Talk:Binomial theorem

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The lead begins with a formula, this is not good for new readers.--Cronholm144 20:11, 21 May 2007 (UTC)

1 Simple derivation section
2 Incorrect?
3 propose adding stats application
4 simplification
5 What if x = 0 ?
6 Any Complex Number
7 Newton's generalized binomial theorem
8 Pascal's Triangle
9 Is this correct?
10 k is not specified

[edit] Simple derivation section

While adding the "simple derivation" section I noticed that "inline" math elements look very cramped in Firefox (and possibly other browsers as well). Is there some good way to fix this (other than separate all math elements from the text, which would probably clutter things up)? Ulfalizer 21:16, 12 June 2007 (UTC)

[edit] Incorrect?

Shouldn't it be:

$(x+y)^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^k$

So the x and y terms descend and ascend in the correct order?

For example:
Let n = 3

$(x+y)^3=\sum_{k=0}^3{3 \choose k}x^{3-k}y^k$

$(x+y)^3={3 \choose 0}x^{3-0}y^0+{3 \choose 1}x^{3-1}y^1+{3 \choose 2}x^{3-2}y^2+{3 \choose 3}x^{3-3}y^3$

$(x+y)^3=1x^{3-0}y^0+3x^{3-1}y^1+3x^{3-2}y^2+1x^{3-3}y^3\,$

$(x+y)^3=1x^3y^0+3x^2y^1+3x^1y^2+1x^0y^3\,$

$(x+y)^3=1x^3+3x^2y^1+3x^1y^2+1y^3\,$

$(x+y)^3=x^3+3x^2y+3xy^2+y^3\,$

Let x = 2 and y = 4

$(2+4)^3=2^3+3(2)^24+3(2)4^2+4^3\,$

$(6)^3=8+48+96+64\,$

$216=216\,$

—Preceding unsigned comment added by 74.134.125.183 (talk • contribs)

They're both the same

It's not hard to see why

$\sum_{k=0}^n{n \choose k}x^{n-k}y^k$

must be exactly the same thing as

$\sum_{k=0}^n{n \choose k}x^ky^{n-k}.$

Just try it, the way you do with your examples above. Michael Hardy 03:06, 5 December 2006 (UTC)

They're the same but...

It should be written

$(x+y)^n=\sum_{k=0}^n{n \choose k}x^{n-k}y^k$

since that is standard notation. also some of the examples and the proof start with the x term first while newton's generalization start with the y term. they should at least be written in one standard way. Heycheckitoutyo 04:11, 29 January 2007 (UTC)

Why is that any more standard than the other way? Both are equally correct; you'll find both forms, depending on the specific textbook you're using. —Lowellian (reply) 03:02, 27 May 2007 (UTC)

--> I agree, revert back to the way it was, since, well it was that way first. x^k * y^(n-k) --JRK, unregistered —Preceding unsigned comment added by 192.197.54.136 (talk) 18:28, 6 November 2007 (UTC)

[edit] propose adding stats application

Let p be the probability of a discrete event taking place. The probability of the event not taking place is 1-p. Let 1-p = q. Then in a series of n trials the probabilty of p taking place r times is

${n \choose r}(p^r)(q^{n-r})$

—Preceding unsigned comment added by 212.159.75.167 (talk • contribs) 17:12, 3 January 2007

This is all covered in a separate article titled binomial distribution. Michael Hardy 00:58, 4 January 2007 (UTC)

[edit] simplification

"whenever n is any non-negative integer"

could read

"when n is a natural number"

—Preceding unsigned comment added by 212.159.75.167 (talk • contribs) 17:16, 3 January 2007

Unfortunately some mathematicians define "natural number" to mean positive integer (0 is not included) and others (especially logicians and set-theorists) define it to mean nonnegative integer (0 is included). So it's ambiguous. Michael Hardy 00:59, 4 January 2007 (UTC)

whenever n is any non-negative integer

this sentence is useless, since the factorial is defined for all complex numbers, except for the negative integers (in which case it is ssaid to be (unsigned) infinity.

—Preceding unsigned comment added by 134.184.49.146 (talk • contribs) 05:07, 2 March 2007

[edit] What if x = 0 ?

if x is 0, the left side of the equation turns into yⁿ but the left side goes to 0.... that doesn't make sense Fresheneesz 22:07, 10 January 2007 (UTC)

When using the binomial theorem it is customary to define

00

to be equal to 1 (see Exponentiation#Zero_to_the_zero_power).

