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In mathematics, the binomial series is the Taylor series of the function (1 + x) α at x = 0, where α is a complex number. Explicitly,

\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}

and the binomial series is the power series on the right hand side of (1), expressed in terms of the (generalized) binomial coefficients

 {\alpha \choose k} := \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!}.

Contents

Special cases

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 formula.

The following variant holds for arbitrary complex β, but is especially useful for handling negative integer exponents in (1):

\frac{1}{(1-z)^{\beta+1}} = \sum_{n=0}^{\infty}{n+\beta \choose n}z^n.

To prove it, substitute x = −z in (1) and apply a binomial coefficient identity.

Convergence

Conditions for convergence

Whether (1) converges depends on the values of the complex numbers α and x. More precisely:

(i) If |x| < 1, the series converges for any complex number α.

(ii) If |x| = 1, the series converges absolutely if and only if either Re(α) > 0 or α = 0.

(iii) If |x| = 1 and x ≠ −1, the series converges if and only if Re(α) > −1.

Identities to be used in the proof

The following hold for any complex number α:

{\alpha \choose 0} = 1,
 {\alpha \choose k+1} = {\alpha\choose k}\,\frac{\alpha-k}{k+1}, \qquad\qquad(1)
 {\alpha \choose k-1} + {\alpha\choose k} = {\alpha+1 \choose k}. \qquad\qquad(2)

When α is not a nonnegative integer, a useful asymptotic relationship for the binomial coefficients is, in Landau notation:

 {\alpha \choose k} = \frac{(-1)^k} {\Gamma(-\alpha)k^ {1+\alpha} } \,(1+o(1)),

as k → ∞. This is essentially equivalent to Euler's definition of the Gamma function:

 \Gamma(z) = \lim_{k \to \infty} \frac{k! \; k^z}{z \; (z+1)\cdots(z+k)}, \,\qquad

and implies immediately the coarser asymptotics

 {\alpha \choose k} =\; O\left(\frac {1} {k^{1+\operatorname{Re}\,\alpha}}\right), \qquad\qquad(3)

as k → ∞, which is sufficient for our needs. The simple bound (3) may also be obtained by means of elementary inequalities (see the addendum below).

Sketch of proof

To prove (i), apply the ratio test and use formula (1) above to show that whenever α is not a nonnegative integer, the radius of convergence is exactly 1. The absolute convergence (ii) follows from formula (3), by comparison with

 \sum_{k=1}^\infty \; \frac {1} {k^s}, \qquad

with s = 1 + Re(α). To prove (iii), first use formula (2) to obtain

(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},

and then use (ii) and formula (3) 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,

 \left|{\alpha \choose k}\; x^k \right| \geq 1.

Summation of the binomial series

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'(x) = α u(x) with initial data u(0) = 1. The unique solution of this problem is the function u(x) = (1 + x)α, 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)α.

History

The first results concerning convergence of the binomial series were discovered by Sir Isaac Newton, so it is therefore sometimes referred to as Newton's binomial theorem. Later, Niels Henrik Abel treated the subject in a memoir.

Addendum: elementary bounds on the coefficients

In order to keep the whole discussion within elementary methods, one may derive the asymptotics (3) proving the inequality

\left|{\alpha \choose k} \right|\leq\frac {M}{k^{1+\mathrm{Re}\,\alpha}},\qquad\forall k\geq1

with

M:= \exp\left(|\alpha|^2 +\mathrm{Re}\, \alpha \right)

as follows. By the inequality of arithmetic and geometric means

\left|{\alpha \choose k} \right|^2=\prod_{j=1}^k \left|1-\frac{1+\alpha}{j}\right|^2 \leq \left( \frac{1}{k}\sum_{j=1}^{k} \left|1-\frac{1+\alpha}{j}\right|^2 \right)^k.

Using the expansion

\textstyle |1-\zeta|^2=1-2\mathrm{Re}\,\zeta +|\zeta|^2

the latter arithmetic mean writes

\frac{1}{k}\sum_{j=1}^{k} \left|1-\frac{1+\alpha}{j}\right|^2= 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)\ .

To estimate its kth power we then use the inequality

\left(1+\frac{r}{k}\right)^k\leq \mathrm{e}^r,

that holds true for any real number r as soon as 1 + r/k ≥ 0. Moreover, we have elementary bounds for the sums:

\sum_{j=1}^k \frac{1}{j}\leq1+\log k; \qquad \sum_{j=1}^k \frac{1}{j^2} \leq 2.

Thus,

\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 )} }

with

M:=\exp\left(|\alpha|^2+\mathrm{Re}\,\alpha\right), \,

proving the claim.

See also








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