NUMERICAL ANALYSIS 10TH EDITION BY
BURDEN PRACTICE EXAM 2026 PREMIUM
QUESTIONS AND ACCURATE ANSWERS ONE
HUNDRED PERCENT CORRECT
◉ Fixed Point Convergence.
Answer: If g is continuously differentiable, g(r)=r, and ‖g'(r)‖<1,
then the fixed point iteration converges linearly with rate S to the
fixed point r.
S=‖g'(r)‖
◉ Newton's Method.
Answer: Approximates the root of non-linear functions by finding
the root of the tangent line.
In the second iteration, plug in the approximation from the first
iteration as "x_n".
Repeat recursively.
◉ Newton's Method Convergence.
,Answer: Newton's Method converges quadratically to a root x if
f'(r)≠0 at rate M.
M=lim(e_(i+1))/e_i²
◉ Derivation of Newton's Method.
Answer: f'(x_0)(x-x_0)=0-f(x_0)
x-x_0=-f(x_0)/f'(x_0)
x=x_0-f(x_0)/f'(x_0)
◉ Secant Method.
Answer: Approximates the root of the function by taking the secant
line through the last two approximation. The new guess is the point
in which the secant line intersects the x-axis.
Start with two initial approximations p₀ and p₁. The next guess, p₂, is
the x-intercept of the line joining (p₀,f(p₀) and (p₁,f(p₁).
The approximation p₃ is the x-intercept of the line joining (p₁,f(p₁))
and (p₂,f(p₂)), and so on.
◉ Method of False Position.
, Answer: Combination of the Secant Method and the Bisection
Method. It uses the Bisection Method to ensure that the two
previous approximations used for the secant line have opposite
signs.
Choose initial guesses p₀ and p₁ with opposite sides (p₀*p₁ < 0).
The next guess, p₂, is the x-intercept of the line joining (p₀,f(p₀) and
(p₁,f(p₁).
Use p₀ and p₂ or p₁ and p₂ as your next guesses. Use p₀ and p₂ if
f(p₀) and f(p₂) have opposite signs. Else, use the other option.
Continue until f(p_n)=0.
◉ Muller's Method.
Answer: A root-finding method that draws a parabola through the
three previous approximations and looks at where it intersects the
x-axis. It generally intersects at 0 or 2 points.
If there are two points, then the point nearest the last point is
chosen as the newest approximation. If there are 0 points, then there
are complex solutions.
BURDEN PRACTICE EXAM 2026 PREMIUM
QUESTIONS AND ACCURATE ANSWERS ONE
HUNDRED PERCENT CORRECT
◉ Fixed Point Convergence.
Answer: If g is continuously differentiable, g(r)=r, and ‖g'(r)‖<1,
then the fixed point iteration converges linearly with rate S to the
fixed point r.
S=‖g'(r)‖
◉ Newton's Method.
Answer: Approximates the root of non-linear functions by finding
the root of the tangent line.
In the second iteration, plug in the approximation from the first
iteration as "x_n".
Repeat recursively.
◉ Newton's Method Convergence.
,Answer: Newton's Method converges quadratically to a root x if
f'(r)≠0 at rate M.
M=lim(e_(i+1))/e_i²
◉ Derivation of Newton's Method.
Answer: f'(x_0)(x-x_0)=0-f(x_0)
x-x_0=-f(x_0)/f'(x_0)
x=x_0-f(x_0)/f'(x_0)
◉ Secant Method.
Answer: Approximates the root of the function by taking the secant
line through the last two approximation. The new guess is the point
in which the secant line intersects the x-axis.
Start with two initial approximations p₀ and p₁. The next guess, p₂, is
the x-intercept of the line joining (p₀,f(p₀) and (p₁,f(p₁).
The approximation p₃ is the x-intercept of the line joining (p₁,f(p₁))
and (p₂,f(p₂)), and so on.
◉ Method of False Position.
, Answer: Combination of the Secant Method and the Bisection
Method. It uses the Bisection Method to ensure that the two
previous approximations used for the secant line have opposite
signs.
Choose initial guesses p₀ and p₁ with opposite sides (p₀*p₁ < 0).
The next guess, p₂, is the x-intercept of the line joining (p₀,f(p₀) and
(p₁,f(p₁).
Use p₀ and p₂ or p₁ and p₂ as your next guesses. Use p₀ and p₂ if
f(p₀) and f(p₂) have opposite signs. Else, use the other option.
Continue until f(p_n)=0.
◉ Muller's Method.
Answer: A root-finding method that draws a parabola through the
three previous approximations and looks at where it intersects the
x-axis. It generally intersects at 0 or 2 points.
If there are two points, then the point nearest the last point is
chosen as the newest approximation. If there are 0 points, then there
are complex solutions.
