| x | f[] | f[,] | f[,,] | f[,,,] |
|---|
| 0 | 0.75 | | | |
| 1.5 | 0.65 | F[x1,x2] = x2−x1F[x2]−F[x1]=1.5−00.65−0.75=−151 | | |
| 4.5 | 1.0 | F[x2,x3] = x3−x2F[x3]−F[x2]=4.5−1.51.0−0.65=607 | F[x1,x2,x3] = x3−x1F[x2,x3]−F[x1,x2]=4.5−07/60−(−1/15)=27011 | |
| $7.5$ | $0.8$ | $F[x_3,x_4]$ = $\frac{F[x_4]-F[x_3]}{x_4-x_3} = \frac{0.8 - 1.0}{7.5 - 4.5} = -\frac{1}{15}$ | $F[x_2,x_3,x_4]$ = $\frac{F[x_3,x_4]-F[x_2,x_3]}{x_4-x_2} =\frac{-1/15 - 7/60}{7.5 -1.5} = -\frac{11}{360}$ | $F[x_1,x_2,x_3,x_4]$ = $\frac{F[x_2,x_3,x_4]-F[x_1,x_2,x_3]}{x_4-x_1} = \frac{-11/360-11/270}{7.5 - 0} = -\frac{77}{8100}$
| 8.9 | 0.5 |F[x4,x5] = x5−x4F[x5]−F[x4]=8.9−7.50.5−0.8=−143 | F[x3,x4,x5] = x5−x3F[x3,x4]−F[x4,x5]=8.9−4.5(−3/14)−(−1/15)=−92431 | F[x2,x3,x4,x5] = x5−x2F[x3,x4,x5]−F[x4,x5,x6]=8.9−1.5−31/924−(−11/360)=−20512883 |
| 10 | 0.3 | F[x5,x6] = x6−x5F[x6]−F[x5]=10−8.90.3−0.5=−112 | F[x4,x5,x6] = x6−x4F[x5,x6]−F[x4,x5]=10−7.5(−2/11)−(−3/14)=771 | F[x3,x4,x5,x6] = x6−x3F[x4,x5,x6]−F[x3,x4,x5]=10−4.51/77−(−31/924)=508243 |
| f[,,,,] | f[,,,,,] |
|---|
| F[x1,x2,x3,x4,x5] = x5−x1F[x2,x3,x4,x5]−F[x1,x2,x3,x4]=8.9−0−83/205128−(−77/8100)=410768820420071 | |
| F[x2,x3,x4,x5,x6] = x6−x2F[x3,x4,x5,x6]−F[x2,x3,x4,x5]=10−1.543/5082−(−83/205128)=1917946820005 | F[x1,x2,x3,x4,x5,x6] = x6−x1F[x2,x3,x4,x5,x6]−F[x1,x2,x3,x4,x5] = 10−020005/19179468−(420071/410768820)=192034423350391687 |
| an | F[] | Value | Approx Value | xn |
|---|
| a0 | F[x1] | 0.75 | 0.75 | 0 |
| a1 | F[x1,x2] | −151 | −0.066667 | 1.5 |
| a2 | F[x1,x2,x3] | 27011 | 0.040741 | 4.5 |
| a3 | F[x1,x2,x3,x4] | −810077 | −0.0095062 | 7.5 |
| a4 | F[x1,x2,x3,x4,x5] | 410768820420071 | 0.001023 | 8.9 |
| a5 | F[x1,x2,x3,x4,x5,x6] | −19203442335027170723 | −0.000141 | 10 |
P6(x)=0.75+x(−0.066667+(x−1.5)(0.040741+(x−4.5)(−0.0095062+(x−7.5)(0.001023+(x−8.9)(−0.000141+(x−10))))))
Gc=1.61
https://atozmath.com/CONM/NumeDiff.aspx?q=DD&q1=1%600%2C1.5%2C4.5%2C7.5%2C8.9%2C10%600.75%2C0.65%2C1.0%2C0.8%2C0.5%2C0.3%600%60DD&dp=10&tm=R&do=1#tblSolution