# Polynomial expansion

1. Mar 23, 2006

### heartless

Hello
I'm trying to do some research on polynomial expansion however my math isn't that good to do high level calculations, proofs and so on. A while ago my friend came up with a formula for finding number of terms in any expansion of a polynomial to the y power. For example for
(a+b+c+d)^4 = 35 terms
Formula: n-y-1Cy for n = number of terms in paranthases.
Can anyone help me, give some clues on how to proof this formula or show me a proof?
Thanks!

2. Mar 23, 2006

### arildno

What is y? What is C? Are you sure you haven't omitted any parantheses here?

3. Mar 23, 2006

### Nimz

Basically, what you're doing in your example to get the 35 terms is picking 4 objects from a collection of 4, allowing repetition. In general, you have $(a_1 + a_2 + \dots + a_n)^y$. Are you saying that the formula is $\binom{n-y-1}{y}$?

4. Mar 23, 2006

### heartless

Sorry for bad explanation and wrong formula after all,
for any polynomial (with number of terms n) to the yth power we may find number of terms after expansion by
n+y-1 C y
C means 'choose' for example aCb = a!
-----------
(a-b)! b!

for polynomial (a+b+c+d)^4 where number of terms is 4 (a, b, c, d) we may find number of terms after expansion by 4+3C4 which is 7C4 = 35
Thanks for all the help.

5. Mar 23, 2006

### Nimz

nCr is the same as $$\binom{n}{r}$$ Just different notation.

6. Mar 24, 2006

### geniusprahar_21

Clearly, each term in the expansion of $$(a_{1} + a_{2} + \dots + a_{n})^y$$ will have degree y.
Let an arbitary term be $$a_{1}^{p}a_{2}^{q}a_{3}^r\dots$$ then according to the condition p+q+r....=y.
Thus, find the number of integer solutions of the above equation. thats your required number of terms.

7. Mar 24, 2006

### hypermonkey2

This is kind of interesting in that it might imply a quick solution to a question such as "how many pairs of integers satisfy a+b=100" and beyond, no?

8. Mar 26, 2006

### geniusprahar_21

well, there is a quick solution to your question hypermonkey2, though i do not know the proof...

the number of solutions to $$a_1x_1 + a_2x_2 + \dots + a_nx_n = p \ where \ b_i \leq x_i \leq c_i \mbox{for}\ 1 \leq i \leq n$$ is given by the coefficient of $$t^n$$ in the expression

$$\prod ( (t^{a_i})^{b_i} + (t^{a_i})^{b_i + 1} + (t^{a_i})^{b_i + 2} + \dots + (t^{a_i})^{c_i})\where \ 1 \leq i \leq n$$

this however also includes solutions so that two $$x_i$$ may be equal. for distinct solutions introduce dummy variables so that condition of distinctness is removed. introduction of a dummy variable can be extended to number of solutions of the equation $$a_1x_1 + a_2x_2 + \dots + a_nx_n \leq p$$.

Last edited: Mar 27, 2006
9. Mar 29, 2006

### robert Ihnot

After spending some time trying to figure out what this is about, I think I see what the problem is. Let p = the power and n=number of linear terms, C = the combinations, then the number of terms in

$$(\sum n)^p=(p+n-1) C (p).$$

For p=1 it is trivial, and easy enough for p=2; I'll go ahead and show it for all n for p=3.

If n=1, then by the formula we should have 3C3=1, which is correct, and represents (a)^3=a^3, which is one term.

Now if it is true for n terms, which I represent as u and we add one more term, call it b, then we have by the binominal formula:

$$(u+b)^3=u^3+3u^2b+3ub^2+b^3$$

In this case b^3 adds only 1 term. 3ub^2 adds n more terms. 3u^2b adds (n+1)C2. (This by how the squares work) and u^3 by the induction hypothses adds (n+2)C3.

So we have (n+2)(n+1)n/6+(n+1)n/2+n+1 =(n+1)/6{n^2+2n+3n+6}=
(n+1)/6{(n+2)(n+3)}=(n+3)C3. Or the induction is complete for p=3.

Last edited: Mar 29, 2006
10. Mar 7, 2010

### nepenthes9997

C actually means combination, not choose. Choose is only a simpler representative word for combination to help people to understand better.