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微積分：第九章
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calculus-chpt9
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<title>微積分：第九章</title>
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<body><div class="container"><h1 id="第九章">第九章</h1>
<h2 id="第一節">第一節</h2>
<h3 id="序列-函數">序列 &lt;=&gt; 函數</h3>
<p>序列可以視為一個將正整數（或非負整數）映射到值的函數：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1134">
a_0 = 1 \\
a_1 = 2 \\
a_2 = 4 \\
f(x) = 2^x
</script></p>
<p>若序列有極限，代表此對應函數有極限。</p>
<h3 id="基本數列收斂性">基本數列收斂性</h3>
<h4 id="交錯序列">交錯序列</h4>
<p>若序列在兩數字之間交換，則數列發散（<script type="math/tex" id="MathJax-Element-1135">2, 4, 2, 4, 2, 4, ...</script>）。</p>
<h4 id="用函數極限">用函數極限</h4>
<p>數列 <script type="math/tex" id="MathJax-Element-1136">\left\{ a_n \right\} = \left\{\frac{n}{1 - 2n}\right\}</script> 收斂至 <script type="math/tex" id="MathJax-Element-1137">-\frac{1}{2}</script>：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1138">
\lim_{n \to \infty} \frac{n}{1 - 2n} = \lim_{n \to \infty} \cfrac{\cfrac{n}{n}}{\cfrac{1 - 2n}{n}} = \lim_{n \to \infty} \cfrac{1}{\cfrac{1}{n} - 2} = \frac{1}{-2}
</script></p>
<h4 id="用羅必達求函數極限">用羅必達求函數極限</h4>
<p>數列 <script type="math/tex" id="MathJax-Element-1139">\left\{ a_n \right\} = \left\{\frac{n^2}{2^n - 1}\right\}</script> 收斂至 <script type="math/tex" id="MathJax-Element-1140">0</script>：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1141">
\lim_{n \to \infty} \frac{n^2}{2^n - 1} = \lim_{n \to \infty} \frac{2n}{2^n\ln 2} = \lim_{n \to \infty} \frac{2}{2^n\left(\ln 2\right)^2}
</script></p>
<h4 id="用夾擠定理">用夾擠定理</h4>
<h5 id="範例一">範例一</h5>
<p><script type="math/tex; mode=display" id="MathJax-Element-1142">
\{c_n\} = \left\{\left(-1\right)^n\frac{1}{n!}\right\} \\
\frac{-1}{2^n} \le \left(-1\right)^n\frac{1}{n!} \le \frac{1}{2^n} \quad n \ge 4 \\
\lim_{n \to \infty} \frac{-1}{2^n} = \lim_{n \to \infty} \frac{1}{2^n} = 0 \\
\lim_{n \to \infty} \left(-1\right)^n\frac{1}{n!} = 0
</script></p>
<h5 id="範例二">範例二</h5>
<p><script type="math/tex; mode=display" id="MathJax-Element-1143">
\{c_n\} = \left\{\frac{1}{n!}\right\} \\
0 \le \left\{\left(-1\right)^n\frac{1}{n!}\right\} \le \frac{1}{2^n} \quad n \ge 4 \\
\lim_{n \to \infty} \frac{1}{2^n} = 0 \\
\lim_{n \to \infty} \left(-1\right)^n\frac{1}{n!} = 0
</script></p>
<h4 id="絕對值定理">絕對值定理</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1144">
\lim_{n \to \infty} \left\lvert a_n \right\rvert = 0 \implies \lim_{n \to \infty} a_n = 0
</script></p>
<h3 id="尋找數列通式">尋找數列通式</h3>
<p>檢查乘法、加法。</p>
<h3 id="單調數列">單調數列</h3>
<p>若數列遞增或是遞減，則稱為單調（Monotonic）。</p>
<h3 id="有界數列">有界數列</h3>
<p>若數列的值永遠在一個值域之內，則稱為有界（Bounded）。</p>
<p>若一個數列單調且有界，則一定收斂。</p>
<h2 id="第二節">第二節</h2>
<h3 id="無窮級數定義">無窮級數定義</h3>
<p>若 <script type="math/tex" id="MathJax-Element-1145">\left\{ S_n \right\} = \left\{ \sum^{n}_{k=1} a_k \right\}</script> 本身收斂，則無窮級數收斂於 <script type="math/tex" id="MathJax-Element-1146">\sum^{\infty}_{k=1} a_k</script>。</p>
<h3 id="無窮級數範例">無窮級數範例</h3>
<h4 id="12n"><script type="math/tex" id="MathJax-Element-1147">1/2^n</script></h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1148">
\begin{align}
S_n = \sum_{n=1}^{\infty} {\frac{1}{2^n}} &= \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdot \cdot \cdot \\
S_n = \sum_{n=1}^{\infty} {\frac{1}{2^n}} &= \frac{2^{n-1}}{2^n} + \frac{2^{n-2}}{2^n} + \frac{2^{n-3}}{2^n} + \cdot \cdot \cdot \\
