# Interpolation with a new point of $f'$

Background:

(Lagrange Interpolation) Let $$f\in C^{n+1}([a,b])$$ and $$x_0,...,x_n\in[a,b]$$. If they are different there is a unique $$p_n\in\mathcal{P}_n$$ such that $$p_n(x_i)=f(x_i)$$. Also, we have that for each $$x\in[a,b]$$ exist $$\xi_x\in[a,b]$$ 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, $$p\in \mathcal{P}_{n+1}$$ 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 $$p\in\mathcal{P}_{n+2}$$ 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?

• By the claim in "In this case, I can prove...", you know there's a $p\in\mathcal{P}_{(n+1)+1}$ attaining specified $p(x_0),\dots,p(x_n),p(x_{n+1}),p'(x_{n+1}).$ Then you're asking whether there's a $p\in\mathcal{P}_{n+2}$ attaining specified $p(x_0),\dots,p(x_n),p'(x_{n+1}).$ But that's an easier problem! You are free to specify an arbitrary value for $p(x_{n+1}),$ like $p(x_{n+1})=42.$ – Dap Jan 1 at 10:43
• $p\in\mathcal P_{n+1},$ due to corresponding choice of $p(x+1).$ Example in my answer shows how to do it, so the proof is not a problem. – Yuri Negometyanov Jan 6 at 0:14

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 $$p\in\mathcal{P}_{n+2}$$.

Dap’s comment shows how to find such $$p\in \mathcal P_{n+2}$$, but it is natural to look for such $$p\in\mathcal P_{n+1}$$. Pick any $$p_0\in\mathcal P_{n+1}$$ 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 $$p\in\mathcal P_{n+1}$$ 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 $$\lambda\in\Bbb R$$ (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 $$p\in\mathcal P_{n+2}$$ 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 $$\quad p(x) = 12\sin \dfrac \pi6\,x\ \text{ for } x\in \{x_0, x_1, x_2\},\quad \text{and}\quad p'(x_3)=1.$$

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 $$p_3(x)=p\left(x, \dfrac{102}{11}\right)=-\dfrac5{11}x^3+\dfrac{71}{11}x,$$ $$p'_3(x)=-\dfrac{15}{11}x^2+\dfrac{71}{11}.$$ $$p_3(-1)=-6,\quad p_3(0)=0,\quad p_3(1)=6,\quad p'_3(2)=1.$$

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

This mean that $$\boxed{p\in\mathcal P_{n+1}}.$$

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 $$x_{n+1}$$.