# A little query involving $n^{th}$ roots of unity

Today when I was solving some math problems I noticed that many of the questions used the cube roots or the $n^{th}$ roots of unity to solve the questions involving a certain periodicity in the function. For instance the image of a function after every $3$ period difference attains the same value. I also posted a question on MSE having such property but the given links provided by other members did not completely explain me what kind of procedure needs to be applied in such questions. The question I posted was

$P(x)$ is a polynomial of degree $3n$ such that

\begin{eqnarray*} P(0) = P(3) = \cdots &=& P(3n) = 2, \\ P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\ P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\ && P(3n+1) = 730.\end{eqnarray*} Determine $n$.

The link to this question is

Note: And please don't mark this as duplicate because I did not get any satisficatory answer to the question stated above. So I expect a proper explanation for my doubt. Any help would be greatly appreciated.

• used the cube roots or the n^th roots of unity to solve the questions involving a certain periodicity in the function The technique is known as series multisection. I did not get any satisfactory answer to the question stated above There were links posted to full solutions. I suggest you narrow down what was not satisfactory with those. – dxiv Jan 3 '18 at 3:48
• In the links posted by others one of the link is artofproblemsolving.com/community/c6h62982p841989. One of the person in comment says "Before writing down the proof(asuuming it to be correct) an intutive note since i was doing complex no.s these days period three reminds me of $\omega$". What is so special of period 3 and how do we use it in simpler way – Rohan Shinde Jan 3 '18 at 4:03
• He also says that "a bit guess work and manipulation may lead u to conclude that $P(k)-1$ $=$ $\frac{2Ime^{\frac{\pi i}{3}}\omega^{k}}{\sqrt 3}$ " But how did he get this thing. Is this some sort of formula or rule? – Rohan Shinde Jan 3 '18 at 4:09
• Don't think we are looking at the same thing. The aops link posted under the other question points to the answer right below the comment you are quoting from, which answer relies on no guesses or assumptions. That's why, as I said, you'd improve your chances at better answers here if you narrowed down the context, and pointed out the difficulty. – dxiv Jan 3 '18 at 4:23
• In the below comment how did pco in the first step itself gave the expression for $P_k(x)$ Will you please explain me how did he get $P_{k}(x)=\frac{1}{(-1)^{3n-k}k!(3n-k)!}\prod_{i=0,i\neq k}^{i=3n}(x-i)$ the major part I did not get is how did get the part $\frac{1}{(-1)^{3n-k}k!(3n-k)!}$ – Rohan Shinde Jan 3 '18 at 4:28

Following up on a comment, below is the annotated transcript of @pco's (Patrick) proof on aops.

1) For integer $k\in[0,3n]$, the unique polynomial $P_{k}$ of degree $3n$

A polynomial of degree $N$ is uniquely defined by its $N$ roots up to a constant factor, and in this case there is an additional condition to determine the constant factor, therefore the claim of "unique" .

and such that $P_{k}(k)=1$ and $P_{k}(p)=0$ for any integer integer $p\neq k\in[0,3n]$

This is shorthand notation for $p \in \{0, 1, 2, \ldots, 3n\} \setminus \{k\}\,$. Purists will frown ;-)

is: $P_{k}(x)=\frac{1}{(-1)^{3n-k}k!(3n-k)!}\prod_{i=0,i\neq k}^{i=3n}(x-i)$.

Given the roots, $P_{k}$ must be of the form $P_{k}(x)= \lambda_k \, \prod_{i=0,i\neq k}^{3n}(x-i)$ for some constant $\lambda_k$. The remaining condition $P_{k}(k)=1$ gives $\lambda_k = \frac{1}{\prod_{i=0,i\neq k}^{3n}(k-i)}$ $= \frac{1}{\prod_{i=0}^{k-1}(k-i)} \cdot \frac{(-1)^{3n-k}}{\prod_{i=k+1}^{3n}(i-k)}$ $= \frac{1}{(-1)^{3n-k}k!(3n-k)!}$.

