Winter 2016 Solutions for Root Finding Practice Problems CS371

Winter 2016
1.
Solutions for Root Finding Practice Problems
CS371/AM242
[16 marks]
a) [2 marks] 𝑓(𝑥) = (𝑥 + √𝑥) ∗ (20 − 𝑥 + √20 − 𝑥) − 155.55
b) [8 marks] Note, with f(x) as above, we have that
1
1
𝑓 ′ (𝑥) = (1 +
) (20 − 𝑥 + √20 − 𝑥) − (𝑥 + √𝑥) (1 +
)
2√𝑥
2√20 − 𝑥
% fn.m
function [ val ] = fn( x )
% Calculates (x+sqrt(x))*(20-x + sqrt(20-x)) - 155.55
y = 20-x;
val = (x+sqrt(x)) * (y + sqrt(y)) - 155.55;
end
% fnprime.m
function [ val ] = fnprime( x )
% Calculates the derivative of fn(x)
y = 20-x;
val1 = (1+1/(2*sqrt(x)))* (y + sqrt(y));
val2 = (x+sqrt(x))*(1 + 1/(2*sqrt(y)));
val = val1 - val2;
end
% a5q1b_soln.m
% choose initial guesses (x0=1, x0=9. x0=another value (5 here)
initial = [1,9,5];
% for each initial value, perform steps of Newton's method until the
% value of fn at x is <= 1.0e-6
for i = 1:3
x = initial(i);
fx = fn(x);
while (abs(fx) > 1.0e-6)
% display the value of x (and its function value)
info = ['x= ' num2str(x) ' f(x)= ' num2str(fx)]; disp(info)
% calculate the next iterate
fx = fn(x);
fprime = fnprime(x);
x = x - fx/fprime;
end
% display the final iterate
info = ['x= ' num2str(x) ' f(x)= ' num2str(fx)]; disp(info)
disp(blanks(1))
end
1
Winter 2016
Solutions for Root Finding Practice Problems
CS371/AM242
My results:
>> a1q5b_soln
x= 1 f(x)= -108.8322
x= 4.3172 f(x)= -108.8322
x= 6.0608 f(x)= -29.9344
x= 6.486 f(x)= -4.9311
x= 6.5127 f(x)= -0.27575
x= 6.5128 f(x)= -0.0010825
x= 6.5128 f(x)= -1.694e-08
x=
x=
x=
x=
x=
x=
x=
x=
9 f(x)= 16.2495
3.3844 f(x)= 16.2495
5.7048 f(x)= -47.4534
6.4338 f(x)= -9.2546
6.5119 f(x)= -0.81793
6.5128 f(x)= -0.0092484
6.5128 f(x)= -1.2358e-06
6.5128 f(x)= -5.6843e-14
x=
x=
x=
x=
x=
x=
5 f(x)= -18.9838
6.2721 f(x)= -18.9838
6.5048 f(x)= -2.5485
6.5128 f(x)= -0.082198
6.5128 f(x)= -9.7225e-05
6.5128 f(x)= -1.3665e-10
c) [5 marks] Bisection method – note that f(x) is continuous over the
i. [1 mark] f(x) is continuous over the interval [1,9], and f(1) = -108.83 < 0 and f(9) =
16.2495 > 0. By the Intermediate Value Theorem, there exists a root of f(x) in the
interval [1,9].
ii. [4 marks] Each iteration of the bisection method halves the length of the interval, so
after k iterations 1
1 𝑘
1 𝑘
1 𝑘−3
|𝑎𝑘 − 𝑏𝑘 | = |𝑎𝑘−1 − 𝑏𝑘−1 | = ⋯ = ( ) |𝑎0 − 𝑏0 | = ( ) 8 = ( )
2
2
2
2
So, we are looking for k such that
1 𝑘−3
( )
≤ 10−6 → 2𝑘−3 ≥ 106 → log 2 2𝑘−3 ≥ log 2 106 → 𝑘 − 3 ≥ 19.93 …
2
That is, we need k ≥22.93… or about 23 iterations.
2
Winter 2016
Solutions for Root Finding Practice Problems
CS371/AM242
2. [16 marks] Fixed point techniques with f(x) = x4 + 2x2 – x - 3
a) [2 marks] x is a fixed point of g(x) if and only if
(𝑥 + 3 − 𝑥 4 )
(𝑥 + 3 − 𝑥 4 )
𝑥= √
↔ 𝑥2 =
↔ 2𝑥 2 = 𝑥 + 3 − 𝑥 4 ↔ 𝑥 4 + 2𝑥 2 − 𝑥 − 3 = 0
2
2
i.e. if and only if x is a root of f(x).
b) [6 marks] My Matlab code to produce the first 20 iterations of the fixed point method with
g:
% g.m
function [ y ] = g( x )
%Produces sqrt((x+3-x^4)/2)
y = sqrt( (x+3-x^4) / 2 );
end
% a1q6b_soln.m
x = 1;
for i = 1:20
disp(num2str(x))
x = g(x);
end
disp(num2str(x))
My results for those 20 iterations:
>> a1q6b_soln
1
1.2247
0.99367
Note that the iterates are oscillating back and forth between a value
1.2286
around between 1.22 and 1.25 and a second value between 0.99 and .95.
