Problem 1:
Write a function that computes the area of a circle.
Calculate_Area <- function(r) {
# This function calculates the area of a circle
area <- pi * r^2
return(area)
}
radii <- c(5, 15, 200)
Calculate_Area(radii)[1] 78.53982 706.85835 125663.70614Problem 2:
Write a function that computes the average absolute value. \(\frac{\lvert x \rvert + \lvert y \rvert}{2}\).
Average_Absolutely <- function(x, y) {
# This function calculates the average absolute value.
# (|x| + |y|) / 2
absolute_average <- (abs(x) + abs(y)) / 2
return(absolute_average)
}
example_x <- c(-3:2)
example_y <- c(-3:2)
Average_Absolutely(example_x, example_y)[1] 3 2 1 0 1 2Problem 3:
Write a function to count the number of odd integers in a given vector x. Test your function of the vectors: c(1, 3, 5, 6, 8, 9), seq(1, 99, 3) and seq(1, 5000, 7).
# Returns the amount of odd integers in a given vector
Count_Odds <- function(x) {
odds <- length(which(x %% 2 > 0))
return(odds)
}
ex_1 <- c(1, 3, 5, 6, 8, 9)
ex_2 <- seq(1, 99, 3)
ex_3 <- seq(1, 5000, 7)
cat("The amount of odds in the first vector are", Count_Odds(ex_1), "\n")The amount of odds in the first vector are 4 cat("The amount of odds in the second vector are", Count_Odds(ex_2), "\n")The amount of odds in the second vector are 17 cat("The amount of odds in the third vector are", Count_Odds(ex_3), "\n")The amount of odds in the third vector are 358 Problem 4:
Use ifelse or switch to return “even” or “odd” depending whether or not each element of the vector x is divisible by 2.
# This function finds the remainder of various integers and assigns them even or odd depending on said remainder.
Even_or_Odd <- function(x) {
y <- x%%2
if (y == 0){
return("even")}
else{
return("odd")}
}
x_1 <- c(1, 3, 5, 6, 8, 9)
x_2 <- 1:20
sapply(x_1, Even_or_Odd)[1] "odd" "odd" "odd" "even" "even" "odd" sapply(x_2, Even_or_Odd) [1] "odd" "even" "odd" "even" "odd" "even" "odd" "even" "odd" "even"
[11] "odd" "even" "odd" "even" "odd" "even" "odd" "even" "odd" "even"Problem 5:
We wish to compute the product of 3, 4 or 5 digits, and then add 42. Write a function to complete this task. Test your function on the vectors: c(3,5,7), c(3,5,7,9) and c(5,6,7,8,9).
sqornshellous_zeta <- function(x) {
# This function does what it says on the tin. That is, to say, takes the product of a vector and adds the answer to life, the universe, and everything.
# This function only works for vectors made up of integers, floats and logicals.
product_value <- prod(x)
final_product_value <- product_value + 42
return(final_product_value)
}
ex_51 <- c(3,5,7)
ex_52 <- c(3,5,7,9)
ex_53 <- c(5,6,7,8,9)
sqornshellous_zeta(ex_51)[1] 147sqornshellous_zeta(ex_52)[1] 987sqornshellous_zeta(ex_53)[1] 15162Problem 6:
The sum formula for the finite geometric series \(h(x,n) = \sum_{i=0}^{n} x^{i} = 1 + x + x^2 + ... + x^n, x \neq 1\) is given by:
\[ h(x,n) = \begin{cases} \frac{1 - x^{n+1}}{1-x} & \text{for $x \neq 1$} \\ n & \text{for $x = 1$} \end{cases} \]
Write a function to compute this formula. Use your function to compute: h(0.7, 57), h(4.2, 9), h(-1.2, 81) and h(1, 342).
piecewise_example <- function(x,n) {
if (x == 1){return(n)}
else{return({(1 - x^(n+1)) / (1-x)})}
}
piecewise_example(0.7, 57)[1] 3.333333piecewise_example(4.2, 9)[1] 533755.9piecewise_example(-1.2, 81)[1] -1413967piecewise_example(1, 342)[1] 342Problem 7.
