伯努利数是伯努利在做等幂和的时候定义的，之后又是如何被定义到上述的幂级数展开式的呢
通常，大多数这类问题都是有不显然的原因的，而且关于Bernoulli数在各种不同场合的出现是有非常大量的文献以及讨论的。&br&&br&下面是一个很知名的heuristic，能够几乎完全给题主的问题一个解答。如果你在读Concrete Mathematics的话跳到9.5章，如果没在读这是一个简单的介绍：&br&考虑随便一个什么函数&img src=&///equation?tex=f%28x%29& alt=&f(x)& eeimg=&1&&，我们可以定义一个离散的差分算子&img src=&///equation?tex=%5CDelta& alt=&\Delta& eeimg=&1&&：&br&&img src=&///equation?tex=%28%5CDelta+f%29%28n%29%3Df%28n%2B1%29-f%28n%29& alt=&(\Delta f)(n)=f(n+1)-f(n)& eeimg=&1&&&br&在离散的状况下这几乎就是你能定义的导数，但是不全是.&br&&br&假如差分算子有逆算子，就只能是求和&img src=&///equation?tex=%28%5CSigma+f%29%28n%29+%3D+%5Csum_%7Bk%3Dm%7D%5E%7Bn-1%7Df%28k%29& alt=&(\Sigma f)(n) = \sum_{k=m}^{n-1}f(k)& eeimg=&1&&，这样一来我们就有：&br&&img src=&///equation?tex=%28%5CSigma%5CDelta+f%29%28n%29+%3D+%5Csum_%7Bk%3Dm%7D%5E%7Bn-1%7D%5CDelta+f%28k%29%3Df%28m%2B1%29-f%28m%29%2B%5Cldots%2Bf%28n%29-f%28n-1%29%3Df%28n%29-f%28m%29& alt=&(\Sigma\Delta f)(n) = \sum_{k=m}^{n-1}\Delta f(k)=f(m+1)-f(m)+\ldots+f(n)-f(n-1)=f(n)-f(m)& eeimg=&1&&&br&这个&img src=&///equation?tex=f%28m%29& alt=&f(m)& eeimg=&1&&有点烦人，就当作m不变好了，就像变上限积分一样把它当作常数.&br&&br&所以，用比较疯狂的写法，我们可以认为&img src=&///equation?tex=%5CSigma+%5CDelta%3D1& alt=&\Sigma \Delta=1& eeimg=&1&&，也就是说&img src=&///equation?tex=%5CSigma+%3D+%5Cfrac%7B1%7D%7B%5CDelta%7D& alt=&\Sigma = \frac{1}{\Delta}& eeimg=&1&&。用中文说，意思就是&b&求和是插分的逆运算&/b&。下面我们就来换个角度看这个“逆”&img src=&///equation?tex=%5Cfrac%7B1%7D%7B%5CDelta%7D& alt=&\frac{1}{\Delta}& eeimg=&1&&的问题：&br&&br&泰勒公式告诉我们，&img src=&///equation?tex=f%28n%2B1%29%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty+%5Cfrac%7Bf%5E%7B%28k%29%7D%28n%29%7D%7Bk%21%7D& alt=&f(n+1)=\sum_{k=0}^\infty \frac{f^{(k)}(n)}{k!}& eeimg=&1&&，这当然通常是错的，但是我们只是在进行heuristic，所以并不重要。这个公式是说，如果我们把导数算子写成&img src=&///equation?tex=%28Df%29%28x%29%3Df%27%28x%29& alt=&(Df)(x)=f'(x)& eeimg=&1&&，泰勒公式就让我们相信，&img src=&///equation?tex=f%28n%2B1%29%3D%5Csum_%7Bk%3D0%7D%5E%5Cinfty+%5Cfrac%7BD%5Ek+f%7D%7Bk%21%7D%28n%29%3De%5ED+f%28n%29& alt=&f(n+1)=\sum_{k=0}^\infty \frac{D^k f}{k!}(n)=e^D f(n)& eeimg=&1&&.&br&&br&可是这有什么用呢？把这个奇怪的&img src=&///equation?tex=f%28n%2B1%29%3De%5EDf+%28n%29& alt=&f(n+1)=e^Df (n)& eeimg=&1&&带入到插分的定义中，我们可以看到&img src=&///equation?tex=%5CDelta+f+%28n%29%3D+%28e%5ED-1%29+f%28n%29& alt=&\Delta f (n)= (e^D-1) f(n)& eeimg=&1&&, 也就是说&img src=&///equation?tex=%5CDelta+%3D+e%5ED-1& alt=&\Delta = e^D-1& eeimg=&1&&，这也就告诉我们&br&&img src=&///equation?tex=%5Cfrac%7B1%7D%7B%5CDelta%7D%3D%5Cfrac%7B1%7D%7BD%7D%5Ccdot%5Cfrac%7BD%7D%7Be%5ED-1%7D& alt=&\frac{1}{\Delta}=\frac{1}{D}\cdot\frac{D}{e^D-1}& eeimg=&1&&&br&看到那个熟悉的Bernoulli数的生成函数了吧？