       
R. 2. Examples of use (more)
Introduction Use as calculator Other calculi Figures Graphs Statistics More
This guide is just an introduction to the main commands, to get confident with the
software and to illustrate some of the activities that can be faced with it. It is
an extended version of this. At the bottom there is an index.
# (A) At startup, the program has a line beginning with >. After this symbol I can
# type an "input", followed by "enter key", and get a possible "output", written in
# blue, preceded by [1]. With some "commands" I can have more lines of output; in
# these cases the following lines are preceded by [2], [3], ... A simple example:
> 10/4
# [1] 2.5
# In the examples we will do we will not put > so that the commands can be copied from
# here and put directly into R
# The parts that follow # (like these) are for those who read; they can be copied to R
# but do not give rise to any execution: in R the texts preceded by # are just comments.
# You can change size and style of the font, and other aspects (for example, to use the
# projector): see.
# To use many commands (indicated by point (C) down) you must first write to R the
# following line. You should copy it and paste it immediately:
source("http://macosa.dima.unige.it/r.R")
# A simple way to terminate the current session without using the mouse consists in
# typing quit()
USE as CALCULATOR
# (B) R can be used as an excellent calculator; the division is "/", the multiplication
# "*", a power elevation, such as 6 to 2, is indicated by 6^2, where "^" indicates that
# the 2 should be thought as if it was raised.
6+2
# [1] 8
62; 26; 6*2; 6/2; 6^2
# [1] 4
# [1] 4
# [1] 12
# [1] 3
# [1] 36
# As you have seen, I can run multiple commands separated by ;. I get the results
# one below the other.
# The results can have many or infinite digits; only a few are shown:
5000/12
# [1] 416.6667
# 5000/12 results 416.66666 ... but the program displays only 7 digits rounding the
# final figure: 416.66666... > 416.6667 (it is closer to 5000/12 than 416.6666).
# (C) In addition to + * / ... with R you can calculate the values of other functions,
# which relate outputs to inputs. Generally the inputs are enclosed in brackets. Some
# examples, round, sqrt or rad2, rad3:
round(5000/12, 3); round(5000/12, 1); round(5000/12, 0); round(5000/12, 2)
# [1] 416.667
# [1] 416.7
# [1] 417
# [1] 400
# round(5000/12, 3) rounds to the 3rd digit after ".", round(5000/12, 1) to 1st digit,
# round(5000/12, 0) to integers, rounds(5000/12, 2) to hundreds. For example, if I
# have a cost of 5000 euros to divide between 12 people I calculate 5000/12, rounding
# the result to the cents, that is to the 2nd digit after ".":
round(5000/12, 2)
# [1] 416.67
sqrt(4); sqrt(3); sqrt(3)^2
# [1] 2
# [1] 1.732051
# [1] 3
# sqrt acts for "square root". The side of a square that must have the area of 3 m^2
# must be 1.73 m, ie 173 cm.
# Instead of sqrt you can use rad2. You can use rad3 to calculate the cubic root.
# The side (in cm) of a volume cube of 500 cm^3 (half liter):
rad3(500)
# [1] 7.937005
# (D) For fractional calculations and to represent finite or periodic numbers in the
# form of fractions it may be useful the command fraction (or "frazio"). Examples:
fraction(1.3333333333333); fraction(0.125); fraction(10/4); fraction(15/60+1/2)
# 4/3 1/8 5/2 3/4
# To get a ratio in percentage form, eg 23/54 as 42.6% or 43%, I do:
23/54*100; round(23/54*100,1); round(23/54*100,0)
# 42.59259 42.6 43
# To write large or small numbers you can use the socalled exponential notation, as
# you do with the calculator, using eg e3 to mean a multiplication for 10^3 and e3 to
# mean a division for 10^3, ie a multiplication for 10^3.
# R uses the same notation to indicate large and small numbers.
# Here are some ways to write 100 thousand and 2 tenths of a thousand:
100000; 100*1000; 100e3; 0.0002; 2e4
# 1e+05 1e+05 1e+05 2e04 2e04
# If you have to calculate 15+27 and divide the result by twice the sum of 36 and 25,
# ie calculate what at school is sometimes written (12+27)/[2*(36+25)], you need to be
# careful: in math the parentheses have different meanings than those given at school:
# square brackets, [ and ],and braces, { and }, are used to indicate particular
# mathematical objects. Software, and mathematicians, write a term like the previous
# using only round parentheses: (12+27)/(2*(36+25)). They eventually change the size:
# (12+27)/(2*(36+25)). Using R you can improve readability by adding white spaces:
(12+27) / ( 2*(36+25) )
# [1] 0.3196721
# In fact, I can put all the spaces I want between brackets, numbers, and operations.
# (E) Here is how to perform the division with the rest between two integers and how
# to list the positive integers for which an integer can be divisible (its divisors).
div(13, 4)
# [1] 3 1
# 3 with remainder 1; in fact 13 = 4*3 + 1
divisors(12) # or: divisori(12)
# [1] 1 2 3 4 6 12
USE as CALCULATOR  Developments
# (E_1) Tree(1),…,Tree(17) allow you to compare how to write mathematical expressions
# on your computer on a single line, how to write them normally, and how to represent
# them with a "tree" without the use of brackets. Knowing how to switch between a
# representation and the other is important to understanding the meaning of formulas.
Trees()
#If you use Tree(1) or Tree(2) ... or Tree(18) you see expressions
#in the usual way, in "one line", as a "tree"
Tree(8)
# (E_2) The results are displayed with 7 digits. If they are integer and have less
# than 13 digits they are displayed exactly:
1.23456*7.654321; 123456*7654321
# [1] 9.449719
# [1] 944971853376
# divInt(M,N,d) computes the division M by N with d decimal digits of the quotient, and
# the remainder. An example:
divInt(450,29, 40)