S_n = \sum_{n=1}^{\infty} {\frac{1}{2^n}} &= \frac{\sum_{k = 0}^{n - 1}{2^{k}}}{2^n} = \frac{2^n-1}{2^n}=1
\end{align}
</script></p>
<h4 id="裂項和telescroping-series">裂項和（Telescroping Series）</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1149">
\sum_{n=1}^{\infty} \frac{1}{n\left(n+1\right)} = \sum_{n=1}^{\infty}{\left( \frac{1}{n} - \frac{1}{n + 1} \right)} = 1 - \frac{1}{n + 1}
</script></p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1150">
\sum_{k = 1}^{n} \left( a_k - a_{k - 1} \right) = a_n - a_0 \\
a_n \to c \implies \sum_{k=1}^{\infty} = c - a_0
</script></p>
<h4 id="用裂項和求解">用裂項和求解</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1151">
S_n = \sum_{n = 1}^{\infty} \frac{2}{4n^2 - 1} \\
\frac{2}{4n^2 - 1} = \frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{1}{2n-1}-\frac{1}{2n+1} \\
\lim_{n\to \infty} S_n = 1
</script></p>
<h4 id="等比級數geometric-series">等比級數（Geometric Series）</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1152">
\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r} \quad 0 < \left\lvert r \right\rvert < 1
</script></p>
<p>若 <script type="math/tex" id="MathJax-Element-1153">\left\lvert r \right\rvert \ge 1</script> 則發散。</p>
<h4 id="用等比級數求解">用等比級數求解</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1154">
\begin{align}
0.\overline{08} &= \frac{8}{10^2} + \frac{8}{10^4} + \cdot\cdot\cdot \\
&= \sum_{n = 0}^{\infty} \left( \frac{8}{10^2} \right) \left( \frac{1}{10^2} \right) ^ n \\
&= \cfrac{\cfrac{8}{10^2}}{1 - \cfrac{1}{10^2}} = \frac{8}{99}
\end{align}
</script></p>
<h3 id="基本定理">基本定理</h3>
<p><script type="math/tex; mode=display" id="MathJax-Element-1155">
\lim_{n \to \infty} S_n = c \implies \lim_{n \to \infty} a_n = 0 \\
\lim_{n \to \infty} a_n \neq 0 \implies S_n \; \mathrm{diverges} \\
</script></p>
<h4 id="基本定理應用">基本定理應用</h4>
<p><script type="math/tex" id="MathJax-Element-1156">\left\{2^n\right\}</script> 級數發散。</p>
<p><script type="math/tex" id="MathJax-Element-1157">\left\{ \frac{n!}{2n!+1} \right\}</script> 級數發散。</p>
<p><script type="math/tex" id="MathJax-Element-1158">\left\{\frac{1}{n}\right\}</script> 級數無法確認。</p>
<h2 id="第三節">第三節</h2>
<h3 id="積分試驗法">積分試驗法</h3>
<p>若 <script type="math/tex" id="MathJax-Element-1159">f\left(x\right) \; x \ge 1</script> 為正、連續且遞減，而 <script type="math/tex" id="MathJax-Element-1160">a_n = f\left(n\right)</script>：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1161">
\sum_{n = 1}^{\infty} a_n \quad \text{and} \quad \int_{1}^{\infty} f\left(x\right) dx \\
\text{either both converge or both diverge.}
</script></p>
<h4 id="證明">證明</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1162">
\sum_{i = 2}^{n} f\left(i\right) \le \int_{1}^{n} f\left(x\right)dx \le \sum_{i = 1}^{n - 1} f\left(i\right) \\
S_n - f \left( 1 \right) \le \int_{1}^{n} f\left(x\right)dx \le S_{n-1} \\
\text{Assume} \; \int_{1}^{n} f\left(x\right)dx \; \text{converges to} \; L \\
S_n - f\left(1\right) \le L \implies S_n \le L + f\left(1\right) \; \text{for} \; n \ge 1 \\
\text{Consequently} \; \left\{S_n\right\} \; \text{is bounded and monotonic, converges.} \\
\text{Assume} \; \int_{1}^{n} f\left(x\right)dx \; \text{diverge to} \; \infty \\
\text{By} \; S_{n-1} \ge \int_{1}^{n} f\left(x\right)dx \text{, } \; S_{n-1} \;\text{diverges}
</script></p>
<h4 id="範例一-1">範例一</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1163">
\begin{align}
f\left(n\right) = a_n &= \frac{n}{n^2 + 1} \\