在这个背景下，题主好奇的&img src=&///equation?tex=%5Cfrac%7Bx%7D%7Be%5Ex-1%7D& alt=&\frac{x}{e^x-1}& eeimg=&1&&几乎给出了差分算子的逆。&br&&br&用Bernoulli数作为系数展开：&br&&img src=&///equation?tex=%5Cfrac%7B1%7D%7B%5CDelta%7D%3D%5Cfrac%7B1%7D%7BD%7D%5Csum_%7Bk%3D0%7D%5E%5Cinfty+B_k+%5Cfrac%7BD%5Ek%7D%7Bk%21%7D%3D%5Cfrac%7B1%7D%7BD%7D%2B%5Csum_%7Bk%3D1%7D%5E%5Cinfty+B_k+%5Cfrac%7BD%5E%7Bk-1%7D%7D%7Bk%21%7D& alt=&\frac{1}{\Delta}=\frac{1}{D}\sum_{k=0}^\infty B_k \frac{D^k}{k!}=\frac{1}{D}+\sum_{k=1}^\infty B_k \frac{D^{k-1}}{k!}& eeimg=&1&&&br&这个结论是什么意思呢？我们把&img src=&///equation?tex=%5Cfrac%7B1%7D%7B%5CDelta%7D+f%28n%29%3D%5B%5Cfrac%7B1%7D%7BD%7D%2B%5Csum_%7Bk%3D1%7D%5E%5Cinfty+B_k+%5Cfrac%7BD%5E%7Bk-1%7D%7D%7Bk%21%7D%5Df%28n%29& alt=&\frac{1}{\Delta} f(n)=[\frac{1}{D}+\sum_{k=1}^\infty B_k \frac{D^{k-1}}{k!}]f(n)& eeimg=&1&&的具体形式写下来看：&br&&img src=&///equation?tex=%5Csum_%7Bl%3Dm%7D%5E%7Bn-1%7Df%28l%29%3D%5Cint_m%5En+f%28x%29dx%2B%5Csum_%7Bk%3D1%7D%5E%5Cinfty+B_k+%5Cfrac%7Bf%5E%7B%28k-1%29%7D%28n%29-f%5E%7B%28k-1%29%7D%28m%29%7D%7Bk%21%7D& alt=&\sum_{l=m}^{n-1}f(l)=\int_m^n f(x)dx+\sum_{k=1}^\infty B_k \frac{f^{(k-1)}(n)-f^{(k-1)}(m)}{k!}& eeimg=&1&&&br&&br&嘿！这看上去是个非常好的公式，虽然不知道对不对。&br&那么我们用个例子看看这个公式对不对：&br&&img src=&///equation?tex=%5Csum_%7Bl%3D1%7D%5E%7Bn-1%7Dl%3D%5Cint_1%5En+xdx%2B+B_1%28n-1%29%3D%5Cfrac%7B1%7D%7B2%7D%28n%5E2-1%29-%5Cfrac%7B1%7D%7B2%7D%28n-1%29%3D%5Cfrac%7B1%7D%7B2%7Dn%28n-1%29& alt=&\sum_{l=1}^{n-1}l=\int_1^n xdx+ B_1(n-1)=\frac{1}{2}(n^2-1)-\frac{1}{2}(n-1)=\frac{1}{2}n(n-1)& eeimg=&1&&&br&好像还不错。&br&&br&但是很遗憾，很多情况下（除了多项式，也许？），第二个无穷和不收敛，所以这个公式不太对。参见Concrete Mathematics p457。 不过，我们计算出来的形式恰好是Euler最初得到的形式. 正确的形式称为Euler-Maclaurin求和公式：&br&&img src=&///equation?tex=%5Csum_%7Bl%3Dm%7D%5E%7Bn-1%7Df%28l%29%3D%5Cint_m%5En+f%28x%29dx%2B%5Csum_%7Bk%3D1%7D%5EpB_k+%5Cfrac%7Bf%5E%7B%28k-1%29%7D%28n%29-f%5E%7B%28k-1%29%7D%28m%29%7D%7Bk%21%7D%2BR_p& alt=&\sum_{l=m}^{n-1}f(l)=\int_m^n f(x)dx+\sum_{k=1}^pB_k \frac{f^{(k-1)}(n)-f^{(k-1)}(m)}{k!}+R_p& eeimg=&1&&&br&那个余项&img src=&///equation?tex=R_p& alt=&R_p& eeimg=&1&&中有Bernoulli多项式和我们的函数的p阶导数乘积的积分。这个余项是非常重要的。&br&关于这个公式，如下的链接是一个比较不错的故事书：&br&Dances between continuous and discrete:
