Questions in category: 级数 (Infinite Series)
分析 >> 数学分析 >> 级数 [58]
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1. 求 $\sqrt[n]{a}$ 的近似值.

Posted by haifeng on 2025-05-04 17:04:28 last update 2025-05-04 17:36:50 | Answers (1) | 收藏


设 $a > 0$, $n$ 为正整数, 求 $\sqrt[n]{a}$ 的近似值.


记 $f_n(a)=\sqrt[n]{a}$, 则

\[f_n(1)=1,\quad f_n(\frac{1}{a})=\frac{1}{f_n(a)}.\] 

 因此, 不妨设 $a>1$. 对任意 $b > 0$, 

\[
f_n(a)=\sqrt[n]{a}=\sqrt[n]{\frac{b^n a}{b^n}}=\frac{1}{b}\sqrt[n]{b^n a}.
\]

对于此正数 $b$, 若存在正数 $c$, 使得 $b^n a=c^n+\varepsilon_n$,  则

\[
f_n(a)=\frac{1}{b}\sqrt[n]{b^n a}=\frac{1}{b}\sqrt[n]{c^n+\varepsilon_n}=\frac{c}{b}\sqrt[n]{1+\frac{\varepsilon_n}{c^n}}.
\]

其中

\[
\sqrt[n]{1+\frac{\varepsilon_n}{c^n}}=(1+\frac{\varepsilon_n}{c^n})^{\frac{1}{n}}
\]

可以根据$(1+x)^{\alpha}$的幂级数展开公式

\[
(1+x)^{\alpha}=1+\alpha x+\frac{\alpha\cdot(\alpha-1)}{2!}x^2+\cdots+\frac{\alpha\cdot(\alpha-1)\cdots(\alpha-m+1)}{m!}x^m+\cdots
\]

进行计算, 这里 $|x| < 1$.

因此, 这里要求 $\varepsilon_n$ 满足 $-c^n <\varepsilon_n < c^n$. 这提供了对 $\sqrt[n]{a}$ 求近似值的一个算法.

 


因此, 第一步找到合适的正数 $b$ 和 $c$, 使得

\[
b^n a=c^n+\varepsilon_n,\quad -1<\frac{\varepsilon_n}{c^n} < 1.
\]

 

 

问题3260

2. 求 $\sum\limits_{n=1}^{2025}\dfrac{1}{n}$.

Posted by haifeng on 2025-01-01 10:27:13 last update 2025-01-01 10:29:20 | Answers (0) | 收藏


>> sum(1/n,n,1,2025)
3352115173648121679965699723941806011419256032990001580679374066690635673990482812901066841801271700388461801784908206217585317799633152724908259332304706360667629003001878183870001502333131118953156208104510058340758878080529641441201716363352192028833735191870444274605947562228855478417708123228043222925440662852462584114305132232316224160037542393428357718483447246524366563705810946798526160237613544464080103850027344804033681561449946186235974194757998829650891355805575132229337279301965017590049027653927293172545766749041705342226163383199329865662055989214659751298242651094008119371952958988906806974245062479233867613031394033135182845292094602805062760059724143163494258112593774902629969655527145841877321341399104407521097942837957157992931796288710213008913639933511534848063729899574408318729848360044462087027654119019039418442189449511365822541979799467791|409254318735421372874527513713809849729135059223982154744686910391244881751394287792064717538513501909093551073331105901230174815690870254022597927796335026690598911658288578341662007257908728685152761650582748164977706081859581162971618587292791162956045102947117094844113969603097500003652004069652979380609518121519013531548885927853149410724755210783428395488365803795111980933624569399933400969497574721790682373349842950967442946949750794524393431836515674176033082971320894446597931131516693687546474012255765758283940350898733234638365606173236127739051742487602390618454018049479926800452950875570239516276819871570674576851400791100528044895551810310590444123249637327298475483510172556378112704981184569797419716363712359628331775989191307093772005846389088398901747521145065831525586976740612039273283797156150977660434534795320274675464655132985044254833228160000
------------------------

 

>> :mode numerical
Switch into numerical calculating mode.
e.g., 1/2+1/3 will return 0.83333333
 
>> setprecision(100)
Now the precision is: 100
 
------------------------
 
>> sum(1/n,n,1,2025)
8.1907875377003144144384170229842614128313271548100196142275290713705512794801417745101431392419948684
------------------------
 

 

3. 判断下列级数的敛散性.

Posted by haifeng on 2024-12-26 10:33:22 last update 2024-12-26 10:34:40 | Answers (1) | 收藏


判断下列级数的敛散性.

(1)    $\sum\limits_{n=1}^{\infty}\dfrac{(-1)^n}{\sqrt{n+1}+(-1)^n}$

 

 

(2)    $\sum\limits_{k=2}^{\infty}\biggl(\dfrac{1}{\sqrt{2k}-1}-\dfrac{1}{\sqrt{2k-1}}\biggr)$

4. 判断下列级数的敛散性.

Posted by haifeng on 2024-12-19 16:00:11 last update 2024-12-19 16:00:11 | Answers (1) | 收藏


判断下列级数的敛散性.

