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16

Linear functions (degree $1$ polynomials) $p(x)$ have constant first differences $(\Delta^1f)(x)=f(x)-f(x-1)$, quadratic functions have constant second differences $(\Delta^2f)(x)=(\Delta^1f)(x)-(\Delta^1f)(x-1)$, and so on. You can use this fact to find $f(0)$ as shown below. First, from the known values, calculate the forward differences (red) from left to ...

7

There is a conceptually simpler way: do some linear algebra and polynomial arithmetic. Instead of a complex linear system, solve $4$ much simpler linear systems. Namely, solve the following problems: find polynomials $p(x), q(x), r(x), s(x)$ such that: \begin{align} (a)\enspace&\begin{cases}p(3)=1\\p(4)=0\\p(5)=0\\p(6)=0\end{cases} &(b)\enspace&...

7

There is this theorem: Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| \to \infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$. It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem. See this question.

6

Frankly, Lagrange interpolation is mostly just useful for theory. Actually computing with it requires huge numbers and catastrophic cancellations. In floating point arithmetic this is very bad. It does have some small advantages: for instance, the Lagrange approach amounts to diagonalizing the problem of finding the coefficients, so it takes only linear time ...

5

Hint Let $g(x)=(x+1)f(x)-x$. This is a polynomial with degree $n+1$ and you know its roots, namely $x=0,1,2, \ldots ,n$. Thus $$g(x)=Ax(x-1)(x-2) \dotsb (x-n),$$ where $A$ can be determined to be $A=\frac{(-1)^{n+1}}{(n+1)!}$

5

There is a unique polynomial of degree $\leq$ (number of points - 1) going through the points. Lagrange and Newton both produce such a thing, therefore the same thing.

5

You can directly find out the polynomial $f$ by considering it according as the $x$-values available: Let $f(x)=a_0+a_1(x-3)+a_2(x-3)(x-4)+a_3(x-3)(x-4)(x-5)$ for real constants $a_0,a_1,a_2,a_3$. Note that we don't need to take into account the value $x=6$ as this is already a cubic polynomial. Then, $f(3)=2\Rightarrow a_0=2$ $f(4)=4\Rightarrow a_1=2$ $... 5 For the$n+1$points$\{\left(x_0,f(x_0)\right),...,\left(x_n,f(x_n)\right)\}$the Lagrange interpolating polynomial is defined as $$P(x)=\sum_{j=0}^nP_j(x)=\sum_{j=0}^n f(x_j)\prod_{k=0\\{k\ne j}}^{n}\frac{x-x_k}{x_j-x_k}$$ Now since the$x_j=j,\; (j=0,1,...,n)$the product can be simplified. Let$p_n(x):=x(x-1)\cdots(x-n)then \begin{align}\prod_{k=0\\{... 4 Given |P(x)|<1\ \forall\ x<1\ \ \ \ \ -----(I) If a\ne0 then \displaystyle\lim_{x\to-\infty}|P(x)|=\infty, this violates (I), hence a=0. Similarly for b and c, if b\ne0 then \displaystyle\lim_{x\to-\infty}|P(x)|=\infty, this violates (I), hence b=0, if c\ne0 then \displaystyle\lim_{x\to-\infty}|P(x)|=\infty, this violates (... 4 The basic Lagrange interpolation with many points is primarily of theoretical interest. Polynomials are pretty well-behaved and well-understood functions, and it is sometimes useful to know while conducting a proof that there exists a well-behaved function that goes through any finite number of points you care to specify. In practice, however, Lagrange ... 4 is there anything wrong with the unique parabola going through those points? I'll assume you mean three points in general position (no two are colinear.) In general, you can take ax^2+bx+c=y, and use all three points to determine a 3 \times 3 system of linear equations to solve for coefficients. In a similar fashion, you could take (x-a)^2+(y-b)^2=r^2... 3 In short, this is product rule. If you are comfortable with the notation for the product rule of N factors (see e.g. Wikipedia), denoted by "(\text{PR})" below, you may calculate:\begin{align*}D_x[(x-x_0)\ldots(x-x_n)]\rvert_{x=x_j}&=D_x[\prod_{i=1}^n (x-x_i)]\rvert_{x=x_j}\\ &\overset{\text{(PR)}}{=}\sum_{i=0}^n\Bigg( \underbrace{D_x[x-x_i]\... 3 Let me just quickly provide the reference to the original question (It was posted by me and I had some problems to derive the formula: Link) So we start with \begin{align} L_j'(x) = \sum_{l\not = j} \frac{1}{x_j-x_l}\prod_{m\not = (j,l)} \frac{x-x_m}{x_j-x_m} \end{align} as you already mentioned, the interesting part is the product \begin{align} \prod_{m\... 3 Actually if\beta$is not one of your$x_j$'s then it's impossible. Indeed, the two polynomials$P$and$P'$have degree$n-1$(at most) and take the same value on all$x_j$for$j\neq i$(namely,$y_j$), so if they both vanished at$\beta$they would have the same value at$n$points, and thus they would be equal, which is not the case since they have ... 3 It can be easily proven using partial fraction expansion. Since this expansion is entailed by Lagrange's interpolation formula for the numerator, it may give you an interesting view of the problem. For$0 \leqslant m \leqslant n$we have the expansion \frac{t^{m}}{t(t+1)\dotsm(t+n)}=\sum_{k=0}^{n}{\frac{c_{k}}{t+k}}.\label{eq}\tag{E} \end{... 3$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}...