Euler’s summation formula&br&&a href=&///?target=https%3A//www.math.nmsu.edu/%7Edavidp/euler2k2.pdf& class=& external& target=&_blank& rel=&nofollow noreferrer&&&span class=&invisible&&https://www.&/span&&span class=&visible&&math.nmsu.edu/~davidp/e&/span&&span class=&invisible&&uler2k2.pdf&/span&&span class=&ellipsis&&&/span&&i class=&icon-external&&&/i&&/a&&br&&br&Bernoulli数本身是个非常复杂的话题，它出现在（但不限于）如下的场合：&br&1. 解析数论里&img src=&///equation?tex=%5Czeta& alt=&\zeta& eeimg=&1&&函数在正偶数的值&br&2. von Staudt-Clausen定理以及代数数论里的Kummer's criterion&br&3. 矩阵里Campbell-Baker-Hausdorff公式：&img src=&///equation?tex=%5Clog+%28e%5EA+e%5EB%29& alt=&\log (e^A e^B)& eeimg=&1&&的展开，参见&a href=&///?target=http%3A//www.math.harvard.edu/%7Eshlomo/docs/lie_algebras.pdf& class=& external& target=&_blank& rel=&nofollow noreferrer&&&span class=&invisible&&http://www.&/span&&span class=&visible&&math.harvard.edu/~shlom&/span&&span class=&invisible&&o/docs/lie_algebras.pdf&/span&&span class=&ellipsis&&&/span&&i class=&icon-external&&&/i&&/a&&br&4. 正切、余切函数的泰勒展开&br&5. 拓扑 里的Todd class&br&6. 某个我不懂的东西：&a href=&///?target=http%3A//ocw.mit.edu/courses/mathematics/18-238-geometry-and-quantum-field-theory-fall-2002/lecture-notes/sec5.pdf& class=& external& target=&_blank& rel=&nofollow noreferrer&&&span class=&invisible&&http://&/span&&span class=&visible&&ocw.mit.edu/courses/mat&/span&&span class=&invisible&&hematics/18-238-geometry-and-quantum-field-theory-fall-2002/lecture-notes/sec5.pdf&/span&&span class=&ellipsis&&&/span&&i class=&icon-external&&&/i&&/a&&br&7. Mathoverflow上的&a href=&///?target=http%3A//mathoverflow.net/questions/61252/why-do-bernoulli-numbers-arise-everywhere/87599& class=& external& target=&_blank& rel=&nofollow noreferrer&&&span class=&invisible&&http://&/span&&span class=&visible&&mathoverflow.net/questi&/span&&span class=&invisible&&ons/61252/why-do-bernoulli-numbers-arise-everywhere/599&/span&&span class=&ellipsis&&&/span&&i class=&icon-external&&&/i&&/a&&br&和这个讨论&a href=&///?target=https%3A//golem.ph.utexas.edu/category/2008/02/metric_spaces.html%23c014936& class=& external& target=&_blank& rel=&nofollow noreferrer&&&span class=&invisible&&https://&/span&&span class=&visible&&golem.ph.utexas.edu/cat&/span&&span class=&invisible&&egory/2008/02/metric_spaces.html#c014936&/span&&span class=&ellipsis&&&/span&&i class=&icon-external&&&/i&&/a&&br&&br&希望有帮助
