source("http://macosa.dima.unige.it/r.R") # If I have not already loaded the library ---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- # # Th symbolInfrepresents theinfinite. The "value"Infis assumed when a number # exceeds a certain value.factorial(170); factorial(171)#7.257416e+306 Inf # Similarly, the value0is assumed when the absolute value of a number is very small. # For details:.Machine#h <- function(x) exp(-x) Plane(0,10, -0,1); graph1(h, 0,10, "brown") integral(h,0, Inf)#1 # Thelimitof h as x -> Inf is 0:h(1); h(10); h(100); h(Inf)#0.3678794 4.539993e-05 3.720076e-44 0 #Note: thelimitof f(x)as x -> Infcan be estimated withf(Inf)not in all cases. # It is good to evaluate f(x) as x grows to control. Ex .:f <- function(x) (1+1/x)^x; f(Inf)#1 # whereas the limit ise:n <- 1:8; f(10^n)#2.593742 2.704814 2.716924 2.718146 2.718268 2.718280 2.718282 2.718282 # Thelimitsof q(x) as q approaches 1 from the right and from the left can be differentq <- function(x) (x^2-1)/(x^2-2*x+1); Plane(-2,2, -15,15); graph2(q,-2,2, "red")q(1-1E-2);q(1-1E-5);q(1-1E-10)#-199 -199999 -Infq(1+1E-2);q(1+1E-5);q(1+1E-10)#201 200001 Inf # # There aresequences[ie functions F whose domain is the set 0,1,2,3,… or the set # -2,-1,0,1,… or 6,7,8,9,…] such that F(n) can not be expressed with a simple term. # Consider for example #^{n}√((n+1)·(n+2)·...·(n+n))/n for n = 1, 2, 3, … # Let's see how to study thelimitof F(n) as n that tends to infinity. First of all, # I write the term in the easiest way to make calculations. In this particular case I # have a product of terms that grow as both in value and quantity. I should rewrite # F(n) as^{n}√(n+1)·^{n}√(n+2)·...·^{n}√(n+n)/n # I translate this intoR, using aforloop to express the product of multiple terms: F = function(n) {x=1; for(i in 1:n) x = x*(n+i)^(1/n); x/n} # Study what's happening. Try with n power of 2. First I also try with n+1, to check # that there are no differences if n is odd (with sequences, especially if there are # some exponentiations, this can happen, even if I could understand that this is not # the case). n=2; c(F(n), F(n+1) ) # 1.732051 1.644141 n=2^2; c(F(n), F(n+1) ) # [1] 1.600543 1.574513 n=2^3; c(F(n), F(n+1) ) # [1] 1.535668 1.528503 All OK, remove F(n+1) n=2^18; F(n) # [1] 1.47152 n=2^19; F(n) # [1] 1.471519 n=2^20; F(n) # [1] 1.471518 n=2^21; F(n) # [1] 1.471518 # I try with WolframAlpha if I can express this term more synthetically. # I get4/e. Verification: 4/exp(1) # [1] 1.471518 OK! # The study of aseries: 1-1/2+1/3-1/4+1/5... # The graph of N -> 1-1/2+ ... 1/Na <- function(n) (-1)^(n+1)/n # a(n): the nth element of the summation N <- function(n) seq(1,n,1) # N = 1 2 ... n S <- function(n) sum(a(N(n))) # sum a(1)+...a(n) NMax <- 20; Plane(0,NMax, 0,1) for(i in 1:NMax) POINT(i,S(i), "brown") for(i in 1:(NMax-1)) polyl(c(i,i+1),c(S(i),S(i+1)), "brown")# What is the value of this series? # I print the sum and the value of the next term that would be added.n <- 10; more(S(n)); a(n+1)#0.645634920634921 0.09090909n <- 100; more(S(n)); a(n+1)#0.688172179310195 0.00990099n <- 1e6; more(S(n)); a(n+1)#0.693146680560195 9.99999e-07> n <- 1e7; more(S(n)); a(n+1)#0.693147130559948 9.999999e-08n <- 1e8; more(S(n));a (n+1)#0.693147175559945 1e-08more(log(2))#0.693147180559945 # #Fourier seriesstudy (see here and here) is easily accomplished # (there are also specificRcommands to study these series). A simple example # (with A1 = A2 = B2 = B3 = A4 = B4 = … = 0) f0 = function(x) 2/3; f1 = function(x) 2*sin(x); f3 = function(x) cos(3*x)/2 F = function(x) f0(x)+f1(x)+f3(x) BF=6.5; HF=2; Plane(0,20, -2,3) graph(F,0,20, "black") graph1(f0,0,20, "brown"); graph1(f1,0,20, "blue"); graph1(f3,0,20, "seagreen") # An example with an infinite number of addends: # 4/π*sin(x) + 4/π*sin(3x)/3 + 4/π*sin(5x)/5 + ...p <- function(x) {f <- 0; for(i in 0:n) f <- f+sin((2*i+1)*x)/(2*i+1); f <- f*4/pi} BF=6.5; HF=2.3 n=0; graph1F(p, -5,10, "black") n=1; graph1F(p, -5,10, "blue") n=2; graph1(p, -5,10, "red") n=3; graph1(p, -5,10, "seagreen") BF=6.5; HF=2.3 n=20; graph1F(p, -5,10, "brown") BF=6.5; HF=2.3 n=2000; graphF(p, -5,10, "brown")# # To get a F function from its Fourier series (or polynomials) you can simply use # WolframAlpha (go here and type "fourier series") # In the case in question, if I put infunction to expand# Piecewise[ { {-1, -pi < x < 0}, {1, 0 < x < pi} } ] # (Is the step represented above, which is then repeated periodically) # and inorder1, 3 e 21, I get the following graphs and the corresponding polynomials # For example, for order = 3 I get the polynomial considered above: # If we represent the amplitude of the sinusoids, not in function of "x" as done # initially, but in function of the frequency, we get vertical segments. # Here is what we get for the Fourier polynomial consisting of four addends: Plane(0,0.2, -1.3,1.3) segm(1/(2*pi),0, 1/(2*pi),4/pi, "black") segm(1/(2*pi*3),0, 1/(2*pi*3),4/(pi*3), "blue") segm(1/(2*pi*5),0, 1/(2*pi*5),4/(pi*5), "red") segm(1/(2*pi*7),0, 1/(2*pi*7),4/(pi*7), "seagreen") # This is the so-calledspectrumof the polynomial. # The concept of "spectrum" is used in all situations where a function can be # represented as a sum of sinusoidal functions: the red function, sum of the three # sinusoids represented, and the corresponding spectrum: BF=3.5; HF=2.3 g=function(t) sin(1200*pi*t) h=function(t) 2*sin(800*pi*t) k=function(t) 3*sin(400*pi*t) a=-0.005; b=0.005; Piano(a,b, -5,5) graph2(g, a,b, "black"); graph2(h, a,b, "blue"); graph2(k, a,b, "green3") f=function(t) g(t)+h(t)+k(t); graph2(f, a,b, "red") # Plane(0,800, -5,5) Segm(600,0, 600,1, "black"); Segm(400,0, 400,2, "blue") Segm(200,0, 200,3, "green3") # the freq. are 1/period, ie 1200π/(2π) = 600, …Other examples of use