Solutions to exercises

1. Intro to R

Exercise

Create your own code block below and run a math operation.

pi * 2
[1] 6.283185

Exercise

Examine the help file of the log() function. How can we compute the the base-10 logarithm of my_object? Your code:

Code 
# setup: these steps were executed before the exercise
my_object <- 10
  1. Examine the log() function.
?log
  1. Compute the base-10 logarithm of my_object.
log(my_object, base = 10)
[1] 1
# alternative:
log10(my_object)
[1] 1

Exercise

Obtain the maximum value of water content per 100g in the data. Your code:

Code 
# setup: these steps were executed before the exercise
my_character_vector <- c("Apple", "Orange", "Watermelon", "Banana")
my_data_frame <- data.frame(fruit = my_character_vector,
                            calories_per_100g = c(52, 47, 30, 89),
                            water_per_100g = c(85.6, 86.8, 91.4, 74.9))
my_data_frame
max(my_data_frame$water_per_100g)
[1] 91.4

2. Tidy data analysis I

Code 
# setup: these steps were executed before the exercises
library(tidyverse)
trump_scores <- read_csv("data/trump_scores_538.csv")

Exercise

Select the variables last_name, party, num_votes, and agree from the data frame. Your code:

trump_scores |> 
  select(last_name, party, num_votes, agree)
# A tibble: 122 × 4
   last_name  party num_votes agree
   <chr>      <chr>     <dbl> <dbl>
 1 Alexander  R           118 0.890
 2 Blunt      R           128 0.906
 3 Brown      D           128 0.258
 4 Burr       R           121 0.893
 5 Baldwin    D           128 0.227
 6 Boozman    R           129 0.915
 7 Blackburn  R           131 0.885
 8 Barrasso   R           129 0.891
 9 Bennet     D           121 0.273
10 Blumenthal D           128 0.203
# ℹ 112 more rows
# alternative
trump_scores |> 
  select(last_name, party:agree)
# A tibble: 122 × 4
   last_name  party num_votes agree
   <chr>      <chr>     <dbl> <dbl>
 1 Alexander  R           118 0.890
 2 Blunt      R           128 0.906
 3 Brown      D           128 0.258
 4 Burr       R           121 0.893
 5 Baldwin    D           128 0.227
 6 Boozman    R           129 0.915
 7 Blackburn  R           131 0.885
 8 Barrasso   R           129 0.891
 9 Bennet     D           121 0.273
10 Blumenthal D           128 0.203
# ℹ 112 more rows

Exercise

  1. Add a new column to the data frame, called diff_agree, which subtracts agree and agree_pred. How would you create abs_diff_agree, defined as the absolute value of diff_agree? Your code:

  2. Filter the data frame to only get senators for which we have information on fewer than (or equal to) five votes. Your code:

