Here are the Bisection Method formulas xm = (xl+xu)/2 I'm not convinced that you understand what the above means. X L - Lower (left) endpoint of an interval x M - Midpoint of an interval x U - Upper (right) endpoint of an interval a) If f(x L)*f(x M) 0, the graph of the function does not cross the x-axis between x L and x M, so we should look in the other half of the interval - in [x M, x U].

Programming exercises in fortran. Demonstrate your. Class, write a brief report summarizing the numerical methods used in exercises 4-7. X x to find the root which is closest to the point x = 3.3. B) Modify the program to use the secant method. Using this program, find the root of the equation.

If so, USE THE SAME VALUE FOR x U (i.e., don't change x U), but reset x L to x M. Your code should NOT include x U = x U. At each step for a) or b), we are shortening the interval by half its length, so that we eventually find the root.

C) If f(x L)*f(x M) = 0 then either f(x L) = 0 or f(x M). There's probably an assumption that f(x L) ≠ 0 and f(x U) ≠ 0, but you didn't show it in the attachment you posted.

Contents • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • [ ] with Ada. Text_Io; use Ada.

Text_Io; procedure Roots_Of_Function is package Real_Io is new Ada. Translation of: print_roots = (f, begin, end, step ) -># Print approximate roots of f between x=begin and x=end, # using sign changes as an indicator that a root has been # encountered. X = begin y = f (x ) last_y = y cross_x_axis = ->(last_y 0 ) or (last_y >0 and y # Smaller steps produce more accurate/precise results in general, # but for many functions we'll never get exact roots, either due # to imperfect binary representation or irrational roots. Step = 1 / 256 f1 = (x ) ->x *x *x - 3 *x *x + 2 *x print_roots f1, - 1, 5, step f2 = (x ) ->x *x - 4 *x + 3 print_roots f2, - 1, 5, step f3 = (x ) ->x - 1.5 print_roots f3, 0, 4, step f4 = (x ) ->x *x - 2 print_roots f4, - 2, 2, step output >coffee roots.coffee ----- Root found at 0 Root found at 1 Root found at 2 ----- Root found at 1 Root found at 3 ----- Root found at 1.5 ----- Root found near -1.4140625 Root found near 1.41796875 [ ]. Output: Root found at 8.6218e-16 Root found at 1.000016 Root found at 1.998914 [ ]% Implemented by Arjun Sunel - module (roots ). - export ( [main / 0 ] ).

Main ( ) ->F = fun ( X ) ->X * X * X - 3 * X * X + 2 * X end, Step = 0.001,% Using smaller steps will provide more accurate results Start = - 1, Stop = 3, Sign = F ( Start ) >Download Lagu Tema Hp Sony Ericsson W880i. 0, X = Start, while ( X, Step, Start, Stop, Sign, F ). While ( X, Step, Start, Stop, Sign, F ) ->Value = F ( X ), if Value == 0 ->% We hit a root: format ( 'Root found at ~p~n', [ X ] ), while ( X + Step, Step, Start, Stop, Value >0, F ); ( Value% We passed a root: format ( 'Root found near ~p~n', [ X ] ), while ( X + Step, Step, Start, Stop, Value >0, F ); X >Stop ->: format ( ' ); true ->while ( X + Step, Step, Start, Stop, Value >0, F ) end.

Output: Root found near 8.6218e-16 Root found near 1.000016 Root found near 2.998915 ok [ ] PROGRAM ROOTS_FUNCTION!VAR E,X,STP,VALUE,S%,I%,LIMIT%,X1,X2,D FUNCTION F(X) F=X*X*X-3*X*X+2*X END FUNCTION BEGIN X=-1 STP=1.0E-6 E=1.0E-9 S%=(F(X)>0) PRINT('VERSION 1: SIMPLY STEPPING X') WHILE XLIMIT% THEN PRINT('ERROR: FUNCTION NOT CONVERGING') EXIT END IF D=(X2-X1)/(F(X2)-F(X1))*F(X2) IF ABS(D). Output: Root (to 12DP) Max. Error 0.08 1e-06 1.16 1e-06 2.34 1e-06 Note that the roots found are all near misses because fractional numbers that seem nice and 'round' in decimal (such as 10^-6) often have some rounding error when represented in binary.

To increase the chances of finding exact integer roots, try using an integer start value with a step value that is a power of two. -- Main procedure print ( 'Root (to 12DP) tMax. Error n' ) for _, r in pairs (root (f, - 1, 3, 2 ^- 10 ) ) do print ( string.format ( '%0.12f', r.val ), r.err ) end. Output: findRoots(#f, -1, 3, 0.0001) Root found at 0 Root found at 1 Root found at 2 findRoots(#f, -1.000001, 3, 0.0001) Root near 9.0713e-005 Root near 1.000099 Root near 2.000099 [ ] If the equation is a polynomial, we can put the coefficients in a vector and use roots: a = [ 1, - 3, 2, 0 ]; r = (a );% let's print it for = 1: 3 n = (a, r ( ) ); ( 'x%d =%f (%f',, r ( ), n ); if (n!= 0.0 ) ( ' not' ); endif ( ' exact) n' ); endfor Otherwise we can program our (simple) method.