HW3 - 1 secant2(inline'x^3 2*x^2-3*x-1-2-3,9.5*10^-11,20...

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1. secant2(inline( 'x^3+2*x^2-3*x-1'),-2,-3,9.5*10^-11,20) secant2(inline( 'x^3+2*x^2-3*x-1'),-2,-3,9.4*10^-11,20) print out secant2.m To obtain the approximation p6, the secant method went through five iterations and  evaluated a total of six functions. 2. a.  newton(inline( '27*x^4+162*x^3-180*x^2+62*x-7'),inline( '62 - 360*x + 486*x^2 +  108*x^3  ' ),0) 10 iterations bisect(inline( '27*x^4+162*x^3-180*x^2+62*x-7'),0,1) 10 iterations Bisect converges faster. Multiplicity of the zero, where x=1/3: f(1/3)=27*x^4+162*x^3-180*x^2+62*x-7=0 f’(1/3) = 62 - 360*x + 486*x^2 + 108*x^3  =0 f’’(1/3) = -360 + 972*x + 324*x^2  =0 f’’’(1/3) = 972 + 648*x   = 1188 Thus, since f(1/3) = f’(1/3) = f’’(1/3) = 0, but f’’’(1/3) = 1188   0, the root of  multiplicity for this zero is 3.   Also, the apparent order of convergence is… b.   (i.) secant2(inline( '(1+log(x)-x)/(1/x - 1)'),.5,2,9.4*10^-11,20) zero is around x=1,  1.618  α≈ has/has 
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