8.2.3. Model Independent Calibration

Calibrating correlations via model independent treatments permits using GRBs as distance indicators, although the calibration is made by means of other standard candles. The idea of model-independent calibrations, however, enables getting the luminosity distance *d*<sup>L</sup> without a priori postulating the background cosmology, healing de facto the circularity problem.

A nice possibility consists of relating distances with model-independent quantities written in terms of a Taylor series expansion of the scale factor. Thus, we first notice

$$H(z) = \left\{ \frac{d}{dz} \left[ \frac{d\_\mathcal{L}(z)}{1+z} \right] \right\}^{-1} \tag{44}$$

and then we consider the following expansions:

$$d\_L(z) = \frac{z}{H\_0} \sum\_{n=0}^{N} \frac{\mathfrak{a}\_n}{n!} z^n \tag{45}$$

$$H(z) = \sum\_{m=0}^{M} \frac{H\_m}{m!} z^m \tag{46}$$

where *α<sup>n</sup>* are the coefficients of the luminosity distance expansion and *H<sup>m</sup>* are the coefficients of the Hubble rate expansion. Thus, baptizing the cosmographic set, *q*0, *j*0,*s*0, as the present values of the following quantities:

$$q(t) = -\frac{1}{a} \frac{d^2 a}{dt^2} \left[\frac{1}{a} \frac{da}{dt}\right]^{-2}, \quad j(t) = +\frac{1}{a} \frac{d^3 a}{dt^3} \left[\frac{1}{a} \frac{da}{dt}\right]^{-3}, \quad s(t) = +\frac{1}{a} \frac{d^4 a}{dt^4} \left[\frac{1}{a} \frac{da}{dt}\right]^{-4} \tag{47}$$

where the scale factor *a* has been considered, with the requirement *a* ≡ (1 + *z*) −1 . The above quantities are named *deceleration*, *jerk*, and *snap* parameters, respectively, we formally have

$$d\_{\mathcal{L}}^{(4)} \simeq \frac{z}{H\_0} \left(\mathfrak{a}\_0 + \mathfrak{a}\_1 z + \mathfrak{a}\_2 \frac{z^2}{2} + \mathfrak{a}\_3 \frac{z^3}{6}\right). \tag{48}$$

The coefficients in Equations (45) and (46), say *α<sup>i</sup>* ≡ *αi*(*q*0, *j*0,*s*0) and *H<sup>i</sup>* ≡ *Hi*(*q*0, *j*0,*s*0), can be determined directly with data, without considering a cosmological model a priori. This treatment is known with the name of *cosmography* or *cosmokinetics*, i.e., the part of cosmology that reconstructs the universe's kinematics model-independently. Thus, at *z* = 0, we have

$$a\_1 = \frac{1}{2}(1 - q\_0) \tag{49}$$

$$\mathfrak{a}\_2 = -\frac{1}{6}(1 - q\_0 - 3q\_0^2 + j\_0) \tag{50}$$

$$\alpha\_3 = \frac{1}{24}(2 - 2q\_0 - 15q\_0^2 - 15q\_0^3 + 5j\_0 + 10q\_0j\_0 + s\_0) \tag{51}$$

and

$$H\_1 = 1 + q\_0 \tag{52}$$

$$H\_2 = \frac{1}{2}(j\_0 - q\_0^2) \tag{53}$$

$$H\_3 = \frac{1}{6} \left[ -3q\_0^2 - 3q\_0^3 + j\_0(3 + 4q\_0) + s\_0 \right]. \tag{54}$$

Although powerful, the above formalism suffers from shortcomings due to the convergence at higher redshifts25, i.e., the high GRB redshifts are very far from *z* = 0. In other words, the standard cosmographic approach fails to be predictive if one employs data at higher redshift domains, which is exactly the case of GRBs.

Healing the convergence problem leads to a high-redshift cosmography. In this respect, several strategies have been suggested. For instance, one could (1) extend the limited convergence radii of Taylor series by changing variables of expansion, using the so-called *auxiliary variables*, or (2) changing the mathematical technique in which the expansions are performed, i.e., involving expansions different from Taylor ones, etc.

In the case of auxiliary variables, one employs a tricky method in which the expansion variable is reformulated as a function of the redshift itself, but with particular convergence properties. In other words, we cosmic quantities are rewritten in a more complicated function of the redshift *z*, namely *y*. Changing the redshift variable from *z* to *y* modifies accordingly the convergence radius. Formally speaking, we write *y* ≡ F(*z*) [9], where we assume F(*z*) a generic function of the redshift. The function F(*z*) is properly chosen from physical prime principles. All F(*z*) prototypes, however, might fulfill a few mathematical conditions:


The first guarantees that at *z* = 0, our time, even *y* is zero. The second that the auxiliary variable does not diverge; otherwise, the convergence problem would still persist. In addition, a further requirement is helpful in constructing F(*z*):

3. F(*z*) *<sup>z</sup>*=−<sup>1</sup> < ∞.

The former condition enables *y* to converge at a future time, as well as *z*. Using these hints toward the formulation of F(*z*), we can suggest a couple of well-consolidated examples of F(*z*):

$$y\_1 = \frac{z}{1+z} \tag{55}$$

*y*<sup>2</sup> = arctan *z* . (56)

The second possibility is offered by rational approximations, where the expansion is thought to be a rational function, instead of a polynomial. This guarantees to optimize the Taylor series with rational approximants that better approach large *z* than the Taylor series, guaranteeing mathematical stability of the new series if data points exceed *z* = 0. Among all the possible choices, here the attention is given to the Padé polynomials, firstly introduced in Ref. [170]. This technique of approximations turns out to be a bookkeeping device to keep the calculations manageable for the cosmography convergence issue. Thus, provided we have Taylor expansions of *f*(*z*) under the form *f*(*z*) = ∑ ∞ *i*=0 *ciz i* , with *c<sup>i</sup>* = *f* (*i*) (0)/*i*!, it is possible to obtain the (*n*, *m*) Padé approximant by

$$P\_{n,m}(z) = \left(\sum\_{i=0}^{n} a\_i z^i\right) \left(1 + \sum\_{j=1}^{m} b\_j z^j\right)^{-1} \tag{57}$$

and requires that *b*<sup>0</sup> = 1. Furthermore, it is important that *f*(*z*) − *Pn*,*m*(*z*) = O(*z n*+*m*+1 ) and the coefficients *b<sup>i</sup>* come from solving the homogeneous system of linear equations ∑ *m j*=1 *b<sup>j</sup> cn*+*k*+*<sup>j</sup>* = −*b*<sup>0</sup> *cn*+*<sup>k</sup>* , valid for *k* = 1, . . . , *m*. Once *b<sup>i</sup>* are known, *a<sup>i</sup>* can be obtained using the formula *a<sup>i</sup>* = ∑ *i k*=0 *bi*−*<sup>k</sup> ck* . Just for an example, we report the (2, 1) Padé polynomial as

$$P\_{2,1}(z) = \frac{z}{H\_0} \left\{ \frac{6(q\_0 - 1) + [q\_0(8 + 3q\_0) - 5 - 2j\_0]z}{2q\_0(3 + z + 3q\_0z) - 2(3 + z + j\_0z)} \right\}.\tag{58}$$
