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| {{Calculus |Series}}
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| In [[mathematics]], the '''binomial series''' is the [[Taylor series]] at ''x'' = 0 of the function ''f'' given by ''f''(''x'') ''='' (1 + ''x'')<sup> ''α''</sup>, where {{nowrap|''α'' ∈ ''ℂ''}} is an arbitrary [[complex number]]. Explicitly,
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| :<math>\begin{align} (1 + x)^\alpha &= \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k \qquad\qquad\qquad (1) \\ &= 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!} x^2 + \cdots, \end{align}</math>
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| and the binomial series is the [[power series]] on the right hand side of (1), expressed in terms of the (generalized) [[binomial coefficient]]s
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| :<math> {\alpha \choose k} := \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!}. </math>
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| == Special cases ==
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| If α is a nonnegative integer ''n'', then the (''n'' + 1)th term and all later terms in the series are 0, since each contains a factor (''n'' − ''n''); thus in this case the series is finite and gives the algebraic [[binomial theorem|binomial formula]].
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| The following variant holds for arbitrary complex ''β'', but is especially useful for handling negative integer exponents in (1):
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| :<math>\frac{1}{(1-z)^{\beta+1}} = \sum_{k=0}^{\infty}{k+\beta \choose k}z^k.</math>
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| To prove it, substitute ''x'' = −''z'' in (1) and apply a binomial coefficient identity. | |
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| == Convergence ==
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| === Conditions for convergence ===
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| Whether (1) converges depends on the values of the complex numbers {{math|''α''}} and {{math|''x''}}. More precisely:
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| {{ordered list|type=lower-roman
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| |1= If {{math| {{abs|''x''}} < 1}}, the series converges [[absolute convergence|absolutely]] for any complex number α.
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| |2= If x = −1, the series converges absolutely [[if and only if]] either {{math|Re(α) > 0}} or {{math|α {{=}} 0}}.
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| |3= If {{math| {{abs|''x''}} {{=}} 1}} and {{math|''x'' ≠ −1}}, the series converges if and only if {{math|Re(α) > −1}}.
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| |4= If {{math| {{abs|''x''}} > 1}}, the series diverges, unless {{math|''α''}} is a non-negative integer (in which case the series is finite).
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| }}
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| Assume now that <math> \alpha </math> is not a non-negative integer and that <math> |x| = 1 </math>. We make the following additional observations, which follow from the ones above:
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| * If {{math|Re(α) > 0}}, the series converges absolutely.
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| * If {{math|-1 < Re(α) ≤ 0}}, the series converges [[conditional convergence|conditionally]] if {{math|''x'' ≠ −1}} and diverges if {{math|''x'' {{=}} −1}}.
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| * If {{math|Re(α) ≤ -1}}, the series diverges.
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| === Identities to be used in the proof ===
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| The following hold for any complex number α:
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| :<math>{\alpha \choose 0} = 1,</math>
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| :<math> {\alpha \choose k+1} = {\alpha\choose k}\,\frac{\alpha-k}{k+1}, \qquad\qquad(2) </math>
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| :<math> {\alpha \choose k-1} + {\alpha\choose k} = {\alpha+1 \choose k}. \qquad\qquad(3) </math>
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| Unless α is a nonnegative integer (in which case the binomial coefficients vanish as ''k'' is larger than α), a useful [[asymptotic analysis|asymptotic]] relationship for the binomial coefficients is, in [[Landau notation]]:
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| :<math> {\alpha \choose k} = \frac{(-1)^k} {\Gamma(-\alpha)k^ {1+\alpha} } \,(1+o(1)), \quad\text{as }k\to\infty. \qquad\qquad(4)</math>
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| This is essentially equivalent to Euler's definition of the [[Gamma function]]:
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| :<math>
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| \Gamma(z) = \lim_{k \to \infty} \frac{k! \; k^z}{z \; (z+1)\cdots(z+k)}, \qquad
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| </math>
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| and implies immediately the coarser bounds
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| :<math> \frac {m} {k^{1+\operatorname{Re}\,\alpha}}\le \left|{\alpha \choose k}\right| \le \frac {M} {k^{1+\operatorname{Re}\,\alpha}}, \qquad\qquad(5) </math>
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| for some positive constants ''m'' and ''M'', which are in fact sufficient for our needs. The simpler bounds (5) may also be obtained by means of elementary inequalities (see the [[#Addendum: elementary bounds on the coefficients|addendum]] below for the latter inequality).
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| === Proof ===
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| To prove (i) and (v), apply the [[ratio test]] and use formula (2) above to show that whenever ''α'' is not a nonnegative integer, the [[radius of convergence]] is exactly 1. Part (ii) follows from formula (5), by comparison with the [[Harmonic series (mathematics)#P-series|p-series]]
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| :<math> \sum_{k=1}^\infty \; \frac {1} {k^p}, \qquad </math>
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| with ''p'' = 1 + Re(α). To prove (iii), first use formula (3) to obtain
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| :<math>(1 + x) \sum_{k=0}^n \; {\alpha \choose k} \; x^k =\sum_{k=0}^n \; {\alpha+1\choose k} \; x^k + {\alpha \choose n} \;x^{n+1}, </math>
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| and then use (ii) and formula (5) again to prove convergence of the right-hand side when Re(α) > −1 is assumed. On the other hand, the series does not converge if |''x''| = 1 and Re(α) ≤ −1, because in that case, for all ''k'',
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| :<math> \left|{\alpha \choose k}\; x^k \right| \geq 1,</math>
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| completing the proof of (iii). Also, the identity above, for ''x=-1'' and with ''α+1'' in place of ''α'' writes
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| :<math>\sum_{k=0}^n \; {\alpha\choose k} \; (-1)^k = {\alpha-1 \choose n} \;(-1)^n, </math>
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| whence (iv) follows using (5) again.
