Ken Ward's Mathematics Pages

Series Binomial Theorem

Series Contents

Page Contents

1. Binomial Theorem
2. Notation
3. "n" as a nonnegative integer
4. Proof of the Binomial Theorem
1. Proof when n and k are positive integers
2. Proof when r is any real number
5. Defining the Binomial Coefficients
6. Sum of Binomial Coefficients
7. Convergence

Binomial Theorem

The theorem is called binomial because it is concerned with a sum of two numbers (bi means two) raised to a power. Where the sum involves more than two numbers, the theorem is called the Multi-nomial Theorem. The Binomial Theorem was first discovered by Sir Isaac Newton.

Notation

We can write a Binomial Coefficient as:
[0.1]

Similarly, combinations can be written:
[0.2]

"n" as a nonnegative integer

The Binomial Theorem or Formula, when n is a nonnegative integer and k=0, 1, 2...n is the kth term, is: [1.1]

When k>n, and both are nonnegative integers, then the Binomial Coefficient is zero. This explains why the above series appears to terminate. That is, it has (n+1) terms.

It can also be written:
[1.2]
Where n and k are references to numbers in Pascal's Triangle, is called a Binomial Coefficient and is read as "n over k".  When it is a combination, it may be read as "n choose k".

Proof of the Binomial Theorem

The Binomial Theorem was stated without proof by Sir Isaac Newton (1642-1727). The Swiss Mathematician, Jacques Bernoulli (Jakob Bernoulli) (1654-1705), proved it for nonnegative integers. Leonhart Euler (1707-1783) presented a faulty proof for negative and fractional powers. Finally, the Norwegian mathematician, Niels Henrik Abel (1802-1829), proved the theorem.

Proof when n and k are positive integers

When n and k are nonnegative integers, then the Binomial Coefficients in:
[1.2, repeated]

can be considered combinations, and read "n choose k", as appropriate. Sir Isaac Newton just rote the formula down in his notebook, without proof, perhaps because he thought the formula was self-evident.

Looking at 1.2, we choose the number of x's, and first we choose none at all. n choose 0 is 1.

Similarly, we choose 1, 2, ... n x's.

When we choose 1 from n, this is the combination n choose 1, which we know is n. When we consider the x-squared term, we choose 2 x's from n, and the formula for this is n(n-1)/2. The combinations appear to prove the theorem.

Another approach, is to assume
[2.1]
where the subscripts are simply for labelling the factors.

The first term of this expansion will be aⁿ, and the coefficient is 1. We will think about choosing x's. So the first terms, which uses all the a's and none of the x's is the number of ways we can choose 0 x's from n, which is only one way.

The next term is the x term. We need to choose 1 x from n possibilities, and this is n ways. The coefficient of x is therefore n. This will be of the power (n-1). The sum of the powers of the n's and a's will always be n.

The x² term, is made up of choosing one x and then choosing another (so they make x-squared). One way to think of this is to pick the first x from the first factor, and then we have a choice of (n-1) factors for the next x. We can choose the first x in n ways and there are (n-1) other choices for the next x. There are therefore n(n-1)/2! ways of choosing two x's and this is the x² coefficient. The factorial 2 comes about in the following way. Consider factors 1 and 2. We can choose an x from factor 1 and then one from factor 2, or we can choose an x from factor 2 and one from factor 1. Clearly, these are effectively the same so we are double counting. So we divide by factorial 2. (This is the number of permutations of 2 objects).

The x3 term is made up of three x's. We have a choice of n x's for the first x, but once we have chosen, we have only (n-1) x's remaining for the second x. And for the third x we have (n-2) choices. This means we can choose three x's in n(n-1)(n-2)/3! ways and this is the coefficient of x³. The factorial 3 makes allowance for identical permutations. With three things, we can arrange them in 6 ways, but, for our purposes, these three x's, irrespective of the order of selection, are the same combinations. For instance, we could choose three x's from the first three factors in 6 ways: 123, 132, 213, 231, 312, 321, but they all make up the same x³!

In general, we can select k x's from n factors in n(n-1)(n-2)... (n-k+1)/k! ways. And this is the coefficient of the general term.

We have proved the Binomial Theorem for nonnegative integers n and k, essentially by creating its terms and showing they are the same as the terms claimed for the Binomial Theorem.

Proof when r is any real number

We now prove the Binomial Theorem when the power r, is any real number: positive, negative, rational, irrational, fractional.. any real number.
I have use (1+x) instead of (a+x) for simplicity, and because we often use the Binomial Theorem in this way. If there is an a, we simply take it out of the brackets.
The following proof uses simple calculus, and the proof rests on the truth of simple calculus. It assumes that a Binomial Expansion can be written as:
[4.1]
where r is a real number and k is an integer. ak are the coefficients of the expansion.