—Preceding unsigned comment added by 213.200.162.130 (talk • contribs) 12:54, 13 January 2007

[edit] Any Complex Number

where r can be any complex number (in particular r can be any real number, not necessarily positive and not necessarily an integer)

I believe that this line, through the use of "in particular", is sort of confusing. It makes it sound like r being in the reals, not necessarily positive, and not necessarily an integer, are, together, a sufficient condition for r to be a complex number (focused on the ones with Im(r)=/= 0, of course). I believe it should be changed.

—Preceding unsigned comment added by 72.189.6.76 (talk • contribs) 03:28, 21 February 2007

I did not write that original text, but I think the text sounds fine and is not confusing. The language used is fairly standard for mathematical writing. —Lowellian (reply) 20:18, 23 May 2007 (UTC)

The sentence should read, "Where r can be any complex number", The brackets are unneccessary. In the act of saying that r can be any complex number, we are basically saying that r can be any value at all, excluding perhaps infinity. A real number is still a complex number, just as much as a square is a rectangle, any logically inclined person can see that. -Glooper—Preceding unsigned comment added by 220.233.184.203 (talk • contribs)

I find the language of the article crystal-clear and that of our anonymous commentator confusing. The words "in particular" mean something. In this case, they mean every real number is a complex number. Our anonymous commentator seems to think they mean every complex number is a real number. Our anonymous commentator needs a dictionary. Michael Hardy 00:19, 7 July 2007 (UTC)

[edit] Newton's generalized binomial theorem

Isaac Newton generalized the formula to other exponents by considering an infinite series:

${(x+y)^r=\sum_{k=0}^\infty {r \choose k} x^k y^{r-k} \quad\quad\quad(2)}$

where r can be any complex number (in particular r can be any real number, not necessarily positive and not necessarily an integer), and the coefficients are given by

$\begin{align} {r \choose k} &{}= {1 \over k!}\prod_{n=0}^{k-1}(r-n)=\frac{r(r-1)(r-2)\cdots(r-(k-1))}{k!}=\frac{r!}{k!\,(r-k)!}. \end{align}$

This is the same as \frac{r!}{k!\,(r-k)!} factorials are defined for ALL complex numbers, except for negative integers

This comment needs to be added because some people like to remove relevant information, because they like (I do not know for what reason whatsoever) to deny the definition of non-integer factorials!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

The following comment is a shame for mathematics:

This negative comment about "not the same as..." seems to be needed. People keep coming along and completing this formula with this expression involving factorials, missing the point of this section.

Bombshell 18:47, 8 March 2007 (UTC)

Yes, but what if r or r-k is a negative integer? In particular, r-k will be a negative integer for some k whenever r is an integer, so if you use \frac{r!}{k!\,(r-k)!}, it's not actully a generalization of the original formula. 75.33.224.226 08:20, 13 October 2007 (UTC)

[edit] Pascal's Triangle

I had this explained to me through the showing of Pascal's Traingle of coefficients, and then the relevent button on the calculator (nCR). Would it perhaps be beneficial if someone showed the traingle, and it how can correspond with a binomial expansion (or for it to be made clearer in the article and not just a reference at the end for it)?

Ginger Warrior 20:36, 25 May 2007 (UTC) Ginger Warrior

I have a question. Im not to familiar for binomial theorems and was wondering if anyone knew the formula for Fto the power (k) (X)=? with the equation F(x) = f(x)g(x). Thank you.—Preceding unsigned comment added by 69.113.236.28 (talk • contribs)

[edit] Is this correct?

In the article, in the generalized binomial theorem section, it defines the generalized binomial coefficients by:

$\begin{align} {r \choose k} &{}= {1 \over k!}\prod_{n=0}^{k-1}(r-n)=\frac{r(r-1)(r-2)\cdots(r-(k-1))}{k!}. \end{align}$

In case k = 0, this is a product of no numbers at all and therefore equal to 1, and in case k = 1 it is equal to r, as the additional factors (r − 1), etc., do not appear.

Another way to express this quantity is

${r \choose k}=\frac{(-1)^k}{k!}(-r)_k$

Is the last line correct? (-r)_k denotes a falling power, so it seems like it should be

${r \choose k}=\frac{(r)_k}{k!}=\frac{(-1)^k}{k!}(-r)^{(k)}$

Perhaps this is just a matter of the notation the original author was using was (r)_k to denote a rising power, but it disagrees with the linked page so should probably be consistent. 75.33.224.226 08:29, 13 October 2007 (UTC)

this is cunfuzeling.—Preceding unsigned comment added by Chicken rule (talk • contribs)

[edit] k is not specified

What is k? —Preceding unsigned comment added by 84.203.7.90 (talk) 10:58, 15 May 2008 (UTC)