Answer to Question #225409 in Quantitative Methods for doraid

Question

Approximate f(0.05) use the following data and the Newton forward-difference formula

x 0.0 0.2 0.4 0.6 0.8

f(x) 1.00000 1.22140 1.49182 1.82212 2.22554

b. use the Newton backward-difference formula to approximate f(0.65)

c.Use stirlings formula to approximate f(0.43).

Answer 1

a)

The linear function is 

$f0 + (x-x0)(f1-f0)\div(x1-x0) \\ =1 + (x-0)(1.22140-1)\div(0.2) \\ =1 + (0.2214)x\div(0.2) \\ =1 + 1.107x$

 

So f(0.05) is approximately$1 + 1.107(0.05) = 1.105535$

b)

Likewise the backwards difference results in :

 $f1 - (x-x1)(f1-f0)\div(x1-x0) \\ =1.82212 - (x-0.6)( 2.22554-1.82212)\div(-0.2)\\ =1.82212 + (x-0.6)(2.0171)\\ So\\ x=0.65\\ 1.922975$

c)

$h=0.-0=0.2$

Taking X₀=0.4 then $p=\frac{X-X_0}{h}=\frac{X-0.4}{2}$

$X=043\\P=\frac{X-X_)}{h}=\frac{0.43-0.4}{0.2}=0.15$

$Y_0=1.49182,\Delta Y_0=0.3303,\Delta ^2Y_{-1}=0.05988,\Delta ^3Y_{-1}=0.01324,\Delta ^4Y_{-2}=0.00238$

striling formula is

$Y_P=Y_0+P.\frac{\Delta Y_0+\Delta Y_{-1}}{2}+\frac{p^2}{2}\Delta ^2Y_{-1}+\frac{p(p^2-1^2)}{3}.\frac{\Delta ^3Y_{-1}+\Delta ^3Y_{-2}}{2}+\frac{p^2(p^2-1^2)}{4}\Delta ^4Y_{-2}$

$Y_{0.15}=1.49182+(0.15).\frac{0.3303+0.27042}{2}+\frac{0.0225}{2}.(0.05988)+\frac{(0.15)(0.225-1)}{6}.\frac{0.01324+0.01086}{2}+\frac{(0.0225)(0.225-1)}{24}.(0.00238)$

$Y_{0.15}=1.49182+0.045054+0.00067365-0.0002944719-0.0000002181\\Y_{015}=1.53725$