Page 76 - Applied Numerical Methods Using MATLAB

PROBLEMS 65
evaluate the following two formulas (for the roots of a quadratic
equation) that are mathematically equivalent and plot the values of the
second root of each pair. Noting that the true values are not available
and so the shape of solution graph is only one practical basis on which
we can assess the quality of numerical solutions, tell which is better in
terms of resisting the loss of significance.
1 √

(i) x 1 ,x 2 = (−b ∓ sign(b) b − 4ac) (P1.20.2a)
2
2a
1 √ c/a

2
(ii) x 1 = (−b − sign(b) b − 4ac), x 2 = (P1.20.2b)
2a x 1
16
14
(c) For 100 values of x over the interval [10 ,10 ], evaluate the follow-
ing two expressions that are mathematically equivalent, plot them, and
based on the graphs, tell which is better in terms of resisting the loss
of significance.
√
2
(i) y = 2x + 1 − 1 (P1.20.3a)
2x 2
(ii) y = √ (P1.20.3b)
2
2x + 1 + 1
−9
(d) For 100 values of x over the interval [10 ,10 −7.4 ], evaluate the fol-
lowing two expressions that are mathematically equivalent, plot them,
and based on the graphs, tell which is better in terms of resisting the
loss of significance.
√ √
(i) y = x + 4 − x + 3 (P1.20.4a)
1
(ii) y = √ √ (P1.20.4b)
x + 4 + x + 3

(e) On purpose to find the value of (300 125 /125!)e −300 , type the following
statement into the MATLAB command window.

>>300^125/prod([1:125])*exp(-300)
What is the result? Is it of any help to change the order of multipli-
cation/division? As an alternative, make a routine which evaluates the
expression

λ k −λ
p(k) = e for λ = 300 and an integer k (P1.20.5)
k!
in a recursive way, say, like p(k + 1) = p(k) ∗ λ/k and then, use the
routine to find the value of (300 125 /125!)e −300 .