(1)    $\sum\limits_{n=1}^{\infty}\dfrac{(-1)^n}{n-\ln n}$

5. 讨论下面级数的敛散性

Posted by haifeng on 2024-04-25 21:28:14 last update 2024-04-25 21:28:14 | Answers (0) | 收藏


讨论下面级数的敛散性

\[
\sum_{n=1}^{\infty}\frac{1}{n^{1+\frac{1}{n}}}
\]

6. 利用 $(1+x)^{\alpha}$ 的展开式计算一些幂.

Posted by haifeng on 2023-12-30 11:24:59 last update 2023-12-30 12:46:36 | Answers (0) | 收藏


$(1+x)^{\alpha}$ 展开成幂级数为

\[
(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\cdots+\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!}x^n+\cdots
\]

这里 $x\in(-1,1)$.  在端点 $\pm 1$ 的敛散性与具体的 $\alpha$ 有关.


计算 $10^{0.9181246048}$, 注意上面级数的收敛区间为 $(-1,1)$, 因此下面的计算方法是错误的.

\[
\begin{split}
10^{0.9181246048}&=(1+9)^{0.9181246048}\\
&=1+0.9181246048\times 9+\frac{0.9181246048(0.9181246048-1)}{2!}9^2\\
&\qquad+\frac{0.9181246048(0.9181246048-1)(0.9181246048-2)}{3!}9^3+\cdots
\end{split}
\]

我们需要将 $10^{\alpha}$ 转换为 $(2^{\alpha})^3\cdot(1.25)^{\alpha}$ 这样的形式. 即使是 $2^{\alpha}$, 有时也有很大的误差, 因为上面的级数在端点 $\pm 1$ 处不一定收敛. 因此需要转换为 $(1+x)^{\alpha}$ 的形式, 其中 $x\in(-1,1)$.


下面使用 Sowya 进行计算.

将下面的代码保存到 code/power.sc 中

//a^t =(1+x)^t
//(1+x)^t = 1+t*x+t*(t-1)/(2!)*x^2+...+t*(t-1)*(t-2)*...*(t-n+1)/(n!)*x^n+...
//make sure x is in (-1,1)

fun power(a,t)
{
    var n=10;
    var x;
    x=a-1;
    var sum=1;
    var coeff, x_pow;
    coeff=1;
    x_pow=1;

    for(var i=0; i< n; i=i+1)   {
        coeff=coeff*(t-i)/(i+1);
        x_pow=x_pow*x;
        sum=sum+coeff*x_pow;
    }
    return sum;
}


启动 Sowya, 并进入到 clox 编程模式,

>> :mode clox

加载 power.sc 文件
> load(code\power.sc)

测试 $2^3$
> print power(2,3);
8>

计算 $2^{3.2}$

> print power(2,3.2);
2243569763|244140625>

 

7. 如果级数 $\sum\limits_{n=1}^{\infty}u_n$ 收敛, 而且 $\lim\limits_{n\rightarrow\infty}\dfrac{v_n}{u_n}=1$, 则能否判断级数 $\sum\limits_{n=1}^{\infty}v_n$ 也收敛? 如若不能, 请举出反例.

Posted by haifeng on 2023-12-27 12:57:43 last update 2023-12-27 12:57:43 | Answers (1) | 收藏


如果级数 $\sum\limits_{n=1}^{\infty}u_n$ 收敛, 而且 $\lim\limits_{n\rightarrow\infty}\dfrac{v_n}{u_n}=1$, 则能否判断级数 $\sum\limits_{n=1}^{\infty}v_n$ 也收敛? 如若不能, 请举出反例.

8. 用柯西审敛准则证明下列级数收敛.

Posted by haifeng on 2023-12-27 12:54:01 last update 2023-12-27 12:54:12 | Answers (1) | 收藏


用柯西审敛准则证明下列级数收敛.

1.    $\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}$

9. 研究级数 $\sum\limits_{n=1}^{\infty}\sin(\alpha n+\beta)$ 的敛散性, 其中 $\alpha$, $\beta$ 为常数.

Posted by haifeng on 2023-12-17 23:47:54 last update 2023-12-17 23:47:54 | Answers (0) | 收藏


研究级数 $\sum\limits_{n=1}^{\infty}\sin(\alpha n+\beta)$ 的敛散性, 其中 $\alpha$, $\beta$ 为常数.

 

 

 

梅加强《数学分析》P.278, 习题 7.

10. 设 $a_1=\frac{1}{2}$, $2na_{n+1}=(n+1)a_n$, 记 $S_n=\sum_{k=1}^{n}a_k$, 求 $S_n$.

Posted by haifeng on 2023-07-05 14:26:23 last update 2023-07-05 14:26:23 | Answers (2) | 收藏


设 $a_1=\frac{1}{2}$, $2na_{n+1}=(n+1)a_n$, 记 $S_n=\sum_{k=1}^{n}a_k$, 求 $S_n$.

 

 

题目来源:   用和不用北太天元做清华大学2018领军数学第24题的差别

 

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