Solving $45x-3795x^3 +95634x^5 - \cdots + 945x^{41}-45x^{43}+x^{45} = N$?

Solution 1:

That should be the expression of $2 \sin (45\, \theta)$ in terms of $2\sin \theta=x$, that is $$2\sin(45 \arcsin \frac{x}{2})$$ The full expression is $$x^{45} - 45 x^{43} + 945 x^{41} - 12300 x^{39} + 111150 x^{37} - 740259 x^{35} + 3764565 x^{33} -\\- 14945040 x^{31} + 46955700 x^{29} - 117679100 x^{27} + 236030652 x^{25} - 378658800 x^{23} +\\+ 483841800 x^{21} - 488494125 x^{19} + 384942375 x^{17} - 232676280 x^{15} + 105306075 x^{13} -\\- 34512075 x^{11} + 7811375 x^9 - 1138500 x^7 + 95634 x^5 - 3795 x^3 + 45 x$$

As for hints, for any $n$ odd the expansion of $2\sin(n \arcsin \frac{x}{2})$ starts (as it should ) with $n x$ and ends with $(-1)^{\frac{n-1}{2}}(- n x^{n-2}+x^n )$. That must have been a strong hint for Viete. The explicit expansion for general odd $n$ can be obtained using Chebyshev polynomials.

Added:If $n$ is a parameter then the Taylor expansion of $2 \sin(n \arcsin\frac{x}{2})$ has the first few terms

$$n x - \frac {n(n^2-1)}{24} x^3 + \frac{n(n^2-1)(n^2-9)}{1920} x^5 +\mathcal{O}(x^7)$$ ( a hypergeometric series in fact). If $n$ is an odd integer then the Taylor series is in fact a polynomial of degree $n$ with integral coefficients, but not otherwise. Let's check the values for $n=45$: $$\frac{45(45^2-1)}{24}=3795\\ \frac{45(45^2-1)(45^2-9)}{1920}=95634$$

In fact the Taylor series is $$2\sin(n \arcsin \frac{x}{2})=\sum_{k\ge 0} (-1)^k \frac{n}{2k+1}\binom{\frac{n-1}{2}+k}{2k}x^{2k+1}$$ a polynomial if $n$ is an odd integer.

The way Viete calculated $\phi_{n}(x)=2\sin(n \arcsin\frac{x}{2})$ for $n=45$ is probably by noticing that $\phi_{m}\circ\phi_n=\phi_{mn}$, and starting with $\phi_3$ and $\phi_5$.

Solution 2:

Perhaps, it is this:

${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}\\= \color{blue}{\displaystyle\sum_{r=0, 2r+1\le n}(-1)^r\dbinom{n}{2r+1}\cos^{n-2r-1}\theta \sin^{2r+1}\theta} $

Proof:

$(\cos \theta+ i\sin \theta)^n = \cos n\theta + i\sin n\theta$

Writing the binomial expansion of LHS we get:

$\dbinom{n}{0}\cos^n \theta + i\dbinom{n}{1}\cos^{n-1}\theta\sin \theta -\dbinom{n}{2}\cos^{n-2}\theta \sin^2\theta - i\dbinom {n}{3}\cos^{n-3}\theta \sin \theta+\dbinom{n}{4}\cos^{n-4}\theta\sin ^4 \theta+ i\dbinom{n}{5}\cos^{n-5}\theta\sin \theta$

Comparing real and imaginary parts, ${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}$... which is the desired result.