1.9 Numerical Methods

    Cards (98)

    • Iterative processes are suitable for problems without exact solutions
    • Steps to find the square root of 5 using the iterative formula xn+1=x_{n + 1} =12(xn+ \frac{1}{2} (x_{n} +5xn) \frac{5}{x_{n}})

      1️⃣ Start with an initial guess x0=x_{0} =2 2
      2️⃣ Calculate x1=x_{1} =12(2+52)= \frac{1}{2} (2 + \frac{5}{2}) =2.25 2.25
      3️⃣ Calculate x2=x_{2} =12(2.25+52.25)= \frac{1}{2} (2.25 + \frac{5}{2.25}) =2.2361 2.2361
      4️⃣ Continue until successive values are within an acceptable margin
    • Iterative processes cannot refine solutions to meet specific accuracy requirements.
      False
    • What do numerical methods provide when high precision is not critical?
      Close approximations
    • Numerical methods are applicable to a wide range of problem types, including nonlinear equations
    • Iterative processes are only suitable for problems with exact analytical solutions.
      False
    • What is the initial step in an iterative process for numerical methods?
      Initial guess
    • Iterative processes allow for accuracy control by refining solutions to meet specific requirements
    • Match the root finding method with its description:
      Bisection ↔️ Halves interval until root is found
      Newton-Raphson ↔️ Uses derivative to converge
      Secant ↔️ Does not require derivative
    • The bisection method always converges if f(a)f(b) < 0.
    • Why might the Newton-Raphson method not converge?
      f′(x)=f'(x) =0 0
    • The secant method converges faster than the bisection method.
    • What does the Newton-Raphson method use to find roots?
      Function and derivative
    • What is the iterative formula for the Newton-Raphson method?
      xn+1=x_{n + 1} =xn−f(xn)f′(xn) x_{n} - \frac{f(x_{n})}{f'(x_{n})}
    • The Secant method uses the formula xn+1=x_{n + 1} =xn−f(xn)(xn−xn−1)f(xn)−f(xn−1) x_{n} - \frac{f(x_{n})(x_{n} - x_{n - 1})}{f(x_{n}) - f(x_{n - 1})} and does not require derivative calculation
    • The Newton-Raphson method requires the derivative of the function f′(x)f'(x) to find roots.
    • Steps to apply the Newton-Raphson method
      1️⃣ Start with an initial guess x0x_{0}.
      2️⃣ Use the iterative formula to find xn+1x_{n + 1}.
      3️⃣ Iterate until ∣xn+1−xn∣|x_{n + 1} - x_{n}| is within a specified tolerance.
    • What is the initial guess for finding the root of f(x) = x^{3} - 2x - 5</latex> using Newton-Raphson?
      x0=x_{0} =2 2
    • After two iterations, the approximate root of f(x)=f(x) =x3−2x−5 x^{3} - 2x - 5 using Newton-Raphson is 2.0945681
    • The Newton-Raphson method may not converge if f′(x)=f'(x) =0 0 or the initial guess is poor.
    • What are numerical methods designed to find?
      Approximate solutions
    • Numerical methods are essential when dealing with equations that cannot be solved analytically
    • Numerical methods provide exact solutions for all mathematical problems.
      False
    • What is the stopping criterion for an iterative process?
      Acceptable error threshold
    • What are numerical methods designed to find?
      Approximate solutions
    • An iterative process starts with an initial guess and improves through each iteration
    • Iterative processes are suitable for problems without exact solutions.
    • What iterative formula can be used to find the square root of5?
      xn+1=x_{n + 1} =12(xn+ \frac{1}{2} (x_{n} +5xn) \frac{5}{x_{n}})
    • An iterative process refines approximate solutions through repeated application of a formula or algorithm
    • Iterative processes are suitable for problems without exact solutions.
    • Steps to find the square root of 5 using the iterative formula
      1️⃣ Start with x0=x_{0} =2 2.
      2️⃣ Iterate: x1=x_{1} =2.25 2.25.
      3️⃣ Iterate: x2=x_{2} =2.2361 2.2361.
      4️⃣ Continue until successive values are within an acceptable margin.
    • Match the root-finding method with its description:
      Bisection ↔️ Halve the interval until the root is found.
      Newton-Raphson ↔️ Use the formula xn+1=x_{n + 1} =xn−f(xn)f′(xn) x_{n} - \frac{f(x_{n})}{f'(x_{n})}.
      Secant ↔️ Use the formula xn+1=x_{n + 1} =xn−f(xn)(xn−xn−1)f(xn)−f(xn−1) x_{n} - \frac{f(x_{n})(x_{n} - x_{n - 1})}{f(x_{n}) - f(x_{n - 1})}.
    • The bisection method always converges if f(a)f(b)<0f(a)f(b) < 0.
    • The Newton-Raphson method converges quickly if the initial guess is close to the root
    • Steps to find the root of f(x)=f(x) =x3−2x−5 x^{3} - 2x - 5 using Newton-Raphson

      1️⃣ Start with x0=x_{0} =2 2.
      2️⃣ Apply the formula xn+1=x_{n + 1} =xn−f(xn)f′(xn) x_{n} - \frac{f(x_{n})}{f'(x_{n})}.
      3️⃣ Iterate until successive values are within an acceptable margin.
    • What is the iterative formula used in the Newton-Raphson method?
      xn+1=x_{n + 1} =xn−f(xn)f′(xn) x_{n} - \frac{f(x_{n})}{f'(x_{n})}
    • The Newton-Raphson method converges quickly if the initial guess is close to the root
    • The Newton-Raphson method requires the calculation of the derivative of the function.
    • What is the formula used in the Secant method?
      xn+1=x_{n + 1} =xn−f(xn)(xn−xn−1)f(xn)−f(xn−1) x_{n} - \frac{f(x_{n})(x_{n} - x_{n - 1})}{f(x_{n}) - f(x_{n - 1})}
    • The Secant method does not require the calculation of the derivative
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