# [1] 15 5 1 7 2 4 1 3 7 9 3 1 0 3 4 4 8 2 7 5 8 6 2 0 6 8
# [27] 9 6 5 5 1 7 2 4 1 3 7 9 3
# [1] 10
# The quotient has at most N1 digits. In this case they are exactly 28 (the remainder
# is 10).
Other CALCULI
# (F) In a formula like area = length*width (area of the rectangle) area, length and
# width are variables, ie the names to which values can be replaced. Also in R I can
# use variables. I have to be careful about the size of the letters: Width, WIDTH and
# width are three different variables. Example:
length = 10; width = 5
length*width
# [1] 50
length*Width
# Error: "Width" not found
# I could also put the result of length*width into another variable and then look
# for the value of it:
area = length*width
area
# [1] 50
# (G) If I have to do the same calculation referring to several numbers, such as 5, 10
# and 3, I can put these into a collection of numbers, ie put them in c(5,10,3) and
# make calculations with this new object:
c(5,10,3)*2
# [1] 10 20 6
# There are also calculations that are made specifically on the collections of objects;
# eg If I want to sort the data I put them as input of the sort function and get the
# output as data in ascending order:
sort( c(5,10,3) )
# [1] 3 5 10
# I can extract a number from a collection by pointing  between [ and ]  the place:
joe = c(5,10,3); joe[1]; joe[2]
# [1] 5
# [1] 10
# Let's see an example of use. I put in height_pupils the heights in cm of some pupils
# of a class (I use "_" to combine "height" and "pupils" in one variable, but I could
# use eg HeightPupils). Then I put them in order:
height_pupils = c(118, 135, 127, 119, 121, 124, 130, 132, 128, 122, 121, 123, 127)
sort(height_pupils)
# [1] 118 119 121 121 122 123 124 127 127 128 130 132 135
# Other calculations I can do (we see the outputs immediately):
data = height_pupils; min(data); max(data); length(data); sum(data)
# [1] 118
# [1] 135
# [1] 13
# [1] 1627
# The meaning of commands is obvious. I copied "height_pupils" into "data" so that I
# had a shorter name to write.
# NOTE If I press the arrow key "up" [^] I see the previous commands again, if I press
# the arrow key "down" [v] I see the subsequent commands: this is convenient for
# recalling or modifying them. Try to use them.
Other CALCULI  Developments
# (G_1) To assign a value to a variable, as well as =, you can use an arrow, that is
# < or >.
Price < 9; 7 > Cost; Profit = Price  Cost; Profit
# [1] 2
FIGURES
# (H) Let's see some basic commands to build geometric figures. Let's go through
# examples.
PLANE(1,10, 1,10)
# A window opens where the left figure appears: a "squared paper" with numbers that
# represent, as in a geographic map, the coordinates of the various points. The point
# whose horizontal and vertical coordinates are 7 and 4 appears red. It was plotted
# with the command:
POINT(7,4, "red")
# The smallest point (blue) was obtained with:
Point(2,4, "blue")
# By moving the mouse over the title bar of the graphics window (see below) I can move
# it. By moving over edges or angles I can resize it (see above on the right).
# To delete a window I do not need, I click the red button X .
# New "planes" appear in overlapping windows. To see the old ones I can move the new
# windows.
# The initial window was wide and high about 3 inches. If I type BF and HF (base and
# height of the window frame) I get:
BF;HF
# [1] 3
# [1] 3
# I can change the size of the window. Example:
BF=4; HF=2; PLANE(50,50, 0,50)
# If I do not change BF and HF the size remains these. Horizontal coordinates range (at
# least) from 50 to 50. Vertical coordinates range (at least) from 0 to 50. To plot
# the new windows in the original dimensions I type:
BF=3; HF=3
# I write 4,2 to indicate the point having the coordinates 4 and 2. With the following
# commands I draw in blue the circle whose center is 4,2 and radius is 3 and in black
# the center. Then I draw the circle that passes through the 3 points whose first
# coordinates I put in x and the second in y. I draw the circle in red and the points
# in black. Instead of "black" I can use number 1.
PLANE(1,10, 1,10)
circle(4,2, 3, "blue"); POINT(4,2,"black")
x = c(5, 8, 9); y = c(9, 3, 7.5)
circle3p(x,y, "red"); POINT(x,y, 1)
# The center of the red circle is also drawn. I got it and the radius as follows:
center3p; radius3p
# [1] 6.076923 5.788462
# [1] 3.387291
POINT(6.076923, 5.788462, "orange")
# Here are some color names [enter Colors() to see others]:
# 1 black 2 grey 3 white 4 brown 5 orange 6 red 7 pink
# 8 magenta 9 blue 10 cyan 11 yellow 12 green 13 seagreen
# (I) Let's see some other figure: lines, angles, ...
PLANE(0,10, 0,10)
# I plot a few points (figure on the left):
POINT(1,9,"blue"); POINT(4,9, "blue"); POINT(0,6,"brown"); POINT(4,8, "brown")
POINT(3,5,"magenta"); POINT(7,4, "magenta"); POINT(2,1, "red"); POINT(7,2,"red")
# then (on the right) two segments, a halfline and a straight line for two points (the
# latter are only partially traced):
segm(1,9, 4,9, "blue"); segm(0,6, 4,8, "brown")
halfline(3,5, 7,4, "magenta") # the halfline starting from 3,5, passing through 7,4
line2p(2,1, 7,2, "red") # the line passing through 2,1 and 7,2
# Let's see some other simple figure:
PLANE(0,10, 0,10)
A = c(0,6); B = c(2,10); C = c(5,4)
POINT(A[1],A[2], "blue"); POINT(B[1],B[2], "blue"); POINT(C[1],C[2], "blue")
line(A[1],A[2], B[1],B[2], "blue"); line(A[1],A[2], C[1],C[2], "blue")
# I have plotted points and traced thin segments with the command line (instead of
# segm). I have stored every point (A, B, C) in a variable, as collection  see (G) 
# of his coordinates.
# With point_point I calculate the distance of A from B, with inclination the
# inclination of AB e AC. With angle I can measure the angle CAB directly.
point_point(A[1],A[2], B[1],B[2])
# [1] 4.472136
inclination(A[1],A[2], B[1],B[2])
# [1] 63.43495
inclination(A[1],A[2], C[1],C[2])
# [1] 21.80141
angle(C,A,B)
# [1] 85.23636
# Then with mid2p I traced the middle point of AB
mid2p(A[1],A[2], B[1],B[2], "red")
# In the figure on the right I traced the axis of AB, ie the line perpendicular to AB
# and passing through its middle point.
perp2p(A[1],A[2], B[1],B[2], "red")
# (J) Let's see how to draw polyline and polygons. With polyline I trace the line
# formed by the segments joining the points with the indicated coordinates. In the
# following example the first point coincides with the last one (in this case the
# polyline is a polygon). leng calculates its length.
PLANE(3,9,2,10)
x = c(3,2,8,8,3); y = c(1,1,9,1,1)
polyline(x,y, "blue"); POINT(x,y, "red")
leng(x,y)
# [1] 36.18034
# If I want to calculate the length of the first 3 segments, between 1^ and 4^ points:
leng(x[1:4],y[1:4])
# [1] 25
# With polyC I close the polyline by coloring the polygon it delimits.
PLANE(3,9,2,10); polyC(x,y,"yellow")
# With polyC I can not rewrite the starting point: x=c(3,2,8,8), y=c(1,1,9,1)
# With color 0 the polygon is transparent
# At the yellow polygon I added a red rectangle, obtained with:
A = c(0,2, 2,0, 0); B = c(4,4, 7,7, 4); polyline(A,B, "red")
FIGURES  Developments
# (J_1) It's easy calculate the area of polygon. In the previous case:
x = c(3,2,8,8); y = c(1,1,9,1)
areaPol(x,y)
# [1] 35
See here to calculate the area of Italy.
