Gauss's and Stirling's Formulas

In case of equidistant tabular points a convenient form for interpolating polynomial can be derived from Lagrange's interpolating polynomial. The process involves renaming or re-designating the tabular points. We illustrate it by deriving the interpolating formula for 6 tabular points. This can be generalized for more number of points. Let the given tabular points be $x_0, x_1=x_0+h, x_2=x_0-h,x_3=x_0+2h,x_4=x_0-2h, x_5=x_0+3h.$ These six points in the given order are not equidistant. We re-designate them for the sake of convenience as follows: $x_{-2}^*=x_4, x_{-1}^*=x_2, x_{0}^*=x_0, x_{1}^*=x_1, x_{2}^*=x_3,x_{3}^*=x_5.$ These re-designated tabular points in their given order are equidistant. Now recall from remark (12.3.3) that Lagrange's interpolating polynomial can also be written as :

$$f(x) \approx f(x_{0})+\delta[x_0,x_1](x-x_{0})+\delta[x_0,x_1,x_2] (x-x_0)(x-x_1)$$

$$+ \delta[x_0,x_1,x_2,x_3] (x-x_0)(x-x_1)(x-x_2)$$

$$+ \delta[x_0,x_1,x_2,x_3,x_{4}](x-x_0)(x-x_1)(x-x_2)(x-x_{3})$$

$$+ \delta[x_0,x_1,x_2,x_3,x_{4},x_5](x-x_0)(x-x_1)(x-x_2)(x-x_{3})(x-x_4),$$

which on using the re-designated points give:

$$f(x) \approx f(x_{0}^*)+\delta[x_0^*,x_1^*](x-x_{0}^*)+\delta[x_0^*,x_1^*,x_{-1}^*] (x-x_0^*)(x-x_1^*)$$

$$+ \delta[x_0^*,x_1^*,x_{-1}^*,x_2^*] (x-x_0^*)(x-x_1^*)(x-x_{-1}^*)$$

$$+ \delta[x_0^*,x_1^*,x_{-1}^*,x_2^*,x_{-2}^*] (x-x_0^*)(x-x_1^*)(x-x_{-1}^*)(x-x_2^*)$$

$$+ \delta[x_0^*,x_1^*,x_{-1}^*,x_2^*,x_{-2}^*,x_3^*] (x-x_0^*)(x-x_1^*)(x-x_{-1}^*)(x-x_2^*)(x-x_{-2}^*).$$

Now note that the points $x_{-2}^*, x_{-1}^*, x_{0}^*, x_{1}^*, x_{2}^*$ and $x_{3}^*$ are equidistant and the divided difference are independent of the order of their arguments. Thus, we have

$$\delta[x_0^*,x_1^*] = \frac{\Delta y_0^*}{h}, \hspace{.25in} \del... ...x_1^*,x_{-1}^*] = \delta[x_{-1}^*,x_0^*,x_1^*]=\frac{\Delta^2 y_{-1}^* }{2h^2},$$

$$\delta[x_0^*,x_1^*,x_{-1}^*,x_2^*] = \delta[x_{-1}^*,x_0^*,x_1^*,x_2^* ]=\frac{\Delta^3 y_{-1}^* }{3!h^3},$$

$$\delta[x_0^*,x_1^*,x_{-1}^*,x_2^*,x_{-2}^*] = \delta[x_{-2}^*, x_{-1}^*,x_0^*,x_1^*,x_2^* ]=\frac{\Delta^4 y_{-2}^* }{4!h^4},$$

$$\delta[x_0^*,x_1^*,x_{-1}^*,x_2^*,x_{-2}^*,x_3^*] = \delta[x_{-2}^*,x_{-1}^*,x_0^*,x_1^*,x_2^*,x_3^* ]=\frac{\Delta^5 y_{-2}^* }{5!h^5},$$

where $y_i^*=f(x_i^*)$ for $i= -2, -1, 0, 1, 2.$ Now using the above relations and the transformation $x=x_0^*+hu,$ we get

$$f(x_0^*+hu) \approx y_{0}^*+\frac{\Delta y_0^*}{h}(hu)+\frac{\Delta^2 y_{-1}^* }{2h^2} (hu)(hu-h)+\frac{\Delta^3 y_{-1}^* }{3!h^3} (hu)(hu-h)(hu+h)$$

$$+ \frac{\Delta^4 y_{-2}^* }{4!h^4} (hu)(hu-h)(hu+h)(hu-2h)$$

$$+ \frac{\Delta^5 y_{-2}^*}{5!h^5}(hu)(hu-h)(hu+h)(hu-2h)(hu+2h).$$

Thus we get the following form of interpolating polynomial

$$f(x_0^*+hu) \approx y_{0}^*+u\Delta y_0^*+u(u-1)\frac{\Delta^2 y_{-1}^* }{2!}+u(u^2-1)\frac{\Delta^3 y_{-1}^* }{3!}$$

$$+ u(u^2-1)(u-2) \frac{\Delta^4 y_{-2}^* }{4!}+ u(u^2-1)(u^2-2^2) \frac{\Delta^5 y_{-2}^* }{5!} .$$ (12.4.1)

