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)

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.

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.