  3. Filter the data frame to only get Democrats who agreed with Trump in at least 30% of votes. Your code:

  1. Add a new column to the data frame, called diff_agree, which subtracts agree and agree_pred. How would you create abs_diff_agree, defined as the absolute value of diff_agree? Your code:
trump_scores |> 
  mutate(diff_agree = agree - agree_pred) |> 
  select(last_name, matches("agree")) # just for clarity
# A tibble: 122 × 4
   last_name  agree agree_pred diff_agree
   <chr>      <dbl>      <dbl>      <dbl>
 1 Alexander  0.890      0.856    0.0336 
 2 Blunt      0.906      0.787    0.120  
 3 Brown      0.258      0.642   -0.384  
 4 Burr       0.893      0.560    0.333  
 5 Baldwin    0.227      0.510   -0.283  
 6 Boozman    0.915      0.851    0.0634 
 7 Blackburn  0.885      0.889   -0.00308
 8 Barrasso   0.891      0.895   -0.00389
 9 Bennet     0.273      0.417   -0.144  
10 Blumenthal 0.203      0.294   -0.0910 
# ℹ 112 more rows
trump_scores |> 
  mutate(abs_diff_agree = abs(agree - agree_pred)) |> 
  select(last_name, matches("agree")) # just for clarity
# A tibble: 122 × 4
   last_name  agree agree_pred abs_diff_agree
   <chr>      <dbl>      <dbl>          <dbl>
 1 Alexander  0.890      0.856        0.0336 
 2 Blunt      0.906      0.787        0.120  
 3 Brown      0.258      0.642        0.384  
 4 Burr       0.893      0.560        0.333  
 5 Baldwin    0.227      0.510        0.283  
 6 Boozman    0.915      0.851        0.0634 
 7 Blackburn  0.885      0.889        0.00308
 8 Barrasso   0.891      0.895        0.00389
 9 Bennet     0.273      0.417        0.144  
10 Blumenthal 0.203      0.294        0.0910 
# ℹ 112 more rows
  1. Filter the data frame to only get senators for which we have information on fewer than (or equal to) five votes. Your code:
trump_scores |> 
  filter(num_votes <= 5)
# A tibble: 5 × 8
  bioguide last_name    state party num_votes agree agree_pred margin_trump
  <chr>    <chr>        <chr> <chr>     <dbl> <dbl>      <dbl>        <dbl>
1 H000273  Hickenlooper CO    D             2   0       0.0302        -4.91
2 H000601  Hagerty      TN    R             2   0       0.115         26.0 
3 K000377  Kelly        AZ    D             5   0.2     0.262          3.55
4 L000571  Lummis       WY    R             2   0.5     0.225         46.3 
5 T000278  Tuberville   AL    R             2   1       0.123         27.7
  1. Filter the data frame to only get Democrats who agreed with Trump in at least 30% of votes. Your code:
trump_scores |> 
  filter(party == "D" & agree >= 0.3)
# A tibble: 11 × 8
   bioguide last_name state party num_votes agree agree_pred margin_trump
   <chr>    <chr>     <chr> <chr>     <dbl> <dbl>      <dbl>        <dbl>
 1 D000607  Donnelly  IN    D            83 0.542      0.833        19.2 
 2 H001069  Heitkamp  ND    D            84 0.548      0.915        35.7 
 3 J000300  Jones     AL    D            68 0.353      0.845        27.7 
 4 K000383  King      ME    D           129 0.372      0.441        -2.96
 5 M001170  McCaskill MO    D            83 0.458      0.830        18.6 
 6 M001183  Manchin   WV    D           129 0.504      0.893        42.2 
 7 N000032  Nelson    FL    D            83 0.434      0.568         1.20
 8 R000608  Rosen     NV    D           136 0.346      0.604        -2.42
 9 S001191  Sinema    AZ    D           135 0.504      0.398         3.55
10 T000464  Tester    MT    D           129 0.302      0.805        20.4 
11 W000805  Warner    VA    D           129 0.349      0.401        -5.32

Exercise

Arrange the data by diff_pred, the difference between agreement and predicted agreement with Trump. (You should have code on how to create this variable from the last exercise). Your code:

trump_scores |> 
  mutate(diff_agree = agree - agree_pred)  |> 
  arrange(diff_agree)
# A tibble: 122 × 9
   bioguide last_name state party num_votes agree agree_pred margin_trump
   <chr>    <chr>     <chr> <chr>     <dbl> <dbl>      <dbl>        <dbl>
 1 T000464  Tester    MT    D           129 0.302      0.805       20.4  
 2 J000300  Jones     AL    D            68 0.353      0.845       27.7  
 3 M001183  Manchin   WV    D           129 0.504      0.893       42.2  
 4 B000944  Brown     OH    D           128 0.258      0.642        8.13 
 5 M001170  McCaskill MO    D            83 0.458      0.830       18.6  
 6 H001069  Heitkamp  ND    D            84 0.548      0.915       35.7  
 7 D000607  Donnelly  IN    D            83 0.542      0.833       19.2  
 8 B001230  Baldwin   WI    D           128 0.227      0.510        0.764
 9 F000457  Franken   MN    D            55 0.236      0.495       -1.52 
10 R000608  Rosen     NV    D           136 0.346      0.604       -2.42 
# ℹ 112 more rows
# ℹ 1 more variable: diff_agree <dbl>

Exercise

Obtain the maximum absolute difference in agreement with Trump (the abs_diff_agree variable from before) for each party.

trump_scores |> 
  mutate(abs_diff_agree = abs(agree - agree_pred)) |> 
  summarize(max_abs_diff = max(abs_diff_agree),
            .by = party)
# A tibble: 2 × 2
  party max_abs_diff
  <chr>        <dbl>
1 R            0.877
2 D            0.503