# The value of π ("pi"), ie the area of a circle of radius 1, and the length of a
# semicircle of radius 1 (it is rounded with 7 digits):
pi
# [1] 3.141593
# The square all of whose sides are tangents to this circle has area 2·2 = 4
# If the radius is 2, 3, … area and semicircunference of the circle are π·4, π·9, …
# and π·2, π·3, …
# (J_2) I can print the coordinates of any point of the screen. If I type:
PLANE(3,6,4,5)
xy(2)
# and click the point marked with a cross (see below), I obtain:
# [1] 4.7 1.3
# They are rounded to 2 digits. If I want to round to 3 digits I type xy(3)
# (J_3) With arrow and dart you can draw arrows. With dirArrow you obtains direction
# and length. See below and the following commands:
arrow(2,3, 2,5, "red"); arrow(2,5, 4,4, "red")
dart(2,2, 2,4, "black"); dart(2,4, 4,3, "black")
dirArrow(2,2, 2,4)
# dir. = 26.56505° leng. = 4.472136
dirArrow1(2,2, 2,4); dirArrow2(2,2, 2,4)
# [1] 26.56505 [1] 4.472136
# With Arrow and Dart instead of the 2nd point I can put direction and length:
Arrow(2,1, 26.56505, 4.472136, "blue")
Dart(2,0, 26.56505, 4.472136, "seagreen")
# With Direction and Directio the point of the arrow is not drawn:
Direction(0,3, 0,4, "brown"); Direction(0,3, 60,4, "orange")
Directio(0,3, 30,4, "black")
Directionx; Directiony; POINT(Directionx, Directiony, "blue")
# [1] 3.464102 [1] 1
# In substance, Direction(a,b, D, R) draws a radius width center in (a,b), direction
# D and lenght R
# With arroW and ArroW I have an intermediate thickness between "dart" and "arrow".
# (J_4) line_line(x,y, p,q, a,b, u,v) is the intersection of the line through x,y and
# p,q and the line through a,b and u,v. The coordinates of it are in solut[1],solut[2].
# See below, on the left:
PLANE(3,9,2,10)
POINT(8,1, 1); POINT(4,10, 1); line2p(8,1, 4,10, "seagreen")
POINT(2,1, 1); POINT(8,7, 1); POINT(2,1, 1); POINT(8,9, 1)
# We remember that instead of "black" we can use 1
line2p(2,1, 8,7, "brown"); line2p(2,1, 8,9, "blue")
line_line(2,1, 8,7, 2,1, 8,9)
# [1] "rette parallele/parallel lines"
line_line(8,1, 4,10, 2,1, 8,9)
# [1] 5.551020 5.734694
POINT(solut[1],solut[2],"red")
# point_inclina(x,y, d, col) draws the line through x,y with the direction of d degrees
# whereas point_incl draws a thin line.
# line_line2(x,y, d1, a,b, d2) is the intersection of the line through x,y with
# direction d1 and the line through a,b and direction d2. The coordinates of it are in
# solut[1],solut[2]. point_line(x,y, a,b,u,v) is the distance between the point x,y
# and the line through a,b and u,v. point_line2(x,y, a,b,u,v) supplies the coordinates
# of the closest point by x,y of the line.
PLANE(3,9,2,10)
POINT(1,5, 1); point_inclina(1,5, 80, "magenta"); point_incl(1,5, 30, 1)
POINT(6,5, "blue"); point_incl(6,5, 20, "blue")
line_line2(1,5, 80, 6,5, 20)
# [1] 1.480053 2.277483
POINT(solut[1],solut[2], "brown")
point_line(6,5, solut[1],solut[2], 1,5)
# [1] 6.893654
point_line2(6,5, solut[1],solut[2], 1,5)
# [1] 0.7889242 6.1970705
POINT(solut[1],solut[2], "seagreen")
line(solut[1],solut[2], 6,5, "red")
# If you use noPrint=1 before line_line or line_line2 the value is not printed (they
# are in solut).
# Another picture:
# While perp2p traces the axis of a segment, perp3p(A,B,C,D, H,K) traces the segment
# perpendicular to the line passing through (A,B) and (C,D) that starts from (H,K).
PLANE(3,9,1,11)
POINT(1,2,"brown"); POINT(7,9,"brown")
line(1,2, 7,9, "brown")
POINT(8,3, "seagreen")
noPrint=1; perp3p(1,2, 7,9, 8,3, "seagreen")
POINT(solut[1],solut[2],"red")
# If I don't use "noPrint" I also get the coordinates of the "red" point:
# [1] 4.592920 6.893805
# symm(H,K, A,B,C,D) gives the point symmetric to (H,K) with respect to the line
# passing through (A,B) and (C,D) (the blue point is symmetric to the green point)
Q = symm(8,3, 1,2,7,9); Q
# [1] 1.185841 10.787611
POINT(Q[1],Q[2], "blue")
# symm2(H,K, A,B,d) gives the point symmetric to (H,K) with respect to the line passing
# through (A,B) with the direction of d degrees. Here's how to get the same point:
d = inclination(1,2, 7,9); d
# [1] 41.18593
U = symm2(8,3, 7,9, d); POINT(U[1],U[2],"blue")
# (J_5) I can easily plot a point on a circle giving the angle that expresses the
# direction in which it is reachable from the center. With circleA(x,y, r, D, col) I
# plot the point of the circle with center x,y and radius r in the direction D. With
# circleA2(x,y, r, D) I print the coordinates. See the picture on the left.
# I can use also xrot(D) and yrot(D), which give the coordinates of the point of the
# circle with center 0,0 and radius 1 in the direction D.
PIANO(4,7, 3,8)
circle(0,4, 3, "blue")
circleA(0,4, 3, 45, "orange"); circleA2(0,4, 3, 45)
# [1] 2.12132 6.12132
ang=c(120,240); circleA(0,4, 3, ang, "red")
0+xrot(45)*3; 4+yrot(45)*3
# [1] 2.12132 [1] 6.12132
# If the circle is specified with 3 points I use the commands circle3pA and circle3pA2:
x=c(2,4,7); y=c(2,3,1); POINT(x,y,"black")
circle3p(x,y, "red")
ang=c(90,180,45); circle3pA(x,y, ang, "blue")
circle3pA2(x,y, ang)
# [1] 4.22727273 1.26399891 6.32262374 2.91781928 0.04545455 2.14080556
type(4.6,3.7,"P"); type(0.5,0.6,"Q")
