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  • 牛顿二项式与 e 级数

    复习一下数学, 找一下回忆.

    先是从二项式平方开始:

    (a+b)^{2} = a^{2} + 2ab + b^2

    其实展开是这样的:

    (a+b)^{2} = aa + ab + ba + bb

    再看立方:

    (a+b)^{3} = a^3 + 3a^{2}b + 3ab^2 + b^3
    (a+b)^{3} = aaa + aab + aba + baa + bba + bab + abb + bbb

    通过排列组合的方式标记, 于是:

    (a+b)^3 = {3 choose 0}a^3 b^0 + {3 choose 1} a^{2}b^1 + {3 choose 2} a^{1}b^2 + {3 choose 3} a^{0} b^{3}

    通过数学归纳法可以拓展:

    (a+b)^n = {n choose 0}a^n b^0 + {n choose 1}a^{n-1}b^1 + {n choose 2}a^{n-2}b^2 + cdots + {n choose n-1}a^1 b^{n-1} + {n choose n}a^0 b^n

    使用求和简写可得:

    (a+b)^n = sum_{k=0}^n {n choose k}a^{n-k}b^k = sum_{k=0}^n {n choose k}a^{k}b^{n-k}

    e 级数

    数学常数 e (The Constant e – NDE/NDT Resource Center) 的定义爲下列极限值:

    e = lim_{n	oinfty} left(1 + frac{1}{n}
ight)^n.

    使用二项式定理能得出

    left(1 + frac{1}{n}
ight)^n = 1 + {n choose 1}frac{1}{n} + {n choose 2}frac{1}{n^2} + {n choose 3}frac{1}{n^3} + cdots + {n choose n}frac{1}{n^n}.

    第 k 项之总和为

    {n choose k}frac{1}{n^k} ;=; frac{1}{k!}cdotfrac{n(n-1)(n-2)cdots (n-k+1)}{n^k}

    因为 n → ∞,右边的表达式趋近1。 因此

    lim_{n	oinfty} {n choose k}frac{1}{n^k} = frac{1}{k!}.

    由于序列的极限可以相加, 所以 e 可以表示为:

    e = sum_{k=0}^inftyfrac{1}{k!}=frac{1}{0!} + frac{1}{1!} + frac{1}{2!} + frac{1}{3!} + cdots.

    计算情况:

    e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + …

    As an example, here is the computation of e to 22 decimal places:

    1/0! = 1/1 = 1.0000000000000000000000000
    1/1! = 1/1 = 1.0000000000000000000000000
    1/2! = 1/2 = 0.5000000000000000000000000
    1/3! = 1/6 = 0.1666666666666666666666667
    1/4! = 1/24 = 0.0416666666666666666666667
    1/5! = 1/120 = 0.0083333333333333333333333
    1/6! = 1/720 = 0.0013888888888888888888889
    1/7! = 1/5040 = 0.0001984126984126984126984
    1/8! = 1/40320 = 0.0000248015873015873015873
    1/9! = 1/362880 = 0.0000027557319223985890653
    1/10! = 1/3628800 = 0.0000002755731922398589065
    1/11! = 1/39916800 = 0.0000000250521083854417188
    1/12! = 1/479001600 = 0.0000000020876756987868099
    1/13! = 1/6227020800 = 0.0000000001605904383682161
    1/14! = 1/87178291200 = 0.0000000000114707455977297
    1/15! = 1/1307674368000 = 0.0000000000007647163731820
    1/16! = 1/20922789887989 = 0.0000000000000477947733239
    1/17! = 1/355687428101759 = 0.0000000000000028114572543
    1/18! = 1/6402373705148490 = 0.0000000000000001561920697
    1/19! = 1/121645101098757000 = 0.0000000000000000082206352
    1/20! = 1/2432901785214670000 = 0.0000000000000000004110318
    1/21! = 1/51091049359062800000 = 0.0000000000000000000195729
    1/22! = 1/1123974373384290000000 = 0.0000000000000000000008897
    1/23! = 1/25839793281653700000000 = 0.0000000000000000000000387
    1/24! = 1/625000000000000000000000 = 0.0000000000000000000000016
    1/25! = 1/10000000000000000000000000 = 0.0000000000000000000000001

    For more information on e, visit the the math forum at mathforum.org
    The sum of the values in the right column is 2.7182818284590452353602875 which is “e.”

    Reference: The mathforum.org

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  • 原文地址:https://www.cnblogs.com/veis/p/6940840.html
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