S_n &= \sum_{n=1}^{\infty} \frac{n}{n^2 + 1} \\
\because f\left(n\right) \gt 0 & \; \text{and continuous, and } f'\left(x\right) < 0 \text{ for } x > 1 \\
\therefore & \text{ We can use Integral Test.} \\
\end{align}
</script></p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1164">
\begin{align}
\int_{1}^{\infty}\frac{x}{x^2 + 1} dx &= \frac{1}{2} \int_{1}^{\infty}\frac{1}{x^2 + 1}dx^2 \\
&= \lim_{t \to \infty} \left[ \ln\left(x^2 + 1\right) \right]^{t}_{1} \\
&= \infty
\end{align}
</script></p>
<h4 id="範例二-1">範例二</h4>
<p>Apply the Integral Test to the series <script type="math/tex" id="MathJax-Element-1165">\sum_{n = 1}^{\infty}\frac{1}{n^2 + 1}</script>：</p>
<p>The function is positive, and continuous for <script type="math/tex" id="MathJax-Element-1166">x \ge 1</script>, and its deriviative is negative for <script type="math/tex" id="MathJax-Element-1167">x \gt 1</script>.</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1168">
\int_{1}^{\infty} \frac{1}{x^2 + 1} dx = \lim_{t \to \infty} \left[ \arctan x \right]^{b}_{1} = \frac{\pi}{4}
</script></p>
<h4 id="範例三">範例三</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1169">
\sum_{n = 2}^{\infty} \frac{1}{n\ln n}
</script></p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1170">
\because f\left(x\right) \gt 0 \text{ for } x \ge 1 \text{, and is continuous. Its derivative } f'\left(x\right) \text{ is negative for } x \gt 1\\
\therefore \text{We can apply the Integral Test.}
</script></p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1171">
\begin{align}
\int_{2}^{\infty} \frac{1}{n\ln n} dn &= \int_{2}^{\infty} \frac{1/n}{\ln n} dn \\
&= \int_{2}^{\infty} \frac{1}{\ln n} d\ln n \\
&= \ln \ln n \vert^{\infty}_{2}
\end{align}
</script></p>
<h3 id="特殊級數">特殊級數</h3>
<p>黎曼 zeta 函數：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1172">
\zeta\left(s\right) = \sum_{n = 1}^{\infty} \frac{1}{n^s}
</script></p>
<p>此級數又稱為 p 級數，在 <script type="math/tex" id="MathJax-Element-1173">s \gt 1</script> 時收斂，<script type="math/tex" id="MathJax-Element-1174">s \le 1</script> 時發散。</p>
<p>若 <script type="math/tex" id="MathJax-Element-1175">s = 1</script>，此級數稱為調和級數（Harmonic Series）；<script type="math/tex" id="MathJax-Element-1176">s = 2</script> 稱為巴塞爾問題（Basel Problem），答案為 <script type="math/tex" id="MathJax-Element-1177">\frac{\pi^2}{6}</script>；<script type="math/tex" id="MathJax-Element-1178">s = 3</script> 稱為阿培里常數。</p>
<h2 id="第四節">第四節</h2>
<h3 id="直接比較法direct-comparison-test">直接比較法（Direct Comparison Test）</h3>
<p>若有兩數列 <script type="math/tex" id="MathJax-Element-1179">\left\{a_n\right\}, \left\{b_n\right\}</script> 以及其無窮級數 <script type="math/tex" id="MathJax-Element-1180">A_\infty, B_\infty</script>，且符合 <script type="math/tex" id="MathJax-Element-1181">0 \lt a_n \lt b_n</script>：</p>
<p>若 <script type="math/tex" id="MathJax-Element-1182">B_\infty</script> 收斂，則 <script type="math/tex" id="MathJax-Element-1183">A_\infty</script> 收斂；若 <script type="math/tex" id="MathJax-Element-1184">A_\infty</script> 發散，則 <script type="math/tex" id="MathJax-Element-1185">B_\infty</script> 發散。</p>
<h3 id="極限比較法limit-comparison-test">極限比較法（Limit Comparison Test）</h3>
<p>若有兩數列 <script type="math/tex" id="MathJax-Element-1186">\left\{a_n\right\}, \left\{b_n\right\}</script> 符合 <script type="math/tex" id="MathJax-Element-1187">\lim_{n \to \infty} \frac{a_n}{b_n} = L</script> 且 <script type="math/tex" id="MathJax-Element-1188">L</script> 是一個非無窮正數，則兩無窮級數 <script type="math/tex" id="MathJax-Element-1189">A_\infty, B_\infty</script> 收斂性必相同。</p>