# I wrote the word "P" and "Q" on the picture with the command type.
# circle_circle(x,y,r1, a,b,r2) finds the eventual intersections between the circle
# with center x,y and radius r1 and the circle with center a,b and radius r2. The
# figure in the center is obtained with:
PIANO(4,7, 3,8)
circle(0,4, 3, "brown"); circle(0,1, 2, "red")
circle_circle(0,4,3, 0,1,2)
# [1] 0 1
P=circle_circle(0,4,3, 0,1,2); POINT(P[1],P[2], "blue")
circle(4,1, 1.5,"seagreen"); circle_circle(0,4,3, 4,1,1.5)
# [1] "troppo distanti/too distant"
circle(4,6, 2,"blue"); circle_circle(0,4,3, 4,6,2)
# [1] 2.012660 6.224679 2.987340 4.275321
Q=circle_circle(0,4,3, 4,6,2); Q[1];Q[2];Q[3];Q[4]
# [1] 2.01266 [1] 6.224679 [1] 2.98734 [1] 4.275321
# To get the figure above on the right we use the commands polylineR(a,b, R,N, D, col)
# and polyRC(a,b, R,N, D, col). The first one draws the regular polygon with center
# a,b, radius R, N sides and one vertex in the direction D; the sides have the colour
# col. The second one cause the interior have the colour col. polyR draws thin sides.
PLANE(1,5, 2,4)
POINT(2,1, "blue"); circle(2,1, 3, "red")
polylineR(2,1, 3, 5, 90, "blue"); polyR(2,1, 2, 5,90, "brown")
polyRC(4,3, 1, 4, 0, "orange")
# (J_6) With ruler, goniometer and compass I can build a triangle if I know the three
# sides or two sides and the inclusive angle or two angles and the inclusive side. Here
# I can build it using triSSS, triSAS and triASA commands. With SIDES and ANGELS I have
# the measurements of sides and angles opposite them, with TRIANGLE() I have the image.
triSSS(5,5,5)
SIDES; ANGLES; TRIANGLE()
# [1] 5 5 5 [1] 60 60 60
triSAS(5,90,5)
SIDES; ANGLES; TRIANGLE()
# [1] 5.000000 5.000000 7.071068 [1] 45 45 90
triASA(30,8,45)
SIDES; ANGLES; TRIANGLE()
# [1] 5.856406 4.141105 8.000000 [1] 45 30 105
# You can use also triASS (1 angle and the next 2 sides) that can produce two outcomes:
triASS(30,2,1); SIDES; ANGLES; TRIANGLE()
# [1] 2.000000 1.000000 1.732051
# [1] 90 30 60
triASS(30,2,1.5); SIDES; ANGLES; TRIANGLE()
# [1] "Then use triASS2(...) too"
# [1] 2.000000 1.500000 2.850085
# [1] 41.81031 30.00000 108.18969
triASS2(30,2,1.5); SIDES; ANGLES; TRIANGLE()
[1] 2.0000000 1.5000000 0.6140168
[1] 138.18969 30.00000 11.81031
# If you want to change the size of the window in advance, first use commands BFt=...
# and HFt=...
# The incentre command lets you easily trace the center
# of the circle inscribed in a triangle (the incentre
# of the triangle). An example:
PLANE(0,5, 0,5); x=c(0,4,5,0); y=c(0,0,5,0)
C = incentre(x,y); polyline(x,y,"blue")
POINT(C[1],C[2],"brown")
circle(C[1],C[2], C[2], "red")
line(0,0,C[1],C[2],"red")
line(4,0,C[1],C[2],"red")
line(5,5,C[1],C[2],"red")
# The incentre is the intersection of the bisectors: 

A = angle(c(4,0),c(0,0),c(5,5)); Directio(0,0, A/2, 10, "brown")
# I get a line that crosses the incentre, and (0,0)
B = angle(c(5,5),c(4,0),c(0,0)); B
# 101.3099 The angle B, and the direction from B to C:
BC = dirArrow1(4,0, 5,5); BC
# 78.69007 The line that crosses the incentre and (4,0)
Directio(4,0, BC+B/2, 10, "brown")  
# How to determine the tangent lines to a circle (of centre
# C1,C2 and radius R) from a point (P1,P2) with the command
# P_tan(P1,P2, C1,C2,R):
PLANE(0,6, 0,6)
POINT(5,5, "blue")
circle(3,2, 1.5, "red")
Q = P_tan(5,5, 3,2,1.5); Q
# 2.211213 3.275858 4.481095 1.762603
l2p(5,5, Q[1],Q[2], "blue")
l2p(5,5, Q[3],Q[4], "blue")
# l2p draw thinner lines than line2p  
GRAPHS
# (K) Let's see how to make a graph of a phenomenon that varies with another. Plane
# (with only one uppercase letter, "P") represents the "x" and "y" with scales that
# can be different.
# Example: average consumption of kg of fresh fruit per year by an Italian.
year = c(1880,1890,1900,1910,1920,1930,1940,1950,1960,1970,1980)
fruit = c(19, 21, 23, 28, 31, 27, 23, 37, 65, 88, 79)
# max is useful for choosing the "Plane": I find the maximum is 88, so I change y
# between 0 and 90
max(fruit)
# [1] 88
BF=4; HF=2.5
Plane(1880,1980, 0,90)
polyline(year,fruit, "blue"); POINT(year,fruit,"brown")
# I also put two inscriptions along the axes with these commands:
abovex("year"); abovey("fruit")
# I can also graph the functions in which "x" and "y" are linked by a formula
# Ex: the conversion of the temperature in degrees Celsius to that in Fahrenheit.
Fa = function(Ce) 32 + 72/40*Ce
Fa(30); Fa(110)
# 22 230
Plane(30,110, 22,230)
graph(Fa, 30,110, "brown")
abovey("°F"); abovex("°C")
POINT(100,Fa(100),"blue"); POINT(0,Fa(0),"blue")
# I got the chart on top left.
Fa(100)
# [1] 212
# To find the temperature in °F that corresponds to 100°C I simply calculated Fa(100).
# To find what is the temperature k in °C that corresponds to 100°F (see the graph on
# the right) I have to solve the problem: "for which k Fa(k) = 100?". I could do it
# "by hand". Or I can proceed using the solution command as follows:
solution(Fa,100, 20,60)
# [1] 37.77778
# I put Fa,100 because I must find the solution of the problem:
# for what input Fa is worth 100?
# I put 20,60 to indicate two values among which is the solution, derived from the
# graph. 37.777… is therefore the value of k.
# I can proceed in the same way for any other value.
# The computer does nothing but speed up a process that I could do directly with zoom,
# to find more precisely the °C degrees to which it corresponds to 100°F
# If you want to see how the solutions are automatically searched, click here.
# NOTE. In more complicated situations I have to look for the solutions graphically
# before using "solution": see K_3.
GRAPHS  Developments
# (K_1) If you want shut down all open graphics devices type:
graphics.off()