Write a function that computes \(f(n) = n!\) using the formula \(n! n(n-1,)(n-2) ... 2 \cdot 1\) and does not use factorial(). For a given integer \(n\), the function prints out:
- n! n> 0
- 1 for n=0
- NaN for n<0 Test on n=5, 15, 0, -8
better_factorial <- function(n){
# Here I have if statements that return 1 if n=0 and a NaN if n<0. I put these at the start to catch them as soon as possible
if (n == 0){return(1)}
if (n < 0){return(NaN)}
return_value <- 1
# This while loop takes a value which is set to 1 and multiplies it by a variable "n" that gets progressively smaller by 1 with every loop. This loop stops once n is no longer greater than 1.
while (n > 1) {
return_value <- return_value * n
n <- n-1
}
return(return_value)
}
cat("5! is", better_factorial(5), "\n")5! is 120 cat("15! is", better_factorial(15), "\n")15! is 1.307674e+12 cat("0! is", better_factorial(0), "\n")0! is 1 cat("-8! is", better_factorial(-8), "\n")-8! is NaN Problem 8
Write a function that prints out the 1st n Lucas numbers for a given n. Test your function for n = 10, 25.
lucas_func <- function(n){
#First I need to make sure no negative values are plugged in here.
if (n < 1){
return(NaN)
}
# Then we just need to setup the first two known values of this sequence. That is, 2 and 1.
if (n == 1){
return(2)
}
if (n == 2){
return(1)
}
#Essentially my game plan is to have a and b add with each other to create a variable "c". After doing that we reassign some values. The key here is the counter. It goes up by 1 at the end of the loop and when it is no longer less than n the loop stops.
a <- 2
b <- 1
counter <- 2
results_list <- c(2, 1) # We create a small vector here with our two given values.
while (counter < n) {
c <- a + b
a <- b
b <- c
counter <- counter + 1
results_list <- append(results_list, c) #This makes that small little vector useful!
}
return(results_list)
}
test_val <- c(10, 25)
sapply(test_val, lucas_func)[[1]]
[1] 2 1 3 4 7 11 18 29 47 76
[[2]]
[1] 2 1 3 4 7 11 18 29 47 76
[11] 123 199 322 521 843 1364 2207 3571 5778 9349
[21] 15127 24476 39603 64079 103682Problem 9
tribo_func <- function(n){
#Essentially my gameplan here is the exact same as problem 8, I just add in another variable.
if (n < 1){
return(NaN)
}
if (n == 1){
return(0)
}
if (n == 2){
return(0)
}
if (n == 3){
return(1)
}
a <- 0
b <- 0
c <- 1
counter <- 3
results_list <- c(0, 0, 1)
while (counter < n) {
d <- a + b + c
a <- b
b <- c
c <- d
counter <- counter + 1
results_list <- append(results_list, d)
}
return(results_list)
}
test_val2 <- c(10, 25)
sapply(test_val2, tribo_func)[[1]]
[1] 0 0 1 1 2 4 7 13 24 44
[[2]]
[1] 0 0 1 1 2 4 7 13 24 44
[11] 81 149 274 504 927 1705 3136 5768 10609 19513
[21] 35890 66012 121415 223317 410744Problem 10
Using the code for the Fibonacci sequence given in class, generate a vector containing every 3rd term of the 1st 30 terms. Then verify using R that is vector contains only even numbers. Theorem: Every third Fibonacci number is even.