<h4 id="範例">範例</h4>
<p>測試 <script type="math/tex" id="MathJax-Element-1190">\sum_{n = 1}^{\infty} \frac{1}{an + b} \; , \; a \gt 0 , \; b \gt 0</script> 的收斂性。</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1191">
a_n = \frac{1}{an + b} \\
b_n = \frac{1}{n} \\
\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{1}{an + b}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{an + b} = \frac{1}{a}
</script></p>
<h4 id="極限方法">極限方法</h4>
<table>
<thead>
<tr>
<th align="center">Given Series</th>
<th align="center">Comparison Series</th>
<th align="center">Conclusion</th>
</tr>
</thead>
<tbody><tr>
<td align="center"><script type="math/tex" id="MathJax-Element-1192">\sum_{n = 1}^{\infty}\frac{1}{3n^2-4n+5}</script></td>
<td align="center"><script type="math/tex" id="MathJax-Element-1193">\sum_{n = 1}^{\infty}\frac{1}{n^2}</script></td>
<td align="center">Both series converge.</td>
</tr>
<tr>
<td align="center"><script type="math/tex" id="MathJax-Element-1194">\sum_{n = 1}^{\infty}\frac{1}{\sqrt{3n-2}}</script></td>
<td align="center"><script type="math/tex" id="MathJax-Element-1195">\sum_{n = 1}^{\infty}\frac{1}{n^2}</script></td>
<td align="center">Both series diverge.</td>
</tr>
<tr>
<td align="center"><script type="math/tex" id="MathJax-Element-1196">\sum_{n = 1}^{\infty}\frac{n^2-10}{4n^5+n^3}</script></td>
<td align="center"><script type="math/tex" id="MathJax-Element-1197">\sum_{n = 1}^{\infty}\frac{1}{n^3}</script></td>
<td align="center">Both series converge.</td>
</tr>
<tr>
<td align="center"><script type="math/tex" id="MathJax-Element-1198">\sum_{n = 1}^{\infty}\frac{n2^n}{4n^3+1}</script></td>
<td align="center"><script type="math/tex" id="MathJax-Element-1199">\sum_{n = 1}^{\infty}\frac{2^n}{n^2}</script></td>
<td align="center">Both series diverge.</td>
</tr>
</tbody></table>
<h2 id="第五節">第五節</h2>
<h3 id="交錯級數">交錯級數</h3>
<p>若 <script type="math/tex" id="MathJax-Element-1200">\left\{a_n\right\} \gt 0</script>，則我們可以這樣定義交錯級數：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1201">
\sum_{n = 1}^{\infty} \left( -1 \right)^n a_n \quad\text{or}\quad\sum_{n = 1}^{\infty} \left( -1 \right)^{n+1} a_n
</script></p>
<h3 id="交錯級數測試alternating-series-test">交錯級數測試（Alternating Series Test）</h3>
<p>交錯級數若符合以下特性則會收斂：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1202">
\lim_{n \to \infty}a_n = 0 \\
a_{n+1} \le a_n, \text{for all } n
</script></p>
<h4 id="範例一-2">範例一</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1203">
\sum_{n=1}^{\infty} \left(-1\right)^{n+1}\frac{1}{n} \\
\lim_{n \to \infty} a_n = 0 \\
\frac{1}{n+1} \le \frac{1}{n} \\
\text{Converges}
</script></p>
<h4 id="範例二-2">範例二</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1204">
\sum_{n=1}^{\infty} \frac{n}{\left(-2\right)^{n-1}} \\
\frac{n}{\left(-2\right)^{n-1}} = \left(-1\right)^{n+1}\frac{n}{2^{n-1}} \\
a_n = \frac{n}{2^{n-1}}\\
\lim_{n \to \infty}a_n = 0 \\
a_n = \frac{n}{2^{n-1}} = \frac{2n}{2^n} \\
a_{n+1} = \frac{n+1}{2^n} \\
\frac{2n}{2^n} \ge \frac{n+1}{2^n},\;\text{for $n \ge 1$}
</script></p>
<h3 id="交錯級數餘項alternating-series-remainder">交錯級數餘項（Alternating Series Remainder）</h3>
<p>若以數列的前 <script type="math/tex" id="MathJax-Element-1205">N</script> 項估計級數和，其誤差值將小於等於第 <script type="math/tex" id="MathJax-Element-1206">N+1</script> 項。</p>
<h3 id="絕對以及條件收斂">絕對以及條件收斂</h3>