# Instead of graph you can use graphF(g, a,b, col) that (automatically) traces the
# graph of the function g between a and b in the colour col in a suitable window.
# If I refer to the foregoing function Fa:
graphF(Fa, 35,105, "blue")
POINT(0,Fa(0),"red"); POINT(100,Fa(100),"red")
# If I use the color 0 line and graphs are dashed. I can change the color width coldash
coldash = "red"
segm(100,0, 100,Fa(100), 0); segm(0,Fa(100), 100,Fa(100), 0)
type(10,220,"F"); type(105,15,"C")
line(40,100, 110,100, 0)
x=solution(Fa,100, 40,110); POINT(x,100, "brown")
coldash = "brown"; segm(x,0, x, 100, 0); type(x,20, 37.8)
# The graph on the right is obtained with:
PLANE(3,3, 1,5)
f1 = function(x) x^2; f2 = function(x) sqrt(x); f3 = function(x) x
graph(f1, 3,3, "black"); graph(f2, 3,3, "seagreen")
# Between 3 and 0 the graph of f2 is not drawn
coldash="red"; graph(f3, 3,3, 0)
# (K_2) The program automatically traces the graph of a function for a large number of
# points. I can trace it for a fixed number of points, if I put it in NpF.
g = function(N) N*3; NpF = 6; graphF(g, 0,5, "brown")
# In the center, the graph whether as a collection of 6 points or a continuous line:
graphF(g, 1,6, "yellow"); NpF = 6; graph(g, 0,5, "red")
# On the right, with NPF 6 small rings have been added.
NPF = 6; graph(g, 0,5, "black")
# If I use NpsF the segments joining xaxis are added
NpsF = 4; graphF(g, 1,5, "blue"); NpsF = 3; graph(g, 0,4, "red")
# (K_3) How to find the points where the graph of a function has a bump? We use the
# command maxmin (or minmax) giving the name of the function and an interval where it
# has a bumb as input:
V < function(x) (x1/3)^21; graphF(V, 4,4, "red")
maxmin(V, 2,2)
# [1] 0.3333333
V( maxmin(V,2,2) )
# [1] 1
POINT(1/3,1, "blue") # The figure on the left:
# If you want to see how the bump is automatically searched, click here.
# We have already seen (K) how to solve an equation. Another example:
# solve (x1/3)^21 = 5 (ie V(x) = 5) for x.
# I see on the graph that the solution is between 2 and 4.
coldash="red"; segm(4,5, 4,5, 0)
solution(V, 5, 2,4)
# [1] 2.782823
POINT( solution(V,5, 2,4), 5, "black")
# In this case (if we are able) we can solve the equation "by hand":
# (x1/3)^21 = 5 I shift 1 to the right (subtraction > addition)
# (x1/3)^2 = 5+1 I make calculations
# (x1/3)^2 = 6 The inverse of "^2" is the square root function
# x1/3 = √6 I move 1/3 to the right
# x = √6+1/3 I make calculations (ie with computer)
# x = 2.782823
# If the solution of the equation corresponds to a point where the graph reaches a
# maximum or a minimum, as in the following case, the maxmin command seen above must
# be used to find the solution. Example: 28*x99*x^2+14*x^349*x^4 = 2
# I look for x such that 28*x99*x^2+14*x^349*x^42 = 0.
T = function(x) 28*x99*x^2+14*x^349*x^42
graphF(T, 5,5, "blue")
graphF(T, 1,1, "blue")
# The graph does not climb over the xaxis. I cannot use "solution".
maxmin(T,1,1)
# [1] 0.1428571
fraction( maxmin(T,1,1) )
# [1] 1/7
T( 1/7 )
# [1] 1.94289e16
# Because of the approximations, I do not obtain exactly 0.
POINT(1/7, 0, "red")
STATISTICS
# (L) The first statistical calculations, which can be seen from the early years of
# school, are cross histograms. At a later time one can use the square or millimeter
# paper. Only then does it make sense to use the computer, to automate some of the
# process that we have learned to deal with "by hand". Here are some examples.
# The inhabitants (in thousands) of the North, the Center and the South (of Italy):
N = 25755; C = 12068; S = 20843; Italia = c(N,C,S)
names = c("N","C","S")
BarNames = names; Bar(Italia)
PieNames = names; Pie(Italia)
StripNames = names; Strip(Italia)
# yellow,cyan,... % 43.90107 20.57069 35.52824
# yellow,cyan,... N C S
# To get rounded percentage distribution  see (C)  I can do:
distribution(Italia, 0)
# 44 21 36
distribution(Italia, 2)
# 43.9 20.57 35.53 # if I want, then I write 43.90 instead of 43.9
distribution(Italia, 1)
# 40 20 40
# (M) These were already classified data. Let's see how to analyze individual data.
# As we see in the following example, we observe that commands can be broken over
# multiple rows. Here are the lengths of many broad beans (ie bean seeds) collected by
# a 12yearold class. All lines must be copied and pasted, from "beans = c(" to the
# final line )".
beans = c(
1.35,1.65,1.80,1.40,1.65,1.80,1.40,1.65,1.85,1.40,1.65,1.85,1.50,1.65,1.90,
1.50,1.65,1.90,1.50,1.65,1.90,1.50,1.70,1.90,1.50,1.70,1.90,1.50,1.70,2.25,
1.55,1.70,1.55,1.70,1.55,1.70,1.60,1.70,1.60,1.75,1.60,1.75,1.60,1.80,1.60,
1.80,1.60,1.80,1.60,1.80,1.00,1.55,1.70,1.75,1.30,1.55,1.70,1.75,1.40,1.60,
1.70,1.75,1.40,1.60,1.70,1.80,1.40,1.60,1.70,1.80,1.40,1.60,1.70,1.80,1.40,
1.60,1.70,1.80,1.40,1.60,1.70,1.80,1.40,1.60,1.70,1.80,1.40,1.60,1.70,1.80,
1.45,1.60,1.70,1.80,1.50,1.60,1.70,1.80,1.50,1.60,1.70,1.85,1.50,1.60,1.70,
1.85,1.50,1.60,1.75,1.90,1.50,1.60,1.75,1.90,1.50,1.65,1.75,1.90,1.55,1.65,
1.75,1.95,1.55,1.65,1.75,2.00,1.55,1.65,1.75,2.30,1.35,1.65,1.80,1.40,1.65,
1.80,1.40,1.65,1.85,1.40,1.65,1.85,1.50,1.65,1.90,1.50,1.65,1.90,1.50,1.65,
1.90,1.50,1.70,1.90,1.50,1.70,1.90,1.50,1.70,2.25,1.55,1.70,1.55,1.70,1.55,
1.70,1.60,1.70,1.60,1.75,1.60,1.75,1.60,1.80,1.60,1.80,1.60,1.80,1.60,1.80,
1.00,1.55,1.70,1.75,1.30,1.55,1.70,1.75,1.40,1.60,1.70,1.75,1.40,1.60,1.70,
1.80,1.40,1.60,1.70,1.80,1.40,1.60,1.70,1.80,1.40,1.60,1.70,1.80,1.40,1.60,
1.70,1.80,1.40,1.60,1.70,1.80,1.40,1.60,1.70,1.80,1.45,1.60,1.70,1.80,1.50,
1.60,1.70,1.80,1.50,1.60,1.70,1.85,1.50,1.60,1.70,1.85,1.50,1.60,1.75,1.90,
1.50,1.60,1.75,1.90,1.50,1.65,1.75,1.90,1.55,1.65,1.75,1.95,1.55,1.65,1.75,
2.00,1.55,1.65,1.75,2.30
)
# We have already seen in (G) how to do some first analysis:
min(beans); max(beans); length(beans)
# 1 2.3 260
# A convenient way (it can also be made without a computer) to organize the data is
# the use of a kind of cross histogram, with the stem command. Example:
stem(beans)