When x=0, then a₀=1

We differentiate and get:
[4.2]
When x=0, a₁=r

Differentiating again and setting x=0
[4.3]
a2=r(r-1)/2

It is evident that we are extracting the Binomial Coefficients. To cut a long story short, let us differentiate the function k times
[4.4]
Setting x=0 and rearranging:
[4.5]

Taking the general term (4.5), we show that the left-hand side equals:
[4.6]

And the right-hand side is the Binomial Theorem! We have therefore proved the Binomial Theorem for all real numbers, so we can legitimately use it with positive and negative and fractional r, and we are no longer limited to integers.  The method of expanding (1+x)^r is known as a Maclaurin Expansion. The Binomial Theorem is so versatile that x can even be a complex number, with a non-vanishing imaginary part!

Defining the Binomial Coefficients

When n and k are nonnegative integers, we can define the Binomial Coefficients as:
[6.1]
k is positive and less than n. If k is less than zero, we have negative factorials, which we haven't defined (see using the Gamma Function to define factorials, for a method to define factorials for fractions and complex numbers. Factorials of the negative integers do not exist.) When k is greater than n, [6.1] is zero, as expected. (This is what makes the Binomial Expansion with n as a nonnegative integer terminate after n+1 terms!)

When r is a real number, not equal to zero, we can define this Binomial Coefficient as:
[6.2]
When r is zero, [6.2] gives zero instead of 1, so we restrict [6.2] to r≠0.  We define B(n,0) as 1.

If the Binomial Coefficient is also a combination (n and r are positive integers), then we can use the rules of combinations.

Sum of Binomial Coefficients

We can write the binomial theorem as:

Where n is a positive integer, and k is a nonnegative integer, 0, 1, ..., n and is the term number.
If we let a=b=1, we find (1+1)ⁿ=2ⁿ is the sum of the terms, because the powers of a and b are all 1, and only the coefficients remain. Naturally, the values of the coefficients are not changed by the values of a and b, so the sum of the coefficients is always 2ⁿ, whatever the values of a and b.

If we write a=1 and b=-1, then (1-1)ⁿ=0. We also notice that the even powers of b will be positive and the odd powers will be negative.

The sum of the terms of a binomial expansion equals the sum of the even terms (and the even powers of b), k=0, 2, etc plus the sum of the odd terms, k=1, 3, 5, etc:


Because when a=1 and b=-1, the odd terms and the even terms cancel out, and their coefficients are therefore equal, we have:


The coefficients do not change with the values of a and b, so the sum of the coefficients of the odd terms is always equal to the sum of the coefficients of the even terms, when n is a positive integer.

Because the sum of the coefficients of the even terms equals the sum of the coefficients of the odd ones, and because the total sum is 2ⁿ, we have the sum of the even terms and the sum of the odd terms are both equal to half the total sum:
■

Convergence

When the Binomial Expansion is finite, when r is a nonnegative integer, then the series is always convergent, being the finite sum of finite terms. It is when the series is infinite that we need to question the when it converges.
 According to the ratio test for series convergence a series converges when:
[7.1]
It diverges when:
[7.2]
And the result is indeterminate when:
[7.3]
For the Binomial Theorem the ratio of the kth and the k-1 terms is:
[7.4]

Applying the ratio test when the Binomial Expansion is an infinite series (r is not a nonnegative integer), we find the limit is
[7.5]

That is, |x|. The Binomial Theorem converges when |x|<1.

When |x| is 1, the ratio test does not advise us on its status.
For example:
[7.6]

This is an infinite series. When x=1, the left-hand side is 1/2 and the right-hand side is 1-1+1-1+... Yet it appears to be divergent, in the sense of meaning not convergent (that is, it does not converge to a single finite value), because it seems to oscillate between -1, 0 and 1. (It seems that the answer 1/2 may be correct, however) When x=-1, the left-hand side is 1/0, which is infinite, and the right-hand side is 1+1+1... which is also clearly infinite. In this case, the series clearly diverges. When |x|=1, we need to examine these cases very carefully. The simple answer, however, is that when |x|=1, the binomial series is indeterminate, so discussing the value when x=1 is meaningless.1/(1+1) is a half, but we cannot obtain this from the Binomial Theorem.

The conclusion here is that when the binomial series is infinite (n is negative or fractional), then it converges when |x|<1






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Faster Arithmetic - by Ken Ward