x5 2 0 x1 1 x3SolutionNewton method xn1xnfxnfxn fxx52 fx5

x^5 + 2 = 0, x1 = ?1

x3=??

Newton method:

x_n+1=x_n-f(x_n)/f\'(x_n)

f(x)=x⁵+2

f\'(x)=5x⁴

f(-1)=1

f\'(-1)=5

x₂=-1-(1/5)=-6/5

f(-6/5)= -0.48832

f\'(-6/5)=10.36800

x₃=-6/5-(-0.48832/10.36800)=-1.1529

$x^5 + 2 = 0, x1 = ?1 x3=??SolutionNewton method: xn+1=xn-f(xn)/f\'(xn) f(x)=x5+2 f\'(x)=5x4 f(-1)=1 f\'(-1)=5 x2=-1-(1/5)=-6/5 f(-6/5)= -0.48832 f\'(-6/5)=10.36$

x5 2 0 x1 1 x3SolutionNewton method xn1xnfxnfxn fxx52 fx5

Solution

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