0.98751
1.2322
Convergence does not appear to be happening (as is possible with fixed
0.98159
point functions if the derivative of g is not well-behaved in the region of a
1.2356
0.97596
root. In this case, g' is undefined when x+3-x4 = 0.
1.2387
0.97068
1.2415
0.96579
1.2441
0.9613
1.2465
0.95723
1.2485
0.95357
1.2504
0.95032
3
Winter 2016
c)
Solutions for Root Finding Practice Problems
CS371/AM242
[2 marks] x is a fixed point of h(x) if and only if:
3𝑥 4 + 2𝑥 2 + 3
= 𝑥 ↔ 3𝑥 4 + 2𝑥 2 + 3 = 4𝑥 4 + 4𝑥 2 − 𝑥 ↔ 𝑥 4 + 2𝑥 2 − 𝑥 − 3 = 0
4𝑥 3 + 4𝑥 − 1
i.e. if and only if f(x) = 0.
d) [6 marks] Matlab code is virtually identical to that in part b, with just h replacing g.
% h.m
function [ y ] = h( x )
%Produces (3x^4+2x^2+3)/(4x^3+4x-1)
numer = 3*x^4 + 2*x^2 + 3;
denom = 4*x^3 + 4*x - 1;
y = numer/denom;
end
% a1q6b_soln.m
x = 1;
for i = 1:20
disp(num2str(x))
x = h(x);
end
disp(num2str(x))
Which leads to the following results:
>> a1q6d_soln
1
1.1429
This fixed point algorithm converges *very* quickly to a fixed point of h, i.e. a root of f.
1.1245
1.1241
A little background: In the area of the root of f nearby x0=1 (as identified here, around
1.1241
1.1241), g'(x) < -2 (i.e. absolute value > 1) but abs(h(x)) < 1. So, these results support
1.1241
the convergence theory we developed for the fixed point algorithms.
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
1.1241
4
Winter 2016
Solutions for Root Finding Practice Problems
1
1
2
𝑥𝑘
3. [10 marks] The iteration 𝑥𝑘+1 = 𝑥𝑘 +
1
1
CS371/AM242
, is used to find a fixed point of the function
𝑔(𝑥) = 2 𝑥 + 𝑥 on the interval [√2, ∞).
a) [5 marks] The function x is continuous for over (0, ∞) - the only discontinuity is when
x=0, which is not the range we are considering. If x >0 then g(x) >0, so the value of g
1
1
is in the same range. We see that √2 is a fixed point of g since 𝑔(√2) = 2 √2 + =
√2
√2
2
+
√2
2
= √2.
1
1
The derivative of g(x) is 𝑔′(𝑥) = 2 − 𝑥 2 . For any real nonzero x, g'(x)< ½ . Also, for
1
1
x≥√2, 𝑥 2 ≤ 2 so g'(x) ≥ 0. In particular, we see that |g'(x)|<1 on the interval [√2, ∞).
So, by the convergence theorem for fixed point algorithms, the given iteration will
converge to x* = √2.
b) [5 marks] From (a), we know that lim𝑘 𝑥𝑘 = √2. Now, we just want to know the rate of
convergence of the iteration. Fixed point algorithms are known to converge at least
linearly (when the converge), so it is at least linear. However, we can show that the
convergence in this case is quadratic.
Consider the ratio of errors of consecutive iterates:
𝑒𝑘+1
𝑒𝑘2
𝑥𝑘2 + 2 − 2√2𝑥
1
1
𝑥
+
−
√2
𝑘
𝑥𝑘+1 − 𝑥
𝑥𝑘+1 − √2
1
2
𝑥𝑘
2𝑥𝑘
=
=
= 2
= 2
=
2
∗
2
(𝑥𝑘 − 𝑥 )
𝑥𝑘 − 2√2𝑥 + 2 𝑥𝑘 − 2√2𝑥 + 2 2𝑥𝑘
(𝑥𝑘 − √2)
∗
So,
|𝑒𝑘+1 |
1
1
= lim
=
2
𝑘→∞ |𝑒𝑘 |
𝑘→∞ 2𝑥𝑘
2√2
lim
1
Therefore, convergence is quadratic (q=2) and the asymptotic error constant λ=2 2.
√
5

Download Report