Fibonacci <- function(n){
fib <- numeric(n)
# Create a vector of length n
fib[1] <- fib[2] <- 1
# Store F 1 and F 2 in the vector
for (i in 3:n) fib[i] = fib[i-1] + fib[i-2]
cat("The first", n, "Fibonacci numbers are", "\n")
print(fib)}
# Here I set fib_30 to the vector created when 30 is put into the fibonacci function.
fib_30 <- Fibonacci(30)The first 30 Fibonacci numbers are
[1] 1 1 2 3 5 8 13 21 34 55
[11] 89 144 233 377 610 987 1597 2584 4181 6765
[21] 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040# From here I filter it out using a sequence. The sequence starts at the third position and moves up by 3.
fib_thirds <- fib_30[seq(3, length(fib_30), 3)]
cat("Every third value in the Fibonacci sequence up to the thirtieth are", fib_thirds, "\n")Every third value in the Fibonacci sequence up to the thirtieth are 2 8 34 144 610 2584 10946 46368 196418 832040 # To check if all the values here are even we can actually use our old Even_or_Odd function!
sapply(fib_thirds, Even_or_Odd) [1] "even" "even" "even" "even" "even" "even" "even" "even" "even" "even"Problem 11
Using the code for the Fibonacci sequence given in class, generate a vector containing every 4th term of the 1st 30 terms. Then verify using R that every number in this vector is divisible by three. Theorem: Every fourth Fibonacci number is divisible by three.
Fibonacci <- function(n){
fib <- numeric(n)
# Create a vector of length n
fib[1] <- fib[2] <- 1
# Store F 1 and F 2 in the vector
for (i in 3:n) fib[i] = fib[i-1] + fib[i-2]
cat("The first", n, "Fibonacci numbers are", "\n")
print(fib)}
# Here I set fib_30 to the vector created when 30 is put into the fibonacci function.
fib_30 <- Fibonacci(30)The first 30 Fibonacci numbers are
[1] 1 1 2 3 5 8 13 21 34 55
[11] 89 144 233 377 610 987 1597 2584 4181 6765
[21] 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040# From here I filter it out using a sequence. The sequence starts at the fourth position and moves up by 4.
fib_fourth <- fib_30[seq(4, length(fib_30), 4)]
cat("Every fourth value in the Fibonacci sequence up to the thirtieth are", fib_fourth, "\n")Every fourth value in the Fibonacci sequence up to the thirtieth are 3 21 144 987 6765 46368 317811 # To see if all of these are divisible by three we just slightly tweak our even or odd function!
divisible_by_three <- function(x) {
y <- x%%3
if (y == 0){
return("TRUE")}
else{
return("FALSE")}
}
sapply(fib_fourth, divisible_by_three)[1] "TRUE" "TRUE" "TRUE" "TRUE" "TRUE" "TRUE" "TRUE"Problem 12
Generate a sequence of ratios with the Fibonacci numbers. List the 1st 30 terms of this sequence and verify the sequence appears to converge to 1.618034.
fibonacci_ratio <- function(n){
fib_vector <- Fibonacci(n+1) #I add the +1 here to prevent things from getting wacky when we reach the end of the final loop.
ratio_vector <- c()
for (i in 0:n) ratio_vector[i] = fib_vector[i+1] / fib_vector[i] #The n on this line could be replaced with n-1 if you remove the +1 from the n a few lines up.
return(ratio_vector)
}
fibonacci_ratio(30)The first 31 Fibonacci numbers are
[1] 1 1 2 3 5 8 13 21 34
[10] 55 89 144 233 377 610 987 1597 2584
[19] 4181 6765 10946 17711 28657 46368 75025 121393 196418
[28] 317811 514229 832040 1346269 [1] 1.000000 2.000000 1.500000 1.666667 1.600000 1.625000 1.615385 1.619048
[9] 1.617647 1.618182 1.617978 1.618056 1.618026 1.618037 1.618033 1.618034
[17] 1.618034 1.618034 1.618034 1.618034 1.618034 1.618034 1.618034 1.618034
[25] 1.618034 1.618034 1.618034 1.618034 1.618034 1.618034