Chebyshev cube root

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In mathematics, in the theory of special functions, the Chebyshev cube root is a particular inverse function of the Chebyshev polynomial of third degree. It is analogous to the cube root function is the inverse of the third power. It can be used to solve the cubic equation.

[edit] Motivation

The cube root function is in some respects not a well-behaved function, or one convenient for the purposes of finding the roots of a cubic equation. While cube roots are well-known and traditional, it is possible to use other algebraic functions to determine the roots, and avoid some of the problems of cube roots. The cube root function has a branch singularity at zero, as a result of which the real cube root function does not extend nicely to a complex cube root function. Moreover, when using cube roots to find the roots of a polynomial with three real roots one must take the roots of complex numbers, which introduces complex numbers into a situation which does not, in fact, require them.

One can get around these problems by using Chebyshev cube roots in place of ordinary cube roots. The polynomial $C 3 = x 3 - 3 x$ is the third Chebyshev polynomial normalized to obtain a monic polynomial. The Chebyshev cube root is then defined as a (suitably chosen) root (depending on t) of the polynomial equation

$x^3 - 3x = t \ .$

The polynomial $C 3 (x)$ satisfies the third order addition relations

$\, 2\,\cos(3x)= C_3(2 \cos x)$

and (as $\, 2\cos(ix)=2\cosh(x)$ )

$2\,\cosh(3x)=C_3(2\cosh x) \ .$

If t is represented as $t = 2cos y$ , then the polynomial equation $x 3 - 3 x = t$ can now be transformed into

$t=2\,\cos y=C_3(2\cos (y/3)) \ .$

The function $C_{1\over3}(t)$ is then defined as (a branch of) the algebraic function of the third order which transforms $2cos(x)$ into $2cos(x / 3)$ . It is given (inverting the relation $t = 2cos(x)$ to $x = arccos(t / 2)$ ) as

$C_{1\over3}(t) = 2 \,\operatorname{cos}\left(\operatorname{arccos}\left({t\over2}\right)/3\right) ,$

if t lies in the real interval [−2, 2]. If t lies in the interval $[2,\infty]$ , then the Chebyshev root is given as

$C_{1\over3}(t) = 2 \,\operatorname{cosh}\left(\operatorname{arccosh}\left({t\over2}\right)/3\right) \ .$

The branch is uniquely defined by the value at $t = 0$ , which is $2\, \operatorname{cos}\left(\operatorname{arccos}(0)/3\right)=2\, \operatorname{cos}(\pi/6)=\sqrt{3}$ , corresponding to the positive solution of $x 3 - 3 x = x (x 2 - 3) = 0$ .

This procedure is precisely analogous to the definition of the cube root in terms of logarithms and exponentials, with arccosh(x/2) resp. arccos(x/2) in the place of ln(x), and 2cosh(x) resp. 2cos(x) in the place of exp(x). The Chebyshev cube root can be constructed as an analytic function on the cut plane $\mathbb{C}\setminus [-\infty,-2]$ and is the unique branch of the algebraic function $C_{1\over3}(t)$ with this property. In the domain $D_1 := \{z \in \mathbb{C}\, | \, \Re{z}>2\}$ it can be defined as

$C_{1\over3}(t)= 2\,\operatorname{cosh}\left(\operatorname{arccosh}\left({t\over2}\right)/3\right)$

where $\operatorname{arccosh}(z/2)=\ln{{z+\sqrt{z^2-4}}\over 2},$ using the branch of the logarithm which is real on the positive real line and the branch of the square root which is positive on the real axis. On the domain $D_2 :=\mathbb{C}\setminus{\{[-\infty,-2] \cup [2,\infty]\}}$ it can be defined as

$C_{1\over3}(t)= 2 \,\operatorname{cos}\left(\operatorname{arccos}\left({t\over2}\right)/3\right)$ ,

where $\operatorname{arccos}(z/2)={\pi \over 2}+i\ln{{iz+\sqrt{4-z^2}}\over 2} \ .$ Both D₁ and D₂ are simply-connected domains in $\mathbb{C}$ on which the functions $\operatorname{arccos}(z)$ and $\operatorname{arccosh}(z)$ are well-defined analytic functions (because the square roots $\sqrt{\pm (z^2-4)}$ exist as analytic functions on D₁ resp. D₂ and the argument functions ${z+\sqrt{z^2-4}}\over 2$ and ${iz+\sqrt{4-z^2}}\over 2$ of the logarithm do not vanish on each domain). Both (partially overlapping) definitions of the Chebyshev cube root on the domains D₁ and D₂ can be put together to define the Chebyshev cube root unambiguously as an analytic function on the larger domain $D= \mathbb{C}\setminus [-\infty,-2]$ . In fact, if one approaches the critical value $t = 2$ from either the left or the right on the real axis the value of each representative will tend to 2. Because $x = 2$ is a simple root of the polynomial $x 3 - 3 x - 2$ the branch of the Chebyshev root (defined as the algebraic function F(t)=2+G(t) satisfying