<h4 id="絕對收斂定理absolute-convergence-theorem">絕對收斂定理（Absolute Convergence Theorem）</h4>
<p>若 <script type="math/tex" id="MathJax-Element-1207">\sum\left\lvert a_n\right\rvert</script> 收斂，<script type="math/tex" id="MathJax-Element-1208">\sum a_n</script> 也收斂。</p>
<h5 id="範例-1">範例</h5>
<p><script type="math/tex; mode=display" id="MathJax-Element-1209">
\sum_{n=1}^{\infty}\frac{\sin n}{n^2} \\
\sum_{n=1}^{\infty}\left\lvert\frac{\sin n}{n^2}\right\rvert \le \sum_{n=1}^{\infty}\left\lvert\frac{1}{n^2}\right\rvert \\
\text{All three series converge.}
</script></p>
<h4 id="定義">定義</h4>
<p><script type="math/tex" id="MathJax-Element-1210">\sum\left\lvert a_n\right\rvert</script> 以及 <script type="math/tex" id="MathJax-Element-1211">\sum a_n</script> 都收斂，稱為絕對收斂。</p>
<p><script type="math/tex" id="MathJax-Element-1212">\sum\left\lvert a_n\right\rvert</script> 發散，但是 <script type="math/tex" id="MathJax-Element-1213">\sum a_n</script> 收斂，稱為條件收斂。</p>
<p>若級數絕對收斂，改變項的順序，不會改變級數和。</p>
<p>若級數條件收斂，改變項的順序，可能會改變級數和。</p>
<h5 id="範例-2">範例</h5>
<p><script type="math/tex; mode=display" id="MathJax-Element-1214">
\sum_{n = 1}^{\infty}\left(-1\right)^n\frac{1}{n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} + \cdot\cdot\cdot = \ln2 \\
\sum_{n = 1}^{\infty}\left(-1\right)^n\frac{1}{n} = 1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \cdot\cdot\cdot \\
\sum_{n = 1}^{\infty}\left(-1\right)^n\frac{1}{n} = \left(1 - \frac{1}{2}\right) - \frac{1}{4} + \left(\frac{1}{3} - \frac{1}{6}\right) - \frac{1}{8} + \frac{1}{5} - \cdot\cdot\cdot \\
\sum_{n = 1}^{\infty}\left(-1\right)^n\frac{1}{n} = \frac{1}{2} - \frac{1}{4} + \frac{1}{6}- \cdot\cdot\cdot \\
\sum_{n = 1}^{\infty}\left(-1\right)^n\frac{1}{n} = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{2} + \frac{1}{3}-\cdot\cdot\cdot\right) = \frac{1}{2}\ln 2\\
</script></p>
<h2 id="第六節">第六節</h2>
<h3 id="比例測試ratio-test">比例測試（Ratio Test）</h3>
<p><script type="math/tex; mode=display" id="MathJax-Element-1215">
\lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \lt 1 \implies \sum a_n \text{ converges absolutely} \\
\lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert \gt 1 \implies \sum a_n \text{ diverges} \\
\lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert = 1 \text{ draws no conclusion}
</script></p>
<h4 id="範例-3">範例</h4>
<p>檢查 <script type="math/tex" id="MathJax-Element-1216">\sum _ {n = 0} ^ {\infty} {\frac{2^n}{n!}}</script> 收斂性：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1217">
a_n = \frac{2^n}{n!} \\
a_{n + 1} = \frac{2 \times 2^n}{\left( n + 1 \right) n! } \\
\lim_{n \to \infty} {2 \over {n + 1}} = 0
</script></p>
<p>級數 <script type="math/tex" id="MathJax-Element-1218">\sum _ {n = 0} ^ {\infty} {\frac{2^n}{n!}}</script> 收斂</p>
<h3 id="開根測試root-test">開根測試（Root Test）</h3>
<p><script type="math/tex; mode=display" id="MathJax-Element-1219">
\lim_{n \to \infty} \sqrt[n]{\left\lvert a_n \right\rvert} \le 1 \implies \sum a_n \text{ converges absolutely} \\
\lim_{n \to \infty} \sqrt[n]{\left\lvert a_n \right\rvert} \gt 1 \implies \sum a_n \text{ diverges} \\
\lim_{n \to \infty} \sqrt[n]{\left\lvert a_n \right\rvert} = 1 \text{ draws no conclusion}
</script></p>
<h4 id="範例-4">範例</h4>
<p>檢查 <script type="math/tex" id="MathJax-Element-1220">\sum _ {n = 0} ^ {\infty} {\frac{e^{2n}}{n^n}}</script> 收斂性：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1221">
a_n = \frac{e^{2n}}{n^n} = \left( \frac{e^{2}}{n}\right) ^ n \\