3

As $\omega - z$ divides $p(\omega) - p(z)$, this quotient is actually a polynomial: $$\frac{p(\omega) - p(z)}{\omega - z} = R(\omega,z) = \sum_{k=0}^{n-1} R_k(\omega)z^k.$$ Then, $$P(z) = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\frac{p(\omega) − p(z)}{\omega − z}d\omega = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\left(\sum_{k=0}^{n-1} R_k(\... 3 If L[x_0, \ldots, x_n; f] means a polynomial of degree  \le n that has the same values as f at x_0, \ldots, x_n, then yes, this is true. 3 In your notation, p(x) is a Lagrange interpolation of degree n-1, hence q(x) is of degree n-1 > 3. Therefore, if q(x) is not the zero polynomial, it can have at most n-1 roots for q(x) = 0. On the other hand, because p(\alpha_i) = \alpha_i^3, you already know that$$\{\alpha_i\mid I = 1,2, \dots, n\}$$are all roots of q(x). Hence, if ... 3 Here is the answer I sketched in the comments, in more detail. Background. Let \mathbb{N}=\left\{ 0,1,2,\ldots\right\} . Fix a field \mathbb{K}. Let V be a \mathbb{K}-vector space. (In your setting, V=\mathbb{K}, but there is nothing gained from this specialization.) If \lambda\in\mathbb{K} is nonzero and if v\in V, then \dfrac{v}{\lambda}... 2 Yes, with some modification, we can use the Lagrange interpolation method. First, note that (y+z)(z+x)(x+y)=(x+y+z)(yz+zx+xy)-xyz. As x+y+z=1, yz+zx+xy=2, and xyz=3, (y+z)(z+x)(x+y)=1\cdot 2-3=-1. The required expression is then$$\begin{align} \sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\right)} =\left(\frac{1}{(y+z)(z+x)(x+y)}\...

2

Here are some thoughts about the problem where I discuss how one would go about if one wants to use the Lagrange interpolation method and some obstacles to applying it. I also give an example of a related problem where the method can easily be applied. I don't consider this a full answer, but it's too long for a comment and might be useful. Added: It ...