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| == Summation of the binomial series ==
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| The usual argument to compute the sum of the binomial series goes as follows. Differentiating term-wise the binomial series within the convergence disk |''x''| < 1 and using formula (1), one has that the sum of the series is an [[analytic function]] solving the ordinary differential equation (1 + ''x'')''u''<nowiki>'</nowiki>(''x'') = α ''u''(''x'') with initial data ''u''(0) = 1. The unique solution of this problem is the function ''u''(''x'') = (1 + ''x'')<sup>α</sup>, which is therefore the sum of the binomial series, at least for |''x''| < 1. The equality extends to |''x''| = 1 whenever the series converges, as a consequence of [[Abel's theorem]] and by continuity of (1 + ''x'')<sup>α</sup>.
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| == History ==
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| The first results concerning binomial series for other than positive-integer exponents were given by Sir [[Isaac Newton]] in the study of areas enclosed under certain curves. Extending work by [[John Wallis]] who calculated such areas for ''y'' = (1 − ''x''<sup>2</sup>)<sup>''n''</sup> with ''n'' = 0, 1, 2, 3, ... he considered fractional exponents. He found for such exponent ''m'' that (in modern formulation) the successive coefficients ''c''<sub>''k''</sub> of (−''x''<sup>2</sup>)<sup>''k''</sup> are to be found by multiplying the preceding coefficient by <math>\tfrac{m-(k-1)}k</math> (as in the case of integer exponents), thereby implicitly giving a formula for these coefficients. He explicitly writes the following instances<ref>[http://www.jstor.org/pss/2305028 The Story of the Binomial Theorem, by J. L. Coolidge], ''The American Mathematical Monthly'' '''56''':3 (1949), pp. 147–157. In fact this source gives all non-constant terms with a negative sign, which is not correct for the second equation; one must assume this is an error of transcription.</ref>
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| :<math>(1-x^2)^{1/2}=1-\frac{x^2}2-\frac{x^4}8-\frac{x^6}{16}\cdots</math>
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| :<math>(1-x^2)^{3/2}=1-\frac{3x^2}2+\frac{3x^4}8+\frac{x^6}{16}\cdots</math>
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| :<math>(1-x^2)^{1/3}=1-\frac{x^2}3-\frac{x^4}9-\frac{5x^6}{81}\cdots</math>
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| The binomial series is therefore sometimes referred to as [[Binomial theorem#Newton's generalized binomial theorem|Newton's binomial theorem]]. Newton gives no proof and is not explicit about the nature of the series; most likely he verified instances treating the series as (again in modern terminology) [[formal power series]].{{citation needed|date=September 2012}} Later, [[Niels Henrik Abel]] treated the subject in a memoir, treating notably questions of convergence.
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| == Elementary bounds on the coefficients ==
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| In order to keep the whole discussion within elementary methods, one may derive the asymptotics (5) proving the inequality
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| :<math>\left|{\alpha \choose k} \right|\leq\frac {M}{k^{1+\mathrm{Re}\,\alpha}},\qquad\forall k\geq1</math>
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| with
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| :<math>M:= \exp\left(|\alpha|^2 +\mathrm{Re}\, \alpha \right)</math>
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| as follows. By the [[inequality of arithmetic and geometric means#The inequality|inequality of arithmetic and geometric means]]
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| :<math>\left|{\alpha \choose k} \right|^2=\prod_{j=1}^k \left|1-\frac{1+\alpha}{j}\right|^2
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| \leq \left( \frac{1}{k}\sum_{j=1}^{k} \left|1-\frac{1+\alpha}{j}\right|^2 \right)^k. </math>
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| Using the expansion
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| :<math>\textstyle |1-\zeta|^2=1-2\mathrm{Re}\,\zeta +|\zeta|^2</math>
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| the latter arithmetic mean writes
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| :<math>\frac{1}{k}\sum_{j=1}^{k} \left|1-\frac{1+\alpha}{j}\right|^2=
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| 1+\frac{1}{k}\left(- 2(1+\mathrm{Re}\,\alpha) \sum_{j=1}^{k}\frac{1}{j}+|1+\alpha|^2\sum_{j=1}^{k}\frac{1}{j^2}\right)\ .</math>
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| To estimate its ''k''th power we then use the inequality
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| :<math>\left(1+\frac{r}{k}\right)^k\leq \mathrm{e}^r,</math>
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| that holds true for any real number ''r'' as soon as 1 + ''r''/''k'' ≥ 0. Moreover, we have elementary bounds for the sums:
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| :<math>\sum_{j=1}^k \frac{1}{j}\leq1+\log k; \qquad \sum_{j=1}^k \frac{1}{j^2} \leq 2.</math>
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| Thus,
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| :<math>\left|{\alpha \choose k} \right|^2\leq \exp\left(- 2(1+\mathrm{Re}\,\alpha )(1+\log k) +2|1+\alpha|^2 \right) = \frac{M^2}{k^{2(1+\mathrm{Re}\,\alpha )} }</math>
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| with
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| :<math>M:=\exp\left(|\alpha|^2+\mathrm{Re}\,\alpha\right), \,</math>
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| proving the claim.
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| ==See also==
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| *[[Binomial theorem#Newton's generalized binomial theorem|Binomial theorem]]
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| *[[Table of Newtonian series]]
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| ==References==
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| {{Reflist}}
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| [[Category:Calculus]]
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| [[Category:Factorial and binomial topics]]
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| [[Category:Mathematical series]]
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| [[Category:Complex analysis]]
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| [[Category:Real analysis]]
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