Representing Functions as Power Series · 将函数表示为幂级数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| power series/ˈpaʊə ˈsɪəriːz/ | 幂级数 | mì jí shù |
Reusing known series to build new ones
- You don't have to compute every Taylor series from derivatives. Manipulate the ones you know.
- Start from the geometric series $\frac{1}{1-x}=\sum x^n$ or the $e^x,\sin,\cos$ series.
- Substitute, multiply, differentiate, or integrate to represent a new function as a power series 幂级数.
- It's algebra on series — fast and clever.
重用已知级数造出新级数
- 你不必从导数计算每个泰勒级数。变形你已知的那些。
- 从几何级数 $\frac{1}{1-x}=\sum x^n$ 或 $e^x,\sin,\cos$ 级数出发。
- 用代换、相乘、求导或积分把一个新函数表示成幂级数。
- 这是对级数做代数——快又巧。
Substitution and multiplication
- Substitute an expression for $x$: replace $x$ with $-x^2$ in $\frac1{1-x}$ to get $\frac{1}{1+x^2}=\sum(-x^2)^n=\sum(-1)^n x^{2n}$.
- Multiply a series by a power of $x$ or a constant: $x\cdot e^x=\sum\frac{x^{n+1}}{n!}$.
- Each move keeps a valid power series (adjust the interval of convergence for substitutions).
- No new derivatives needed — just plug and combine.
代换与相乘
- 代换一个表达式给 $x$:在 $\frac1{1-x}$ 中把 $x$ 换成 $-x^2$,得 $\frac{1}{1+x^2}=\sum(-x^2)^n=\sum(-1)^n x^{2n}$。
- 把级数乘以 $x$ 的幂或常数:$x\cdot e^x=\sum\frac{x^{n+1}}{n!}$。
- 每一步都保持一个有效的幂级数(代换要调整收敛区间)。
- 不需要新导数——只需代入并组合。
Built from the geometric series · 基于几何级数构建
y = a / (1 − b·x)
Substituting $-x^2$ into · 生成 $\tfrac{1}{1-x}=\sum x^n$ builds the series for $\tfrac{1}{1+x^2}$ — no new derivatives needed. · 将 $-x^2$ 代入 $\tfrac{1}{1-x}=\sum x^n$ 构建 $\tfrac{1}{1+x^2}$ 的级数 —— 无需新导数。
Substituting $u=-x^2$ into · 生成 $\tfrac{1}{1-u}=\sum u^n$ gives $\tfrac{1}{1+x^2}=$ · 将 $u=-x^2$ 代入 $\tfrac{1}{1-u}=\sum u^n$ 得到 $\tfrac{1}{1+x^2}=$
$(-x^2)^n=(-1)^n x^{2n}$.
Which moves generate new power series from known ones? · 哪些操作可从已知级数生成新的幂级数？
Substitute, scale/multiply, differentiate, integrate — not random reordering. · 代入、缩放/乘法、求导、积分——而非随机重排。
Differentiating and integrating term by term
- Inside its interval of convergence, a power series can be differentiated or integrated term by term.
- Integrate $\frac{1}{1+x^2}=\sum(-1)^n x^{2n}$ to get $\arctan x=\sum\frac{(-1)^n x^{2n+1}}{2n+1}$.
- Differentiate a series to get the series of the derivative.
- These moves generate famous series (like $\arctan$ and $\ln(1+x)$) from simpler ones.
逐项求导与积分
- 在收敛区间内,幂级数可以逐项求导或逐项积分。
- 对 $\frac{1}{1+x^2}=\sum(-1)^n x^{2n}$ 积分,得 $\arctan x=\sum\frac{(-1)^n x^{2n+1}}{2n+1}$。
- 对级数求导,得到导数的级数。
- 这些手法从更简单的级数生成著名级数(如 $\arctan$ 和 $\ln(1+x)$)。
Within its interval of convergence, a power series can be differentiated and integrated term by term. · 在其收敛区间内，幂级数可以逐项微分和积分。
A valid move inside the interval of convergence. · 收敛区间内的有效操作。
Integrating $\sum(-1)^n x^{2n}$ term by term gives the series for... · 逐项积分 $\sum(-1)^n x^{2n}$ 得到...的级数
$\int\tfrac{1}{1+x^2}\,dx=\arctan x$.
A powerful toolkit
- Substitute, scale, multiply, differentiate, integrate — five moves, endless new series.
- This is often far faster than computing high derivatives directly.
- The new series inherits (roughly) the radius of convergence of the one you started from.
- Recognizing which move to apply is the real skill.
一套强大的工具箱
- 代换、缩放、相乘、求导、积分——五招,无穷多新级数。
- 这常比直接计算高阶导数快得多。
- 新级数(大致)继承你出发那个的收敛半径。
- 认出该用哪一招才是真本事。
Integrating a power series term by term introduces a constant ____, fixed by a known value. · 逐项积分幂级数会引入一个常数 ____, 由已知值确定。
E.g. $\arctan 0=0$ fixes $C=0$. · 例如：$\arctan 0=0$ 固定 $C=0$。
Substituting $x\to -x^2$ can change the interval of convergence. · 代入$x\to -x^2$可能会改变收敛区间。
Re-derive the interval in the new variable. · 在新变量中重新推导该区间。
Term-by-term differentiation and integration are valid only within the interval of convergence (and integrating adds a $+C$ — pin it down with a known value, e.g. $\arctan 0=0$). When you substitute (like $x\to -x^2$), the interval of convergence changes too — re-derive it from the new variable.
逐项求导和积分仅在收敛区间内有效(积分要加 $+C$——用一个已知值确定它,如 $\arctan 0=0$)。当你代换(如 $x\to -x^2$)时,收敛区间也会改变——从新变量重新推导它。
Find a power series for $\dfrac{1}{1+x^2}$.
- Start from $\dfrac{1}{1-u}=\sum u^n$ and substitute $u=-x^2$.
- $\dfrac{1}{1+x^2}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^n x^{2n}=1-x^2+x^4-\cdots$ (for $|x|<1$).
求 $\dfrac{1}{1+x^2}$ 的幂级数。
- 从 $\dfrac{1}{1-u}=\sum u^n$ 出发,代入 $u=-x^2$。
- $\dfrac{1}{1+x^2}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^n x^{2n}=1-x^2+x^4-\cdots$(当 $|x|<1$)。
To represent a function as a power series, manipulate a known series: substitute, multiply/scale, or differentiate/integrate term by term (valid within the interval of convergence; integration adds $+C$). E.g. substituting into $\frac{1}{1-x}$ and integrating yields the series for $\frac{1}{1+x^2}$ and $\arctan x$.
要把函数表示成幂级数,变形一个已知级数:代换、相乘/缩放,或逐项求导/积分(在收敛区间内有效;积分要加 $+C$)。例如代入 $\frac{1}{1-x}$ 并积分,得到 $\frac{1}{1+x^2}$ 与 $\arctan x$ 的级数。