Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number.

Question

There was a problem in a book:

Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.

It's not a hard question, but I find a special sequence:

$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=y^3-3y \\ x^5+\dfrac{1}{x^5}=\left(x+\dfrac{1}{x}\right)^5-5\left(x^3+\dfrac{1}{x^3}\right)-10\left(x+\dfrac{1}{x}\right)=y^5-5\left(y^3-3y\right)-10y=y^5-5y^3+5y\\x^7+\dfrac{1}{x^7}=\left(x+\dfrac{1}{x}\right)^7-7\left(x^5+\dfrac{1}{x^5}\right)-21\left(x^3+\dfrac{1}{x^3}\right)-35\left(x+\dfrac{1}{x}\right)=y^7-7\left(y^5-5y^3+5y\right)-21\left(y^3-3y\right)-35y=y^7-7y^5+14y^3-7y$

I find that the coefficient has some relationship between the Pascal Triangle, such as $y^7-7y^5+14y-7y=y^7-7 \binom{2}{0}y^5+7\binom{2}{1}y^3-7\binom{2}{2}y$. That's strange but funny! However, I can't really prove this, or show that it is false. Hope there is someone who can answer me. Thank you!

Answer 1

I don't see a clear pattern in the coefficients but there is a recurrence: $$ y_{n+2} = y_2 y_n -y_{n-2} = (y^2-2)y_n -y_{n-2} $$ where $$ y_n = x^n+\dfrac{1}{x^n} $$

Answer 2

Let $x=e^{iz}$. We want to express $x^n+\frac{1}{x^n}=2\cos(nz)$ in terms of $x+\frac{1}{x}=2\cos(z)$, which can be done through Chebyshev polynomials of the first kind:

$$ x^n+\frac{1}{x^n}=2\cos(nz) = 2\,T_n(\cos z) = 2\, T_n\left(\frac{x+\frac{1}{x}}{2}\right).$$

Answer 3

Let $s_1:=x+\dfrac1x$. Then

$$x=\frac{s_1\pm\sqrt{s_1^2-4}}2$$ and

$$s_n=\left(\frac{s_1+\sqrt{s_1^2-4}}2\right)^n+\left(\frac{s_1-\sqrt{s_1^2-4}}2\right)^n.$$

Then by the Binomial development, after cancellation of the odd terms

$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\binom n{2k} s_1^{n-2k}\left(s_1^2-4\right)^{k}.$$

Then, developing again,

$$s_n=\frac1{2^{n-1}}\sum_{2k=0}^n\sum_{j=0}^k\binom n{2k}\binom kj(-4)^{k-j} s_1^{n-2k+2j}$$

and you can regroup the terms by equal powers of $s_1$.

Answer 4

Krechmar's `A problem book in Algebra' gives an identity if $x+z=p$ and $xz=q$, then $$x^n+z^n=p^n-\frac{n}{1}p^{n-2} q+\frac{n(n-3)}{1.2}p^{n-4} q^2+...+(-1)^k \frac{n(n-k-1)(n-k-2)....(n-2k+1)}{k!} p^{n-2k} q^k+...$$ Take $z=1/x$ then $q=1$ and letting $x+\frac{1}{x}=y$ we can write $$f_n(x)=x^n+\frac{1}{x^n}=\sum_{k=0}^{n/2} A_{n,k} ~y^{n-2k},...(1)$$ where $$A_{n,k}=(-1)^k \frac{n(n-k-1)(n-k-2)....(n-2k+1)}{k!}....(2)$$ We can re-write $A_{n,0}=1$ and $$A_{n,k}=(-1)^k\frac{n}{k}{n-k-1 \choose k-1}, ~ 0<k <n/2 ......(3)$$

Fpr $n=5$ we get $A_{5,0}=1, A_{5,1}= -5, A_{5,2}=5.$ For $n=7$ we get $A_{7,0}=1, A_{7,1}=-7, A_{7,2}= 14, A_{7,3}=-7$ For $n=8$, we get $A_{8,0}=1, A_{8,1}=-8, A_{8,2}= 20, A_{8,3}=-16, A_{8,4}=2.$ So $$x^8+1/x^8=y^8-8y^6+20y^4-16y^2+2,~ y=x+1/x.$$ Similarly $$x^9+1/x^9=y^9-9y^7+27y^5-30y^3+9y.$$

Finally, we would like to assert that the expansion (1) along with (2) or (3) is the required generalization which works for both even and odd $n$.