HW. 4-13               Courtney Taylor

Question 2

Series[f[a+(h/2)],{h,0,4}]

Series[f[a+h],{h,0,4}]

Series[f'[a+h],{h,0,4}]

(1/2)(f[a]+Series[f[a+h],{h,0,4}])-(h/8)(Series[f'[a+h],{h,0,4}]-f'[a])-Series[f[a+(h/2)],{h,0,4}]

Question 3

l0[x_]:=(((x-1)(x-2)(x-3))/((0-1)(0-2)(0-3)))

l1[x_]:= (((x-0)(x-2)(x-3))/((1-0)(1-2)(1-3)))

l2[x_]:= (((x-0)(x-1)(x-3))/((2-0)(2-1)(2-3)))

l3[x_]:= (((x-0)(x-1)(x-2))/((3-0)(3-1)(3-2)))

g[x_]=0*D[D[l0[x],x],x]+1*D[D[l1[x],x],x]+5*D[D[l2[x],x],x]+13*D[D[l3[x],x],x]

g[0]

Question 6
Part A

(Series[f[a+h],{h,0,6}]-Series[f[a-h],{h,0,6}])

(2 f'[a] h+(1/3) Derivative[3][f][a] (h^3)+(1/60) Derivative[5][f][a] (h^5))/(2h)

d[h_]:=(f'[a] +(1/6) Derivative[3][f][a] (h^2)+(1/120) Derivative[5][f][a] (h^4))

d[2h]

Simplify[3d[h]-2d[(3/2)h]]

Part B

((3.9-3.1)/(1.2-1))

((4-3.1)/(1.3-1))

3(4)-2(3)

Other Question Involving the Richard’s Extrapolation, Second Derivatives, and Centered Difference Formula:

re[x_]:=((f[a+h]-2f[a]+f[a-h])/(h^2))=f’’[a]

((Series[f[a+h],{h,0,6}]-2 f[a]+Series[f[a-h],{h,0,6}])/(h^2))

re[h_]:=f''[a]+(1/12) Derivative[4][f][a] (h^2)+(1/360) Derivative[6][f][a] (h^4)

re[h]

re[2h]

Simplify[4*re[h]-re[2h]]

Simplify[(4/3)*re[h]-(1/3)re[2h]]

re1[h_]:=(4/3) *((f[a+h]-2f[a]+f[a-h])/(h^2))- (1/3)*((f[a+2h]-2f[a]+f[a-2h])/(4 (h^2)))