Create programs (remember to hand in a hard copy of your code) for the Bisection

Question

Create programs (remember to hand in a hard copy of your code) for the Bisection, Secant, and Newton's methods to find the root of , where

a. Use the bisection method with  and  as starting values. Give the first four iterates showing the intervals, , and the function value at the midpoint, .

,  ,   
,  ,   
,  ,   
,  ,   

Continue the iteration until the solution is accurate to a tolerance of

The number of iterations is   
The approximate root is  

b. Use the secant method with  and  as starting values. Give the first four iterates showing the updated value of the root, , and the function value at this point, .

,   
,   
,   
,   

Continue the iteration until the solution is accurate to a tolerance of

The number of iterations is   
The approximate root is  

c. Use Newton's method with  as a starting value. Give the first four iterates showing the updated value of the root, , and the function value at this point, .

,   
,   
,   
,   

Continue the iteration until the solution is accurate to a tolerance of

The number of iterations is   
The approximate root is  

Explanation / Answer

1) secant

6dgt 10dgt 14dgt 18dgt 22dgt 26dgt 30dgt 34dgt 38dgt 42dgt 46dgt 50dgt n x accuracy x f(x) 2 1.775 2 -0.421309955000589816 3 1.8875 2 -0.002700162166624038 4 1.94375 1.9 0.241544925675145248 5 1.915625 1.9 0.11634861486049395 6 1.9015625 1.9 0.05606820867211787 7 1.89453125 1.89 0.02649655862895399 8 1.891015625 1.89 0.011851523729519266 9 1.8892578125 1.89 0.004564036061743092 10

conitue


45 1.88815405096463281 1.8881540509646 -6.5904E-14 46 1.8881540509646456 1.8881540509646 -1.3074E-14 47 1.88815405096465199 1.88815405096465 1.3338E-14 48 1.88815405096464879 1.88815405096465 1.38E-16 49 1.88815405096464719 1.88815405096465 -6.472E-15

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