3. (B) Program for Newton’s backward interpolation.
| ||
| y=[0.2588190 0.3420201 0.4226183 0.5 0.5735764 0.6427876]; | ||
| h=5 | ||
| c=1; | ||
| for i=1:5 | ||
| d1(c)=y(i+1)-y(i); | ||
| c=c+1; | ||
| end | ||
| c=1; | ||
| for i=1:4 | ||
| d2(c)=d1(i+1)-d1(i); | ||
| c=c+1 | ||
| end | ||
| c=1; | ||
| for i=1:3 | ||
| d3(c)=d2(i+1)-d2(i); | ||
| c=c+1; | ||
| end | ||
| c=1; | ||
| for i=1:2 | ||
| d4(c)=d3(i+1)-d3(i); | ||
| c=c+1; | ||
| end | ||
| c=1; | ||
| for i=1:1 | ||
| d5(c)=d4(i+1)-d4(i); | ||
| c=c+1 ; | ||
| end | ||
| d=[d1(5) d2(4) d3(3) d2(2) d1(1)]; | ||
| x0=38; | ||
| pp=1; | ||
| y_x=y(6); | ||
| p=(x0-x(6))/h; | ||
| for i=1:5 | ||
| pp=1; | ||
| for j=1:i | ||
| pp=pp*(p-(j-1)) | ||
| end | ||
| y_x=y_x+(pp*d(i))/factorial(i); | ||
| end | ||
| printf('value of function at %f is:%f',x0,y_x); | ||
3. (B) Program for Newton’s backward interpolation.
Reviewed by admin
on
December 24, 2019
Rating:
No comments: