微积分表

微分表

`f(x)` `f^'(x)` `f(x)` `f^'(x)`
`sin x` `cos x` `sinh x` `cosh x`
`cos x` `-sin x` `cosh x` `sinh x`
`tan x` `1/(cos^2 x)` `tanh x` `1/(cosh^2 x)`
`cot x` `- 1/(sin^2 x)` `coth x` `- 1/(sinh^2 x)`
`arcsin x` `1/sqrt(1-x^2)` `"arsh "x` `1/sqrt(x^2+1)`
`arccos x` `- 1/sqrt(1-x^2)` `"arch "x` `1/sqrt(x^2-1)`
`arctan x` `1/(1+x^2)` `"arth "x` `1/(1-x^2)`

积分表

`f^'(x)` `f(x)`
`tan x` `-ln |cos x|`
`cot x` `ln |sin x|`
`sec x` `ln |sec x + tan x|`, `-1/2 ln {:(1-sinx)/(1+sinx):}`
`csc x` `ln |csc x - cot x|`, `1/2 ln {:(1-cosx)/(1+cosx):}, ln |tan{:x/2:}|`
`1/sqrt(a^2-x^2)` `arcsin{:x/a:}`, `-arccos{:x/a:}`
`1/(a^2+x^2)` `1/a arctan{:x/a:}`
`tanh x` `ln("e"^x+"e"^-x)`, `ln cosh x`
`coth x` `ln |"e"^x-"e"^-x|`, `ln |sinh x|`
`1/cosh x` `2 arctan "e"^x`, `arctan sinh x`
`1/sinh x` `ln {:|"e"^x-1|/("e"^x+1):}`, `1/2 ln {:(coshx-1)/(coshx+1):}`, `ln {:(coshx-1) / |sinhx|:}`
`1/sqrt(x^2+a^2)` `ln (x+sqrt(x^2+a^2))`, `"arsh "{:x/a:}`
`1/sqrt(x^2-a^2)` `ln |x+sqrt(x^2-a^2)|`, `sgn x * "arch "{:|x|/a:}`
`1/(x^2-a^2)` `1/(2a) ln |(x-a)/(x+a)|`, `-1/a "arth "(x/a)^(sgn(a-|x|))`

`sinh x = ("e"^x - "e"^-x)/2` `quad cosh x = ("e"^x + "e"^-x)/2`, `quad tanh x = ("e"^x - "e"^-x)/("e"^x + "e"^-x)`;
`"arsh "x = ln (x+sqrt(x^2+1))`, `quad x in RR`,
`"arch "x = ln (x+sqrt(x^2-1))`, `quad x ge 1`,
`"arth "x = 1/2 ln ((1+x)/(1-x))`, `quad -1 lt x lt 1`.

