Find the number of terms in $$\left(x+\frac{1}{x}+x^2 + \frac{1}{x^2}\right)^{15}$$

I dont have any clue on how to start this problem. Terms can be formed from various combinations! Thank you!


Actually answer provided is $61$, so it seems all powers are present.

  • $\begingroup$ Relevant: en.wikipedia.org/wiki/Multinomial_theorem $\endgroup$ – JavaMan Jul 2 '17 at 4:09
  • $\begingroup$ Before or after combining like terms? $\endgroup$ – pjs36 Jul 2 '17 at 4:10
  • $\begingroup$ @pjs36 After combining! $\endgroup$ – akhmeteni Jul 2 '17 at 4:11
  • $\begingroup$ @JavaMan will all powers be present? $\endgroup$ – akhmeteni Jul 2 '17 at 4:11
  • $\begingroup$ @NairitSarkar an example of power that would be absent? $\endgroup$ – akhmeteni Jul 2 '17 at 4:15

$$\left(x+\frac{1}{x}+x^2 + \frac{1}{x^2}\right)^{15}$$ $$= \left(\frac{x^{3}+ x + x^4 + 1}{x^{2}}\right)^{15}$$ $$= \left(\frac{1 + x+ x^3 + x^4}{x^{2}}\right)^{15}$$ $$= x^{-30} \cdot\left(1 + x+ x^3 + x^4\right)^{15}$$

Consider the expansion of $\left(1 + x+ x^3 + x^4\right) ^ {15}$.

We have the following characteristics:

  1. The maximum power of $x$ in the expansion is $60$
  2. Each term will have a power which is formed as follows: let there be 15 slots and we fill them with numbers from the set $\{0, 1, 3, 4\}$ which are the powers of $x$ in the expression. Can we construct sets such the sum of numbers in each set will be able to represent all numbers from 0..60?

The answer to (2) seems to be in the affirmative. The proof is a simple exercise. Consider all of the $15$ slots to be filled with the value $4$. That will give rise to $60$. If you want to have construct numbers from $56 - 59$, it can be done as follows:

$$60: 15 \cdot 4 + 1 : 15\, slots$$ $$59: 14 \cdot 4 + 1 \cdot 3 : 15\, slots$$ $$58: 13 \cdot 4 + 2 \cdot 3 : 15\, slots$$ $$57: 14 \cdot 4 + 1 \cdot 1 : 15\, slots$$ $$56: 14 \cdot 4 + 1 \cdot 0 : 15\, slots$$

We can provide a simple inductive proof based on the above reasoning. Hence we will have all powers in the expansion from $\{0..60\}$ and hence the answer is $\boxed{61}$.

Side Note: This problem is quite similar to a problem on weights, and we could be sure that in such cases where we have a good set of weights and a large enough number of slots, we could represent a large number of values. However, if you consider weights of the form $\{1, 3, 5\}$ and you want to represent values from $\{0..15\}$, i.e., the numbers in the expansion of $\left(x + x^3 + x^9\right)^3$ will not have all values (since $4$ for example is not representable as such a sum.

  • $\begingroup$ Nice answer! I was sure that leaving it as $x+\frac{1}{x}+x^2 + \frac{1}{x^2}$ is the way to go, because $x+\frac{1}{x}$ and $x^2 + \frac{1}{x^2}$ cancel so nicely when raised to powers. $\endgroup$ – Ovi Jul 2 '17 at 5:20
  • $\begingroup$ We can also say that in $(x+x^2+x^3+x^4)^{15}$, $x^{4k}$ is present, taking only $x^4$ for creating $x^{4k}$. Then replace one $x^4$ with $x^3$ so we have $x^{4k−1}$ present! similarly $x^{4k−2}$ and $x^{4k−3 }$ are also present! Is this right? $\endgroup$ – akhmeteni Jul 2 '17 at 5:27
  • $\begingroup$ @akhmeteni yes, that is essentially the inductive proof. $\endgroup$ – user1952500 Jul 2 '17 at 5:32
  • $\begingroup$ @Ovi thanks for your kind words! $\endgroup$ – user1952500 Jul 2 '17 at 5:32

The "top term" will be $x^{30}$ and the "bottom term" will be $x^{-30}$. In between there will be terms involving $x^{29}$, $x^{28}$ etc., with various coefficients. There seems no reason to think any coefficients will vanish.

  • $\begingroup$ Why wont they vanish? Because they are binary? i mean 1,2 as powers? $\endgroup$ – akhmeteni Jul 2 '17 at 4:13
  • $\begingroup$ +1. It's the obvious answer unless some coefficient vanishes out which isn't the case. Just 'for completeness' you should add something like $\left[30-\left(-30\right)\right] + 1 = \color{#f00}{61}$. $\endgroup$ – Felix Marin Jul 3 '17 at 2:14

We have $$ f(x)=(x^{-2}+x^{-1}+x^{1}+x^{2})^{15} = \sum_{k=-30}^{30}c_k x^k \tag{1}$$ where $c_k$ is given by the number of ways for representing $k$ as $-2a-b+c+2d$ with $a,b,c,d\in\mathbb{N}$ an $a+b+c+d=15$. We clearly have $c_k=c_{-k}$ by swapping $a$ and $d$, $b$ and $c$. By direct inspection every coefficient of $g(x)=(x^{-2}+x^{-1}+x^{1}+x^{2})^{3}$ is positive, hence the same holds for the coefficients of $f(x)=g(x)^5$. This gives $$ \forall k\in[-30,30],\qquad c_k > 0 \tag{2}$$ hence there are $61$ terms.

In general, if $a_1,a_2,\ldots,a_{n}$ with $n\geq 2$ is a sequence of distinct integers, the Laurent series of $(x^{a_1}+x^{a_2}+\ldots+x^{a_n})^m $ has positive coefficients for any $m\in\mathbb{N}$ large enough.


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