\lim_{n \to \infty} \sqrt[n]{\left\lvert \left( \frac{e^{2}}{n}\right) ^ n \right\rvert} = 0
</script></p>
<p>級數 <script type="math/tex" id="MathJax-Element-1222">\sum _ {n = 0} ^ {\infty} {\frac{e^{2n}}{n^n}}</script> 收斂</p>
<h3 id="測試級數的方法">測試級數的方法</h3>
<ol>
<li>最後一項是否趨近 <script type="math/tex" id="MathJax-Element-1223">0</script>，若否，級數發散</li>
<li>級數是否特殊：幾何、p 級數、裂項和、交錯和</li>
<li>可以使用積分法、開根法或是比例法測試</li>
<li>可以與特殊級數比較</li>
</ol>
<h3 id="方法">方法</h3>
<p><img src="https://doc-10-4k-docs.googleusercontent.com/docs/securesc/ha0ro937gcuc7l7deffksulhg5h7mbp1/ce3t9kekl5k96360v8h7hgi1jd0ndsh6/1433880000000/15787471668676610797/*/0B2yT2KVwX0BUZjRDUDBSZmM1Z0E?e=download" alt="方法整理" title=""></p>
<h2 id="第七節">第七節</h2>
<h3 id="泰勒展開式">泰勒展開式</h3>
<p>將函數 <script type="math/tex" id="MathJax-Element-1224">f\left(x\right)</script> 在 <script type="math/tex" id="MathJax-Element-1225">x = c</script> 點展開：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1226">
f\left(x\right) = \sum_{n = 0}^{\infty} {\frac{f^{\left(n\right)}\left(c\right)}{n!}} \left(x - c\right) ^ n
</script></p>
<h3 id="泰勒多項式趨近">泰勒多項式趨近</h3>
<p>若三次趨近某函數：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1227">
P_3\left(x\right) = 1 + x + \frac{1}{2} x^2 + \frac{1}{3!}x^3
</script></p>
<p>若 <script type="math/tex" id="MathJax-Element-1228">c = 0</script>，也可以稱為麥克勞林多項式（Maclaurin Polynomial）。</p>
<h3 id="泰勒多項式的餘式">泰勒多項式的餘式</h3>
<p><script type="math/tex; mode=display" id="MathJax-Element-1229">
f\left(x\right) = P_n\left(x\right) + R_n\left(x\right) \\
\text{Error} = \left\lvert R_n\left(x\right) \right\rvert = \left\lvert f\left(x\right) - P_n\left(x\right) \right\rvert
</script></p>
<h3 id="泰勒定理及拉格朗日於餘項">泰勒定理及拉格朗日於餘項</h3>
<p>若函數 <script type="math/tex" id="MathJax-Element-1230">f\left(x\right)</script> 可在一個包含 <script type="math/tex" id="MathJax-Element-1231">c</script> 的區間 <script type="math/tex" id="MathJax-Element-1232">I</script> 被微分 <script type="math/tex" id="MathJax-Element-1233">n + 1</script> 次，則對於區間 <script type="math/tex" id="MathJax-Element-1234">I</script> 內的每一個 <script type="math/tex" id="MathJax-Element-1235">x</script>，存在一個 <script type="math/tex" id="MathJax-Element-1236">z</script> 介於 <script type="math/tex" id="MathJax-Element-1237">c</script> 以及 <script type="math/tex" id="MathJax-Element-1238">x</script> 之間符合：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1239">
f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \cdot\cdot\cdot + \frac{f^{(n)}(c)}{n!}(x-c)^n + R_n(x) \\
R_n(x) = \frac{f^{n+1}(z)}{(n+1)!}(x-c)^{n+1}
</script></p>
<p>經過推導：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1240">
\left\lvert R_n(x) \right\rvert \le \frac{\lvert x-c\rvert ^ {n+1}}{(n+1)!} \text{ max } \left\lvert f^{n+1}(z) \right\rvert
</script></p>
<p>若 <script type="math/tex" id="MathJax-Element-1241">n = 0</script>，可得均值定理。</p>
<h2 id="第八章">第八章</h2>
<h3 id="冪級數">冪級數</h3>
<p><script type="math/tex; mode=display" id="MathJax-Element-1242">
\sum_{0}^{\infty} a_nx^n \\
\sum_{0}^{\infty} a_n(x-c)^n
</script></p>
<p>後者稱為中心在 <script type="math/tex" id="MathJax-Element-1243">c</script> 的級數。</p>
<h3 id="收斂區間及收斂半徑radius-and-interval-of-convergence">收斂區間及收斂半徑（Radius and Interval of Convergence）</h3>
<p>冪級數 <script type="math/tex" id="MathJax-Element-1244">f(x) = \sum_{0}^{\infty} a_n(x-c)^n</script> 的收斂有三種可能：</p>
<ol>
<li>當 <script type="math/tex" id="MathJax-Element-1245">x = c</script> 收斂</li>
