Question 1.3

sin((exp(1)^(-10))^(.5))

## [1] 0.006737896

G <- function(N){
N <- sample(1:1000000, 1)
print((N*(N+1)/2)^2)
}

log(exp(3))

## [1] 3

log(exp(7))

## [1] 7

log(exp(300))

## [1] 300

#After various calculations, this resulted to be the correct answer. It makes #sense, as the logarithm of e^x is always x.

#Question 2.1

n <- 100 
x <- seq(1, n)
sum(x)

## [1] 5050

#The result is 5050. In 1.1, the reuslt was also 5050. This is the equivalent of the sum of the first 100 integers.

#Question 2.2

#The “sum” command is advantegeous because it sums values from a sequence. Our approach in Question 1 was just to assign a value to N. Instead, for this question, we created a list from 1 to N and made R add it up. This can be advntageous when you don’t know the extreme values of a sequence. In that case, the “sum” command can add up all the values present in that sequence without you knowing how many there actually are.

#Question 2.3

x <- seq(from = 0, to = 1, by = 0.1)

#Mean:

(sum(x))/(11)

## [1] 0.5

#Standard Deviation:

((sum((x-0.5)^2)/11)^(.5))

## [1] 0.3162278

#Question 2.4

mean(runif(n=10,min=0,max=1))

## [1] 0.4918936

sd(runif(n=10,min=0,max=1))

## [1] 0.3690243

mean(runif(n=100,min=0,max=1))

## [1] 0.498278

sd(runif(n=100,min=0,max=1))

## [1] 0.2800847

mean(runif(n=1000,min=0,max=1))

## [1] 0.4911496

sd(runif(n=1000,min=0,max=1))

## [1] 0.2906211

mean(runif(n=10000,min=0,max=1))

## [1] 0.4993677

sd(runif(n=10000,min=0,max=1))

## [1] 0.2892717

#Question 3.1

MAT1 = matrix(0, nrow=10, ncol=15)
n <- 1
MAT2= matrix(n, nrow=10, ncol=15)
rnorm(n, mean=2, sd=5)

## [1] 8.361837

#Question 3.2

c ("London", "New York", "Paris", "Shanghai", "New Delhi", "Johannesburg")

## [1] "London"       "New York"     "Paris"        "Shanghai"     "New Delhi"   
## [6] "Johannesburg"

CityNames <- c ("London", "New York", "Paris", "Shanghai", "New Delhi", "Johannesburg")



class(CityNames)

## [1] "character"

#CityNames is in the class of characters. 


length(CityNames)

## [1] 6

#The length is 6



c(21, 30, 26, 34, 38, 29)

## [1] 21 30 26 34 38 29

Temperature <- c(21, 30, 26, 34, 38, 29)




CityTemps1 <- rbind (CityNames, Temperature)

CityTemps2 <- cbind (CityNames, Temperature)



class(CityTemps1)

## [1] "matrix"

class(CityTemps2)

## [1] "matrix"

#The class of each CityTemps object is matrix. 



nrow(CityTemps1)

## [1] 2

ncol(CityTemps1)

## [1] 6

#2 rows 6 columns

nrow(CityTemps2)

## [1] 6

ncol(CityTemps2)

## [1] 2

#6 rows 2 columns




#coerce to dataframe

dataframe1 <- as.data.frame(CityTemps1)
dataframe2 <- as.data.frame(CityTemps2)

class(dataframe1)

## [1] "data.frame"

class(dataframe2)

## [1] "data.frame"

#The CityTemps objects are now dataframes, after double-checking with the class command. 

#At last, the structure

str(dataframe1)

## 'data.frame':    2 obs. of  6 variables:
##  $ V1: Factor w/ 2 levels "21","London": 2 1
##   ..- attr(*, "names")= chr  "CityNames" "Temperature"
##  $ V2: Factor w/ 2 levels "30","New York": 2 1
##   ..- attr(*, "names")= chr  "CityNames" "Temperature"
##  $ V3: Factor w/ 2 levels "26","Paris": 2 1
##   ..- attr(*, "names")= chr  "CityNames" "Temperature"
##  $ V4: Factor w/ 2 levels "34","Shanghai": 2 1
##   ..- attr(*, "names")= chr  "CityNames" "Temperature"
##  $ V5: Factor w/ 2 levels "38","New Delhi": 2 1
##   ..- attr(*, "names")= chr  "CityNames" "Temperature"
##  $ V6: Factor w/ 2 levels "29","Johannesburg": 2 1
##   ..- attr(*, "names")= chr  "CityNames" "Temperature"

str(dataframe2)

## 'data.frame':    6 obs. of  2 variables:
##  $ CityNames  : Factor w/ 6 levels "Johannesburg",..: 2 4 5 6 3 1
##  $ Temperature: Factor w/ 6 levels "21","26","29",..: 1 4 2 5 6 3

Question 1.1

Question 1.2

Question 1.3