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xxx... 
= 2 
We need to find root of f (x)  ln(2)  2ln(x) 
Find Root from Modified Secant Method 
1) Initial x x0  3 
2) Time steps δ  0.0001 
ModifiedSecantMethod(δx0) i  0 
x0  x0 
ε0  100 
εi 10 10  while  
xi1 xi 
f xiδxi 
f xi δ xi      f xi 
 
 
  
εi1 
xi1  xi 
xi1 
 100 
i  i  1 
x 
ε 
 
 
 
x_Numerical ModifiedSecantMethod(δx0)0  
x_Numerical 
3 
0.7437711009 
1.2217389913 
1.400486319 
1.4141474062 
1.4142135641 
1.4142135624 
1.4142135624 
 
 

h(x_Numericaly_Numerical) 

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Mathcad - modified secant method 2

  • 1. xxx... = 2 We need to find root of f (x)  ln(2)  2ln(x) Find Root from Modified Secant Method 1) Initial x x0  3 2) Time steps δ  0.0001 ModifiedSecantMethod(δx0) i  0 x0  x0 ε0  100 εi 10 10  while  xi1 xi f xiδxi f xi δ xi      f xi     εi1 xi1  xi xi1  100 i  i  1 x ε    x_Numerical ModifiedSecantMethod(δx0)0  x_Numerical 3 0.7437711009 1.2217389913 1.400486319 1.4141474062 1.4142135641 1.4142135624 1.4142135624   