Lagrange Error Bound · 拉格朗日误差界
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Lagrange error bound/ˈlæɡreɪndʒ ˈerə baʊnd/ | 拉格朗日误差界 | lā gé lǎng rì wù chā jiè |
Bounding a Taylor polynomial's error
- A Taylor polynomial is only an approximation. How wrong can it be?
- The Lagrange error bound 拉格朗日误差界 gives a guaranteed ceiling on the remainder.
- It uses the next derivative — the first one the polynomial didn't match.
- With it, you can prove an approximation is accurate to a required tolerance.
界定泰勒多项式的误差
- 泰勒多项式只是近似。它能错多少?
- 拉格朗日误差界给余项一个有保证的上限。
- 它使用下一个导数——多项式没匹配的第一个。
- 有了它,你能证明一个近似达到所需精度。
The bound
- The error of the degree-$n$ Taylor polynomial $P_n$ at $x$ satisfies:
-
$$|R_n(x)|\le \frac{M}{(n+1)!}\,\big|x-a\big|^{\,n+1}$$
- Here $M$ is a maximum of $\big|f^{(n+1)}\big|$ on the interval between $a$ and $x$.
- It looks like the next Taylor term, but with the derivative replaced by its worst-case size $M$.
那个界
- $n$ 次泰勒多项式 $P_n$ 在 $x$ 处的误差满足:
-
$$|R_n(x)|\le \frac{M}{(n+1)!}\,\big|x-a\big|^{\,n+1}$$
- 这里 $M$ 是 $\big|f^{(n+1)}\big|$ 在 $a$ 与 $x$ 之间区间上的一个最大界。
- 它看起来像下一个泰勒项,只是把导数换成它的最坏情况大小 $M$。
The gap the bound caps · 界限所覆盖的间隙
y = a·e^{bx}
The Lagrange bound caps the gap between $e^x$ and its Taylor polynomial using the next derivative's maximum $M$. · 拉格朗日余项将$e^x$与其泰勒多项式之间的误差上限由下一阶导数的最大值$M$给出。
The Lagrange error bound is $|R_n(x)|\le$ · 拉格朗日误差界是 $|R_n(x)|\le$
Uses the $(n+1)$th derivative bound and $(n+1)!$. · 使用第 $(n+1)$ 阶导数界和 $(n+1)!$。
Finding the pieces
- $(n+1)$: one more than the polynomial's degree — the first missing term's order.
- $M$: a bound on the $(n+1)$th derivative's absolute value between $a$ and $x$ (use the biggest it gets).
- $|x-a|^{n+1}$: the distance from the center, raised to that power.
- Multiply, divide by $(n+1)!$, and you have a guaranteed error ceiling.
找出各部分
- $(n+1)$: 比多项式次数多一——第一个缺失项的阶。
- $M$: $a$ 与 $x$ 之间第 $(n+1)$ 阶导数绝对值的界(取它能达到的最大)。
- $|x-a|^{n+1}$: 离中心的距离,升到那个幂。
- 相乘,除以 $(n+1)!$,你就有了有保证的误差上限。
For a degree-$n$ polynomial, $M$ bounds which derivative? · 对于 n 次（degree-$n$）多项式，$M$ 界定哪个导数？
One past the degree: the $(n+1)$th derivative. · 超过次数的一次：第 $(n+1)$ 阶导数。
In the bound, $M$ is... · 在界限中，$M$ 是...
A worst-case bound on the next derivative. · 下一个导数的最坏情况界。
Proving accuracy
- To show $P_n$ is within, say, $0.001$: compute the Lagrange bound and check it's $\le 0.001$.
- Because the bound overestimates the true error, if the bound is small, the real error is smaller.
- Larger $n$ shrinks the bound (bigger factorial, more accuracy).
- It's the rigorous companion to Taylor approximation.
证明精度
- 要证明 $P_n$ 在(比如)$0.001$ 以内:算出拉格朗日界并检查它 $\le 0.001$。
- 因为这个界高估真实误差,若界小,真实误差更小。
- 更大的 $n$ 缩小这个界(更大的阶乘,更高的精度)。
- 它是泰勒近似的严格伴侣。
With $M=1.2$, $n=2$, $x=0.1$, $a=0$: $\dfrac{1.2}{3!}(0.1)^3=$ ? (a decimal) · 已知 $M=1.2$, $n=2$, $x=0.1$, $a=0$: $\dfrac{1.2}{3!}(0.1)^3=$ ? (小数)
$\frac{1.2}{6}\cdot0.001=0.0002$.
The Lagrange bound is a guaranteed ceiling, usually larger than the actual error. · 拉格朗日界是一个保证的上限，通常大于实际误差。
If the bound is small, the true error is smaller. · 如果界限很小，真实误差会更小。
Increasing the degree $n$ generally makes the error bound... · 增加次数 $n$ 通常会使误差界...
A bigger factorial shrinks the bound → more accuracy. · 更大的阶乘缩小界限 → 提高精度。
Use the $(n+1)$th derivative and the factorial $(n+1)!$ — one past the polynomial's degree, not $n$. $M$ is the maximum of $|f^{(n+1)}|$ on the interval between $a$ and $x$ (a worst-case bound, so pick a value you can justify is large enough). The bound is a guarantee, usually larger than the actual error.
使用第 $(n+1)$ 阶导数和阶乘 $(n+1)!$——比多项式次数多一,而非 $n$。$M$ 是 $|f^{(n+1)}|$ 在 $a$ 与 $x$ 之间区间上的最大(最坏情况的界,所以选一个你能论证足够大的值)。这个界是保证,通常大于实际误差。
Bound the error of the degree-$2$ Maclaurin polynomial for $e^x$ at $x=0.1$.
- Next derivative: $f'''(x)=e^x$; on $[0,0.1]$, $M=e^{0.1}\le 1.2$.
- $|R_2(0.1)|\le \dfrac{1.2}{3!}\,(0.1)^3=\dfrac{1.2}{6}(0.001)=0.0002$.
- So $1+0.1+\tfrac{0.01}{2}=1.105$ is within $0.0002$ of $e^{0.1}$. ✓
界定 $e^x$ 的 $2$ 次麦克劳林多项式在 $x=0.1$ 处的误差。
- 下一个导数:$f'''(x)=e^x$;在 $[0,0.1]$ 上,$M=e^{0.1}\le 1.2$。
- $|R_2(0.1)|\le \dfrac{1.2}{3!}\,(0.1)^3=\dfrac{1.2}{6}(0.001)=0.0002$。
- 所以 $1+0.1+\tfrac{0.01}{2}=1.105$ 在 $e^{0.1}$ 的 $0.0002$ 以内。✓
The Lagrange error bound: $|R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1}$, where $M$ bounds the $(n+1)$th derivative on the interval between $a$ and $x$. It's the next Taylor term with the derivative replaced by its max — a guaranteed error ceiling for proving an approximation meets a tolerance.
拉格朗日误差界:$|R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1}$,其中 $M$ 界住 $a$ 与 $x$ 之间的第 $(n+1)$ 阶导数。它是把导数换成其最大值的下一个泰勒项——一个有保证的误差上限,用来证明近似满足精度。