$\, F(t)^3-3F(t)-2=9G(t)+6G(t)^2+G(t)^3=0$

and $F (2) = 2$ exists locally as an analytic function in a (sufficiently small) neighbourhood U of $t = 2$ (according to the (complex-analytic ) inverse function theorem) and takes real values if $t=2\pm \epsilon, \,\epsilon >0$ . Then it must coincide (on the intersection $U \cap D_1$ and $U \cap D_2$ ) with each of the two representatives (in terms of arccos z resp. arccosh z) constructed above. Therefore the Chebyshev cube root is in fact an analytic function on the whole of the domain D.

An alternative construction of the Chebyshev cube root in terms of hypergeometric functions is sketched in the next subsection.

[edit] Representation as hypergeometric function

The expression

$2 \,\operatorname{cos}\left(\operatorname{arccos}\left({t\over2}\right)/3\right)=2\,\operatorname{cos}\left({\pi\over 6}-\operatorname{arcsin}\left({t\over2}\right)/3\right)$

can be transformed (using the difference-to-product trigonometric identity for the cosine) into the representation

$\sqrt{3} \,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin }\left(\operatorname{arcsin}\left({t\over2}\right)/3\right) \ .$

For general complex parameter $\lambda \ne 0$ the functions $2\,\operatorname{cos}\,(\lambda\, \operatorname{arcsin}(x/2))$ and $2\,\operatorname{sin}\,(\lambda \,\operatorname{arcsin}(x/2))$ are two linearly independent solutions of the second-order linear differential equation

$\, (4-x^2)y''-xy'+\lambda^2 y=0$

which can be obtained by differentiating the functional relations $\, f(2\sin x)=2\sin(\lambda x)$ resp. $\, f(2\sin x)=2\cos(\lambda x)$ twice with respect to x. The differential equation

$\, (4-x^2)y''-xy'+\lambda^2 y=0$

is equivalent (under the affine substitution $x \mapsto (2-4x)$ ) to the hypergeometric differential equation

$x(1-x) \,y''+{{1-2x}\over 2}\,y'+\lambda^2 y=0$

with parameters $c={1\over 2},\, a=\lambda,\, b=-\lambda$ . According to the general theory of the hypergeometric equation it has (unless c is zero or a negative integer) a uniquely defined solution g which is analytic in x=0 and satisfies $\,g(0)=1$ . It is given by the hypergeometric series (see hypergeometric function)

$\,F(a,b,c;z):=\,_2F_1 (a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n(b)_n}{(c)_n} \, \frac {z^n} {n!} \ .$

Transforming back to the original differential equation one finds a solution $g(x)=F(\lambda,-\lambda,{1\over2} ;{{2-x}\over 4})$ of the differential equation

$\, (4-x^2)y''-xy'+\lambda^2 y=0$

which is analytic at $x = 2$ (unique up to scalar multiple). The representation

$C_{{1\over 3}}(t)= \sqrt{3} \,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)$

obtained above shows that the Chebyshev cube root is a solution of the differential equation

$\, (4-x^2)y''-xy'+\lambda^2 y=0$

for $\lambda={1\over 3}$ which is analytic at $x = 2$ . It must be proportional to the argument-shifted hypergeometric series and thus

$C_{{1\over 3}}(t)=2F({1\over 3},-{1\over 3},{1\over2} ;{{2-t}\over 4}) = \sum_{n=0}^\infty \frac{2}{1-3n} {3n \choose n}\left(\frac{2-t}{27}\right)^n\ ,$

where the last series converges if $| t - 2 | < 4$ . All three roots $r 1, r 2, r 3$ of the equation $x 3 - 3 x - t = 0$ are linear combinations of the two functions $f_1(t)=2\operatorname{sin}\,\left({1\over 3} \operatorname{arcsin}{t\over2}\right)$ and $f_2(t)=2\operatorname{cos}\,\left({1\over 3} \operatorname{arcsin}{t\over2}\right) \ .$ By construction

$r_1=C_{{1\over 3}}(t)=\sqrt{3} \,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin}\,\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)\, ,$

the other two roots are

$r_2=-C_{{1\over 3}}(-t)=-\sqrt{3} \,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin}\,\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)$