Exercise

Draw a column plot with the agreement with Trump of Bernie Sanders and Ted Cruz. What happens if you use last_name as the y aesthetic mapping and agree in the x aesthetic mapping? Your code:

Code 
# setup: this step was executed before the exercise
trump_scores_ss <- trump_scores |> 
  filter(num_votes >= 10)
ggplot(trump_scores_ss |> filter(last_name %in% c("Cruz", "Sanders")),
       aes(y = last_name, x = agree)) +
  geom_col()

# alternative
ggplot(trump_scores_ss |> filter(last_name == "Cruz" | last_name == "Sanders"),
       aes(y = last_name, x = agree)) +
  geom_col()

3. Matrices

Exercise

Get the product of the first three elements of vector \(d\). Write the notation by hand and use R to obtain the number.

\[\overrightarrow d = \begin{bmatrix} 12 & 7 & -2 & 3 & 1 \end{bmatrix}\]

Code 
# setup: these steps were executed before the exercise
vector_d <- c(12, 7, -2, 3, -1)

\[\prod_{i=1}^3 d_i = 12 \cdot 7 \cdot (-2) = -168\]

prod(vector_d[1:3])
[1] -168

Exercise

  1. Calculate \(A + B\) \[A= \begin{bmatrix} 1 & 0 \\ -2 & -1 \end{bmatrix}\]

\[B = \begin{bmatrix} 5 & 1 \\ 2 & -1 \end{bmatrix}\]

  1. Calculate \(A - B\) \[A= \begin{bmatrix} 6 & -2 & 8 & 12 \\ 4 & 42 & 8 & -6 \end{bmatrix}\] \[B = \begin{bmatrix} 18 & 42 & 3 & 7 \\ 0 & -42 & 15 & 4 \end{bmatrix}\]
A1 <- matrix(c(1,-2,0,-1), nrow = 2)
B1 <- matrix(c(5,2,1,-1), nrow = 2)
A1 + B1
     [,1] [,2]
[1,]    6    1
[2,]    0   -2
A2 <- matrix(c(6,4,-2,42,8,8,12,-6), nrow = 2)
B2 <- matrix(c(18,0,42,-42,3,15,7,4), nrow = 2)
A2 - B2
     [,1] [,2] [,3] [,4]
[1,]  -12  -44    5    5
[2,]    4   84   -7  -10

Exercise

Calculate \(2\times A\) and \(-3 \times B\). Again, do one by hand and the other one using R. \[A= \begin{bmatrix} 1 & 4 & 8 \\ 0 & -1 & 3 \end{bmatrix}\] \[ B = \begin{bmatrix} -15 & 1 & 5 \\ 2 & -42 & 0 \\ 7 & 1 & 6 \end{bmatrix}\]

A3 <- matrix(c(1,0,4,-1,8,3), nrow = 2)
2 * A3
     [,1] [,2] [,3]
[1,]    2    8   16
[2,]    0   -2    6
B3 <- matrix(c(-15,2,7,1,-42,1,5,0,6), nrow = 3)
-3 * B3
     [,1] [,2] [,3]
[1,]   45   -3  -15
[2,]   -6  126    0
[3,]  -21   -3  -18

4. Tidy data analysis II

Exercise

  1. Create a dummy variable, d_large_pop, for whether the country-year has a population of more than 1 million. Then compute its mean. Your code:
  2. Which countries are recorded as “Never colonized”? Change their values to other reasonable codings and compute a tabulation with count(). Your code:
Code 
# setup: these steps were executed before the exercise
library(tidyverse)
qog_csv <- read_csv("data/sample_qog_bas_ts_jan23.csv")
qog <- qog_csv
  1. Create the dummy variable d_large_pop.
qog |> 
  mutate(d_large_pop = if_else(wdi_pop >= 1000000, 1, 0)) |> 
  count(d_large_pop) # to check if it went well
# A tibble: 2 × 2
  d_large_pop     n
        <dbl> <int>
1           0   341
2           1   744
  1. Change the coding of “Never colonized” countries to something else, and compute a tabulation with count().
qog |> 
  filter(ht_colonial == "Never colonized") |> 
  count(cname)
# A tibble: 2 × 2
  cname             n
  <chr>         <int>
1 Canada           31
2 United States    31
qog |> 
  mutate(ht_colonial_recoded = case_when(
    cname == "Canada" ~ "French/British",
    cname == "United States" ~ "British",
    .default = ht_colonial
  )) |> 
  count(ht_colonial_recoded)
# A tibble: 6 × 2
  ht_colonial_recoded     n
  <chr>               <int>
1 British               403
2 Dutch                  31
3 French                 31
4 French/British         31
5 Portuguese             31
6 Spanish               558