级数与无穷乘积

幂级数

    设 `|x| lt 1`, 则
  1. `sum_(n ge 0) x^n = 1/(1-x)` (几何级数);
  2. `sum_(n ge 1) x^n/n = -ln(1-x)`;
  3. `sum_(n ge 1, "odd") x^n/n = "arth "x = 1/2 ln|(1+x)/(1-x)|`;
  1. 由等式 `(1-x)(1 + x + cdots + x^n) = 1 - x^(n+1)` 得 `1 + x + cdots + x^n = (1-x^(n+1))/(1-x)`, 两边令 `n to oo` 即得结论.
  2. 由几何级数两边从 `0` 到 `x` 积分即可.
  3. 由 `sum_(n ge 0) t^(2n) = 1/(1-t^2)` 两边从 `0` 到 `x` 积分即可.
收敛域 函数 级数展开
`|x| lt oo` `"e"^x` `sum_(n ge 0) x^n/(n!) = 1 + x + x^2/(2!) + x^3/(3!) + cdots`
`|x| lt 1` `1/(1-x)` `sum_(n ge 0) x^n = 1 + x + x^2 + x^3 + cdots`
`|x| lt 1` `(1+x)^alpha` `sum_(n ge 0) (alpha;n) x^n = 1 + alpha x + (alpha(alpha-1)) / (2!) x^2 + (alpha(alpha-1)(alpha-2))/(3!) + cdots`
`|x| lt 1` `ln(1+x)` `sum_(n ge 0) (-1)^n x^(n+1)/(n+1) = x - x^2/2 + x^3/3 - x^4/4 + cdots`
`|x| lt oo` `sinh x` `sum_(n ge 0) x^(2n+1)/((2n+1)!) = x + x^3/(3!) + x^5/(5!) + x^7/(7!) + cdots`
`|x| lt oo` `cosh x` `sum_(n ge 0) x^(2n)/((2n)!) = 1 + x^2/(2!) + x^4/(4!) + x^6/(6!) + cdots`
`|x| lt 1` `"arth "x` `sum_(n ge 0) x^(2n+1)/(2n+1) = x + x^3/3 + x^5/5 + x^7/7 + cdots`
`|x| lt 1` `arcsin x` `sum_(n ge 0) ((2n-1)!!)/((2n)!!) x^(2n+1)/(2n+1) = x + 1/6 x^3 + 3/40 x^5 + 5/112 x^7 + 35/1152 x^9 + cdots`
`|x| gt 1` `"arch "x` `ln 2x - sum_(n ge 1) (-1)^n ((2n-1)!!)/((2n)!!) x^(-2n)/(2n) = ln 2x - 1/(4x^2) - 3/(32x^4) - 15/(288x^6) - cdots`
`|x| lt 2pi` `x/("e"^x-1)` `sum_(n ge 0) B_n/(n!) x^n = 1 - x/2 + x^2/6 -x^4/30 + 5/66 x^6 - cdots`
`|x| lt pi` `coth x` `1/x sum_(n ge 0) B_(2n)/((2n)!) (2x)^(2n) = 1/x + 1/3 x - 1/45 x^3 + 2/945 x^5 + cdots`
`|x| lt pi/2` `tanh x` `1/x sum_(n ge 1) B_(2n)/((2n)!) (4^n-1)(2x)^(2n) = x - 1/3 x^3 + 2/15 x^5 - 17/315 x^7 + 62/2835 x^9 - 1382/155925 x^11 + cdots`
`|x| lt pi` `csc x` `1/x sum_(n ge 0) (-1)^n B_(2n)/((2n)!) (2-4^n) x^(2n) = 1/x + 1/6 x + 7/360 x^3 + 31/15120 x^5 + 127/604800 x^7 + 73/3421440 x^9 + cdots`
`|x| lt pi/2` `1/(cosh x)` `sum_(n ge 0) E_(2n)/((2n)!) x^(2n) = 1 - 1/2 x^2 + 5/24 x^4 - 61/720 x^6 - cdots`
`"erf"(x)` `2/(sqrt pi) sum_(n ge 0) (-1)^n/(n!) x^(2n+1)/(2n+1)`
  1. 在 `1/sqrt(1-x^2) = sum_(n ge 0) (-1/2; n) (-x^2)^n` `= sum_(n ge 0) ((-1/2)(-3/2) cdots (-(2n-1)/2))/(n!) (-x^2)^n` `= sum_(n ge 0) ((2n-1)!!)/((2n)!!) x^(2n)` 两边逐项积分, 就得到 `arcsin x` 的展式.
  2. 将 `"erf"(x) = 2/sqrt pi int_0^x "e"^(-t^2) dt` 的被积函数展开并逐项积分, 就得到 `"erf"(x)` 的展式.

双阶乘 `n` 为正整数时, `(2n)!! = 2 * 4 * 6 * cdots * 2n`, `quad (2n-1)!! = 1 * 3 * 5 * cdots * (2n-1)`. 特别约定 `0!! = (-1)!! = 1`.

