## 数学代写|数值分析代写numerical analysis代考|Existence of B-Spline

The infinite number of points $\Omega_{\infty}=\left{x_i\right}_{i \in \mathcal{Z}}$ with $x_i<x_{i+1}$ are considered so that
$$\left{\begin{array}{l} i \rightarrow-\infty \Rightarrow x_i \rightarrow-\infty \ i \rightarrow \infty \Rightarrow x_i \rightarrow \infty \end{array}\right.$$
We have to show that for every $i \in \mathbb{Z}$, there is exactly one spline $S \in S_l\left(\Omega_{\infty}\right)$, so that $S(x)=0$ for $x_{i+l+1} \geq x$ and $x<x_i$, and if the normalization condition is also met, then we have:
$$\int_{-\infty}^{\infty} S(x) d x=\int_{x_i}^{x_{i+t+1}} S(x) d x=1$$
Proof: For $x \in\left[x_{i-1}, x_{i+l+2}\right]$, it is clear that the expansion of spline $S$ in terms of one-way bases as
$$S(x)=\sum_{i=0}^l a_i x_i+\sum_{i=1}^{n-1} b_i\left(x-x_i\right)_{+}^l$$ does not include all or part of the basis polynomials because for $x<x_i, S(x)=0$. So, we can say that on the above-mentioned interval:
$$S(x)=\sum_{k=0}^m b_k\left(x-x_{i+k}\right){+}^l$$ which is determined for some values of $m$. Now if we assume $m=l+1$, we have: $$S(x)=0, \quad x \geq x{i+l+1}$$
and also,
$$\left(x-x_{i+k}\right){+}^l=\left(x-x{i+k}\right)^l, \quad k=0,1, \ldots, m$$
Therefore, it can be concluded that the coefficients $b_0, \ldots, b_{l+1}$ are the unique answers of the following non-singular system of equations:
\begin{aligned} &\sum_{k=0}^{l+1} b_k\left(x-x_{i+k}\right){+}^l=0, \quad x \geq x{i+l+1} \ &\left{\begin{array}{l} b_0+b_1+\cdots+b_{l+1}=0 \ b_0 x_i+b_1 x_{i+1}+\cdots+b_{l+1} x_{i+l+1}=0 \ \vdots \ b_0 x_i^l+b_1 x_{i+1}^l+\cdots+b_{l+1} x_{i+l+1}^l=0 \end{array}\right. \end{aligned}

## 数学代写|数值分析代写numerical analysis代考|B-Spline Positivity

Spline $B_{l v}$ has at most $2 l$ zeros in $\left[t_v, t_{v+l+1}\right]$, where each of points $t_v$ and $t_{v+l+1}$ are repeated zeros of $B_{l v}$ with the order of $l$, because $B_{l v} \in c^{l-1}(-\infty, \infty)$. It can be concluded that on the interval $\left(t_v, t_{v+l+1}\right)$, no root can exist. On the interval $\left(t_v, t_{v+l+1}\right), B_{l v}$ is not zero, so if it is negative, it must intersect the $x$ axe at a point, in which case a zero is obtained; therefore, it is positive. In fact, on the interval $\left(t_{v+l}, t_{v+l+1}\right)$, we have:
$$B_{l v}(x)=\left(\prod_{r=v}^{v+l}\left(t_{v+l+1}-t_r\right)\right)^{-1}\left(t_{v+l+1}-x\right)^l>0$$

Now, consider the space $S_l\left(\Omega_n\right)$ of the $l$ degree splines that are defined over $\left[x_0, x_n\right]$. In the defined B-spline set, only $B_{l, n-1}, \ldots, B_{l,-l}$ in $\left[x_0, x_n\right]$ have non-zero values. We show that these $n+l$ spline functions form a basis. So we prove that they are linearly independent.
To prove the linear independence of $B_{l, n-1}, \ldots, B_{l,-l}$, we must prove that if for every $x \in\left[x_0, x_n\right]$
$$S(x)=\beta_{-1} B_{l,-1}(x)+\cdots+\beta_{n-1} B_{l, n-1}(x)=0$$
Then, $\beta_{-l}=\cdots=\beta_{n-1}=0$.
To do this, we add node $x_{-l-1}$, with $x_{-l-1}<x_{-l}$. Using the B-spline definition for $x \in\left[x_{-l-1}, x_{-l}\right]$, we have $S(x)=0$. Now if $S(x)=0$ is zero over $\left[x_0, x_1\right]$, then the number of fundamental zeros in $\left[x_{-l}, x_0\right]$ satisfies the following condition.
$$\tau+v=l<l+1 \Rightarrow r \leq \tau-(v+l+1)<0$$
Therefore, it can be concluded that $S(x)=0$ over $\left[x_{-1}, x_0\right]$, so on the whole interval, we have $S(x)=0$.