Exercise

Calculate the median value of the corruption variable for each region (i.e., perform a grouped summary). Your code:

qog |> 
  summarize(med_corr = median(vdem_corr, na.rm = T), .by = region)
# A tibble: 4 × 2
  region           med_corr
  <chr>               <dbl>
1 Caribbean          0.301 
2 South America      0.531 
3 Central America    0.734 
4 Northern America   0.0505

Exercise

Convert back gdp_long to a wide format using pivot_wider(). Check out the help file using ?pivot_wider. Your code:

Code 
# setup: these steps were executed before the exercise
library(readxl)
gdp <- read_excel("data/wdi_gdp_ppp.xlsx")
gdp_long <- gdp |> 
  pivot_longer(cols = -c(country_name, country_code), 
               names_to = "year", 
               values_to = "wdi_gdp_ppp", 
               names_transform = as.integer) 
gdp_long |> 
  pivot_wider(id_cols = c(country_name, country_code), # can omit in this case too
              values_from = wdi_gdp_ppp, 
              names_from = year) 
# A tibble: 266 × 35
   country_name        country_code   `1990`   `1991`   `1992`   `1993`   `1994`
   <chr>               <chr>           <dbl>    <dbl>    <dbl>    <dbl>    <dbl>
 1 Aruba               ABW           2.03e 9  2.19e 9  2.32e 9  2.48e 9  2.69e 9
 2 Africa Eastern and… AFE           9.41e11  9.42e11  9.23e11  9.19e11  9.35e11
 3 Afghanistan         AFG          NA       NA       NA       NA       NA      
 4 Africa Western and… AFW           5.76e11  5.84e11  5.98e11  5.92e11  5.91e11
 5 Angola              AGO           6.85e10  6.92e10  6.52e10  4.95e10  5.02e10
 6 Albania             ALB           1.59e10  1.14e10  1.06e10  1.16e10  1.26e10
 7 Andorra             AND          NA       NA       NA       NA       NA      
 8 Arab World          ARB           2.19e12  2.25e12  2.35e12  2.41e12  2.48e12
 9 United Arab Emirat… ARE           2.01e11  2.03e11  2.10e11  2.12e11  2.27e11
10 Argentina           ARG           4.61e11  5.04e11  5.43e11  5.88e11  6.22e11
# ℹ 256 more rows
# ℹ 28 more variables: `1995` <dbl>, `1996` <dbl>, `1997` <dbl>, `1998` <dbl>,
#   `1999` <dbl>, `2000` <dbl>, `2001` <dbl>, `2002` <dbl>, `2003` <dbl>,
#   `2004` <dbl>, `2005` <dbl>, `2006` <dbl>, `2007` <dbl>, `2008` <dbl>,
#   `2009` <dbl>, `2010` <dbl>, `2011` <dbl>, `2012` <dbl>, `2013` <dbl>,
#   `2014` <dbl>, `2015` <dbl>, `2016` <dbl>, `2017` <dbl>, `2018` <dbl>,
#   `2019` <dbl>, `2020` <dbl>, `2021` <dbl>, `2022` <dbl>

Exercise

There is a dataset on country’s CO2 emissions, again from the World Bank (2023), in “data/wdi_co2.csv”. Load the dataset into R and add a new variable with its information, wdi_co2, to our qog_plus data frame. Finally, compute the average values of CO2 emissions per capita, by country. Tip: this exercise requires you to do many steps—plan ahead before you start coding! Your code:

Code 
# setup: these steps were executed before the exercise
library(tidyverse)
qog <- read_csv("data/sample_qog_bas_ts_jan23.csv")
gdp <- readxl::read_excel("data/wdi_gdp_ppp.xlsx")

gdp_long <- gdp |> 
  pivot_longer(cols = -c(country_name, country_code),
               names_to = "year",
               values_to = "wdi_gdp_ppp", 
               names_transform = as.integer)

qog_plus <- left_join(qog,
                      gdp_long,
                      by = c("ccodealp" = "country_code",
                             "year"))
  1. Load data (notice the .csv format):
emissions <- read_csv("data/wdi_co2.csv")
Rows: 266 Columns: 35
── Column specification ────────────────────────────────────────────────────────
Delimiter: ","
chr  (2): country_name, country_code
dbl (31): 1990, 1991, 1992, 1993, 1994, 1995, 1996, 1997, 1998, 1999, 2000, ...
lgl  (2): 2021, 2022

ℹ Use `spec()` to retrieve the full column specification for this data.
ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.
  1. Pivot data to long, creating the “wdi_co2” variable:
emissions_long <- emissions |> 
  pivot_longer(cols = -c(country_name, country_code),
               names_to = "year",
               values_to = "wdi_co2",  
               names_transform = as.integer)
  1. Merge-in information to our existing qog_plus data frame:
qog_plus2 <- left_join(qog_plus,
                       emissions_long,
                       by = c("ccodealp" = "country_code",
                              "year"))

  1. Create column for emissions per capita (here we do per 1,000 people).

  2. Summarize information to get mean values at the country level (remember that na.rm = T is always a conscious decision):

qog_plus2 |> 
  mutate(emissions_pc = 1000 * wdi_co2 / wdi_pop) |> 
  summarize(emissions_pc_country = mean(emissions_pc, na.rm = T),
            .by = cname)
# A tibble: 35 × 2
   cname               emissions_pc_country
   <chr>                              <dbl>
 1 Antigua and Barbuda                 4.60
 2 Argentina                           3.71
 3 Bahamas (the)                       6.17
 4 Barbados                            4.53
 5 Bolivia                             1.36
 6 Brazil                              1.84
 7 Belize                              1.74
 8 Canada                             15.8 
 9 Chile                               3.64
10 Colombia                            1.54
# ℹ 25 more rows

Exercise

Draw a scatterplot with time in the x-axis and democracy scores in the y-axis. Your code:

ggplot(qog_plus2) + aes(year, vdem_polyarchy) + geom_point()
Warning: Removed 248 rows containing missing values or values outside the scale range
(`geom_point()`).

Exercise

Using your merged dataset from the previous section, plot the trajectories of C02 per capita emissions for the US and Haiti. Use adequate scales.

ggplot(qog_plus2 |> filter(cname %in% c("Haiti", "United States")), 
       aes(x = year, y = 1000 * wdi_co2 / wdi_pop)) +
  geom_line() +
  facet_wrap(~cname, scales = "free_y") +
  labs(x = "Year", y = "CO2 Emissions Per Capita",
       title = "CO2 Emissions Per Capita in Haiti and the United States",
       caption = "Source: World Development Indicators (World Bank, 2023) in QOG dataset.")

5. Functions

Exercise When graphed, vertical lines cannot touch functions at more than one point. Why? Which of the following represent functions?

Figure 1: Vertical line test: examples.

  1. Function ✅

  2. Function ✅

  3. NOT a function 🚫

  4. Function ✅

  5. Function ✅

  6. NOT a function 🚫

  7. Function ✅

  8. NOT a function 🚫

Exercise

Create a function that calculates the area of a circle from its diameter. So your_function(d = 6) should yield the same result as the example above. Your code:

Code 
# setup: these steps were executed before the exercise
circ_area_r <- function(r){
    pi * r ^ 2
}
circ_area_r(r = 3)
[1] 28.27433
circ_area_d <- function(d){
    pi * (d/2) ^ 2
}
circ_area_d(d = 6)
[1] 28.27433

Exercise

Graph the function \(y = x^2 + 2x - 10\), i.e., a quadratic function with \(a=1\), \(b=2\), and \(c=-10\). Next, try switching up these values and the xlim = argument. How do they each alter the function (and plot)?