  1. `B_(2n)` 是 Bernoulli 数. 满足 `B_0 = 1, B_1 = -1/2, B_3 = B_5 = B_7 = cdots = 0`. 事实上由生成函数知道 `x = ("e"^x-1) sum_(k ge 0) B_k x^k/(k!)` `= (sum_(j ge 1) x^j/(j!)) (sum_(k ge 0) B_k x^k/(k!))` `= sum_(n ge 1) x^n/(n!) sum_(k=0)^(n-1) B_k (n; k)`. 等号两边比较系数得 `sum_(k=0)^n (n+1; k) B_k = 0`, `quad n ge 1`.
  2. `E_(2n)` 是 Euler 数, 它是正负号交错的整数: `E_0 = 1`, `E_2 = -1`, `E_4 = 5`, `E_6 = -61`, `E_8 = 1385`, `E_10 = -50521`. 由 `E_(2n)` 的生成函数知道, `1 = cosh x sum_(k ge 0) E_(2k)/((2k)!) x^(2k)` `= (sum_(j ge 0) x^(2j)/((2j)!)) (sum_(k ge 0) E_(2k)/((2k)!) x^(2k))` `= sum_(n ge 0) x^(2n)/((2n)!) sum_(k=0)^n (2n;2k) E_(2k)` 比较等式两边的各次项系数, 得到递推公式: `sum_(k=0)^n (2n;2k) E_(2k) = 0`, `quad n ge 1`.
  3. 三角函数与其对应的双曲三角函数有简单联系 `cos x = cosh("i"x)`, `quad cosh x = cos("i"x)`,
    `sin x = -"i" sinh("i"x)`, `quad sinh x = -"i" sin("i"x)`,
    `tan x = -"i" tanh("i"x)`, `quad tanh x = -"i" tan("i"x)`,
    `cot x = "i" coth("i"x)`, `quad coth x = "i" cot("i"x)`,
    `arccos x = "arch "x("i"x)`, `quad "arch "x = arccos("i"x)`,
    `arcsin x = -"i arsh "x("i"x)`, `quad "arsh "x = -"i"arcsin("i"x)`,
    `arctan x = -"i arth "x("i"x)`, `quad "arth "x = -"i"arctan("i"x)`,     因此只需将一个三角函数的级数从第二项起交错改变符号, 就得到对应的双曲三角函数的级数, 反之亦然. 如 `cos x = sum_(n ge 0) (-1)^n x^(2n)/((2n)!)`,
    `"arsh "x = sum_(n ge 0) (-1)^n ((2n-1)!!)/((2n)!!) x^(2n+1)/(2n+1)`,
    `cot x = 1/x sum_(n ge 0) (-1)^n B_(2n)/((2n)!) (2x)^(2n)`.

[题源 brilliant.org] 求 `(sum_(n=1)^oo n/((2n+1)!))/(sum_(n=1)^oo n/((2n-1)!))`.

将求和写成 `(2/(3!) + 4/(5!) + 6/(7!) + cdots)/ (2/(1!) + 4/(3!) + 6/(5!) + cdots)` 分子等于 `(3-1)/(3!) + (5-1)/(5!) + (7-1)/(7!) + cdots` `= 1/(2!) - 1/(3!) + 1/(4!) - 1/(5!) + 1/(6!) - 1/(7!) + cdots` `= "e"^-1 - 1 + 1/(1!)` `= "e"^-1`. 类似可得分母等于 `"e"`, 于是结果为 `"e"^-2`.

由 digamma 函数的展式得 `sum_(n ge 1)(1/n - 1/(n+s)) = psi(s+1) + gamma`.

`prod_(n ge 1) (1-x^2/n^2) = (sin pi x)/(pi x)`, 两边取对数再求导, 整理得 `sum_(n ge 1) 1/(n^2-x^2) = 1/(2x^2) (1 - pi x cot pi x)`. 容易得到 `sum_(n ge 1) 1/(n^2+1) = pi/2 "coth "pi - 1/2`. 此外可以裂项得到 `sum_(n ge 2) 1/(n^2-1) = 3/4`.

[群友 我是某用户的壹零叁] 把 `x = "i"z` 代入正弦无穷乘积, 变成双曲正弦, 可得 `prod_(n ge 1) (1+1/n^2) = (sinh pi)/pi`.

    求 `f(x) = sum_(n ge 0) x^n/(n!!)` (`0!! = 1`).

`f(0) = 1`, 且 `f` 适合微分方程 `y' = x y + 1`. 解得 `f(x) = sqrt(2/pi) int_(-oo)^x "e"^((x^2-t^2)/2) dt` `= "e"^(x^2/2) ("erf"(sqrt 2 x)+1)`. 利用 Wallis 公式知道, 收敛半径为 `oo`.