# The decimal point is 1 digit(s) to the left of the 
# 10  00
# 11 
# 12 
# 13  0055
# 14  000000000000000000000055
# 15  0000000000000000000000005555555555555555
# 16  0000000000000000000000000000000000000000005555555555555555555555
# 17  0000000000000000000000000000000000000000005555555555555555555555
# 18  00000000000000000000000000000055555555
# 19  000000000000000055
# 20  00
# 21 
# 22  55
# 23  00
# The first phrase ("The decimal …") says that the decimal point is a digit to the left
# of , meaning that 1000, 11, … are for 1.0, 1.1,… Here's how I would represent the
# first six data of "beans" (1.35,1.65,1.80,1.40,1.65,1.80) on the squared paper. The
# computer sorts the digits to the right of  but it's not important to do so.
# The name "stem" comes from the fact that the data thus placed has the shape of leaves
# arranged along a stem. I can command that stem, putting a smaller or larger number
# (as below) of 1, classifies the data differently:
stem(beans, 1.5)
# 10  00
# 10 
# 11 
# 11 
# 12 
# 12 
# 13  00
# 13  55
# 14  0000000000000000000000
# 14  55
# 15  000000000000000000000000
# 15  5555555555555555
# 16  000000000000000000000000000000000000000000
# 16  5555555555555555555555
# 17  000000000000000000000000000000000000000000
# 17  5555555555555555555555
# 18  000000000000000000000000000000
# 18  55555555
# 19  0000000000000000
# 19  55
# 20  00
# 20 
# 21 
# 21 
# 22 
# 22  55
# 23  00
# After learning how to do "by hand" simple histograms, you can use the software:
histogram(beans)
# [the distance of the sides of the grid is 5 %]
# Frequencies and percentage freq.:
# 2, 0,0, 4, 24, 40, 64, 64, 38, 18, 2,0,4
# 0.77,0,0,1.54,9.23,15.38,24.62,24.62,14.62,6.92,0.77,0,1.54
# For other statistics use the command morestat()
# If the classes are many you can get the histogram to the right, first putting:
noClass=1
# As a number indicating where the data are (more or less), I can use the median, which
# is the central value of the data; if the values are 5 I take the 3rd one. If data is
# an even quantity there are two values in the center; usually one takes the first of
# them. In our case:
Median(beans)
# [1] 1.65
# I could also find it by ordering the values  see (G) , which are 260, and taking
# the 130th:
beans1 = sort(beans); beans1[130]
# [1] 1.65
# If data is an odd quantity median gives the mean of the two values in the center; it
# "can" be used (instead of Median), but it can be have a larger number of significant
# digits than data. See here for an alternative.
# I could also find it by ordering the values  see (G) , which are 260, and taking
# the 130th:
beans1 = sort(beans); beans1[130]
# [1] 1.65
# Another used value is the mean, the sum of the data divided by their amount:
sum(beans)/length(beans)
# [1] 1.659231
# I can also get it with the command mean. I should then round off his value:
mean(beans); round( mean(beans),2 )
# [1] 1.659231
# [1] 1.66
# (N) The histogram command automatically chooses the intervals where the data is
# classified. In this way it is easy to build histograms even in basic school.
# If I want, I can build the histogram by choosing the intervals. If I type:
Histogram(beans, 1,2.4, 0.2) # for "beans" see (M)
# (I written in upper case) I obtain the histogram from 1 to 2.4 with 0.2 wide classes
# As suggested by the "histogram" outputs,
# for other statistics use the command morestat(). Let's do it:
morestat()
#
# Min. 1st Qu. Median Mean 3rd Qu. Max.
# 1.000 1.550 1.650 1.659 1.750 2.300
# The brown dots are 5^ and 95^ percentiles
# The red dot is the mean
# If I write noMean=1; morestat() the mean is not shown:
# The outputs of this command are only available at the age of 12 or 13, but can be very
# useful to the teachers:
# minimun, 1^ quartile (demarcation between the first 25% of data and the next ones),
# median (or 2^ quartile or 50^ percentile), mean, 3^ quartile (or 75^ percentile),
# maximum.
# This diagram, that has the shape of a "box with whiskers", is called boxplot.
# If I want only the boxplot I can use statistics(beans) [statistic(…) does not even
# print the values]
# Another command is boxPlot. With the command out we can exclude data that is distant
# from other observations (the outliers). With affine we can compare different data in
# the same scale. See here.
# Incidentally, we did not talk about mode. It's not a fundamental concept and it's not
# easy to focus, because it depends on how you choose the intervals where you want to
# classify the data: in the basic school it is better not to mention it, and present it
# only in the higher schools when the histograms are approximated with curves. In cases
# like that showed in (L) we can use the term "the most frequent class" rather than
# "mode".
STATISTICS  Developments
# (N_1) Some other graphs. besides those presented in (L):
N = 25755; C = 12068; S = 20843; Italia = c(N,C,S)
names = c("N","C","S")
noGrid=1; BarNames=names; Bar(Italia) # without grid
BarNames=names; BAR(Italia) # original data
BarNames=names; BarH(Italia) # horizontal bars
PieNames=names; PIE(Italia) # pie widthout grid
PieNames=names; PIEh(Italia) # half a pie
# How to build double histograms using the
# pyramid (or pyramid2) command: see here.  

 
# (N_2) The histograms classify the data in intervals [.,.). See the first example:
d = c(1.6, 2, 2.3, 2.6, 3.1, 3.7, 4, 5)
Histogram(d, 1,6, 1)
# [the distance of the sides of the grid is 5 %]
# Frequencies and percentage freq.:
# 1, 3, 2, 1, 1
# 12.5,37.5,25,12.5,12.5
# For other statistics use the command morestat()
# But if I choose 5 as the right endpoint I obtain the second histogram:
Histogram(d, 1,5, 1)
