_i$) for each ${\alpha }_i$ and ${\beta }_i$.

1. Finding ${\alpha }_i(x)=(a_{i}x+b_i)(L_i(x){)}^2$

• Condition: ${\alpha }_i(x_i)=1$

  • $(a_i{x}_i+b_i)(L_i(x_i){)}^2=1$

  • Since $L_i(x_i)=1$, this gives: $a_i{x}_i+b_i=1$ (Your equation $\stackrel{◯}{1}$)

• Condition: ${\alpha }_i^{\prime }(x_i)=0$

  • First, get the derivative ${\alpha }_i^{\prime }(x)=a_i(L_i(x){)}^2+(a_{i}x+b_i)(2L_i(x)L_i^{\prime }(x))$

  • Plug in $x_i$: ${\alpha }_i^{\prime }(x_i)=a_i(L_i(x_i){)}^2+(a_i{x}_i+b_i)(2L_i(x_i)L_i^{\prime }(x_i))$

  • Sub in $L_i(x_i)=1$ and $a_i{x}_i+b_i=1$: $a_i(1{)}^2+(1)(2\cdot 1\cdot L_i^{\prime }(x_i))=0$

  • This simplifies to $a_i+2L_i^{\prime }(x_i)=0$, which gives:

    • $a_i=-2L_i^{\prime }(x_i)$ (Your purple box)

• Substitute $a_i$ back into $\stackrel{◯}{1}$:

  • $(-2L_i^{\prime }(x_i))x_i+b_i=1$

  • $b_i=1+2x_i{L}_i^{\prime }(x_i)$ (Your other purple box)

2. Finding ${\beta }_i(x)=(c_{i}x+d_i)(L_i(x){)}^2$

Your notes set this up but don’t complete it. Let’s finish the derivation:

• Condition: ${\beta }_i(x_i)=0$

  • $(c_i{x}_i+d_i)(L_i(x_i){)}^2=0$

  • Since $L_i(x_i)=1$, this gives: $c_i{x}_i+d_i=0$

• Condition: ${\beta }_i^{\prime }(x_i)=1$

  • First, get the derivative ${\beta }_i^{\prime }(x)=c_i(L_i(x){)}^2+(c_{i}x+d_i)(2L_i(x)L_i^{\prime }(x))$

  • Plug in $x_i$: ${\beta }_i^{\prime }(x_i)=c_i(L_i(x_i){)}^2+(c_i{x}_i+d_i)(2L_i(x_i)L_i^{\prime }(x_i))$

  • Sub in $L_i(x_i)=1$ and $c_i{x}_i+d_i=0$: $c_i(1{)}^2+(0)(2\cdot 1\cdot L_i^{\prime }(x_i))=1$

  • This simplifies to: $c_i=1$

• Substitute $c_i$ back:

  • $(1)x_i+d_i=0$

  • $d_i=-x_i$

By finding these coefficients, you have successfully derived the Hermite basis polynomials.

Would you like to see how these coefficients $a_i,b_i$ and $c_i,d_i$ are typically simplified into their final, standard forms?