<li>存在一個 <script type="math/tex" id="MathJax-Element-1246">R > 0</script> 使得 <script type="math/tex" id="MathJax-Element-1247">\lvert x - c \rvert < R</script> 時收斂，<script type="math/tex" id="MathJax-Element-1248">\lvert x - c \rvert > R</script> 時發散</li>
<li><script type="math/tex" id="MathJax-Element-1249">x</script> 為任意數皆收斂</li>
</ol>
<h4 id="範例一-3">範例一</h4>
<p>檢查 <script type="math/tex" id="MathJax-Element-1250">f(x) = \sum_{0}^{\infty} n!x^n</script> 的收斂性：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1251">
f(x) = \sum_{0}^{\infty} n!x^n = \sum_{0}^{\infty} n!(x-0)^n \\
f(0) = 1 \text{ ... Converges at } x = 0 \\
\lim_{n \to \infty} \left\lvert \frac{(n+1)!x^{n+1}}{n!x^n} \right\rvert = \lvert x \rvert \lim_{n \to \infty} (n + 1) = \infty
</script></p>
<p>當 <script type="math/tex" id="MathJax-Element-1252"> x = 0 </script> 時收斂，半徑 <script type="math/tex" id="MathJax-Element-1253">0</script>。</p>
<h4 id="範例二-3">範例二</h4>
<p>檢查 <script type="math/tex" id="MathJax-Element-1254">f(x) = \sum_{0}^{\infty} 3(x-2)^n</script> 的收斂性：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1255">
\lim_{n \to \infty} \left\lvert \frac{3(x-2)^{n+1}}{3(x-2)^n} \right\rvert = \lvert x - 2 \rvert
</script></p>
<p>當 <script type="math/tex" id="MathJax-Element-1256">\lvert x - 2 \rvert \lt 1</script> 收斂，半徑 <script type="math/tex" id="MathJax-Element-1257">1</script>。</p>
<h3 id="極點收斂性">極點收斂性</h3>
<p>兩個極點的收斂性有可能不一樣，必須分開檢查：</p>
<p>收斂區間 <script type="math/tex" id="MathJax-Element-1258">[-1, 1)</script> 表示 <script type="math/tex" id="MathJax-Element-1259">x = -1</script> 時收斂，<script type="math/tex" id="MathJax-Element-1260">x = 1</script> 時發散。</p>
<h3 id="冪級數的微分與積分">冪級數的微分與積分</h3>
<p>對於 <script type="math/tex" id="MathJax-Element-1261">f(x) = \sum_{0}^{\infty} a_n(x-c)^n</script> 且當收斂半徑 <script type="math/tex" id="MathJax-Element-1262">R > 0</script> 表示在區間 <script type="math/tex" id="MathJax-Element-1263">(c - R, c + R)</script> 是可微的：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1264">
f'(x) = \sum_{n=1}^{\infty} na_n(x-c)^{n-1} \\
\int f(x)dx = C + \sum_{n=0}^{\infty} a_n \frac{(x-c)^{n+1}}{n + 1}
</script></p>
<h2 id="第九節">第九節</h2>
<h3 id="幾何冪級數">幾何冪級數</h3>
<p>因為 <br>
<script type="math/tex; mode=display" id="MathJax-Element-1265">
\sum_{n = 0}^{\infty} ar^n = \frac{a}{1-r} , \quad 0 < \lvert r \rvert < 1
</script></p>
<p>我們可以改寫</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1266">
\frac{1}{1-x} = \sum_{n = 0}^{\infty} ar^n = \sum_{n = 0}^{\infty} x^n = 1 + x + x^2 + \cdot\cdot\cdot, \; \lvert x\rvert < 1
</script></p>
<p>或是</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1267">
\frac{1}{1-x} = \frac{1}{2-(x+1)} = \frac{12}{1-[(x+1)/2]} = \frac{a}{1-r} \\
\frac{1}{1-x} = \sum_{n=0}^{\infty} \frac{1}{2} \left( \frac{x + 1}{2} \right) ^ n, \; \lvert x + 1 \rvert < 2
</script></p>
<h4 id="範例一-4">範例一</h4>
<p>找出 <script type="math/tex" id="MathJax-Element-1268">f(x) = \frac{4}{x+2}</script> 中心在 <script type="math/tex" id="MathJax-Element-1269">0</script> 的幾何級數：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1270">
\frac{4}{2 + x} = \frac{2}{1-(-x/2)} = \frac{a}{1-r} \\
a = 2 \\
r = - \frac{x}{2} \\
\frac{4}{x + 2} = \sum_{n = 0}^{\infty}ar^n = \sum_{n = 0}^{\infty}2\left( - \frac{x}{2}\right)^n \\
\left\lvert \frac{x}{2} \right\rvert < 1
</script></p>
<h4 id="範例二-4">範例二</h4>