# [the distance of the sides of the grid is 5 %]
# Frequencies and percentage freq.:
# 1, 3, 2, 2
# 12.5,37.5,25,25
# For other statistics use the command morestat()
# If I have the lenghts of the beans  see (L)  in order:
# 1, 1, 1.3, 1.3, 1.35, 1.35, 1.4, 1.4, 1.4, 1.4, 1.4, ...
# I can use rep(x,n) to repeat every datum n times:
beans = c(rep(1,2), rep(1.3,2), rep(1.35,2), rep(1.4,22), rep(1.45,2), rep(1.5,24),
rep(1.55,16), rep(1.6,42), rep(1.65,22), rep(1.7,42), rep(1.75,22), rep(1.8,30),
rep(1.85,8), rep(1.9,16), rep(1.95,2), rep(2,2), rep(2.25,2), rep(2.3,2))
# To study integer data I can use Histogram command: I must supply the intervals
# centered on integer values. The study of the cast of a die made of thin cardboard:
die < c(
5,5,4,3,3,1,6,2,5,6,2,5,3,6,5,6,3,2,1,6,1,6,6,3,5,2,1,5,2,5,5,3,6,6,4,6,6,5,6,2,
1,6,6,3,2,2,5,6,3,2,6,6,6,4,2,3,6,6,6,2,6,4,5,4,6,2,3,5,6,6,2,2,1,5,5,3,5,3,5,6,
2,1,4,3,6,6,6,4,3,4,6,6,4,5,4,4,2,2,6,4,4,6,2,5,3,6,6,3,3,2,2,6,1,4,6,5,3,3,4,1,
6,4,6,6,2,5,5,1,5,2,6,5,2,2,4,1,2,2,6,1,6,5,5,6,1,3,3,4,5,5,5,6,4,4,6,3,6,3,6,6,
4,2,6,6,6,1,4,5,6,5,6,5,5,5,6,6,2,1,6,3,6,6,5,3,5,2,2,4,6,6,5,2,5,5,6,6,5,2,6,1)
# The outcomes are 1,2,3,4,5,6. I choose 11/2 and 6+1/2 as extremes:
Histogram(die, 0.5,6.5, 1)
# Frequencies and percentage freq.:
# 16, 32, 25, 23, 40, 64
# 8, 16, 12.5, 11.5, 20, 32
mean(die); median(die) # the mean is not 3.5: the die is not balanced
# [1] 4.155
# [1] 5
# To obtain the histogram with stem I must use the parameter scale=s to change the
# scale of the plot (s>1 or s<1) (s=2 will cause the plot to be roughly twice as long
# as the default)
stem(die,scale=0.5)
The decimal point is at the 
1  0000000000000000
2  00000000000000000000000000000000
3  0000000000000000000000000
4  00000000000000000000000
5  0000000000000000000000000000000000000000
6  0000000000000000000000000000000000000000000000000000000000000000
# With the option "width=0" I obtain the lengths of the bars:
stem(die,scale=1/2,width=0)
# The decimal point is at the 
# 1  +16
# 2  +32
# 3  +25
# 4  +23
# 5  +40
# 6  +64
# (N_3) To study data already in class intervals we use the command histoclas: see here
# (also for the command percentile).
# (N_4) Other diagrams:
# See here. We use pointO, pointV and pointD to draw the points in various forms.
MORE
# (O) Let's look at some other tool, which can be useful to the teacher.
# Given three positive integers such as 90, 120 and 150, it is easy to find the
# greatest common divisor (GCD):
divisors(90); divisors(120); divisors(150)
# [1] 1 2 3 5 6 9 10 15 18 30 45 90
# [1] 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120
# [1] 1 2 3 5 6 10 15 25 30 50 75 150
# Obviously it is 30. I can find it even with a specific command:
numbers = c(90,120,150); GCD(numbers)
# [1] 30
# I could also use the command primes(N) that expresses N as the product of "prime"
# numbers (ie integers greater than 1 that are multiple only of 1 and itself):
primes(90); primes(120); primes(150)
# [1] 2 3 3 5
# [1] 2 2 2 3 5
# [1] 2 3 5 5
# The maximum number for which 2*3*3*5 and 2*2*2*3*5 and 2*3*5*55 are divisible is:
2*3*5
# [1] 30
# How to find the least common multiple (LCM) of the same numbers:
LCM(numbers)
# [1] 1800
# Wishing, I could find it by taking every prime number that is in primes(90),
# primes(120) or primes(150) with exponent equal to the maximum number of times
# that appears in one of them:
2^3 * 3^2 * 5^2
# [1] 1800
# Primes(M,N) print the prime numbers between M and N
n=1; Primes(n,n+500)
# [1] 1 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
# [22] 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179
# [43] 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293
# [64] 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431
# [85] 433 439 443 449 457 461 463 467 479 487 491 499
n=5000; Primes(n,n+300)
[1] 5003 5009 5011 5021 5023 5039 5051 5059 5077 5081 5087 5099 5101 5107 5113 5119 5147
[18] 5153 5167 5171 5179 5189 5197 5209 5227 5231 5233 5237 5261 5273 5279 5281 5297
# (P) Some graphic tools.
# You can display the time with clock(); after
# 10 sec the time is printed.
clock()
# [1] 09 15 32
# With clock2 you can display the time you want
clock2(12,25,40)  
# An "empty" pie and a goniometer (the goniometer is the circle divided into degrees (°),
Pie(0); gonio() # ie into 360 equal parts)
GONIO() # A quarter of a goniometer
Strip(0) # A strip
The rectangle of millimeter paper shown on the
right is obtained with mmpaper1()
(50 mm × 50 mm)
With mmpaper2() and mmpaper3() I have
other rectangle:
100 mm × 50 mm and 50 mm × 100 mm
(for other millimeter sheets,
see the next document)
To add axes I can use:
segm(0,50,0,50,"black"); segm(0,50,0,50,"black")
or for ex. segm(50,5, 50,5, "blue") if I want
that x axis is blue and above 5 mm  
# In a new window, I can build a check paper (m columns, n rows) with SQUARES(m,n)
# With SQUARESC(m,n,col) I can modify the color.
BF=4.5; HF=1.5; SQUARES(15,5)
BF=2; HF=1.5; SQUARES(6,5,"red"); POINT(4,1, "blue")
# With BF=3; HF=3; SQUARES(10,10) I have 10×10 squares with side 5 mm long.
#
# If you have enlarged the graphic window (for example with BF=6; HF=4), with
# RATIO(tot1, tot2) you can represent the proportional transformation of data that
# connects tot1 and tot2. Start with the command INPUT(datum) or OUTPUT(datum). If you
# then use rATIO you can represent other data in the same window.
INPUT=20; RATIO(60,12)
# I obtain the left figure.
INPUT=50; RATIO(100,60)
INPUT=80; rATIO(100,60); OUTPUT=20; rATIO(100,60)