Code 
# setup: these steps were executed before the exercise
library(ggplot2)
  1. Graph \(y = x^2 + 2x - 10\).
ggplot() +
  stat_function(fun = function(x){x^2 + 2*x - 10},
                xlim = c(-5, 5))

  1. Switch up the values and the xlim = argument.
ggplot() +
  stat_function(fun = function(x){-3*x^2 - 6*x + 9},
                xlim = c(-10, 10))

Exercise

We’ll briefly introduce Desmos, an online graphing calculator. Use Desmos to graph the following function \(y = 1x^3 + 1x^2 + 1x + 1\). What happens when you change the \(a\), \(b\), \(c\), and \(d\) parameters?

(we’ll show how to do this in R here, but you could use Desmos)

  1. Graph \(y = 1x^3 + 1x^2 + 1x + 1\).
ggplot() +
  stat_function(fun = function(x){x^3 + x^2 + x + 1},
                xlim = c(-10, 10))

  1. Switch up the values.
ggplot() +
  stat_function(fun = function(x){-2*x^3 + 4*x^2 + 8*x + 16},
                xlim = c(-10, 10))

Exercise

Solve the problems below, simplifying as much as you can. \[log_{10}(1000)\] \[log_2(\dfrac{8}{32})\] \[10^{log_{10}(300)}\] \[ln(1)\] \[ln(e^2)\] \[ln(5e)\]

log10(1000)
[1] 3
log2(8/32)
[1] -2
10^(log10(300))
[1] 300
log(1)
[1] 0
log(exp(2))
[1] 2
log(5*exp(1))
[1] 2.609438

Exercise

Compute g(f(5)) using the definitions above. First do it manually, and then check your answer with R.

Code 
# setup: these steps were executed before the exercise
f <- function(x){x ^ 2}
g <- function(x){x - 3}

\[f(5) = 5^2 = 25\] \[g(25) = 25 - 3 = 22\]

g(f(5)) # no pipeline approach
[1] 22
f(5) |> g() # pipeline approach
[1] 22

6. Calculus

Exercise

  1. Use the slope formula to calculate the rate of change between 5 and 6.
  2. Use the slope formula to calculate the rate of change between 5 and 5.5.
  3. Use the slope formula to calculate the rate of change between 5 and 5.1.
(6^2 - 5^2) / (6 - 5)
[1] 11
(5.5^2 - 5^2) / (5.5 - 5)
[1] 10.5
(5.1^2 - 5^2) / (5.1 - 5)
[1] 10.1

Exercise

Use the differentiation rules we have covered so far to calculate the derivatives of \(y\) with respect to \(x\) of the following functions:

  1. \(y = 2x^2 + 10\)
  2. \(y = 5x^4 - \frac{2}{3}x^3\)
  3. \(y = 9 \sqrt x\)
  4. \(y = \frac{4}{x^2}\)
  5. \(y = ax^3 + b\), where \(a\) and \(b\) are constants.
  6. \(y = \frac{2w}{5}\)

  1. \(4x\) (sum rule, constant rule, coefficient rule, power rule)

  2. \(20x^3-2x^2\) (sum rule, coefficient rule, power rule)

  3. \(\frac{-9}{2\sqrt x}\) (power rule)

  4. \(-\frac{8}{x^3}\) (coefficient rule, power rule)

  5. \(3ax^2\) (sum rule, constant rule, coefficient rule, power rule)

  6. \(0\) (constant rule)

Exercise

Compute the following:

  1. \(\frac{d}{dx}(10e^x)\)
  2. \(\frac{d}{dx}(ln(x) - \frac{e^2}{3})\)

  1. \(10e^x\) (coefficient rule, exponent rule)

  2. \(\frac{1}{x}\) (difference rule, constant rule, logarithm rule)

Exercise

Use the differentiation rules we have covered so far to calculate the derivatives of \(y\) with respect to \(x\) of the following functions:

  1. \(x^3 \cdot x\)
  2. \(e^x \cdot x^2\)
  3. \((3x^4-8)^2\)

  1. \(4x^3\) (power rule)

  2. \(e^x x^2 + 2xe^x\) (product rule, exponent rule, power rule)

  3. \(24x^3(3x^4-8)\) (chain rule, difference rule, constant rule, power rule)

Exercise

Take the partial derivative with respect to \(x\) and with respect to \(z\) of the following functions. What would the notation for each look like?

