Binomial series

In mathematics, the binomial series is a generalization of the polynomial that comes from a binomial formula expression like $(1+x)^{n}$ for a nonnegative integer $n$ . Specifically, the binomial series is the MacLaurin series for the function $f(x)=(1+x)^{\alpha }$ , where $\alpha \in \mathbb {C}$ and $|x|<1$ . Explicitly,

{\begin{aligned}(1+x)^{\alpha }&=\sum _{k=0}^{\infty }\;{\binom {\alpha }{k}}\;x^{k}=1+\alpha x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+{\frac {\alpha (\alpha -1)(\alpha -2)}{3!}}x^{3}+\cdots \end{aligned}}

(1)

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

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

Note that if $α$ is a nonnegative integer $n$ then the $x n + 1$ term and all later terms in the series are $0$ , since each contains a factor of $(n - n)$ . Thus, in this case, the series is finite and gives the algebraic binomial formula.

YouTube Encyclopedic

1/5
Views:
1 680 094
1 965 358
452 124
5 220
90 118

Transcription

Voiceover:It doesn't take long to realize that taking higher and higher powers of binomials can get painful, but let's just work through a few just to realize how quickly they get painful. If we take the binomial a plus b, it's a binomial because it has two terms right over here, let's take that to the 0 power. Anything that's non-zero to the 0 power, that's just going to be equal to 1. That wasn't too bad. Now what about a plus b to the 1st power? That's just going to be a plus b. Now what about a plus b squared? Now, if you haven't been practicing taking powers of binomials, you might have been tempted to say a squared plus b squared, but that would be incorrect. If you did that, you should give yourself a very gentle but not overly discouraging slap on the wrist or the brain or something. a plus b squared is not a squared plus b squared. It is a plus b times a plus b. Then if you do this, it will be a times a, which is a squared, plus a times b, which is ab, plus b times a, which is another ab, plus b times b, which is b squared. You have two ab's here, so you could add them together, so it's equal to a squared plus 2ab plus b squared. Now things are going to get a little bit more interesting. What is a plus b to the 3rd power going to be equal to? I encourage you to pause this video and try to figure that out on your own. Well, we know that a plus b to the 3rd power is just a plus b to the 2nd power times another a plus b. Let's just multiply this times a plus b to figure out what it is. I'll do this. Let's see. Let's multiply that times a plus b. I'm just going to multiply it this way. First, I'll multiply b times all of these things. I'll do it in this green color. b times b squared is b to the 3rd power. b times 2ab is 2a squared, so 2ab squared, and then b times a squared is ba squared, or a squared b, a squared b. I'll multiply b times all of this stuff. Now let's multiply a times all this stuff. a times b squared is ab squared, ab squared. a times 2ab is 2a squared b, 2a squared b, and then a times a squared is a to the 3rd power. Now when we add all of these things together, we get, we get a to the 3rd power plus, let's see, we have 1 a squared b plus another, plus 2 more a squared b's. That's going to be 3a squared b plus 3ab squared. 2ab squared plus another ab squared is going to be 3ab squared plus b to the 3rd power. Just taking some of the 3rd power, this already took us a little reasonable amount of time, and so you can imagine how painful it might get to do something like a plus b to the 4th power, or even worse, if you're trying to find a plus b to the 10th power, or to the 20th power. This would take you all day or maybe even longer than that. It would be incredibly, incredibly painful. That's where the binomial theorem becomes useful. What is the binomial theorem? The binomial theorem tells us, let me write this down, binomial theorem. Binomial theorem, it tells us that if we have a binomial, and I'll just stick with the a plus b for now, if I have, and I'm going to try to color code this a little bit, if I have the binomial a plus b, a plus b, and I'm going to raise it the nth power, I'm going to raise this to the nth power, the binomial theorem tells us that this is going to be equal to, and the notation is going to look a little bit complicated at first, but then we'll work through an actual example, is going to be equal to the sum from k equals 0, k equals 0 to n, this n and this n are the same number, of ... I don't want to ... that's kind of a garish color ... of n choose k, n choose k, and we'll review that in a second; this comes straight out of combinatorics; n choose k times a to the n minus k, n minus k, times b, times b to the k, b to the k power. Now this seems a little bit unwieldy. Let's just review, remind ourselves what n choose k actually means. If we say n choose k, I'll do the same colors, n choose k, we remember from combinatorics this would be equal to n factorial, n factorial over k factorial, over k factorial times n minus k factorial, n minus k factorial, so n minus k minus k factorial, let me color code this, n minus k factorial. Let's try to apply this. Let's just start applying it to the thing that started to intimidate us, say, a plus b to the 4th power. Let's figure out what that's going to be. Let's try this. So a, and I'm going to try to keep it