Binomial Expansions and Pascal's TriangleDate: 7/10/96 at 16:58:31 From: Anonymous Subject: Binomial Expansions Can you supply the definition of what a binomial expansion is, where it would be used, why, and how to do one? This would be a great help because I may be able to use it for forecasting. Thanks in advance. Cordially, Anthony Fama Date: 7/10/96 at 20:54:23 From: Doctor Pete Subject: Re: Binomial Expansions A *binomial* is a polynomial expression with two terms, like x+y, x^2+1 (x squared plus 1), or x^4-3*x. *Binomial expansion* refers to a formula by which one can "expand out" expressions like (x+y)^5 and (3*x+2)^n, where the entire binomial is raised to some power. Usually, binomial expansion is introduced using a construction called Pascal's Triangle, but I prefer to think of it in terms of something called the *binomial coefficient*, which I'll explain later. First, we'll look at the "generic" binomial x+y, and its powers (x+y)^2, (x+y)^3, ... (x+y)^n. Notice the following: (x+y)^1 = x+y (x+y)^2 = (x+y)(x+y) = x^2+2*x*y+y^2 (x+y)^3 = (x+y)(x+y)^2 = x^3+3*x^2*y+3*x*y^2+y^3 (x+y)^4 = (x+y)(x+y)^3 = x^4+4*x^3*y+6*x^2*y^2+4*x*y^3+y^4 ... What kind of patterns can we see from this? Well, there are two things going on here, namely the powers of x and y, and the coefficients. Let's look at the powers: Power of (x,y) in the (k)th term: k=1 k=2 k=3 k=4 k=5 (x+y)^1: (1,0) (0,1) (x+y)^2: (2,0) (1,1) (0,2) (x+y)^3: (3,0) (2,1) (1,2) (0,3) (x+y)^4: (4,0) (3,1) (2,2) (1,3) (0,4) Clearly, (x+y)^n will have n+1 terms, and we can infer from the above that the (k)th term will have a (n-k+1)th power of x, and a (k-1)th power of y. Now, let's look at the coefficients: Coefficient in the (k)th term: k=1 k=2 k=3 k=4 k=5 (x+y)^1: 1 1 (x+y)^2: 1 2 1 (x+y)^3: 1 3 3 1 (x+y)^4: 1 4 6 4 1 Now, this pattern isn't so clear.... Let's write it like this: Row 0: (1) <I added this just for fun, it'll become 1: 1 1 clear why I did later 2: 1 2 1 3: 1 3 3 1 4: 1 4 6 4 1 Notice that any entry above which isn't 1 is the sum of the two entries diagonally above it: For example, the 6 is 3+3. This is Pascal's Triangle, and it is easy to see how to generate any row: just put 1's along the sides, and add pairs of values from the previous row to get the next. For example, the next row will be 1 5 10 10 5 1, the next after that will be 1 6 15 20 15 6 1, and so on. Is it a coincidence that the (n)th row corresponds to the coefficients of the expansion (x+y)^n? Not at all. Sidenote: Note I added a (0)th row to the triangle; it's not just there to make the figure a pretty triangle, but it actually corresponds to (x+y)^0 = v c1, so it has a valid mathematical interpretation. Finally, we have a systematic approach to finding the expansion of (x+y)^n. Let B(n,k) be the (k)th entry in the (n)th row of Pascal's Triangle. Then (x+y)^n = B(n,1)*x^n + B(n,2)*x^(n-1)*y + ... + B(n,n+1)*y^n, or in summation notation, n+1 \---\ (x+y)^n = > B(n,k)*x^(n-k+1)*y^(k-1) . /---/ k=1 Now, is there an easy way to compute B(n,k)? That is, how can you find B(n,k) without having to write all these rows of Pascal's Triangle? Well, yes, there is, which is precisely why there exists a connection between Pascal's Triangle and the binomial expansion formula. But first, notice that it's easier to start with k = 0 rather than k = 1 in the above formula; so we'll count starting from the (0)th term, rather than the first. Our formula becomes n \---\ (x+y)^n = > C(n,k)*x^(n-k)*y^k , /---/ k=0 where C(n,k) = B(n,k+1). We call C(n,k) a *binomial coefficient*. It has the value n! C(n,k) = -------- , k!(n-k)! where n! (read "n factorial"), is 1*2*3*...*n (so 3! = 1*2*3 = 6, for example, and 0! = 1). In combinatorics, it is also called "the number of combinations of n objects taken k at a time." You can see why I like to think of binomial expansions in terms of binomial coefficients: it's direct and simple. Pascal's Triangle is a useful way to learn about binomial expansion, but is very inconvenient to use. Now, I'll leave you with two exercises, the first easy, the second a bit more difficult: 1) Show that C(n,k) = C(n,n-k). 2) Show that C(n,k) indeed corresponds to the (k)th entry in the (n)th row of Pascal's Triangle. (Remember to count from k=0 and n=0.) Here's a hint: Show that C(n,0) = C(n,n) = 1, since 0!=1; this establishes the "sides" of the triangle. Then show that C(n,k) = C(n-1,k-1) + C(n-1,k) for 1 <= k <= n-1; this establishes the "add the diagonals" property in Pascal's Triangle. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 7/15/96 at 13:56:35 From: Anonymous Subject: Re[4]: Binomial Expansions Your effort is truly exemplary! My background is in EE with an MBA in I.S. I am also a member of the Mensa 1000 and various professional organizations, IEEE, NSPE, NPA, etc... I applaud you on your work and look forward to conjuring up more interesting questions... Hello again, Thanks for your compliments! I suppose I could have provided a stock response, but no, that was an original reply, no rehearsals or quotes. :) As for your question about applications of the Binomial Theorem, I can't really think of any *direct* ways of applying it to real-life situations -- I like to think of it as just another tool (a powerful one at that) in the vast arsenal of manipulations that one uses in problem-solving. You never know when it might be needed in solving a differential equation, or finding the roots of a polynomial. Binomial expansion is used more often than you might think; (x+1)^3 = x^3+3*x^2+3*x+1 is just an example. It's a case where it's easier to remember "1 3 3 1" rather than multiplying out. In essence, the most common use of binomial expansion is in expression simplification; the expansion is usually done when you expect terms to cancel. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/