Fourier 级数

    设 `x in (0, 2pi)`, 则
  1. `sum_(n ge 1) (sin n x)/n = (pi-x)/2`;
  2. `sum_(n ge 1) (cos n x)/n^2 = x^2/4 - (pi x)/2 + pi^2/6`.
  3. `sum_(n ge 1) (sin n x)/n^3 = x^3/12 - (pi x^2)/4 + (pi^2 x)/6`.
    4. 在 1. 中替换 `x |-> pi-x`, 就得到 `x` 在 `|x| lt pi` 上的 Fourier 展开. 类似地, 由 2, 3 分别得 `x^2, x^3` 的 Fourier 展开.
  1. 直接计算 `(pi-x)//2` 的 Fourier 级数即可.
  2. 上题结果在 `[0, x]` 上逐项积分得 `sum_(n ge 1) (1-cos n x)/n^2 = pi/2 x - x^2/4`, 再利用 `sum_(n ge 1) n^-2 = pi^2//6` 即得结论.
  3. 由上题结果在 `[0, x]` 上积分即得.
    类似上题有
  1. `sum_(n ge 1) (cos n x)/n = -ln(2 sin{:x/2:})`;
  2. `sum_(n ge 1) (sin n x)/n^2 = -int_pi^x ln(2 sin{:t/2:}) dt`;
    3. 令 `x = pi-x`, 又得到
  3. `sum_(n ge 1) (-1)^n (cos n x)/n = -ln(2 cos{:x/2:})`;
  4. `sum_(n ge 1) (-1)^(n-1) (sin n x)/n^2 = int_0^x ln(2 cos{:t/2:}) dt`;
    `x = 0, pi` 时, 等式两边等于 `0`; `x = pi/2` 时, 等于 Calatan 常数 `K`.

一般地, `f` 在 `(-pi,pi)` 上的 Fourier 级数可以由以下公式计算: `f(x) = a_0/2 + sum_(n ge 1) (a_n cos n x + b_n sin n x)`,
`a_n = 1/pi int_(-pi)^pi f(x) cos n x dx`, `quad b_n = 1/pi int_(-pi)^pi f(x) sin n x dx`.

收敛域 函数 Fourier 展开
`|x| lt pi` `x` `2 sum_(n ge 1) (-1)^(n-1) (sin n x)/n`
`|x| le pi` `x^2` `pi^2/3 + 4 sum_(n ge 1) (-1)^n (cos n x)/n^2`
`|x| lt pi` `x^3` `sum_(n ge 1) (12-2n^2 pi^2)/n^3 (-1)^n sin n x`
`|x| lt pi` `"sgn "x` `4/pi sum_(n ge 1, "odd") (sin n x)/n`
`|x| le pi` `|x|` `pi/2 - 4/pi sum_(n ge 1, "odd") (cos n x)/n^2`
`|x| le pi` `x sin x` `1 - (cos x)/2 - 2 sum_(n ge 2) (-1)^n (cos n x)/(n^2-1)`
`|x| lt pi` `x cos x` `-(sin x)/2 + 2 sum_(n ge 2) (-1)^n (n sin n x)/(n^2-1)`
`|x| lt oo` `|sin x|` `2/pi - 4/pi sum_(n ge 1, "even") (cos n x)/(n^2 - 1)`
`|x| lt oo` `|cos x|` `2/pi - 4/pi sum_(n ge 1, "even") "i"^n (cos n x)/(n^2 - 1)`
`|x| lt pi` `sinh a x` `-2/pi sinh a pi sum_(n ge 1) (-1)^n (n sin n x)/(n^2+a^2)`
`|x| le pi` `cosh a x` `2/pi sinh a pi(1/(2a) + sum_(n ge 1) (-1)^n (a cos n x)/(n^2+a^2))`
`|x| lt pi` `"e"^(a x)` `2/pi sinh a pi(1/(2a) + sum_(n ge 1) (-1)^n (a cos n x - n sin n x)/(n^2 + a^2))`

提示: 1. `x sin x` 和 `x cos x` 积分时需借助积化和差; 2. `"e"^(a x) = sinh a x + cosh a x`.

在 `x` 的展式中取 `x = pi/2`, 得到 `pi//4 = sum_(n ge 1) (-1)^(n-1)/(2n-1)`; 在 `x^2` 的展式中取 `x = pi`, 得到 `sum_(n ge 1) n^-2 = pi^2//6`.

