Intereting Posts

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Suppose $f,g$ are continuous functions from $X$ to $Y$, where $X,Y$ are topologies, and $f=g$
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What is the half-derivative of zeta at $s=0$ (and how to compute it)?
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The algebraic equations in one variable, in the general case, cannot be solved by radicals. While the basic operations and root extraction applied to the coefficients of the equations of degree $2, 3, 4 $ are sufficient to express the general solution, for arbitrary degree equations, the general solution can be expressed only in terms of much more complicated functions than the root function.

And even more: trigonometric and logarithmic functions are not enough to express the general solution of an algebraic equation of any degree (cf. Beyond the Quartic Equation, Bruce King, p. 57).

However, it is possible to express the general solution of certain equations using trigonometric functions. The quintessential example is the trigonometric solution to the cubic equation in the case of three real roots. In the case of a real root and two complex roots, the general solution is expressed by means of hyperbolic functions.

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Another example equally interesting, but less known, is the trigonometric solution to the quadratic equation.

The De Moivre’s quintic can also be solved by means of trigonometric functions. The change of variable $ x = y – \frac{a}{y}$ will reduce the equation $ x^5+5ax^3+5a^2x+2b = 0$ to $y^{10}+2by^5-a^5$, and consequently

$$x = \sqrt[5]{-b+\sqrt{b^2+a^5}}+\sqrt[5]{-b-\sqrt{b^2+a^5}}$$

If $b^2+a^5 < 0$, then the roots are trigonometrical functions of the coefficients; the change of variable $ x = -2\phi\sqrt{-a}$ reduces the given equation to

$$ 16\phi^5-20\phi^3+5\phi = \frac{b}{a^2\sqrt{-a}}$$

which coincides with $\sin5\theta$. By taking $\phi = \sin \theta$ and $\frac{b}{a^2\sqrt{-a}} = \sin5\theta$, we get

$$ x = -2\sqrt{-a}\sin\Big(\tfrac{1}{5}\arcsin\tfrac{b}{\sqrt{-a^5}}+\tfrac{2\pi k}{5}\Big)$$

for $ k = 0, 1… 4$. There is a similar approach to solve this quintic by means of hyperbolic functions when $b^2+a^5 > 0$. This leads to me to the following question:

Let $P(x) = a_1x^n+a_2x^{n-1}+\dotsb + a_{n-1}x+a_n = 0$ be a

solvableequation with real coefficients. Is always possible to express the roots of $P(x)$ by means of the four elementary operations, extraction of roots and trigonometrical and hyperbolics functions applied to the coefficients $a_1… a_n$?

In fact, this is possible for some solvable equations as shown in the above examples, the case of quadratic, cubic, and a certain quintic equation. But is this a property of all solvable equations? If the answer is yes, how can be expressed the roots of each solvable equation in terms of trigonometric and hyperbolic functions applied to their coefficients?

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The answer is **Yes**, *an equation solvable in radicals always has a trigonometric solution*. We can illustrate it using the quintic but you can extrapolate it to other degrees. A *depressed quintic* $F(x)=0$ (one without the $x^4$ term) has the solution,

$$x = z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\tag1$$

where the $z_i$ are the roots of its quartic *Lagrange resolvent*,

$$(z^2+b_1z-a_1^5)(z^2+b_2z-a_2^5) = 0\tag2$$

If you have its Langrange, then one can easily find $a_1,a_2,b_1,b_2$ and the trigonometric solution to the solvable quintic is,

$x = 2\sqrt{-a_1}\;\cos\Big(\tfrac{1}{5}\,\arccos\big(\tfrac{-b_1}{2\sqrt{-a_1^5}}\big)-\tfrac{2\pi\,m_1}{5}\Big)+2\sqrt{-a_2}\;\cos\Big(\tfrac{1}{5}\,\arccos\big(\tfrac{-b_2}{2\sqrt{-a_2^5}}\big)-\tfrac{2\pi\,m_2}{5}\Big)\tag3$

where $m_1,m_2$ are appropriately chosen integers from $0,1,2,3,4$.

Thus, for the quintic, you generally need ** two** cosine sums and the DeMoivre quintic is just a special case where $z_3=z_4=0$ and one of its cosine sums is equal to zero.

Example.

Let’s use the one in this post,

$$x^5 – 2x^4 + x^3 + x^2 – x + 1=0$$

Its real root is given by,

$$x_1= \frac{1}{5}\Big(2+z_1^{1/5}+z_2^{1/5}+z_3^{1/5}+z_4^{1/5}\Big) = -0.90879\dots$$

where its Lagrange resolvent is,

$$z^4 + 572z^3 + 70444z^2 + 1600203z – 29^5=0$$

or,

$$\Big(z^2+(286+75\sqrt{5})z+\big(\tfrac{3+5\sqrt{5}}{2}\big)^5\Big)\Big(z^2+(286-75\sqrt{5})z+\big(\tfrac{3-5\sqrt{5}}{2}\big)^5\Big)=0\tag4$$

Equating the terms of $(2)$ and $(4)$, one gets $a_1,a_2,b_1,b_2$. By inspection, we find $m_1 = 3$, $m_2 = 4$, so

$$\begin{aligned}

x_1 &= \small{\tfrac{1}{5}\left(2+2\sqrt{-a_1}\;\cos\Big(\tfrac{1}{5}\,\arccos\big(\tfrac{-b_1}{2\sqrt{-a_1^5}}\big)-\tfrac{6\pi}{5}\Big)+2\sqrt{-a_2}\;\cos\Big(\tfrac{1}{5}\,\arccos\big(\tfrac{-b_2}{2\sqrt{-a_2^5}}\big)-\tfrac{8\pi}{5}\Big)\right)}\\

&= -0.90879\dots

\end{aligned}$$

It’s more complicated than the cubic, but is to be expected since the quintic is higher. Formula $(1)$ can be extrapolated to other prime degrees $n$. For composite $n$, they can be eventually decomposed to equations of prime $n$, so the result still holds.

P.S.

A quick way to see the result is that since the $n$th power of any complex number can be turned to polar form,

$$z^{1/n}=r^{1/n}\big(\cos (\alpha) + i \sin (\alpha)\big)^{1/n} = r^{1/n} \big(\cos (\alpha/n) + i \sin (\alpha/n)\big)$$

then $(1)$ already implicitly involves trigonometry, as was pointed out in the comments by Yves Daoust.

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