The Binomial Series
(((1 + x))^p)=(∑_(k = 0)^∞)(({{p}, {k}})) (x^k)
for any real  p  and  | x | < 1

by
K. D.  Stroyan
University of Iowa

copyright 1997 by Academic Press, Inc. - All rights reserved.

Special Mathematica Functions used in this NoteBook

Off[General::spell1];
Needs["Graphics`Colors`"]
seriesGraphs[f_, series_, radius_, numterms_Integer, opns___Rule] :=
	Module[ {functions, graylevels, i, j,n,x,terms},
	Do[
	functions = Table[series[x,n], {n, 0, i}] ;
	AppendTo[functions, f[x]] ;
	graylevels = Table[{GrayLevel[j]}, {j, 6/10, 0, -6/(10 i)}] ; 
	graylevels=AppendTo[graylevels,{AbsoluteThickness[2],Red}];
	Plot[Release[functions], {x, -radius, radius},
		PlotStyle -> graylevels,PlotLabel ->i"terms",
		opns,AspectRatio->Automatic];
	,{i, 1, numterms}];
	Do[
	Plot[Release[{f[x], series[x,n]}],{x, -radius, radius},
		PlotLabel ->n"terms ", opns,
		PlotStyle -> {{AbsoluteThickness[2],Red},{GrayLevel[0]}},
		AspectRatio->Automatic];
	,{n, 1, numterms}];
	];

Needs["Graphics`Colors`"]
powerGraphs[f_,  radius_, numterms_, opns___Rule] :=
	Module[ {functions, graylevels, i, j,n,x,terms},
	Do[
	functions = Table[Normal[Series[f[x],{x,0,n}]], {n, 0, i}] ;
	AppendTo[functions, f[x]] ;
	graylevels = Table[{GrayLevel[j]}, {j, 6/10, 0, -6/(10 i)}] ; 
	graylevels=AppendTo[graylevels,{AbsoluteThickness[2],Red}];
	Plot[Release[functions], {x, -radius, radius},
		PlotStyle -> graylevels, PlotLabel ->i"terms",
		opns,AspectRatio->Automatic];
	,{i, 1, numterms}];
	Do[
	Plot[Release[{Normal[Series[f[x],{x,0,n}]],f[x]}],{x, -radius, radius},
		PlotLabel ->n"terms ", opns, 
		PlotStyle -> {{GrayLevel[0]},{AbsoluteThickness[2],Red}},AspectRatio->Automatic];
	,{n, 1, numterms}];
	];
On[General::spell1];

Notebook Overview

This NoteBook studies the binomial series

(((1 + x))^p)=(∑_(k = 0)^∞)(({{p}, {k}})) (x^k)

References to the Text

This NoteBook goes with Problem 26.1 of the text.

Step-by-step

The Finite Series

In high school you learned how to raise a binomial to a power, in particular  (1+x)^p  can be expanded to 1 + c[1] x + c[2] x^2 + ... + c[p] x^p, when  p  is a positive integer.    The coefficients in these expansions are called "binomial coefficients."  For example,

(((1 + x))^2) = 1 + 2 x + (x^2)(((1 + x))^3) = 1 + 3 x + 3 (x^2) + (x^3)(((1 + x))^4) = 1 + 4 x + 6 (x^2) + 4 (x^3) + (x^4)

When the power is  2, the coefficients in the algebraic expansion are 1, 2, 1.  Mathematica  calls these Binomial[2,0], Binomial[2,1], Binomial[2,2].  When the power is  3, the coefficients are 1, 3, 3, 1.  Mathematica  calls these Binomial[3,0], Binomial[3,1], Binomial[3,2], Binomial[3,3].  When the power is  4, the coefficients are 1, 4, 6, 4, 1.

You may remember some of the tricks of the binomial trade.  One important fact is that

(({{p}, {k}}))= Binomial[p, k]  =  
          ((p (p - 1) (p - 2) ⋯ 
              (p - k + 1))/(k (k - 1) (k - 2) 
              ⋯1))

Here's how to do this with Mathematica:

TableForm[Table[Binomial[n, k], {n, 2, 10}, {k, 0, n}]]

Clear[x, n, bino]; bino[x_, n_] := 1 + (∑_(k = 1)^n) Binomial[p, k] (x^k); TableForm[Table[bino[x, 10], {p, 1, 9}]]

Of course, the previous computation was pretty stupid.  Here's an easier way:

TableForm[Table[Expand[(((1 + x))^p)], {p, 0, 9}]]

Let's ask Mathematica if these are the same:

p = 10; Expand[(((1 + x))^p)] == bino[x, p]

The Infinite Series

Maybe the computation wasn't entirely stupid, let's try  p = 1/2 to see how to expand the square root.

p = (1/2)Expand[(((1 + x))^p)]bino[x, 10]

This is peculiar, let's see more:

p = (1/2)bino[x, 20]

Humm, interesting.  Let's plot:

Clear[f, x, p]; f[x_] := (((1 + x))^p); p = (1/2)radius = 1.25; numterms = 9; Print[bino[x, numterms]]; seriesGraphs[f, bino, radius, numterms, AspectRatio -> 1, 
      PlotRange -> {-1, 3}];

The binomial expansion converges to (1+x)^p for -1 < x < 1.

