newtontoapproxln

newtontoapproxln -...

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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). 
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newtontoapproxln -...

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