积分变换

详细的讨论参见 积分变换.

Laplace 变换

用 Gamma 函数的定义验证表中的第一条公式; 后面几条都是它的推论.

原函数 `f(t)` 像函数 `F(s) = int_0^oo "e"^(-s t) f(t) dt` 收敛域
`t^(p-1) "e"^(z t)` `(Gamma(p))/(s-z)^p, p gt 0` `"Re"s gt "Re"z`
`"e"^(z t)` `1/(s-z)` `"Re"s gt "Re"z`
`t^n` (`n` 是非负整数) `n!/s^(n+1)` `"Re"s gt 0`
`cos a t` `s/(s^2+a^2)` `"Re"s gt 0`
`sin a t` `a/(s^2+a^2)` `"Re"s gt 0`
`(sin a t)/t`, `a gt 0` `arctan(a/s)` 证明
`delta(t)` (单位脉冲函数) `1` 证明见下
`t^(a-1) ln^n t`, `a gt 0`, `n in ZZ^+` `1/s^a sum_(k=0)^n (n;k) Gamma^((n-k))(a)(-ln s)^k`. `"Re"s gt 0`
`ln t` `-(gamma + ln s)/s` `"Re"s gt 0`
  1. `cc L[delta(t)]`. 严格说, `delta(t)` 是广义函数, 我们暂不追究详情; 直观上它是下面函数族当 `epsi to 0` 的极限: `f_epsi(t) = { 1//epsi, if |t| lt epsi; 0, otherwise :}` 于是 `cc L[delta(t)]` `= lim_(epsi to 0) int_0^epsi 1/epsi "e"^(-s t) dt` `= lim_(epsi to 0) 1/(epsi s) (1-"e"^(-epsi s)) = 1`.
  2. `cc L[t^(a-1) ln^n t]`. 由 `cc L[t^(a-1)] = (Gamma(a))/s^a` 两边对 `a` 求 `n` 阶导即得结论. 特别 `a = n = 1` 时 `cc L[ln t] = -(gamma + ln s)/s`

Mellin 变换

`f(x)` `F(s) = int_0^oo x^(s-1) f(x) dx` 收敛域
`"e"^-x` `Gamma(s)` `"Re"s gt 0`
`1/("e"^x-1)` `Gamma(s) zeta(s)` `"Re"s gt 1`
`1/("e"^x+1)` `Gamma(s) eta(s)` `"Re"s gt 1`
`1/(x+a)` `pi/(sin pi s) a^(s-1)` `0 lt "Re"s lt 1`

Fourier 变换

`f(x)` `F(xi) = int_(-oo)^oo f(x) "e"^(-2pi"i"xi x) dx`
`"e"^(-pi x^2)` `"e"^(-pi xi^2)`

连分数

[DLMF, 數心《无限套娃的连分数》, 知乎: 连分数展开的一点公式]

Euler 连分数公式 `sum_(i=0)^n prod_(j=0)^i a_j` `= a_0/(1-) a_1/(1+a_1-) a_2/(1+a_2-) cdots` `a_(n-1)/(1+a_(n-1)-) a_n/(1+a_n)`.

记 `s_k = sum_(i=k)^n prod_(j=k)^i a_j`, 则 `s_k = sum_(i=k)^n a_k prod_(j=k+1)^i a_j` `= a_k(1 + sum_(i=k+1)^n prod_(j=k+1)^i a_j)` `= a_k(1+s_(k+1))`, `k = 0, 1, cdots, n-1`. 因此 `s_0 = a_0 (1+s_1)` `= a_0/(1-s_1//(1+s_1))`,
`s_k/(1+s_k)` `= a_k/(a_k + a_k//s_k)` `= a_k/(a_k + 1//(1+s_(k+1)))` `= a_k/(a_k + 1 - s_(k+1)//(1+s_(k+1)))`. 连续运用上式即得原式成立.

级数化为连分数 用 `u_i` 替换 Euler 连分数中的乘积 `prod_(j=0)^i a_j`, 得到 `sum_(i=0)^n u_i` `= a_0/(1-) a_1/(b_1-) a_2/(b_2-) cdots a_n/b_n`, 其中 `a_0 = u_0`, `b_0 = 1`, `a_n = u_n//u_(n-1)`, `quad b_n = 1 + a_n`, `quad n ge 1`. 故一个各项不为零的级数, 形式上可以表示为连分数, 使得其第 `n` 个部分和等于连分数的第 `n` 项近似.