通常，大多数这类问题都是有不显然的原因的，而且关于Bernoulli数在各种不同场合的出现是有非常大量的文献以及讨论的。下面是一个很知名的heuristic，能够几乎完全给题主的问题一个解答。如果你在读Concrete Mathematics的话跳到9.5章，如果没在读这是一个…
先简单地说一下吧~伯努利数&br&&img src=&///equation?tex=Bn& alt=&Bn& eeimg=&1&&的第一次发现是与下列数列求和的公式有关：&br&&img src=&///equation?tex=%5Csum_%7Bk%3D0%7D%5E%7Bm-1%7D%7Bk%5En%7D%3D0%5En%2B1%5En%2B2%5En%2B..........%2B%28m-1%29%5En+& alt=&\sum_{k=0}^{m-1}{k^n}=0^n+1^n+2^n+..........+(m-1)^n & eeimg=&1&&&br&其中n为固定的任意正整数。&br&&br&这数列的求和公式中，只看最后一项，可以知道左边的通项可以写成：&br&&img src=&///equation?tex=%28m-i%29%5En++& alt=&(m-i)^n
& eeimg=&1&& 其中，&img src=&///equation?tex=i%5Cin+%281%2Cm%29& alt=&i\in (1,m)& eeimg=&1&&&br&&br&下面我们看一个生成函数&br&&img src=&///equation?tex=e%5Ex-1%3D%281%2Bx%2B%5Cfrac%7Bx%5E2%7D%7B2%21%7D%2B%5Cfrac%7Bx%5E3%7D%7B3%21%7D%2B.....%2B%5Cfrac%7Bx%5En%7D%7Bn%21%7D%2B.......+%29-1%3Dx%2B%5Cfrac%7Bx%5E2%7D%7B2%21%7D%2B%5Cfrac%7Bx%5E3%7D%7B3%21%7D%2B......& alt=&e^x-1=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+.....+\frac{x^n}{n!}+....... )-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+......& eeimg=&1&&&br&所以&br&我们有&br&&img src=&///equation?tex=x%3D%28B_%7B0%7D%2BB_%7B1%7Dx%2BB_%7B2%7D%5Cfrac%7Bx%5E2%7D%7B2%21%7D%2BB_%7B3%7D%5Cfrac%7Bx%5E3%7D%7B3%21%7D%2B....+%2BB_%7Bn%7D%5Cfrac%7Bx%5En%7D%7Bn%21%7D%2B........%29%5Ctimes+%28x%2B%5Cfrac%7Bx%5E2%7D%7B2%21%7D%2B%5Cfrac%7Bx%5E3%7D%7B3%21%7D%2B%5Cfrac%7Bx%5E4%7D%7B4%21%7D%2B.......%2B%5Cfrac%7Bx%5En%7D%7Bn%21%7D%29& alt=&x=(B_{0}+B_{1}x+B_{2}\frac{x^2}{2!}+B_{3}\frac{x^3}{3!}+.... +B_{n}\frac{x^n}{n!}+........)\times (x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+.......+\frac{x^n}{n!})& eeimg=&1&&&br&如果记&img src=&///equation?tex=a_%7B0%7D%3D0%2Ca_%7Bn%7D%3D1+%28n%5Cgeq+1%29& alt=&a_{0}=0,a_{n}=1 (n\geq 1)& eeimg=&1&&&br&我们有&br&&img src=&///equation?tex=x%3D%28%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7Ba_%7Bn%7D%7D%7Bn%21%7D%7D+x%5En%29%5Ctimes+%28%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7BB_%7Bn%7D%7D%7Bn%21%7D%7D+x%5En%29& alt=&x=(\sum_{n=0}^{\infty}{\frac{a_{n}}{n!}} x^n)\times (\sum_{n=0}^{\infty}{\frac{B_{n}}{n!}} x^n)& eeimg=&1&&&br&我们再定义一个项数&img src=&///equation?tex=C_%7Bn%7D& alt=&C_{n}& eeimg=&1&&&br&&img src=&///equation?tex=x%3D%28%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bc_%7Bn%7D%7D%7Bn%21%7D%7D+x%5En%29& alt=&x=(\sum_{n=0}^{\infty}{\frac{c_{n}}{n!