color-coded so you know what's going on, a plus b, although it takes me a little bit more time to keep switching colors, but hopefully it's worth it, a plus b. Let's take that to the 4th power. The binomial theorem tells us this is going to be equal to, and I'm just going to use this exact notation, this is going to be the sum from k equals 0, k equals 0 to 4, to 4 of 4 choose k, 4 choose k, 4 choose ... let me do that k in that purple color, 4 choose k of a to the 4 minus k power, 4 minus k power times b to the k power, b to the k power. Now what is that going to be equal to? Well, let's just actually just do the sum. This is going to be equal to, so we're going to start at k equals 0, so when k equals 0, it's going to be 4 choose 0, 4 choose 0, times a to the 4 minus 0 power, well, that's just going to be a to the 4th power, times b to the 0 power. b to the 0 power is just going to be equal to 1, so we could just put a 1 here if we want to, or we could just leave it like that. This is what we get when k equals 0. Then to that, we're going to add when k equals 1. k equals 1 is going to be, the coefficient is going to be 4 choose 1, and it's going to be times a to the 4 minus 1 power, so a to the 3rd power, and I'll just stick with that color, times b to the k power. Well, now, k is 1b to the 1st power. Then to that, we're going to add, we're going to add 4 choose 2, 4 choose 2 times a to the ... well, now k is 2. 4 minus 2 is 2. a squared. I think you see a pattern here. a to the 4th, a to the 3rd, a squared, and then times b to the k. Well, k is 2 now, so b squared, and you see a pattern again. You could say b to the 0, b to the 1, b squared, and we only have two more terms to add here, plus 4 choose 3, 4 choose 3 times 4 minus 3 is 1, times a, or a to the 1st, I guess we could say, and then b to the 3rd power, times a to the 1st b to the third, and then only one more term, plus 4 choose, 4 choose 4. k is now 4. This is going to be our last term right now. We're getting k goes from 0 all the way to 4, 4 choose 4. a to the 4 minus 4, that's just going to be 1, a to the 0, that's just 1, so we're going to be left with just b to the k power, and b is 4 right over here. We're almost done. We've expanded it out. We just need it figure out what 4 choose 0, 4 choose 1, 4 choose 2, et cetera, et cetera are, so let's figure that out. We could just apply this over and over again. So 4 choose 0, 4 choose 0 is equal to 4 factorial over 0 factorial times 4 minus 0 factorial. That's just going to be 4 factorial again. 0 factorial, at least for these purposes, we are defining to be equal to 1, so this whole thing is going to be equal to 1, so this coefficient is 1. Let's see. Let's keep going here. So 4 choose 1 is going to be 4 factorial over 1 factorial times 4 minus 1 factorial, 4 minus 1 factorial, so 3 factorial. What's this going to be? 1 factorial is just going to be 1. 4 factorial is 4 times 3 times 2 times 1. 3 factorial is 3 times 2 times 1. Let me make that clear. 4 times 3 times 2 times 1 over 3 times 2 times 1 is just going to leave us with 4. This right over here is just going to be 4. Then we need to figure out what 4 choose 2 is. 4 choose 2 is going to be 4 factorial over 2 factorial times what's 4 minus ... this is going to be n minus k, 4 minus 2 over 2 factorial. So what is this going to be? Let me scroll over to the right a little bit. This is going to be 4 times 3 times 2 times 1 over 2 factorial is 2, over 2 times 2. This is 2, this is 2, so 2 times 2 is same thing as 4. We're left with 3 times 2 times 1, which is equal to 6. That's equal to 6. Then what is 4 choose 3? I'll use some space down here. So 4 choose 3, 4 choose 3 is equal to 4 factorial over, over 3 factorial times 4 minus 3 factorial, so that's just going to be 1 factorial. Well, we already figured out what that is. That's the same thing as this right over here. You just swap the 1 factorial and the 3 factorial. We already figured out that this is going to be equal to 4. That is equal to 4. 4 choose 4? Well, this is just going to be, let me just do it over here, 4 choose 4 is 4 factorial over 4 factorial times 0 factorial, which is the exact thing we had here, which we figured out was 1. Just like that, we're done. We were able to figure out what a plus b to the 4th power is. It's 1a to the 4th plus 4a to the 3rd b to the 1st plus 6a squared b squared plus 4ab cubed plus b to the 4th. Actually, let me just write that down, since we did all that work. This is equal to a to the 4th plus, plus 4, plus 4a to the 3rd, a to the 3rd b plus, plus 6, plus 6a squared b squared, a squared b squared, plus, plus, plus 4, I think you see a pattern here, plus 4a times b to the 3rd power plus b to the 4th power, plus b to the 4th power. There is an interesting pattern here. There is a symmetry where you have the coefficient, you go 1, 4, 6 for the middle term, and then you go back to 4, and then you go back to 1. Then you also see that pattern, is that you start at a to the 4th, a to the 3rd, a squared, a, and then you could say there is an a to the 0 here, and then you started b to the 0, which we didn't write it, but that's just 1, then b to the 1st, b squared, b to the 3rd, b to the 4th. This is just one application or one example. In future videos, we'll do more examples of the binomial theorem and also try to understand why it works.