We have already shown that $S(x)$ being zero on the whole interval of $\left[x_0, x_n\right]$ is similar to $S(x)$ being zero on the interval $\left[x_{-l}, x_n\right]$. Therefore, for linear independence on the interval $\left[x_0, x_n\right]$, it is enough to examine the linear independence in $\left[x_{-1}, x_n\right]$.
Now, to check the linear independence of $B_{l, n-1}, \ldots, B_{l,-l}$ on the interval $\left[x_{-1}, x_n\right]$, we do this over the subintervals.
If $S(x)=0$ over $\left[x_{-l}, x_{-l+1}\right]$, then $S(x)=\beta_{-l} B_{l,-l}(x)=0$ where $B_{l,-l}(x)>0$ so $\beta_{-l}=0$. Now over $\left[x_{-l+1}, x_{-l+2}\right]$, if $S(x)=\beta_{-l+1} B_{l,-l+1}=0$, then $\beta_{-l+1}=0$. If we continue in this way, the coefficients will be zero on the whole interval and therefue it has lineat independente.

So, we show that B-splines $B_{l, n-1}, \ldots, B_{l,-l}$ form a basis over $\left[x_0, x_n\right]$ for space $S_l\left(\Omega_n\right)$.

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|Existence of B-Spline

$$\left{ 一世→−∞⇒X一世→−∞ 一世→∞⇒X一世→∞\正确的。 在和H一个在和吨○sH○在吨H一个吨F○r和在和r是一世∈从,吨H和r和一世s和X一个C吨l是○n和spl一世n和小号∈小号l(哦∞),s○吨H一个吨小号(X)=0F○rX一世+l+1≥X一个ndX<X一世,一个nd一世F吨H和n○r米一个l一世和一个吨一世○nC○nd一世吨一世○n一世s一个ls○米和吨,吨H和n在和H一个在和: \int_{-\infty}^{\infty} S(x) dx=\int_{x_i}^{x_{i+t+1}} S(x) dx=1 磷r○○F:F○rX∈[X一世−1,X一世+l+2],一世吨一世sCl和一个r吨H一个吨吨H和和Xp一个ns一世○n○Fspl一世n和小号一世n吨和r米s○F○n和−在一个是b一个s和s一个s S(x)=\sum_{i=0}^l a_i x_i+\sum_{i=1}^{n-1} b_i\left(x-x_i\right)_{+}^l d○和sn○吨一世nCl在d和一个ll○rp一个r吨○F吨H和b一个s一世sp○l是n○米一世一个lsb和C一个在s和F○rX<X一世,小号(X)=0.小号○,在和C一个ns一个是吨H一个吨○n吨H和一个b○在和−米和n吨一世○n和d一世n吨和r在一个l: S(x)=\sum_{k=0}^m b_k\left(x-x_{i+k}\right){+}^l 在H一世CH一世sd和吨和r米一世n和dF○rs○米和在一个l在和s○F米.ñ○在一世F在和一个ss在米和米=l+1,在和H一个在和:S(x)=0, \quad x \geq x{i+l+1} 一个nd一个ls○, \left(x-x_{i+k}\right){+}^l=\left(xx{i+k}\right)^l, \quad k=0,1, \ldots, m 吨H和r和F○r和,一世吨C一个nb和C○nCl在d和d吨H一个吨吨H和C○和FF一世C一世和n吨sb0,…,bl+1一个r和吨H和在n一世q在和一个ns在和rs○F吨H和F○ll○在一世nGn○n−s一世nG在l一个rs是s吨和米○F和q在一个吨一世○ns: \begin{对齐} &\sum_{k=0}^{l+1} b_k\left(x-x_{i+k}\right){+}^l=0, \quad x \geq x{i +l+1} \ &\左{ b0+b1+⋯+bl+1=0 b0X一世+b1X一世+1+⋯+bl+1X一世+l+1=0 ⋮ b0X一世l+b1X一世+1l+⋯+bl+1X一世+l+1l=0\正确的。 \end{对齐}$$

## 数学代写|数值分析代写numerical analysis代考|B-Spline Positivity

$$B_{l v}(x)=\left(\prod_{r=v}^{v+l}\left(t_{v+l+1}-t_r\right)\right)^{-1}\left(t_{v+l+1}-x\right)^l>0$$

$$S(x)=\beta_{-1} B_{l,-1}(x)+\cdots+\beta_{n-1} B_{l, n-1}(x)=0$$

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