Radius and Interval of Convergence of Power Series · 幂级数的收敛半径与收敛区间
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| power series/ˈpaʊə ˈsɪəriːz/ | 幂级数 | mì jí shù |
| radius of convergence/ˈreɪdɪəs ɒv kənˈvɜːdʒəns/ | 收敛半径 | shōu liǎn bàn jìng |
| interval of convergence/ˈɪntəvl ɒv kənˈvɜːdʒəns/ | 收敛区间 | shōu liǎn qū jiān |
Where does a power series make sense?
- A power series 幂级数 is an infinite polynomial: $\sum c_n(x-a)^n$ — like a Taylor series with variable $x$.
- It converges for some $x$ and diverges for others. Which $x$?
- The answer is an interval centered at $a$, measured by the radius of convergence 收敛半径 $R$.
- Finding $R$ and the exact endpoints is the goal.
幂级数在哪里有意义?
- 幂级数是一个无穷多项式:$\sum c_n(x-a)^n$——像带变量 $x$ 的泰勒级数。
- 它对某些 $x$ 收敛,对另一些发散。哪些 $x$?
- 答案是一个以 $a$ 为中心的区间,由收敛半径 $R$ 度量。
- 求 $R$ 和确切的端点就是目标。
The radius of convergence
- A power series converges for $|x-a|
- $R$ is the radius: the series converges within distance $R$ of the center $a$.
- Three possibilities: $R=0$ (only at $a$), $R=\infty$ (all $x$), or a finite $R$.
- Find $R$ with the Ratio Test on the terms of the series.
收敛半径
- 幂级数对 $|x-a|
- $R$ 是半径:级数在离中心 $a$ 距离 $R$ 以内收敛。
- 三种可能:$R=0$(仅在 $a$)、$R=\infty$(所有 $x$)、或有限 $R$。
- 用级数各项的比值判别法求 $R$。
The interval of convergence · 收敛区间
A power series converges within radius $R$ of its center — the interval runs from $a-R$ to · 到 $a+R$, endpoints checked separately. · 幂级数在其中心周围半径 $R$ 内收敛——区间从 $a-R$ 到 $a+R$，端点需单独检验。
A power series $\sum c_n(x-a)^n$ converges for... · 幂级数 $\sum c_n(x-a)^n$ 对...收敛
Within radius $R$ of the center $a$. · 在$R$半径内，以$a$为中心。
The radius of convergence $R$ can be... · 收敛半径 $R$ 可以是...
A radius is never negative. · 半径永远不会为负。
Getting R from the Ratio Test
- Apply the Ratio Test: set $\lim\big|\tfrac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\big|<1$ and solve for $|x-a|$.
- That inequality has the form $|x-a|
- The center is $a$; the interval runs from $a-R$ to $a+R$.
- The ratio test does the heavy lifting — the algebra just isolates $|x-a|$.
由比值判别法求 R
- 应用比值判别法:令 $\lim\big|\tfrac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n}\big|<1$,解出 $|x-a|$。
- 那个不等式呈 $|x-a|
- 中心是 $a$;区间从 $a-R$ 到 $a+R$。
- 比值判别法承担繁重工作——代数只是孤立出 $|x-a|$。
For · 支持 $\sum\tfrac{x^n}{n}$, the Ratio Test gives $|x|<1$. What is $R$? · 对于 $\sum\tfrac{x^n}{n}$，比值判别法给出 $|x|<1$。$R$ 是什么？
$R=1$.
The radius of convergence is typically found using the... · 收敛半径通常通过...求得
The Ratio Test on the terms reveals $|x-a|
Check the endpoints separately
- The Ratio Test is inconclusive at the endpoints $x=a\pm R$ (there $L=1$).
- So test each endpoint individually by plugging it in and using another series test.
- The interval of convergence 收敛区间 may include neither, one, or both endpoints.
- Write the interval with the correct open/closed brackets after checking both ends.
分别检查端点
- 比值判别法在端点 $x=a\pm R$ 不确定(那里 $L=1$)。
- 所以逐个检验每个端点:代入并用别的级数判别法。
- 收敛区间可能不含、含一个或含两个端点。
- 检查两端后,用正确的开/闭括号写出区间。
The Ratio Test is inconclusive at the endpoints, which must be tested separately. · 比值判别法在端点处无法得出结论，必须单独检验。
At $x=a\pm R$, $L=1$; use another test. · 在 $x=a\pm R$ 处，$L=1$; 使用其他判别法。
For · 支持 $\sum\tfrac{x^n}{n}$: $x=1$ diverges, $x=-1$ converges. The interval of convergence is... · 对于 $\sum\tfrac{x^n}{n}$: $x=1$ 发散，$x=-1$ 收敛。收敛区间是...
Include $-1$ (converges), exclude $1$ (diverges). · 包含 $-1$ (收敛)，排除 $1$ (发散)。
The Ratio Test gives the radius but is inconclusive at the endpoints — you must test $x=a-R$ and $x=a+R$ separately (they can converge or diverge independently). Don't assume the interval is open or closed; the answer might be $[a-R,a+R)$, $(a-R,a+R]$, etc. Check both ends every time.
比值判别法给出半径,但在端点不确定——你必须分别检验 $x=a-R$ 与 $x=a+R$(它们可以独立地收敛或发散)。别假设区间是开或闭;答案可能是 $[a-R,a+R)$、$(a-R,a+R]$ 等。每次都检查两端。
Find the radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty}\dfrac{x^n}{n}$.
- Ratio Test: $\lim\big|\tfrac{x^{n+1}/(n+1)}{x^n/n}\big|=|x|\cdot\lim\tfrac{n}{n+1}=|x|<1$, so $R=1$.
- Endpoints: $x=1\Rightarrow\sum\tfrac1n$ (diverges); $x=-1\Rightarrow\sum\tfrac{(-1)^n}{n}$ (converges).
- Interval of convergence: $[-1,1)$.
求 $\displaystyle\sum_{n=1}^{\infty}\dfrac{x^n}{n}$ 的收敛半径与区间。
- 比值判别法:$\lim\big|\tfrac{x^{n+1}/(n+1)}{x^n/n}\big|=|x|\cdot\lim\tfrac{n}{n+1}=|x|<1$,所以 $R=1$。
- 端点:$x=1\Rightarrow\sum\tfrac1n$(发散);$x=-1\Rightarrow\sum\tfrac{(-1)^n}{n}$(收敛)。
- 收敛区间:$[-1,1)$。
A power series $\sum c_n(x-a)^n$ converges for $|x-a|
幂级数 $\sum c_n(x-a)^n$ 对 $|x-a|