newtontoapproxln -...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Use Newton interpolation polynomial to approximate   ln(x)  using three data points  x1=1, f(x1)=0,  x2=4, f(x2)=1.386294,  x3=6, f(x3)=1.791759.  Construct divided differences table.  F[x1]=0,  F[x2]=1.386294, F[x3]=1.791759,  F[x1,x2]=(1.386294-0)/(4-1)=0.4620981, F[x2,x3]=(1.791759-1.386294)/(6-4)=0.2027325,  F[x1,x2,x3]=(0.2027325-0.4620981)/(6-1)=-0.0518731  The Newton polynomial is given by  f2(x)= 0+ 0.4620981(x-1)-0.0518731(x-1)(x-4)  (b) Add a fourth point [x4=5; f(x4)=1.609438] to the interpolation above to get an  improved estimation of ln(x).  Amend the divided difference table above  F[x1]=0,  F[x2]=1.386294, F[x3]=1.791759 (old),  F[x1,x2]=(1.386294-0)/(4-1)=0.4620981, F[x2,x3]=(1.791759-1.386294)/(6-4)=0.2027325  (old),  F[x3,x4]=(1-609438-1.791759)/(5-6)=0.1823216 (new),  F[x1,x2,x3]=(0.2027325-0.4620981)/(6-1)=-0.0518731 (old),   F[x2,x3,x4]=(0.1823216 -0.2027325)/(5-4)=-0.02041100 (new),   F[x1,x2,x3,x4]=( -0.02041100 (-0.0518731))/(5-1)=0.007865529 (new). 
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

newtontoapproxln -...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online