Notice that $P_{k}(3n+1)=(-1)^{3n-k}\binom{3n+1}{k}$

$P_{k}(3n+1) = \frac{1}{(-1)^{3n-k}k!(3n-k)!} \prod_{i=0,i\neq k}^{3n}(3n+1-i) = \frac{(-1)^{3n-k}}{k!(3n-k)!} \cdot \frac{\prod_{i=0}^{3n}(3n+1-i)}{3n-k+1} = \frac{(-1)^{3n-k} \, (3n+1)!}{k!(3n-k+1)!}\,$.

2) So the requested polynomial is well defined by its values for 0 to $3n$ and we have : $P(x)=\sum_{k=0}^{k=n}2P_{3k}(x)+\sum_{k=0}^{k=n-1}P_{3k+1}(x)$

The equality holds because each of the terms on the RHS is $1$ at exactly one of $\{0,1,2,\ldots,3n\}$ while $0$ at all others, and that's how the "polynomial is well defined" part should be interpreted. $\,P(x)$ is not otherwise fully determined at this point, since $n$ is yet unknown.

3) We can now compute $S=P(3n+1)$
$S=\sum_{k=0}^{k=n}2(-1)^{3n-3k}\binom{3n+1}{3k}+\sum_{k=0}^{k=n-1}(-1)^{3n-3k-1}\binom{3n+1}{3k+1}$

Let $A=\sum_{k=0}^{k=n}(-1)^{3k}\binom{3n+1}{3k}$, $B=\sum_{k=0}^{k=n}(-1)^{3k+1}\binom{3n+1}{3k+1}$ and $C=\sum_{k=0}^{k=n-1}(-1)^{3k+2}\binom{3n+1}{3k+2}$.
We have :
$S=(-1)^{3n}2A+(-1)^{3n}B+1$

That's just the previous form written for $x=3n+1\,$.

$(1-1)^{3n+1}=0=A+B+C$
$(1-j)^{3n+1}=A+Bj+Cj^{2}$ where $j$ is one one of the complex root of $x^{3}=1$.
$(1-j^{2})^{3n+1}=A+Bj^{2}+Cj$

These follow from the binomial expansion of $(1-x)^{3n+1}$ for $x=1, j, j^2$ respectively, using that $j^3=1, 1+j+j^2=0$. The technique of "extracting" powers in an arithmetic progression from a polynomial (or series) is known as series multisection, as was pointed out to me while posting this.

And : $A=\frac{(1-j)^{3n+1}+(1-j^{2})^{3n+1}}{3}$ and $B=\frac{j^{2}(1-j)^{3n+1}+j(1-j^{2})^{3n+1}}{3}$

Follows from solving the above for $A,B,C$.

$S=(-1)^{3n}2A+(-1)^{3n}B+1=1-j(j-1)^{3n}-j^{2}(j^{2}-1)^{3n}$

If $n$ is odd ($n=2p+1$), we have $S=1+(-1)^{p}3^{3p+2}$
If $n$ is even ($n=2p$), we have $S=1+(-1)^{p}3^{3p}$

$\,j^2=-1-j\,$, and $\,(j^2-1)^{3n}$ $= (j+1)^{3n} \cdot (j-1)^{3n}$ $= (-j^2)^{3n}(j-1)^{3n}=(-1)^{3n}(j-1)^{3n}\,$.

4) In our case, we need to solve $S=730=3^{6}+1$
Clearly $p=2$ and $n=4$

Q.E.D.

• Thanks a lot for such brief explanation :-) – Rohan Shinde Jan 3 '18 at 7:24
• Sarcasmometer evened back now ;-) Glad it helped. – dxiv Jan 3 '18 at 7:31
• That was a good one!! – Rohan Shinde Jan 3 '18 at 8:25