Convergence

Conditions for convergence

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

If $| x | < 1$ , the series converges absolutely for any complex number $α$ .
If $| x | = 1$ , the series converges absolutely if and only if either $Re(α) > 0$ or $α = 0$ , where $Re(α)$ denotes the real part of $α$ .
If $| x | = 1$ and $x \neq -1$ , the series converges if and only if $Re(α) > -1$ .
If $x = -1$ , the series converges if and only if either $Re(α) > 0$ or $α = 0$ .
If $| x | > 1$ , the series diverges except when $α$ is a non-negative integer, in which case the series is a finite sum.

In particular, if $α$ is not a non-negative integer, the situation at the boundary of the disk of convergence, $| x | = 1$ , is summarized as follows:

If $Re(α) > 0$ , the series converges absolutely.
If $-1 < Re(α) \leq 0$ , the series converges conditionally if $x \neq -1$ and diverges if $x = -1$ .
If $Re(α) \leq -1$ , the series diverges.

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}},

(2)

{\alpha  \choose k-1}+{\alpha  \choose k}={\alpha +1 \choose k}.

(3)

Unless $\alpha$ is a nonnegative integer (in which case the binomial coefficients vanish as $k$ is larger than $\alpha$ ), 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)),\quad {\text{as }}k\to \infty .

(4)

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)}},

and implies immediately the coarser bounds

{\frac {m}{k^{1+\operatorname {Re} \,\alpha }}}\leq \left|{\alpha  \choose k}\right|\leq {\frac {M}{k^{1+\operatorname {Re} \alpha }}},

(5)

for some positive constants $m$ and $M$ .

Formula (2) for the generalized binomial coefficient can be rewritten as

{\alpha  \choose k}=\prod _{j=1}^{k}\left({\frac {\alpha +1}{j}}-1\right).

(6)

Proof

To prove (i) and (v), apply the ratio test and use formula (2) above to show that whenever $\alpha$ is not a nonnegative integer, the radius of convergence is exactly 1. Part (ii) follows from formula (5), by comparison with the $p$ -series

\sum _{k=1}^{\infty }\;{\frac {1}{k^{p}}},

with $p=1+\operatorname {Re} \alpha$ . To prove (iii), first use formula (3) 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},

(7)

and then use (ii) and formula (5) again to prove convergence of the right-hand side when $\operatorname {Re} \alpha >-1$ is assumed. On the other hand, the series does not converge if $|x|=1$ and $\operatorname {Re} \alpha \leq -1$ , again by formula (5). Alternatively, we may observe that for all $j$ , ${\textstyle \left|{\frac {\alpha +1}{j}}-1\right|\geq 1-{\frac {\operatorname {Re} \alpha +1}{j}}\geq 1}$ . Thus, by formula (6), for all ${\textstyle k,\left|{\alpha \choose k}\right|\geq 1}$ . This completes the proof of (iii). Turning to (iv), we use identity (7) above with $x=-1$ and $\alpha -1$ in place of $\alpha$ , along with formula (4), to obtain