Similarly using the tabular points $x_0, x_1=x_0-h, x_2=x_0+h,x_3=x_0-2h,x_4=x_0+2h, x_5=x_0-3h,$ and the re-designating them, as $x_{-3}^*,x_{-2}^*, x_{-1}^*, x_{0}^*, x_{1}^*$ and $x_{2}^*,$ we get another form of interpolating polynomial as follows:

$$f(x_0^*+hu) \approx y_{0}^*+u\Delta y_{-1}^*+u(u+1)\frac{\Delta^2 y_{-1}^* }{2!}+u(u^2-1)\frac{\Delta^3 y_{-2}^* }{3!}$$

$$+ u(u^2-1)(u+2) \frac{\Delta^4 y_{-2}^* }{4!}+ u(u^2-1)(u^2-2^2) \frac{\Delta^5 y_{-3}^* }{5!} .$$ (12.4.2)

Now taking the average of the two interpoating polynomials (12.4.1) and (12.4.2) (called GAUSS'S FIRST AND SECOND INTERPOLATING FORMULAS, respectively), we obtain Sterling's Formula of interpolation:

$$f(x_0^*+hu) \approx y_{0}^*+u\frac{\Delta y_{-1}^*+\Delta y_0^*}{2}+u^2\frac{\Delta^2... ...frac { u(u^2-1)}{2}\left[\frac{\Delta^3 y_{-2}^*+\Delta^3 y_{-1}^* }{3!}\right]$$

$$+ u^2(u^2-1)\frac{\Delta^4 y_{-2}^* }{4!}+ \frac{u(u^2-1)(u^2-2^2)}{2} \left[\frac{\Delta^5 y_{-3}^* +\Delta^5 y_{-2}^*}{5!}\right]+\cdots .$$ (12.4.3)

These are very helpful when, the point of interpolation lies near the middle of the interpolating interval. In this case one usually writes the diagonal form of the difference table.

EXAMPLE 12.4.1   Using the following data, find by Sterling's formula, the value of $f(x)= cot(\pi x)$ at $x = 0.225:$

$x$	0.20	0.21	0.22	0.23	0.24
$f(x)$	1.37638	1.28919	1.20879	1.13427	1.06489

Here the point $x=0.225$ lies near the central tabular point $x=0.22.$ Thus , we define $x_{-2}= 0.20, x_{-1}= 0.21, x_0=0.22, x_1=0.23, x_2=0.24,$ to get the difference table in diagonal form as:

$x_{-2}=0.20$	$y_{-2}=1.37638$
		$\Delta y_{-2}= -.08719$
$x_{-1}=.021$	$y_{-1}=1.28919$		$\Delta^2 y_{-2}=.00679$
		$\Delta y_{-1}=-.08040$		$\Delta^3 y_{-2}=-.00091$
$x_0=0.22$	$y_0=1.20879$		$\Delta^2 y_{-1}=.00588$		$\Delta^4 y_{-2}=.00017$
		$\Delta y_{0}=-.07452$		$\Delta^3 y_{-1}= -.00074$
$x_1=0.23$	$y_{1}= 1.13427$		$\Delta^2 y_{0}=.00514$
		$\Delta y_{1}=-.06938$
$x_2=0.24$	$y_2=1.06489$

(here, $\Delta y_0 = y_1-y_0=1.13427-1.20879=-.07452;\Delta y_{-1}=1.20879-1.28919=-0.08040;$ and $\Delta^2y_{-1}=\Delta y_0 - \Delta y_{-1}=.00588,$ etc.).

Using the Sterling's formula with $u=\displaystyle\frac{0.225-0.22}{0.01}= 0.5,$ we get $f(0.225)$ as follows:

$$f(0.225) = 1.20879+0.5\frac{-.08040-.07452}{2}+(-0.5)^2\frac{.00588}{2}$$

$$+ \frac{-0.5(0.5^2-1)}{2}\frac{(-.00091-.00074)}{3!} 0.5^2(0.5^2-1)\frac{.00017}{4!}$$

$$= 1.1708457$$

Note that tabulated value of $cot (\pi x)$ at $x=0.225$ is 1.1708496.

EXERCISE 12.4.2   Compute from the following table the value of $y$ for $x=0.05$ :

$x$	0.00	0.02	0.04	0.06	0.08
$y$	0.00000	0.02256	0.04511	0.06762	0.09007