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Mathematics A
1. Pure Mathematics
1.9 Numerical Methods
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Iterative processes are suitable for problems without exact
solutions
Steps to find the square root of 5 using the iterative formula
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
1
2
(
x
n
+
\frac{1}{2} (x_{n} +
2
1
(
x
n
+
5
x
n
)
\frac{5}{x_{n}})
x
n
5
)
1️⃣ Start with an initial guess
x
0
=
x_{0} =
x
0
=
2
2
2
2️⃣ Calculate
x
1
=
x_{1} =
x
1
=
1
2
(
2
+
5
2
)
=
\frac{1}{2} (2 + \frac{5}{2}) =
2
1
(
2
+
2
5
)
=
2.25
2.25
2.25
3️⃣ Calculate
x
2
=
x_{2} =
x
2
=
1
2
(
2.25
+
5
2.25
)
=
\frac{1}{2} (2.25 + \frac{5}{2.25}) =
2
1
(
2.25
+
2.25
5
)
=
2.2361
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) =
f
′
(
x
)
=
0
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?
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
x_{n} - \frac{f(x_{n})}{f'(x_{n})}
x
n
−
f
′
(
x
n
)
f
(
x
n
)
The Secant method uses the formula
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
x
n
−
f
(
x
n
)
(
x
n
−
x
n
−
1
)
f
(
x
n
)
−
f
(
x
n
−
1
)
x_{n} - \frac{f(x_{n})(x_{n} - x_{n - 1})}{f(x_{n}) - f(x_{n - 1})}
x
n
−
f
(
x
n
)
−
f
(
x
n
−
1
)
f
(
x
n
)
(
x
n
−
x
n
−
1
)
and does not require derivative calculation
The Newton-Raphson method requires the derivative of the function
f
′
(
x
)
f'(x)
f
′
(
x
)
to find roots.
Steps to apply the Newton-Raphson method
1️⃣ Start with an initial guess
x
0
x_{0}
x
0
.
2️⃣ Use the iterative formula to find
x
n
+
1
x_{n + 1}
x
n
+
1
.
3️⃣ Iterate until
∣
x
n
+
1
−
x
n
∣
|x_{n + 1} - x_{n}|
∣
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?
x
0
=
x_{0} =
x
0
=
2
2
2
After two iterations, the approximate root of
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
−
2
x
−
5
x^{3} - 2x - 5
x
3
−
2
x
−
5
using Newton-Raphson is 2.0945681
The Newton-Raphson method may not converge if
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
0
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?
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
1
2
(
x
n
+
\frac{1}{2} (x_{n} +
2
1
(
x
n
+
5
x
n
)
\frac{5}{x_{n}})
x
n
5
)
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
x
0
=
x_{0} =
x
0
=
2
2
2
.
2️⃣ Iterate:
x
1
=
x_{1} =
x
1
=
2.25
2.25
2.25
.
3️⃣ Iterate:
x
2
=
x_{2} =
x
2
=
2.2361
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
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
x_{n} - \frac{f(x_{n})}{f'(x_{n})}
x
n
−
f
′
(
x
n
)
f
(
x
n
)
.
Secant ↔️ Use the formula
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
x
n
−
f
(
x
n
)
(
x
n
−
x
n
−
1
)
f
(
x
n
)
−
f
(
x
n
−
1
)
x_{n} - \frac{f(x_{n})(x_{n} - x_{n - 1})}{f(x_{n}) - f(x_{n - 1})}
x
n
−
f
(
x
n
)
−
f
(
x
n
−
1
)
f
(
x
n
)
(
x
n
−
x
n
−
1
)
.
The bisection method always converges if
f
(
a
)
f
(
b
)
<
0
f(a)f(b) < 0
f
(
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) =
f
(
x
)
=
x
3
−
2
x
−
5
x^{3} - 2x - 5
x
3
−
2
x
−
5
using Newton-Raphson
1️⃣ Start with
x
0
=
x_{0} =
x
0
=
2
2
2
.
2️⃣ Apply the formula
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
x_{n} - \frac{f(x_{n})}{f'(x_{n})}
x
n
−
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?
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
x
n
−
f
(
x
n
)
f
′
(
x
n
)
x_{n} - \frac{f(x_{n})}{f'(x_{n})}
x
n
−
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?
x
n
+
1
=
x_{n + 1} =
x
n
+
1
=
x
n
−
f
(
x
n
)
(
x
n
−
x
n
−
1
)
f
(
x
n
)
−
f
(
x
n
−
1
)
x_{n} - \frac{f(x_{n})(x_{n} - x_{n - 1})}{f(x_{n}) - f(x_{n - 1})}
x
n
−
f
(
x
n
)
−
f
(
x
n
−
1
)
f
(
x
n
)
(
x
n
−
x
n
−
1
)
The Secant method does not require the calculation of the
derivative
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