2

Let's say you have a vector space (which you want to interpolate in) $\Phi$ of dimension $n$. Then, the nodes would be $n$ points $(\xi_1, \ldots, \xi_n)$ in the domain of your basis functions which satisfy the following: for arbitrary given values $(y_1, \ldots, y_n)$, you can find a unique $f \in \Phi$ such that $$f(\xi_i) = y_i \qquad \forall i=1,\... 2 Let \omega:=\exp\left(\frac{2\pi\text{i}}{3}\right)=\frac{-1+\sqrt{-3}}{2}. Define$$\tau(z):=\left(\frac{1-\omega^2}{3}\right)z^2+\left(\frac{1-\omega}{3}\right)z+1\text{ for all }z\in\mathbb{C}\,.$$Note that, for each k\in\mathbb{Z}, we have$$\tau(\omega^k)=\left\{ \begin{array}{ll} 2&\text{if }k\equiv0\pmod{3}\,,\\ 1&\text{if }k\equiv1\pmod{...

2

The OP is equivalent to: given \begin{align} P(x)&=\sum_{i=0}^{3n} a_i x^i \tag{1}\label{1} ,\\ P(3i-2)&=1,\quad i=1,\cdots,n \tag{2}\label{2} ;\\ P(3i-1)&=0,\quad i=1,\cdots,n \tag{3}\label{3} ;\\ P(3i)&=2,\quad i=1,\cdots,n \tag{4}\label{4} ;\\ P(3n+1)&=730 \tag{5}\label{5} ;\\ P(0)&=2 \tag{6}\label{6} , \end{align} determine $n$...

2

Define the polynomials $$L_i(x) = \underset{j \neq i}{\prod_{0 \le j \le n}} \frac{(x-x_j)}{(x_i - x_j)} = \begin{cases} 1 & \text{ if } x = i \\ 0 & \text{ if } x = j \neq i \end{cases}$$ for each $i \in \{0,1,\cdots,3n\}$. It follows that since the $L_i$'s have degree $3n$, we have (I leave the computations up to you) : $$P(x) = \sum_{i=0}^{3n}... 2 Let y_0,\dots,y_n be such that |y_i|<\frac{n!}{2^n} for all i. Let us prove that the interpolating polynomial for (x_i,y_i) cannot have a leading coefficient of 1. I'll work with the case x_i=i+s for a fixed integer s and leave it to you to generalize. We write the interpolating polynomial in Lagrange form:$$P(x)=\sum_{i=0}^n y_i \prod_{j=...

2

Since $P$ has degree $n$ and $P'$ has degree $n-1$ at most, $P-P'$ has degree $n$, so it has $n$ distincts roots at most. Since $P(x_i)=y_i=P'(x_i)$ for $i\in\{1,\dots,n\}$, $P'(\beta) \neq P(\beta)=0$, so $P'$ cannot have $\beta$ has a root.

2

Start with $T_{n-1}(x)$, the Chebyshev Polynomial of the First Kind. From Lagrange Interpolation, we have $$T_{n-1}(x)=\sum_{i=1}^n T_{n-1}(x_i) \prod_{j \not= i} \frac{x-x_j}{x_i-x_j}$$ By Triangle Inequality, we have $$|T_{n-1}(x)| \le \sum_{i=1}^n |T_{n-1}(x_i)| \prod_{j \not= i} |\frac{x-x_j}{x_i-x_j}|$$ Dividing by $x^{n-1}$, we have |\frac{T_{n-1}(...

2

$\forall x\in (0,1)$, let $\displaystyle \phi_{x}(t)=f(t)-p(t)-\frac{f(x)-p(x)}{x^{2}(x-1)^{2}} t^{2}(t-1)^{2}$ where $t\in [0,1]$ Note that $\phi_{x}$ has three roots $t=0,x,1$. $\exists \alpha_{i}\in (0,1)$ such that $\phi_{x}'(\alpha_{i})=0$ and $\alpha_{i} \notin \{0,x,1\}$ for $i=1,2$ But $\phi_{x}'(0)=\phi_{x}'(1)=0$, i.e. $\phi_{x}'$ has four ...

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