}} x^n)& eeimg=&1&&&br&再看看多项式展开，x的系数相等，所以Bn和an系数，下标和为n就可以啦~&br&我们有：&br&&img src=&///equation?tex=c_%7Bn%7D%3D%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%7BC_%7Bn%7D%5E%7Bk%7DB_ka_%7Bn-k%7D+%7D+& alt=&c_{n}=\sum_{k=0}^{n}{C_{n}^{k}B_ka_{n-k} } & eeimg=&1&&&br&因为&br&&img src=&///equation?tex=a_%7B0%7D%3D0%2Ca_%7Bn%7D%3D1+%28n%5Cgeq+1%29& alt=&a_{0}=0,a_{n}=1 (n\geq 1)& eeimg=&1&&&br&所以我们有：&br&&img src=&///equation?tex=c_%7Bn%7D%3D%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%7BC_%7Bn%7D%5E%7Bk%7DB_k%7D+& alt=&c_{n}=\sum_{k=0}^{n-1}{C_{n}^{k}B_k} & eeimg=&1&&&br&&img src=&///equation?tex=x%3D%28%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7Bc_%7Bn%7D%7D%7Bn%21%7D%7D+x%5En%29& alt=&x=(\sum_{n=0}^{\infty}{\frac{c_{n}}{n!}} x^n)& eeimg=&1&&两端关于x的幂相等，所以最后只保留n=1的项。于是，我们有：&br&&br&&img src=&///equation?tex=%5Csum_%7Bk%3D0%7D%5E%7Bn-1%7D%7BC_%7Bn%7D%5E%7Bk%7DB_k%7D%3D0+& alt=&\sum_{k=0}^{n-1}{C_{n}^{k}B_k}=0 & eeimg=&1&&&img src=&///equation?tex=n%3D%282%2C3%2C....%29& alt=&n=(2,3,....)& eeimg=&1&&&br&&br&在上面式子两端加上&img src=&///equation?tex=B_n& alt=&B_n& eeimg=&1&&&br&&br&我们有：&br&&img src=&///equation?tex=%5Csum_%7Bk%3D0%7D%5E%7Bn%7D%7BC_%7Bn%7D%5E%7Bk%7DB_k%7D%3DB_n& alt=&\sum_{k=0}^{n}{C_{n}^{k}B_k}=B_n& eeimg=&1&&&br&&br&这就是&img src=&///equation?tex=B_n& alt=&B_n& eeimg=&1&&的递推公式。&br&&br&令n=2，B0=1&br&B0+2B1+B2=B2&br&由此B1=1/2&br&&br&再带入回去：&br&&img src=&///equation?tex=x%3D%28%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7Ba_%7Bn%7D%7D%7Bn%21%7D%7D+x%5En%29%5Ctimes+%28%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7BB_%7Bn%7D%7D%7Bn%21%7D%7D+x%5En%29& alt=&x=(\sum_{n=0}^{\infty}{\frac{a_{n}}{n!}} x^n)\times (\sum_{n=0}^{\infty}{\frac{B_{n}}{n!}} x^n)& eeimg=&1&&&br&就变成了：&br&&img src=&///equation?tex=%5Cfrac%7Bx%7D%7Be%5Ex-1%7D+%2B%5Cfrac%7Bx%7D%7B2%7D%3D1%2B%28%5Csum_%7Bn%3D2%7D%5E%7B%5Cinfty%7D%7B%5Cfrac%7BB_%7Bn%7D%7D%7Bn%21%7D%7D+x%5En%29& alt=&\frac{x}{e^x-1} +\frac{x}{2}=1+(\sum_{n=2}^{\infty}{\frac{B_{n}}{n!}} x^n)& eeimg=&1&&&br&&br&很容易知道，左边的函数是&img src=&///equation?tex=f%28x%29%3Df%28-x%29& alt=&f(x)=f(-x)& eeimg=&1&&&br&所以，等式右边只有关于&img src=&///equation?tex=x& alt=&x& eeimg=&1&&的偶次项。&br&&img src=&///equation?tex=B_%7B2n-1%7D%3D0& alt=&B_{2n-1}=0& eeimg=&1&&&br&这便是用生成函数的方法来表示数~&br&&br&&br&现在生成函数的方法已经进一步拓展到其他。