\sum _{k=0}^{n}\;{\alpha  \choose k}\;(-1)^{k}={\alpha -1 \choose n}\;(-1)^{n}={\frac {1}{\Gamma (-\alpha +1)n^{\alpha }}}(1+o(1))

as $n\to \infty$ . Assertion (iv) now follows from the asymptotic behavior of the sequence $n^{-\alpha }=e^{-\alpha \log(n)}$ . (Precisely, $\left|e^{-\alpha \log n}\right|=e^{-\operatorname {Re} \alpha \,\log n}$ certainly converges to $0$ if $\operatorname {Re} \alpha >0$ and diverges to $+\infty$ if $\operatorname {Re} \alpha <0$ . If $\operatorname {Re} \alpha =0$ , then $n^{-\alpha }=e^{-i\operatorname {Im} \alpha \log n}$ converges if and only if the sequence $\operatorname {Im} \alpha \log n$ converges ${\bmod {2\pi }}$ , which is certainly true if $\alpha =0$ but false if $\operatorname {Im} \alpha \neq 0$ : in the latter case the sequence is dense ${\bmod {2\pi }}$ , due to the fact that $\log n$ diverges and $\log(n+1)-\log n$ converges to zero).

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 disk of convergence $| x | < 1$ and using formula (1), one has that the sum of the series is an analytic function solving the ordinary differential equation $(1 + x) u'(x) - αu (x) = 0$ with initial condition $u (0) = 1$ .

The unique solution of this problem is the function $u (x) = (1 + x) α$ . Indeed, multiplying by the integrating factor $(1 + x) - α -1$ gives

0=(1+x)^{-\alpha }u'(x)-\alpha (1+x)^{-\alpha -1}u(x)={\big [}(1+x)^{-\alpha }u(x){\big ]}'\,,

so the function $(1 + x) -α u (x)$ is a constant, which the initial condition tells us is $1$ . That is, $u (x) = (1 + x) α$ is the sum of the binomial series for $| x | < 1$ .

The equality extends to $| x | = 1$ whenever the series converges, as a consequence of Abel's theorem and by continuity of $(1 + x) α$ .

Negative binomial series

Closely related is the negative binomial series defined by the MacLaurin series for the function $g(x)=(1-x)^{-\alpha }$ , where $\alpha \in \mathbb {C}$ and $|x|<1$ . Explicitly,

{\begin{aligned}{\frac {1}{(1-x)^{\alpha }}}&=\sum _{k=0}^{\infty }\;{\frac {g^{(k)}(0)}{k!}}\;x^{k}\\&=1+\alpha x+{\frac {\alpha (\alpha +1)}{2!}}x^{2}+{\frac {\alpha (\alpha +1)(\alpha +2)}{3!}}x^{3}+\cdots ,\end{aligned}}

which is written in terms of the multiset coefficient

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

When $α$ is a positive integer, several common sequences are apparent. The case $α = 1$ gives the series $1 + x + x 2 + x 3 + ...$ , where the coefficient of each term of the series is simply $1$ . The case $α = 2$ gives the series $1 + 2 x + 3 x 2 + 4 x 3 + ...$ , which has the counting numbers as coefficients. The case $α = 3$ gives the series $1 + 3 x + 6 x 2 + 10 x 3 + ...$ , which has the triangle numbers as coefficients. The case $α = 4$ gives the series $1 + 4 x + 10 x 2 + 20 x 3 + ...$ , which has the tetrahedral numbers as coefficients, and similarly for higher integer values of $α$ .

The negative binomial series includes the case of the geometric series, the power series^[1]

{\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}

(which is the negative binomial series when

\alpha =1

, convergent in the disc

|x|<1

) and, more generally, series obtained by differentiation of the geometric power series:

{\frac {1}{(1-x)^{n}}}={\frac {1}{(n-1)!}}{\frac {d^{n-1}}{dx^{n-1}}}{\frac {1}{1-x}}

with

\alpha =n

, a positive integer.^[2]

History

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. John Wallis built upon this work by considering expressions of the form $y = (1 - x 2) m$ where $m$ is a fraction. He found that (written in modern terms) the successive coefficients $c k$ of $(- x 2) k$ are to be found by multiplying the preceding coefficient by $m$ − ( $k$ − 1)/ $k$ (as in the case of integer exponents), thereby implicitly giving a formula for these coefficients. He explicitly writes the following instances^[a]

(1-x^{2})^{1/2}=1-{\frac {x^{2}}{2}}-{\frac {x^{4}}{8}}-{\frac {x^{6}}{16}}\cdots

(1-x^{2})^{3/2}=1-{\frac {3x^{2}}{2}}+{\frac {3x^{4}}{8}}+{\frac {x^{6}}{16}}\cdots

(1-x^{2})^{1/3}=1-{\frac {x^{2}}{3}}-{\frac {x^{4}}{9}}-{\frac {5x^{6}}{81}}\cdots

The binomial series is therefore sometimes referred to as Newton's binomial theorem. Newton gives no proof and is not explicit about the nature of the series. Later, on 1826 Niels Henrik Abel discussed the subject in a paper published on Crelle's Journal, treating notably questions of convergence. ^[4]

Footnotes

Notes

^ ^[3] 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.

