| English | 意味 |
|---|---|
| power series/ˈpaʊə ˈsɪəriːz/ | 冪級数 |
| radius of convergence/ˈreɪdɪəs ɒv kənˈvɜːdʒəns/ | 収束半径 |
| interval of convergence/ˈɪntəvl ɒv kənˈvɜːdʒəns/ | 収束区間 |
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.
The radius of convergence
- A power series converges for $|x-a|<R$ and diverges for $|x-a|>R$.
- $R$ 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.
The interval of convergence
A power series converges within radius $R$ of its center — the interval runs from $a-R$ to $a+R$, endpoints checked separately.
A power series $\sum c_n(x-a)^n$ converges for...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Within radius $R$ of the center $a$.
The radius of convergence $R$ can be...
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
A radius is never negative.
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 $|x-a|<R$, revealing the radius directly.
- The center is $a$; the interval runs from $a-R$ to $a+R$.
- The ratio test does the heavy lifting — the algebra just isolates $|x-a|$.
For $\sum\tfrac{x^n}{n}$, the Ratio Test gives $|x|<1$. What is $R$?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$R=1$.
The radius of convergence is typically found using the...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The Ratio Test on the terms reveals $|x-a|<R$.
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.
The Ratio Test is inconclusive at the endpoints, which must be tested separately.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
At $x=a\pm R$, $L=1$; use another test.
For $\sum\tfrac{x^n}{n}$: $x=1$ diverges, $x=-1$ converges. The interval of convergence is...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Include $-1$ (converges), exclude $1$ (diverges).
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.
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: $[-1,1)$.
A power series $\sum c_n(x-a)^n$ converges for $|x-a|<R$, where $R$ is the radius of convergence (found via the Ratio Test). The interval of convergence is $a-R$ to $a+R$, with the endpoints tested separately (the ratio test is inconclusive there) to decide open vs. closed.