  1. \(y = 3xz - x\)
  2. \(x^3+z^3+x^4z^4\)
  3. \(e^{xz}\)

\(\frac{\delta}{\delta x}(3xz - x) = 3z - 1\) (difference rule, coefficient rule, power rule)

\(\frac{\delta}{\delta z}(3xz - x) = 3x\) (difference rule, constant rule, coefficient rule)

\(\frac{\delta}{\delta x}(x^3+z^3+x^4z^4) = 4x^3z^4+3x^2\) (add rule, coefficient rule, power rule)

\(\frac{\delta}{\delta z}(x^3+z^3+x^4z^4) = 4x^4z^3+3z^2\) (add rule, coefficient rule, power rule)

\(\frac{\delta}{\delta x}(e^{xz}) = e^{xz}z\) (chain rule, exponent rule, coefficient rule)

\(\frac{\delta}{\delta z}(e^{xz}) = e^{xz}x\) (chain rule, exponent rule, coefficient rule)

Exercise

Identify the global extrema of the function \(\displaystyle \frac{x^3}{3} - \frac{3}{2}x^2 -10x\) in the interval \([-6, 6]\).

  1. Take the first derivative

\((\frac{x^3}{3} - \frac{3}{2}x^2 -10x)' = x^2 - 3x - 10\) (sum rule, coefficient rule, power rule)

  1. Set the derivative equal to zero and obtain its roots (F.O.C)

$ x^2 - 3x - 10 = (x-5)(x+2)$

\((x-5)(x+2) = 0\)

\(x^*_1 = 5,\, x^*_2=-2\)

  1. Calculate the second derivative and substitute the roots (S.O.C.)

\((x^2 - 3x - 10)' = 2x -3\)

  1. \(2x^*_1-3 = 2\cdot5-3 = 7\) (since it is positive, this is a minimum)

  2. \(2x^*_2-3 = 2\cdot(-2)-3 = -7\) (since it is negative, this is a maximum)

  1. Adjudicate between these critical points or the bounds.

Minimum critical point: \(f(5) = \frac{(5)^3}{3} - \frac{3}{2}(5)^2 -10 \cdot (5) = −45.8\overline{3}\).

Maximum critical point: \(f(-2) = \frac{(-2)^3}{3} - \frac{3}{2}(-2)^2 -10 \cdot (-2) = 11.\overline{3}\).

Lower bound: \(f(-6) = \frac{(-6)^3}{3} - \frac{3}{2}(-6)^2 -10 \cdot (-6) = −66\)

Upper bound: \(f(6) = \frac{(-6)^3}{3} - \frac{3}{2}(-6)^2 -10 \cdot (-6) = −42\)

So we conclude that, for the \([-6, 6]\) interval, the global minimum is at the lower bound (\(x=-6\)) and the global maximum is at the critical point at \(x=-2\).

Exercise

Solve the following indefinite integrals:

  1. \(\int x^2 \, dx\)
  2. \(\int 3x^2\, dx\)
  3. \(\int x\, dx\)
  4. \(\int (3x^2 + 2x - 7\,)dx\)
  5. \(\int \dfrac{2}{x}\,dx\)

  1. \(\frac{x^3}{3} + C\) (power rule)

  2. \(x^3 + C\) (coefficient rule, power rule)

  3. \(\frac{x^2}{2} + C\) (power rule)

  4. \(x^3 + x^2 - 7x + C\) (sum/difference rule, coefficient rule, power rule)

  5. \(2 ln(x) + C\) (coefficient rule, reciprocal rule)

And solve the following definite integrals:

  1. \(\displaystyle\int_{1}^{7} x^2 \, dx\)
  2. \(\displaystyle\int_{1}^{10} 3x^2 \, dx\)
  3. \(\displaystyle\int_7^7 x\, dx\)
  4. \(\displaystyle\int_{1}^{5} 3x^2 + 2x - 7\,dx\)
  5. \(\int_{1}^{e} \dfrac{2}{x}\,dx\)

In the following, FTC stands for the Fundamental Theorem of Calculus

  1. \(114\) (substitute from previous answer, FTC)
  2. \(999\) (substitute from previous answer, FTC)
  3. \(0\) (there is no area between 7 and 7)
  4. \(120\) (substitute from previous answer, FTC)
  5. \(2\) (substitute from previous answer, FTC)