Notice the difficulty at  x = -1 (or Sqrt[0].)  Let's compare values at the worst place on the graph where we can hope for convergence, x = 1:

x = 1.0(Sqrt[1 + x])bino[x, 10]bino[x, 20]

Convergence is better inside:

x = 0.5(Sqrt[1 + x])bino[x, 5]bino[x, 10]

Practice Exercises

Divergence outside |x| < 1:

Plot 20 terms of the binomial series and the function (Sqrt[1 + x]) for -2 < x < 2.  What do you observe?

p = (1/2)radius = 2; numterms = 20; Print[bino[x, numterms]]; seriesGraphs[f, bino, radius, numterms, AspectRatio -> 1, 
      PlotRange -> {-1, 3}];

Termination of Binomial[n,k]

Use the formula

(({{p}, {k}}))= Binomial[p, k]  =  
          ((p (p - 1) (p - 2) ⋯ 
              (p - k + 1))/(k (k - 1) (k - 2) 
              ⋯1))

to show that if  p  is a positive integer, then  Binomial[p,k] = 0 for k > p.

If  p  is not a positive integer, does the sequence

Binomial[p,0], Binomial[p,1], Binomial[p,2], ...

ever end?  What if  p < 0?,  p = 0?

Plot (1+x)^p & bino[x,n] for various  p

Plot the binomial series approximation for several values of p:

p = 0

Explain this graph:

p = 0radius = 2; numterms = 9; Print[bino[x, numterms]]; seriesGraphs[f, bino, radius, numterms, AspectRatio -> 1, 
      PlotRange -> {-1, 3}];

p = -1

Except for being reversed, this graph should look familiar.  What is it?

p = -1radius = 2; numterms = 9; Print[bino[x, numterms]]; seriesGraphs[f, bino, radius, numterms, AspectRatio -> 1, 
      PlotRange -> {-1, 3}];

p = -3/2

p = -(3/2)radius = 2; numterms = 9; Print[bino[x, numterms]]; seriesGraphs[f, bino, radius, numterms, AspectRatio -> 1, 
      PlotRange -> {-1, 3}];

Problem 26.1: The Differential Equation

(0) Uniqueness

Solutions to initial value problems are unique provided the growth law is smooth.  Look up the exact statement of the Existence & Uniqueness Theorem from Chapter 21 and show that the problem

y[0] == 1,
	y'[x]==(p/(1 + x)) y[x]

satisfies the theorem.  (HINT: What is the function  f[x,y]  in the formula  dy/dx = f[x,y]?  Where is f[x,y] smooth?  Does the theorem guarantee anything about  x = -1?)

(1) The Analytical Solution

Show that  y[x] = (((1 + x))^p) is a solution to the above initial value problem.

Clear[x, y, p]; DSolve[{y[0] == 1,(1 + x) y'[x]==p y[x]},{y[x]},x]

(2) Some Algebra

Show that the binomial coefficients given by

(({{p}, {k}}))= Binomial[p, k]  =  
          ((p (p - 1) (p - 2) ⋯ 
              (p - k + 1))/(k (k - 1) (k - 2) 
              ⋯1))

satisfy

(k+1) Binomial[p,k+1] == (p-k) Binomial[p,k]

for any real   p  and any positive integer k.  Use this to show that

(k+1)Binomial[p,k+1] + k Binomial[p,k] == p Binomial[p,k]

Warm-up for (3) & (4):

HINT: The next computation is just a start:

\(Clear[p, \ n, \ x]; \ 
\((1\  + \ x)\)*D[bino[x, \ n], \ x]\)p*bino[x,  n]\(p\  = \ \(-2\); \ 
n\  = \ 7; \ 

    Expand[\((1\  + \ x)\)*D[bino[x, \ n], \ x], \ x]\)Expand[p*bino[x,  n],  x]

\(n\  = \ 10; \ 
Expand[\((1\  + \ x)\)*D[bino[x, \ n], \ x], \ x]\)Expand[p*bino[x,  n],  x]

\(n\  = \ 100; \ 

    Expand[\((1\  + \ x)\)*D[bino[x, \ n], \ x], \ x]\  - \ 
\ \ 
      Expand[p*bino[x, \ n], \ x]\)

(3) Term-by-term Differentiation

Differentiate bino[x,n] with respect to  x  and show

D[bino[x,Infinity],x] ==(∑_(k = 1)^∞) k Binomial[p, k] (x^(k - 1))
	== (∑_(h = 0)^∞)(h + 1) Binomial[p, h + 1] 
          (x^h)

(4) Multiplication

Multiply

(x+1) D[ bino[x,Infinity] , x] =(x + 1) 
        (∑_(h = 0)^∞)(h + 1) Binomial[p, h + 1] 
            (x^h)

and use

(k+1)Binomial[p,k+1] + k Binomial[p,k] == p Binomial[p,k]

to show

(1+x) D[bino[x,Infinity],x] == p bino[x,Infinity]

Why can't you show this for a fixed finite  n?

(5) Equate equal things

Use a converence theorem to prove that the series defining

bino[x,Infinity]

converges for  | x | < 1.  Then use part (0) and the fact that bino[x,Infinity] and (1+x)^p both satisfy the same initial value problem to prove that

(1+x)^p = Sum [ Binomial[p,k] x^k, {k,0,Infinity}]

YOUR EXPLANATION:

(6) Write out bino[x,8]  when  p = 1/2

Clear[x]p = (1/2)bino[x, 8]