无穷乘积化为连分数 利用下式将无穷乘积写为级数: `prod_(n ge 1) (1 + a_n/b_n)` `= 1 + a_1/b_1 + a_2/b_2(1+a_1/b_1) + a_3/b_3(1+a_1/b_1)(1+a_2/b_2) + cdots` 再将级数化为连分数, 得 `1/(1-) (a_1//b_1)/(1+a_1//b_1-)` `(a_2//b_2 (1+b_1//a_1))/(1+a_2//b_2(1+b_1//a_1)-)` `(a_3//b_3 (1+b_2//a_2))/(1+a_3//b_3(1+b_2//a_2)-) cdots`.

    利用 Taylor 展开,
  1. 指数函数 `"e"^x = 1 + 1 * x + 1 * x * x^2/2 + cdots`
    `= 1/(1-) x/(1+x-) (x//2)/(1+x//2-) (x//3)/(1+x//3-) cdots`
    `= 1/(1-) x/(1+x-) x/(2+x-) (2x)/(3+x-) (3x)/(4+x-) cdots`,     特别 `"e" = 1/(1-) 1/(2-) 1/(3-) 2/(4-) 3/(5-) cdots`. 上面的连分数只是幂级数的另一种写法, 其各项截断等于幂级数的各个部分和, 因此其收敛性由幂级数的收敛性保证. 和 `"e"` 的无穷级数一样, 该连分数收敛很快, 取 n = 12 就得到 `"e"` 的近似值 2.718281828703704.
  2. 正弦函数 `sin x = x + x * (-x^2)/3 + x * (-x^2)/3 * (-x^2)/5 + cdots`
    `= x/(1-) ((-x^2)//(2*3))/(1-x^2//(2*3)-) ((-x^2)//(4*5))/(1-x^2//(4*5)-) cdots`
    `= x/(1+) x^2/(2*3-x^2+) (2*3 x^2)/(4*5-x^2+) cdots`
  3. 反正切函数 `arctan x = x + x * (-x^2)/3 + x * (-x^2)/3 * (-3x^2)/5 + cdots`
    `= x/(1-) (-x^2//3)/(1-x^2//3-) (-3x^2//5)/(1-3x^2//5-) cdots`
    `= x/(1+) x^2/(3-x^2+) (3x)^2/(5-3x^2+) (5x)^2/(7-5x^2+) cdots`,     特别 `pi/4 = 1/(1+) 1^2/(2+) 3^2/(2+) 5^2/(2+) cdots`. 和 `arctan` 的展式一样, 上面的连分数收敛很慢. 取 n = 999, 只得到 `pi` 的近似值 3.143584671557847.

连分数的第 `n` 项近似 记 `b_1/(a_1+) b_2/(a_2+) cdots b_n/a_n = y_n/x_n` (不约分), 则分子和分母均满足二阶线性递推关系: `x_0 = 1`, `quad x_1 = a_1`, `quad x_n = a_n x_(n-1) + b_n x_(n-2)`,
`y_0 = 0`, `quad y_1 = b_1`, `quad y_n = a_n y_(n-1) + b_n y_(n-2)`. 上式给出连分数第 `n` 项近似的算法. 反之, 任意两个满足上面递推关系的数列, 补充定义 `x_0, y_0` 后, 它们商可以展开为连分数. 例如 Fibonacci 数列满足 `F_0 = 0`, `F_1 = 1`, `F_n = F_(n-1) + F_(n-2)`, 因此 `F_n/F_(n+1)` `= F_n/(F_n + F_(n-1))` `= 1/(1+F_(n-1)//F_n)` `= cdots` `= 1/(1+) 1/(1+) cdots 1/1`.