7. Probability, statistics, and simulations

Exercise

Compute the probability of seeing between 1 and 10 voters of the candidate in a sample of 100 people.

pbinom(q = 10, size = 100, prob = 0.02) - 
  dbinom(x = 0, size = 100, prob = 0.02)
[1] 0.8673748

Exercise

Evaluate the CDF of \(Y \sim U(-2, 2)\) at point \(y = 1\). Use the formula and punif().

\[A = F(1) = P(Y\leq 1) = 3 \cdot(1/4) = 0.75\]

punif(q = 1, min = -2, max = 2)
[1] 0.75

Exercise

What is the probability of obtaining a value above 1.96 or below -1.96 in a standard normal probability distribution? Hint: use the pnorm() function.

pnorm(-1.96) + (1 - pnorm(1.96))
[1] 0.04999579

Exercise

Compute and plot my_rnorm, a vector with one million draws from a Normal distribution \(Z\) with mean equal to zero and standard deviation equal to one (\(Z\sim N(0,1)\)). You can recycle code from what we did for the uniform distribution!

set.seed(1) # set a seed
my_rnorm <- rnorm(n = 1000000)

ggplot(data.frame(my_rnorm), aes(x = my_rnorm)) +
  geom_histogram(binwidth = 0.25, boundary = 0, closed = "right") +
  scale_x_continuous(breaks = seq(-5, 5, 1), limits = c(-5, 5))

8. Text analysis

Exercise

What score (out of 10) would you give Barbie or Oppenheimer? Write your score in one sentence (e.g., I would give Barbie seven of ten stars.) If you have not seen either, write a sentence about which you would like to see more.

Store that text as a string (string3) and combine it with our existing cat_string to produce a new concatenated string called cat_string2. Finally, count the total number of characters within cat_string2. Your code:

Code 
# setup: these steps were executed before the exercise
library(stringr)
my_string <- "I know people who have seen the Barbie movie 2, 3, even 4 times!"
my_string2 <- "I wonder if they have seen Oppenheimer, too."
cat_string <- str_c(my_string, my_string2, sep = " ")
string3 <- "I would give Barbie 7 out of 10 stars."
string3
[1] "I would give Barbie 7 out of 10 stars."
cat_string2 <- str_c(cat_string, string3, sep = " ")
cat_string2
[1] "I know people who have seen the Barbie movie 2, 3, even 4 times! I wonder if they have seen Oppenheimer, too. I would give Barbie 7 out of 10 stars."
str_length(cat_string2)
[1] 148

Exercise

Look up the lyrics to your favorite song at the moment (no guilty pleasures here!). Then, follow the process described above to count the words: store the text as a string, convert to a tibble, tokenize, and count.

When you are done counting, create a visualization for the chorus using the ggplot code above. Your code:

  1. Store the text as a string.
library(tidytext)
dummy <- c("I been goin' dummy (Huh)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Yeah)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy (Goin' dummy)",
           "I been goin' dummy",
           "Dumbass, I been goin' dummy")
  1. Convert to a tibble.
dummy_df <- tibble(line = 1:9, text = dummy)
dummy_df
# A tibble: 9 × 2
   line text                            
  <int> <chr>                           
1     1 I been goin' dummy (Huh)        
2     2 I been goin' dummy (Goin' dummy)
3     3 I been goin' dummy (Goin' dummy)
4     4 I been goin' dummy (Goin' dummy)
5     5 I been goin' dummy (Yeah)       
6     6 I been goin' dummy (Goin' dummy)
7     7 I been goin' dummy (Goin' dummy)
8     8 I been goin' dummy              
9     9 Dumbass, I been goin' dummy
  1. Tokenize.
dummy_tok <- unnest_tokens(dummy_df, word, text)
  1. Count.
dummy_tok |>
  count(word, sort = TRUE)
# A tibble: 7 × 2
  word        n
  <chr>   <int>
1 dummy      14
2 goin       14
3 been        9
4 i           9
5 dumbass     1
6 huh         1
7 yeah        1
  1. Visualize.
dummy_tok |>
  count(word, sort = TRUE) |>
  mutate(word = reorder(word, n)) |>
  ggplot(aes(n, word)) +
  geom_col()