Csdrhrt

Background:

(Lagrange Interpolation) Let $fin C^{n+1}([a,b])$ and $x_0,...,x_nin[a,b]$. If they are different there is a unique $p_ninmathcal{P}_n$ such that $p_n(x_i)=f(x_i)$. Also, we have that for each $xin[a,b]$ exist $xi_xin[a,b]$ such that

$$f(x)-p_n(x)=dfrac{f^{(n+1)}(xi_x)}{(n+1)!}W_{n+1}(x)$$
where $W_{n+1}(x)=(x-x_0)(x-x_1).dots.(x-x_n)$.

Now, we can consider other problem: find $p(x)$ such that $p(x_i)=f(x_i)$ for $i=0,...,n$ and $p'(x_n)=f'(x_n)$. In this case, I can prove that the additional data $p'(x_n)=f'(x_n)$ is equivalent to interpolate $f$ into one more data, ie, $pin mathcal{P}_{n+1}$ and the bound is similar with n+2.

Intuitively, this fact is because the problem is equivalent that the problem: "find $p$ such that $p(x_i)=f(x_i)$ for $i=0,...,n+1$ with $x_{n+1}=x_n +epsilon$", using that $f'(x_n)sim dfrac{f(x_n)-f(x_{n+1})}{epsilon}$.

This problem can be generalizated (Hermite interpolation) for more $x_j$'s and $f'', f'''$, etc, always in $x_i$ with $i=0,...,n$.

And here is my question:

Consider the problem: find $p$ such that $p(x_i)=f(x_i)$ for $i=0,...,n$ and $p'(x_{n+1})=f'(x_{n+1})$, where $x_{n+1}in [a,b]$ is a new point.

My intuition says that $pinmathcal{P}_{n+2}$ because one (new) point in $f'$ can be replaced with two points in $f$, $x_{n+1}-epsilon$ and $x_{n+1}+epsilon$. Is this correct? Any proof?

I’ll assume that $mathcal P_n$ is the set of all polynomials with real coeffiecients of degree at most $n$.

find $p$ such that $p(x_i)=f(x_i)$ for $i=0,...,n$ and $p'(x_{n+1})=f'(x_{n+1})$, where $x_{n+1}in [a,b]$ is a new point. ... My intuition says that $pinmathcal{P}_{n+2}$.

Dap’s comment shows how to find such $pin mathcal P_{n+2}$, but it is natural to look for such $pinmathcal P_{n+1}$. Pick any $p_0inmathcal P_{n+1}$ such that $p(x_i)=f(x_i)$ for $i=0,dots,n$. For instance, we can put as $p_0$ the unique polynomial of $mathcal P_n$ with this property. Let $pinmathcal P_{n+1}$ be any polynomial such that $p(x_i)=f(x_i)$ for each $i=0,dots,n$. Then $(p-p_0)(x_i)=0$ for each $i$, so $(p-p_0)(x)= lambda W_{n+1}(x)$ for some $lambdainBbb R$ (we recall that $W_{n+1}(x)=(x-x_0)dots (x-x_n)$). It remains to choose $lambda$ to satisfy $p'(x_{n+1})=f'(x_{n+1})$, that is $f'(x_{n+1})-p’_0(x_{n+1})=lambda W’_{n+1}(x_{n+1})$. If $W’_{n+1}(x_{n+1})ne 0$ then there exists a unique such $lambda$. Otherwise we are lucky if $ p’_0(x_{n+1})$ already equals $f'(x_{n+1})$ but have to look for a polynomial $pinmathcal P_{n+2}$ as is described in Dap’s comment, otherwise.

I think that requirements to $f(x)$ is not valuable, because really we have the set of points.

Let us consider the example
$$quad p(x) = 12sin dfrac pi6,x text{ for } xin {x_0, x_1, x_2},quad text{and}quad p'(x_3)=1.$$

Last condition can be provided due to "right" choice of value $p(x_3).$

Denote $mathbf{p(x_3)=t}$ and construct Lagrange interpolatiщn polynomial with
unknown parameter $t.$

Then calculate parameter $t,$ using the
condition to the polynomial derivative in the given point.

The data table is
begin{vmatrix}
x_0 & p_0 & x_1 & p_1 & x_2 & p_2 & x_3 & p_3 & (p'(x_3))\
-1 & -6 & 0 & 0 & 1 & 6 & 2 & t & 1\
end{vmatrix}
Let us construct Lagrange interpolation polynomial
$$p(x,t) = dfrac{x(x-1)(x-2)}{(-1)(-1-1)(-1-2)}(-6)
+dfrac{(x+1)x(x-2)}{(1+1)1(1-2)}6
+dfrac{(x+1)x(x-1)}{(2+1)2(2-1)}t$$
$$p(x,t) = x(x-1)(x-2)-3(x+1)x(x-2) + (x+1)x(x-1)dfrac t6,$$
$$p(x,t)=dfrac t6(x^3-x)+8x-2x^3.$$
The derivative is
$$p'(x,t)=dfrac t6(3x^2-1)+8-6x^2,$$
$$p'(2,t)=dfrac{11}6t-16 = 1,$$
so
$$t=dfrac{102}{11}.$$
This allows to obtain the required polynomial
$$p_3(x)=pleft(x, dfrac{102}{11}right)=-dfrac5{11}x^3+dfrac{71}{11}x,$$
$$p'_3(x)=-dfrac{15}{11}x^2+dfrac{71}{11}.$$
$$p_3(-1)=-6,quad p_3(0)=0,quad p_3(1)=6,quad p'_3(2)=1.$$

Easy to see that one new condition increments required polynomial degree on $1,$
due to the value choice.

This mean that
$$boxed{pinmathcal P_{n+1}}.$$

Remark. Can be used, for example, condition $p(3)=t$ instead $p(x_3)=t.$

Let $p_i$ for $i=1,$...$n+1$ be the usual Lagrange polynomial which is zero at each $x_j$ except $x_i$, where it is 1. Then let $q_i$ be given by $p_i(x) (x- x_{n+1})$ for each $i$. Normalize each $q_i$ up to $q_n$ so they are 1 at $x_i$, and $q_{n+1}$ so that it’s derivative at $x_{n+1}$ is 1. Now each of our polynomials has a double root at $x_{n+1}$ except the last one, whose derivative is 1 there. Take now the linear combination $sum_{i=1}^n a_i q_i + b q_{n+1}$. So this way we need degree $n+2$.

Slightly less elegantly we can do it this way: specify out Lagrange polynomials at points $x_1$ through $x_n$ and then at some other point $y$, just so long as as the the polynomial which is 1 at $y$ does not have derivative 0 at $x_{n+1}$. Then taking linearly combinations of these $n+1$ polynomials we see that we can specify the values at $x_1$ through $x_n$ with the first $n$ coefficients, and then by varying the coefficient of that last polynomial we can control the derivative of the linear combination at $x_{n+1}$.