Polynomial Expansion: Find Number of Terms in Any Expansion

[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!

 

[arildno]

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

 

[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 $\binom{n-y-1}{y}$?

 

[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.

 

[Nimz]

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

 

[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. that's your required number of terms.

 

[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?

 

[geniusprahar_21]

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?

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 $$a_1x_1 + a_2x_2 + \dots + a_nx_n \leq p$$.

 

[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.

 

[nepenthes9997]

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.

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