&br&&br&数学是描述逻辑的语言~比如说，你要证明一个逻辑，把N表示成m个数的平方和，这种表示方法的个数。这种逻辑怎么证明？用其他的工具？实验？都无法来直接的证明。所以，数学就告诉你，我独创了一种语言。比如，我就定义一个数论函数&br&这个函数就是把N表示成m个数的平方和的表法个数~于是，椭圆函数论中一个著名的等式就诞生了~那就是&br&&br&&img src=&///equation?tex=%281%2B2x%2B2x%5E4%2B2x%5E9%2B....%29%5E2%3D1%2B4%28%5Cfrac%7Bx%7D%7B1-x%7D-%5Cfrac%7Bx%5E3%7D%7B1-x%5E3%7D%2B%5Cfrac%7Bx%5E5%7D%7B1-x%5E5%7D-.....+%29& alt=&(1+2x+2x^4+2x^9+....)^2=1+4(\frac{x}{1-x}-\frac{x^3}{1-x^3}+\frac{x^5}{1-x^5}-..... )& eeimg=&1&&&br&&br&有人说，为什么会有左边的公式。你仔细想想&br&&img src=&///equation?tex=%5Csum_%7Bm%3D-%5Cinfty+%7D%5E%7B%5Cinfty%7D+x%5E%7Bm%5E2%7D+%3D1%2B2x%2B2x%5E2%2B2x%5E4%2B2x%5E9%2B......& alt=&\sum_{m=-\infty }^{\infty} x^{m^2} =1+2x+2x^2+2x^4+2x^9+......& eeimg=&1&&&br&&br&这是恒成立的~因为根据级数的定义嘛~&br&这就是解决数学问题的第一步，把常见的逻辑问题“翻译”成为数学语言。而平方和的表法个数，我们用初等数论没法解决，我们考虑引入一个新的东西，比如把平方弄到指数上去。当然你弄到对数上也可以，不过对数做起来不怎么方便，这是第一步。&br&&br&第二步，根据等式寻找逻辑关系。平方表示方法个数。要有&img src=&///equation?tex=m_%7B1%7D%5E2%2Bm_2%5E2& alt=&m_{1}^2+m_2^2& eeimg=&1&&&br&把上面等式两边平方，是不是左边&img src=&///equation?tex=x& alt=&x& eeimg=&1&&的指数就变成了&img src=&///equation?tex=m_%7B1%7D%5E2%2Bm_2%5E2& alt=&m_{1}^2+m_2^2& eeimg=&1&&?&br&而右边，令&img src=&///equation?tex=n%3Dm_%7B1%7D%5E2%2Bm_2%5E2& alt=&n=m_{1}^2+m_2^2& eeimg=&1&&&br&是不是就是表示把n表示成两个平方和相加有几种表示方法了？&br&&br&数学类似这样的问题都可以用生成函数的方法解决~
先简单地说一下吧~伯努利数Bn的第一次发现是与下列数列求和的公式有关：\sum_{k=0}^{m-1}{k^n}=0^n+1^n+2^n+..........+(m-1)^n 其中n为固定的任意正整数。这数列的求和公式中，只看最后一项，可以知道左边的通项可以写成：(m-i)^n 其中，i\in (1,m)下面我…
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display: 'inlay-fix'（1+x）/(2+x^2)^2怎么展开成关于x的幂级数?
写了一会儿只写出x/（2+x^2）^2关于x的幂级数展开,加个1我就不懂了~给你点思路,自己加油吧~对x/（2+x^2）^2积分得-1/2(2+x^2),提项得-1/4(1+x^2/2),即是-1/4乘上1/（1+x^2/2）利用幂级数形式1/（1+x）,将x^2/2代入x,展开为幂级数,乘上-1/4,然后求导,得原式幂级数展开.这是x/（2+x^2）^2幂级数展开,加了个1就复杂了,我懒得想了,你自己加油~期末不挂科!补充一下：1/(2+x^2)^2化为-1/4*(1+x^2/2)^-2,即是-1/4乘上(1+x^2/2)^-2,利用幂级数形式（1+x）^a,令a=-2,x替换为x^2/2,求出幂级数后与上面求的那个相加,就是（1+x）/(2+x^2)^2的关于x的幂级数展开了胡乱算的,不知道对不对,你自己问问老师吧~加油!
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f(x)=（1+x）/(2+x^2)^2利用泰勒级数展开f(x)=f(x0)+f`( x0)(x- x0)+f``( x0)(x-x0)²/2!+f```( x0)(x- x0)³/3!+...fn(x0)(x- x0)n/n!+....令x0=0f(x)=f(0)+f`( 0)x+f``( 0)x²/2!+f```( 0)x³/3!+...fn(0)x^n/n!+....
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函数展开成幂级数函数幂级数展开幂级数的和函数幂级数求和函数幂函数展开
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