MACM 316: CA2 Solved

Description

First graph is part a), second is part b), third is a demonstrative sketch for part e), where purple is g(x) and black is y = x
a) The graph oscillates similarly to the cos(x) function and
fluctuates between f(x) = -1 and +1 on the negative interval
and f(x) = 0 and 2 on the positive interval. There are 8 roots
on the interval. The roots are roughly x = -17.28, -14.14,
-11.00, -7.854, -4.712, -1.609, 3.076, 3.199.
b) Limit as x approaches positive infinity = 1, negative
infinity = 0. Simpler limit functions:
𝑓−(𝑥) = cos(𝑥) , 𝑎𝑛𝑑 𝑓+(𝑥) = cos(𝑥) + 1
all roots of f-(x) = cos (x) is 𝝅
𝟐
+ 𝝅𝒏, where n is any integer.
All roots of f+(x) = cos (x) + 1 are 𝝅 + 𝟐𝝅𝒏
c) first: 𝑥
∗ = −1.60928, 𝑎𝑛𝑑 𝑥− = −1.57080
second: 𝑥
∗ = −4.71231, 𝑎𝑛𝑑 𝑥− = −4.71239
third: 𝑥
∗ = −7.85398, 𝑎𝑛𝑑 𝑥− = −7.85398
𝑓𝑖𝑟𝑠𝑡 𝑒𝑟𝑟𝑜𝑟 = |−1.60928 − (−1.57080)| = 0.03848
There are around 1-2 significant digits which means the
error is high and the first 𝑥− does not approximate 𝑥
∗
well. However, the accuracy increases in the next two
approximations and are around 5-6 digits.
d) first: 𝑥
∗ = 3.07643, 𝑎𝑛𝑑 𝑥+ = 3.14159
second: 𝑥
∗ = 3.19924, 𝑎𝑛𝑑 𝑥+ = 9.42478
third: 𝑥
∗ = 10.0000, 𝑎𝑛𝑑 𝑥+ = 15.7080
fourth: 𝑥
∗ = 16.0000, 𝑎𝑛𝑑 𝑥+ = 21.9911
𝑓𝑖𝑟𝑠𝑡 𝑒𝑟𝑟𝑜𝑟 = |3.07643 − 3.14159| = 0.06516
I chose the initial brackets by looking at the graph and
estimating. The first estimate once again isn’t very
accurate with only 1-2 accurate significant digits.
However since f(x) has an extra root between x = [3,4],
the roots no longer line up as x increases. After the first two positive roots in f(x), however, the function gets close to y =
0 but never crosses meaning there are no more roots. The function f+(x) continues to have roots, however, meaning f+(x)
does not accurately approximate f(x).
e) The root found using the fixed-point method is 𝑥
∗ = 3.07642 so therefore, finding
a fixed point of g(x) is the same as finding a root of f(x) = 0. The error is 0.00001
meaning 5 accurate significant figures. However, this fixed-point is calculated from
both an initial guess of x = -1.5 and x = 3.0 which is unexpected. This is because y = x
and x = g(x) only intersects at x = 3.07642 meaning it will only converge at x =
3.07642. The real problem with using our specific g(x) to find all roots is that it is not
unique and only has 1 intersection with y = x. No matter our initial guess, the fixed-point method always converges to
the single fixed point and can’t find all roots of f(x).