| English | 意味 |
|---|---|
| Lagrange error bound/ˈlæɡreɪndʒ ˈerə baʊnd/ | ラグランジュの誤差上限 |
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.
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 $M$.
The gap the bound caps
y = a·e^{bx}
The Lagrange bound caps the gap between $e^x$ and its Taylor polynomial using the next derivative's maximum $M$.
The Lagrange error bound is $|R_n(x)|\le$
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Uses the $(n+1)$th derivative bound and $(n+1)!$.
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.
For a degree-$n$ polynomial, $M$ bounds which derivative?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
One past the degree: the $(n+1)$th derivative.
In the bound, $M$ is...
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A worst-case bound on the next derivative.
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.
With $M=1.2$, $n=2$, $x=0.1$, $a=0$: $\dfrac{1.2}{3!}(0.1)^3=$ ? (a decimal)
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$\frac{1.2}{6}\cdot0.001=0.0002$.
The Lagrange bound is a guaranteed ceiling, usually larger than the actual error.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
If the bound is small, the true error is smaller.
Increasing the degree $n$ generally makes the error bound...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A bigger factorial shrinks the bound → more accuracy.
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.
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 $e^{0.1}$. ✓
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.