Description
1. Write two R functions which will evaluate polynomials of the form
P(x) = c1 + c2x + · · · + cn−1x
n−2 + cnx
n−1
.
Your functions should take a vector x and the vector (called it coef) of polynomial coefficients
as arguments and they should return the values of the evaluated polynomial, i.e., if x =
(x1, . . . , xn), the return value is a vector of (P(x1), . . . , P(xn)).
(a) The first function is called directpoly. You can use for loop to loop the sum in the
polynomial formula. But you cannot use for loop to calculate for different values of x,
i.e., the computation for x must be vectorized. Do one error checking on coef input and
return a proper error message if its length is less than 2. Test your function with x = 1 : 3
and coef=c(2,17,-3), i.e., c1 = 2, c2 = 17, c3 = −3.
(b) For moderate to large values of n, evaluate of a polynomial at x can be done more efficiently
by using Horner’s rule. This algorithm is:
Step 1: set output=cn;
Step 2: for i = n − 1, . . . , 1, set output=output*x + ci
;
Step 3: return output.
The second function is called hornerpoly by using Horner’s algorithm. Again you cannot
use for loop to calculate for different values of x, i.e., the computation for x must be
vectorized. Do one error checking on coef input and return a proper error message if its
length is less than 2. Test your function with x = 1 : 3 and coef=c(2,17,-3).
(c) Do some timing comparisons of the functions in (a) and (b). Try the following code
system.time(dirpoly(x=seq(-10,10, length=5000000), c(1,-2,2,3,4,6,7,8)))
system.time(hornerpoly(x=seq(-10,10, length=5000000), c(1,-2,2,3,4,6,7,8)))
Run a few times. How faster is the Horner’s algorithm? Comment your findings.
2. Write an R function called my.unif to generate a sequence of uniform pseudo-random numbers.
The argument list should be n—sample size, a—multiplier, c—increment with default value 0,
m—modulus, and x0—seed. The function should use
xi = (axi−1 + c) (mod m), i = 1, . . . , n
to generate a vector of x = (x1, . . . , xn) and return value as x/m. Use your function to
generate 50 uniform pseudo-random numbers with a = 172, c = 13, m = 30307 and initial
seed x0 = 17218. Construct a histogram to see if its distribution is like uniform[0,1]. Also
generate 50 uniform pseudo-random numbers with a = 171, c = 51, m = 32767 and initial seed
x0 = 2020. Construct a histogram to see if its distribution is like uniform[0,1].
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3. Generate 1000 uniform pseudo-random variates using using the runif function, assigning them
to a vector called U. Use set.seed(2020).
(a) Compute the average, variance, and standard deviation of the numbers in U.
(b) Compare your results with true mean, variance, and standard deviation.
(c) Compute the proportion of the values of U that are less than 0.6, and compare with the
probability that a uniform random variable U is less than 0.6.
(d) Construct a histogram of the values of U and comment your findings.
4. Construct two R function for generating random numbers from the following density.
(a) Use the inverse method to implement an R function with inputs n and a = 1. The density
function is
f(x) = x
a
2
e
−x
2/(2a
2
)
, a > 0, x ≥ 0.
Test your function with n = 20 and plot a histogram with n = 1000 and the option
probability=TRUE. Then add the density curve f(x) with color=2 to the plot. Comment
your findings. No looping is allowed.
(b) Use the reject method to implement an R function with inputs n, M, and a = 1, with the
same density given in (a). Notice that the range is fixed from 0 to 5 and M is needed to be
the max value of f(x). Test your function with n = 20 and plot a histogram with n = 1000
and the option probability=TRUE. Then add the density curve f(x) with color=2 to the
plot. Comment your findings. No looping is allowed.
(c) Use system.time function to record the time of generating a vector of 100,000 random
numbers for each function implemented in (a) and (b) respectively. Comment your findings.
5. Let U be a uniform[0,1] random valuable. Use runif function to simulate 10000 values of
uniform[0,1] for U, called u.
(a) Estimate E[U
2
] and construct 95% confidence interval. Compare with the true value and
comment the accuracy.
(b) Since E[U
2
] = E[U
2 + (1 − U)
2
]/2, this suggests that we can use v = (u
2 + (1 − u)
2
)/2 to
estimate E(V ) where V = (U
2+(1−U)
2
)/2. Construct 95% confidence interval. Compare
with the result in (a) and comment your findings.
(c) Similarly, use the fact E[U
2
] = E[(U/2)2 + (1 − U/2)2
]/2 to construct another estimator
and a 95% confidence interval. Compare with the results in (a) and (b) and comment
your findings.
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