# I obtain the right figure.
# Width RATIO(100,tot) or RATIO(tot,100) I can obtain the graphic representation of
# percentages. It is a convenient instrument to proportionally transform data.
MORE  Developments
# (P_1) We can generate regular sequences:
seq(3,10, 1); seq(3,10, 1.5); seq(3,10, len=5)
# [1] 3 4 5 6 7 8 9 10 # from 3 to 10 with increment 1
# [1] 3.0 4.5 6.0 7.5 9.0 # from 3 to 10 with increment 1.5 (10.5 > 10)
# [1] 3.00 4.75 6.50 8.25 10.00 # form 3 to 10 with length (of the sequence) 5
# If the increment is 1 or 1 it is useful employ :
3:10; 3.4:5
# [1] 3 4 5 6 7 8 9 10
# [1] 3.4 2.4 1.4 0.4 0.6 1.6 2.6 3.6 4.6
# (P_2) If I use the command PLANEW, PLANEw and PLANEww, I can plot the figures without
# grid and axes (similarly PlaneW, Planew and Planeww work):
PLANE(1,1, 1,1); circle(0,0, 1, "red")
PLANEW(1,1, 1,1); circle(0,0, 1, "red")
PLANEw(1,1, 1,1); circle(0,0, 1, "red")
PLANEww(1,1, 1,1); circle(0,0, 1, "red")
# Indeed, the circle on the right was obtained with circleC(0,0, 1, "yellow")
# before circle(0,0, 1, "red") to paint a yellowcoloured circle
# How to obtatin only a grid (with equal or no scales), either without axes:
GRID(0,10, 0,10); Grid(0,10, 0,20); GRIDw(0,10, 0,10); Gridw(0,10, 0,20)
# With gridH, gridV, GridH, GridV you can obtain grid as you like:
PLANEw(0,100, 0,100)
coldash="orange" # With coldash I can change the colour of the small grid
# With Coldash I can change the colour of the other grid
gridH( seq(0,100,10) ); gridV( seq(0,100,10) )
GridH( seq(0,100,50) ); GridV( seq(0,100,50) )
# On the right: PLANE(0,15,0,15); Coldash="grey"; GridV(0:15); GridH(0:15) (... see)
# (P_3) I can draw a goniometer centered in x,y with radius r by Gonio(x,y, r)
Gonio(5,6, 5) # See here for the other commands that draw the figure
# See here for drawing the following figures with goniom, goniome (and gonioM,gonioMe):
# By CROSS I can add crosses to the SQUARES
SQUARES(5,4)
CROSS(1,1); CROSS(1,2)
CROSS(2,1); CROSS(2,2); CROSS(2,3); CROSS(2,4)
CROSS(3,1); CROSS(3,2); CROSS(3,3)
CROSS(4,1); CROSS(4,2)
CROSS(5,1); CROSS(5,2); CROSS(5,3); CROSS(5,4)
# By LCROSS I can add labels:
LCROSS(1,"A"); LCROSS(2,"B"); LCROSS(3,"C"); LCROSS(4,"D"); LCROSS(5,"E")
 
# By CIRCLE I can add circles to the SQUARES
SQUARES(5,4)
CROSS(1,1); CIRCLE(1,2)
CROSS(2,1); CROSS(2,2); CIRCLE(2,3); CROSS(2,4)
CROSS(3,1); CIRCLE(3,2); CROSS(3,3)
CROSS(4,1); CROSS(4,2)
CROSS(5,1); CROSS(5,2); CROSS(5,3); CIRCLE(5,4)
LCROSS(1,"A"); LCROSS(2,"B"); LCROSS(3,"C"); LCROSS(4,"D"); LCROSS(5,"E")
 
# By COUNT I can add numbers to the SQUARES
SQUARES(5,4)
COUNT(4)
CROSS(2,1); CROSS(2,2); CROSS(2,3)
CROSS(3,1); CROSS(3,2); CROSS(3,3)
CROSS(5,1); CROSS(5,2); CROSS(5,3); CROSS(5,4)
LCROSS(2,"Fo"); LCROSS(3,"Sw"); LCROSS(4,"Ba"); LCROSS(5,"Vo")  
# I can also easily make thermometers, with the command THERMO. An example:
THERMO(7); THERMO(c(7,22)); THERMO2(c(3,15))
# THERMO(N) realize an empty, N wide thermometer (as if you put BF=N),
# THERMO(c(N,M)) realize an empty, N wide thermometer that marks M degrees
# THERMO2(N) and THERMO2(c(N,M)) carry out thin and horizontal prints (they are useful
# for preparing materials, exercises …
# THERMO(N) e THERMO(c(N,M)) are useful for preparing thermometers to hang in class
# room. Here are the pictures obtained with N=10 (made smaller many times):
# (P_4) The transformations, like that between °C and °F (see), which are obtained with
# a proportional transformation followed by a translation, are called affine maps. We
# can easily convert x by the affinity that transforms the interval [a, b] into the
# interval [c, d] by the command: affine(x, a,b, c,d). Here, how we can redefine Fa:
Fa = function(Ce) affine(Ce, 0,100, 32,212)
Plane(35,110, 5,220)
graph(Fa,40,120, "blue"); POINT(0,32,"brown"); POINT(100,212,"brown")
# The image in the center:
PLANE(1,10, 1,10);
x=c(0,2,4,10); y=c(1,1,1,1); POINT(x,y,"brown")
x1 = affine(x, 0,10, 5,10); POINT(x1,y*3,"blue")
x2 = affine(x, 0,10, 0,5); POINT(x2,y*6,"magenta")
x3 = affine(x, 0,10, 1,4); POINT(x3,y*9,"seagreen")
# On the right, affine transformations of x and of y:
PLANE(0,10, 0,10)
x = c(1,1,7,7,4,1,7,3,0,4); y = c(6,0,0,6,8,6,6,0,2,8); polyline(x,y, "blue")
x1=affine(x, 1,7, 5,9); y1=affine(y, 0,8, 1,10); polyline(x1,y1, "brown")
# You can make figures that are not easy to construct with other software. Here you can
# see how to draw equilateral but not equiangular polygons (with Direction and polyC).
Command INDEX
Introduction A
> source("http://macosa.dima.unige.it/r.R")
Use as calculator B  E
* / ; .
round sqrt rad2
rad3 fraction …e…
div divisors
Developments E1  E2
Trees divInt
Other calculi F  G
= c(…) sort
[ ] min max
length sum
Developments G1
> <
Figures H  J
PLANE BF HF circle
POINT Point circle3p center3p
radius3p Colors segm
halfline line2p line
point_point inclination angle
mid2p perp2p polyline
leng […:…] polyC
Developments J1  J6
areaPol xy arrow
dart dirArrow Direction
line_line solut point_inclina
line_line2 noPrint circleA
xrot yrot circle3pA type
circle_circle polylineR polyRC
triLLL triLAL triALA
incentre P_tan
Graphs K
Plane abovex abovey
function graph solution
Developments K1  K3
graphF coldash NpF NPF NpsF
maxmin
Statistics L  N
Bar Pie Strip
distribution min max
length stem histogram
noClass median mean
Histogram morestat statistics
Developments N1  N4
BAR BarH PIE PIEh rep Histogram
stem/scale histoclas pointO/D/V
More O  P
clock GCD primes
LCM Pie(0) gonio(0)
GONIO(0) Strip(0) mmpaper1/2/3
SQUARES INPUT OUTPUT
r/RATIO
Developments P1  P4
seq : PLANE/W/w/ww Plane/W/w/ww
circleC GRID/w Grid/w gridH/V GridH/V
Gonio CROSS CIRCLE LCROSS COUNT THERMO/2
affine
For other commands see Other examples of use.