Citations

^ George Andrews (2018), "The geometric series in calculus" (PDF), The American Mathematical Monthly, 105 (1): 36–40, doi:10.1080/00029890.1998.12004846
^ Knopp, Konrad (1944), Theory and applications of infinite series, Blackie and Son, §22.
^ Coolidge 1949.
^ Abel 1826.

References

Abel, Niels (1826), "Recherches sur la série 1 + (m/1)x + (m(m − 1)/1.2)x² + (m(m − 1)(m − 2)/1.2.3)x³ + ...", Journal für die reine und angewandte Mathematik, 1: 311–339
Coolidge, J. L. (1949), "The Story of the Binomial Theorem", The American Mathematical Monthly, 56 (3): 147–157, doi:10.2307/2305028, JSTOR 2305028

External links

Weisstein, Eric W. "Binomial Series". MathWorld.
Weisstein, Eric W. "Binomial Theorem". MathWorld.
binomial formula at PlanetMath.
Solomentsev, E.D. (2001) [1994], "Binomial series", Encyclopedia of Mathematics, EMS Press
"How Isaac Newton Discovered the Binomial Power Series". August 31, 2022.

v t e Calculus
Precalculus	Binomial theorem Concave function Continuous function Factorial Finite difference Free variables and bound variables Graph of a function Linear function Radian Rolle's theorem Secant Slope Tangent
Limits	Indeterminate form Limit of a function One-sided limit Limit of a sequence Order of approximation (ε, δ)-definition of limit
Differential calculus	Derivative Second derivative Partial derivative Differential Differential operator Mean value theorem Notation Leibniz's notation Newton's notation Rules of differentiation linearity Power Sum Chain L'Hôpital's Product General Leibniz's rule Quotient Other techniques Implicit differentiation Inverse functions and differentiation Logarithmic derivative Related rates Stationary points First derivative test Second derivative test Extreme value theorem Maximum and minimum Further applications Newton's method Taylor's theorem Differential equation Ordinary differential equation Partial differential equation Stochastic differential equation
Integral calculus	Antiderivative Arc length Riemann integral Basic properties Constant of integration Fundamental theorem of calculus Differentiating under the integral sign Integration by parts Integration by substitution trigonometric Euler Tangent half-angle substitution Partial fractions in integration Quadratic integral Trapezoidal rule Volumes Washer method Shell method Integral equation Integro-differential equation
Vector calculus	Derivatives Curl Directional derivative Divergence Gradient Laplacian Basic theorems Line integrals Green's Stokes' Gauss'
Multivariable calculus	Divergence theorem Geometric Hessian matrix Jacobian matrix and determinant Lagrange multiplier Line integral Matrix Multiple integral Partial derivative Surface integral Volume integral Advanced topics Differential forms Exterior derivative Generalized Stokes' theorem Tensor calculus
Sequences and series	Arithmetico-geometric sequence Types of series Alternating Binomial Fourier Geometric Harmonic Infinite Power Maclaurin Taylor Telescoping Tests of convergence Abel's Alternating series Cauchy condensation Direct comparison Dirichlet's Integral Limit comparison Ratio Root Term
Special functions and numbers	Bernoulli numbers e (mathematical constant) Exponential function Natural logarithm Stirling's approximation
History of calculus	Adequality Brook Taylor Colin Maclaurin Generality of algebra Gottfried Wilhelm Leibniz Infinitesimal Infinitesimal calculus Isaac Newton Fluxion Law of Continuity Leonhard Euler Method of Fluxions The Method of Mechanical Theorems
Lists	Differentiation rules List of integrals of exponential functions List of integrals of hyperbolic functions List of integrals of inverse hyperbolic functions List of integrals of inverse trigonometric functions List of integrals of irrational functions List of integrals of logarithmic functions List of integrals of rational functions List of integrals of trigonometric functions Secant Secant cubed List of limits Lists of integrals
Miscellaneous topics	Complex calculus Contour integral Differential geometry Manifold Curvature of curves of surfaces Tensor Euler–Maclaurin formula Gabriel's horn Integration Bee Proof that 22/7 exceeds π Regiomontanus' angle maximization problem Steinmetz solid

This page was last edited on 28 January 2024, at 12:56

From Wikipedia, the free encyclopedia