Talk:Binomial theorem

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

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

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

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

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

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

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

Let x = 2 and y = 4

$(2+4)^{3}=2^{3}+3(2)^{2}4+3(2)4^{2}+4^{3}\,$

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

$216=216\,$

— 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

$\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}.$

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)

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

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

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

simplification

"whenever n is any non-negative integer"

could read

"when n is a natural number"

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)

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 $0^{0}$ to be equal to 1 (see Exponentiation#Zero_to_the_zero_power).

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.