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Method and examples
Numerical Differentiation using
Newton's Forward, Backward Method
Method
 
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OR
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 x =
  1. x1.41.61.82.02.2
    f(x)4.05524.95306.04967.38919.0250
    and x=1.4
  2. x1.41.61.82.02.2
    f(x)4.05524.95306.04967.38919.0250
    and x=2.2
  3. x0.00.10.20.30.4
    f(x)1.00000.99750.99000.97760.8604
    and x=0.0
  4. x0.00.10.20.30.4
    f(x)1.00000.99750.99000.97760.8604
    and x=0.1
  5. x0.00.10.20.30.4
    f(x)1.00000.99750.99000.97760.8604
    and x=0.3
  6. x0.00.10.20.30.4
    f(x)1.00000.99750.99000.97760.8604
    and x=0.4
f(x) =
x1 = and x2 =
 x =
Step value (h) =  OR  Interval (N) =
  1. `f(x)=2x^3-4x+1`
    x1 = 2 and x2 = 4
    x = 3.5
    Step value (h) = 0.5
    or N = 5
  2. `f(x)=2x^3-4x+1`
    x1 = 2 and x2 = 4
    x = 2.25
    Step value (h) = 0.25
    or N = 8
  3. `f(x)=x^3-x+1`
    x1 = 2 and x2 = 4
    x = 3.5
    Step value (h) = 0.5
    or N = 5
  4. `f(x)=x^3-x+1`
    x1 = 2 and x2 = 4
    x = 2.25
    Step value (h) = 0.25
    or N = 8
  5. `f(x)=x^3+x+2`
    x1 = 2 and x2 = 4
    x = 3.5
    Step value (h) = 0.5
    or N = 5
  6. `f(x)=x^3+x+2`
    x1 = 2 and x2 = 4
    x = 2.25
    Step value (h) = 0.25
    or N = 8
SolutionHelp

 
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