<p>找出 <script type="math/tex" id="MathJax-Element-1271">f(x) = \frac{1}{x}</script> 中心在 <script type="math/tex" id="MathJax-Element-1272">1</script> 的幾何級數：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1273">
f(x) = \frac{1}{x} = \frac{1}{1-(1-x)} = \frac{a}{1-r} \\
a = 1 \\
r = 1 - x \\
\frac{1}{x} = \sum_{n = 0}^{\infty} (1-x)^n = \sum_{n = 0}^{\infty} (-1)^n(x-1)^n \\
\frac{1}{x} = 1 - (x - 1) + (x - 1)^2 - (x - 1)^3 + \cdot\cdot\cdot \\
\text{Converges when } \left\lvert x - 1 \right\rvert \lt 1
</script></p>
<h3 id="冪級數的操作">冪級數的操作</h3>
<p>若兩個冪級數相加，則結果的收斂區間為他們的聯集。</p>
<h4 id="範例-5">範例</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1274">
f(x) = \frac{3x - 1}{x^2 - 1} = \frac{2}{x + 1} + \frac{1}{x - 1} \\
\frac{2}{x + 1} = \frac{2}{1-(-x)} = \sum_{n = 0}^{\infty}2(-1)^nx^n , \; \lvert x \rvert < 1 \\
\frac{1}{x - 1} = \frac{-1}{1-x} = - \sum_{n = 0}^{\infty} x^n , \; \lvert x \rvert < 1
</script></p>
<h3 id="用積分找冪級數">用積分找冪級數</h3>
<p>找 <script type="math/tex" id="MathJax-Element-1275">\arctan x</script> 的冪級數表示法：</p>
<p>因為 <script type="math/tex" id="MathJax-Element-1276">{d\arctan{x} \over dx} = {1 \over 1+x^2}</script>，所以我們這樣寫：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1277">
f(x) = {1 \over 1 + x} \\
f(x^2) = {1 \over 1 + x^2} \\
\arctan x = \int {1 \over 1 + x^2} dx + C \\
\arctan x = \sum_{n = 0}^{\infty} (-1)^n {x^{2n + 1} \over 2n + 1}
</script></p>
<h3 id="用級數找-pi">用級數找 <script type="math/tex" id="MathJax-Element-1278">\pi</script></h3>
<p><script type="math/tex; mode=display" id="MathJax-Element-1279">
\pi = 4\left( 4\arctan {1 \over 5} - \arctan {1 \over 239} \right)
</script></p>
<p>若趨近五項，可得 <script type="math/tex" id="MathJax-Element-1280">\pi \approx 3.1415926</script></p>
<h2 id="第十節">第十節</h2>
<h3 id="泰勒及麥克勞林級數">泰勒及麥克勞林級數</h3>
<p>若 <script type="math/tex" id="MathJax-Element-1281">f(x) = \sum a_n (x - c)^n</script> 對於一個包含 <script type="math/tex" id="MathJax-Element-1282">c</script> 的區間 <script type="math/tex" id="MathJax-Element-1283">I</script> 中的所有 <script type="math/tex" id="MathJax-Element-1284">x</script> 皆成立，則：</p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1285">
a_n = {f^{(n)}(c) \over n!} \\
f(x) = f(c) + f'(c)(x-c) + {f''(c) \over 2!}(x-c)^2 + \cdot\cdot\cdot
</script></p>
<h4 id="證明-1">證明</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1286">
f^{(0)}(x) = a_0 + a_1(x - c) + a_2(x - c)^2 + \cdot\cdot\cdot \\
f^{(1)}(x) = a_1 + 2a_2(x - c) + 3a_3(x - c)^2 + \cdot\cdot\cdot \\
f^{(2)}(x) = 2a_2 + 3!a_3(x - c) + 4 \times 3a_3(x - c)^2 + \cdot\cdot\cdot \\
f^{(n)}(x) = n!a_n + (n + 1)!a_{n+1}(x-c) + \cdot\cdot\cdot
</script></p>
<p><script type="math/tex; mode=display" id="MathJax-Element-1287">
f^{(0)}(x) = 0!a_0 \\
f^{(1)}(x) = 1!a_1 \\
f^{(2)}(x) = 2!a_2 \\
f^{(3)}(x) = 3!a_3 \\
f^{(n)}(x) = n!a_n
</script></p>
<p>同樣的，若 <script type="math/tex" id="MathJax-Element-1288">c = 0</script>，就成為麥克勞林級數。</p>
<h3 id="泰勒級數的收斂性">泰勒級數的收斂性</h3>
<p><script type="math/tex; mode=display" id="MathJax-Element-1289">
\lim R_n = 0 \text{ for all $x$ in the interval $I$} \\
f(x) = \sum_{n = 0}^{\infty} {f^{(n)}(c) \over n!} (x - c)^n
</script></p>
<h4 id="證明-2">證明</h4>
<p><script type="math/tex; mode=display" id="MathJax-Element-1290">
P_n(x) = f(x) - R_n(x) \\
\lim_{n \to \infty} S_n(x) = \lim_{n \to \infty} P_n(x) = f(x) - \lim_{n \to \infty} R_n(x) = f(x)
</script></p>
<h4 id="範例一-5">範例一</h4></div></body>
</html>
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