容易验证边界条件成立. 假设成立 `y_n/x_n = (a_n y_(n-1) + b_n y_(n-2))/(a_n x_(n-1) + b_n x_(n-2))`, 用 `a_n + b_(n+1)//a_(n+1)` 替换公式中的 `a_n` 得 `y_(n+1)/x_(n+1)` `= ((a_n + b_(n+1)//a_(n+1)) y_(n-1) + y_n - a_n y_(n-1)) / ((a_n + b_(n+1)//a_(n+1)) x_(n-1) + x_n - a_n x_(n-1))` `= (a_(n+1) y_n + b_(n+1) y_(n-1))/(a_(n+1) x_n + b_(n+1) x_(n-1))`.

考虑超几何函数 `{::}_2 F_1(a,b";"c";"z)` `= 1 + (a b)/c z + (a(a+1)b(b+1))/(c(c+1)) z^2/(2!) + cdots`, 取 `f_(2n) = {::}_2 F_1(a+n,b+n";"c+2n-1";"z)`,
`f_(2n+1) = {::}_2 F_1(a+n+1,b+n";"c+2n";"z)`,
`n = 0, 1, 2, cdots`, 则 `f_n = f_(n+1) + k_n z f_(n+2)`, 其中 `k_(2n) = ((a-c-n+1)(b+n))/((c+2n-1)(c+2n))`, `quad k_(2n+1) = ((b-c-n-1)(a+n+1))/((c+2n)(c+2n+1))`. 于是 `f_0/f_1` 可以展开为 `f_0/(f_1)` `= (f_1 + k_0 z f_2)/(f_1)` `= 1 + (k_0 z)/(f_1 // f_2)` `= cdots` `= 1 + (k_0 z)/(1+) (k_1 z)/(1+) cdots`. 特别取 `a = 0`, 则 `f_1 = 1`, 我们就得到 `f_0` 的连分数. 鉴于许多初等函数都能写成超几何函数, 如 `arctan z = z {::}_2 F_1(1,1/2";"3/2";"-z^2)`. 我们得到一些初等函数的连分数:

`"e"^z = 1/(1-) z/(color(blue)1+) z/(color(red)2-) z/(color(blue)3+) z/(color(red)2-) z/(color(blue)5+) z/(color(red)2-) cdots`,
`ln(1+z) = z/(1+) z/(2+) z/(3+) (4z)/(4+) (4z)/(5+) (9z)/(6+) (9z)/(7+) cdots`,
`tan z = z/(1-) z^2/(3-) z^2/(5-) z^2/(7-) cdots`, `quad z != (n+1/2)pi, n in ZZ`,
`arctan z = z/(1+) z^2/(3+) (2z)^2/(5+) (3z)^2/(7+) cdots`, `quad "i"z !in (-oo, -1) uu (1, oo)`,
`arcsin z/sqrt(1-z^2) = z/(1-) (1*2 z^2)/(3-) (1*2 z^2)/(5-) (3*4z^2)/(7-) (3*4z^2)/(9-) cdots`, `quad z !in (-oo, -1) uu (1, oo)`.

`tanh x` 的连分数的另一证明: 记 `a_(n,k) = 1//(2n)!(2n+1)(2n+3)cdots(2n+2k-1)`, 则 `a_(n,k-1) - (2k-1) a_(n,k)` `= { a_(n-1,k-1), if n ge 1; 0, if n = 0 :}` 进而有 `s_k - (2k-1)` `:= (sum_(n ge 0) a_(n,k-1) x^(2n))/(sum_(n ge 0) a_(n,k) x^(2n)) - (2k-1)` `= (sum_(n ge 1) a_(n-1,k+1)x^(2n))/(sum_(n ge 0) a_(n,k) x^(2n))` `= x^2(sum_(n ge 0) a_(n,k+1)x^(2n))/(sum_(n ge 0) a_(n,k) x^(2n))` `= x^2 // s_(k+1)`, 即 `s_k = (2k-1) + x^2//s_(k+1)`. 将 `sinh x` 和 `cosh x` 的 Taylor 展式相除, `tanh x = x (1 + x^2/(3!) + x^4/(5!) + cdots)/(1 + x^2/(2!) + x^4/(4!) + cdots)` `= x (sum_(n ge 0) a_(n,1) x^(2n))/(sum_(n ge 0) a_(n,0) x^(2n))` `= x // s_1`, 递推得 `tanh x = x/(1+) x^2/(3+) x^2/(5+) x^2/(7+) cdots`.