The scoping rules for R are the main feature that make it di erent from the original S language The scoping rules determine how a value is associated with a free variable in a function R uses lexical scoping or static scoping A common alternative is dynamic scoping Related to the scoping rules is how R uses the search list to bind a value to
integrate() - adaptive quadrature over a nite or in nite interval 4 Plotting plot() - generic R object plotting par() - set or query graphical parameters
3 Suppose that f: R R is a continuous function such that lim x1 f(x) = 0; lim x1 f(x) = 0: (a) Give a precise statement of what these limits mean (b) Prove that fis bounded on R and attains either a maximum or minimum value (c) Give examples to show that fmay: (i) attain its maximum but not its in mum; (ii) attain both its maximum and
Use R to Compute Numerical Integrals In short, you may use R to nd out a numerical answer to an n-fold integral I To integrate a one-dimensional integral over a nite or in nite interval, use R function integrate For example, nd out ∫ 1 0 1 (x+1) p x dx >## define the integrated function >integrand
Example 100 Consider again the function f: R R, f(x) = 4x 1 We want to know whether each element of R has a preimage Yes, it has, let us see why: we want to show that there exists xsuch that f(x) = 4x 1 = y Given y, we have the relation x= (y+ 1)=4 thus this xis indeed sent to y by f Example 101 Consider again the function g: R R, g(x
Package ‘pracma’ January 23, 2021 Type Package Version 2 3 3 Date 2021-01-22 Title Practical Numerical Math Functions Depends R (>= 3 1 0) Imports graphics, grDevices, stats, utils
Title R/Weka Interface Description An R interface to Weka (Version 3 9 3) Weka is a collection of machine learning algorithms for data mining tasks written in Java, containing tools for data pre-processing, classification, regression, clustering, association rules, and visualization Package 'RWeka' contains the interface code, the
1- Donner l’expression de la tension v en fonction de E; R0 et R 2- Montrer qu’à l’équilibre du pont ( lorsque v = 0 V ), on a : R = R0 3- En utilisant le tableau caractérisant le capteur résistif, exprimer R en fonction de P Déterminer alors la valeur des résistances réglables R0
The revenue function , R(x), is the total revenue realized from the sale of x units of the product The profit function , P(x), is the total profit realized from the manufacturing and sale of the x units of product Formulas: Suppose a firm has fixed cost of F dollars, production cost of c dollars per unit and selling
[PDF] read.csv2 r
[PDF] which dans r
[PDF] opérateur logique r
[PDF] fonction subset r
[PDF] r compter le nombre d'occurence
[PDF] read.table sous r
[PDF] fonction apply r
[PDF] croquis france potentialités et contraintes
[PDF] néon configuration électronique
[PDF] ion carbone formule
[PDF] le territoire français des milieux aménagés des milieux ? ménager
[PDF] ion sulfite
[PDF] exercice enthalpie libre et potentiel chimique
[PDF] palme funeraire
[PDF] ion sodium nombre de charges positives du noyau
Solutions to Sample Questions
Midterm 1: Math 125A, Fall 2012
1.(a) Suppose thatf: (0;1)!Ris uniformly continuous on (0;1). If (xn)
is a Cauchy sequence in (0;1) andyn=f(xn), prove that (yn) is a Cauchy sequence inR. (b) Give a counter-example to show that the result in (a) need not be true iff: (0;1)!Ris only assumed to be continuous.
Solution.
(a) Let >0 be given. Sincefis uniformly continuous on (0;1), there exists >0 such that jxyj< andx;y2(0;1) implies thatjf(x)f(y)j< : Since (xn) is a Cauchy sequence, there existsN2Nsuch that m;n > Nimplies thatjxmxnj< :
It follows that
m;n > Nimplies thatjf(xm)f(xn)j< ; which shows that (f(xn)) is a Cauchy sequence. (b) Suppose thatf(x) = 1=xforx2(0;1) andxn= 1=nforn2N. Thenfis continuous on (0;1) since it is a rational function with nonzero denominator. The sequence (xn) is Cauchy since it converges to 0 and every convergent sequence is Cauchy (or give a direct proof). However, y n=f(xn) =nand jynymj 1 for everym;n2Nwithm6=n; so (yn) is not Cauchy. Remark.Since every Cauchy sequence converges, it follows from this result that we can extend a uniformly continuous functionf: (0;1)!Rto a uniformly continuous functionf: [0;1]!Rby dening f(0) = limn!1f(xn) where (xn) is any sequence in (0;1) such thatxn!0 asn! 1, and f(1) = limn!1f(xn) where (xn) is any sequence in (0;1) withxn!1 asn! 1. (You have to check that the valuesf(0) andf(1) are independent of the choice of sequences to show thatfis well dened.) However, we cannot extend a non-uniformly continuous function on (0;1), such asf(x) = 1=x, to a continuous function on [0;1].
2.(a) State the-denition for a functionf:R!Rto be continuous at
c2R. (b) Dene the oor functionf:R!Rby f(x) = the largest integern2Zsuch thatnx: For example,f(3:14) = 3,f(7) = 7,f(3:14) =4. Determine, with proof, wherefis continuous and where it is discontinuous.
Solution.
(a) A functionf:R!Ris continuous atc2Rif for every >0 there exists >0 such that jxcj< implies thatjf(x)f(c)j< : (b) The oor function is discontinuous at every integerc2Zand continuous at everyc =2Z.
Ifc2Z, dene sequences (xn), (yn) by
x n=c1n ; yn=c+1n
Thenxn!candyn!casn! 1, but for everyn2N
f(xn) =c1; f(yn) =c sof(xn)!c1 andf(yn)!cconverge to dierent limits. The sequential denition of continuity implies thatfis discontinuous atc. (It has a jump discontinuity atc2Z.) Suppose thatc =2Z. Thenn < c < n+ 1 for some integern2Z, and we can dene >0 by = min(cn;n+ 1c): Sincejxcj< implies thatn < x < n+ 1 andf(x) =nfor all such x, we have jxcj< implies thatjf(x)f(c)j= 0: Therefore we can use this >0 for every >0 in the denition of continuity, andfis continuous atc.
3.Suppose thatf:R!Ris a continuous function such that
lim x!1f(x) = 0;limx!1f(x) = 0: (a) Give a precise statement of what these limits mean. (b) Prove thatfis bounded onRand attains either a maximum or minimum value. (c) Give examples to show thatfmay: (i) attain its maximum but not its inmum; (ii) attain both its maximum and minimum.
Solution.
(a) The statement limx!1f(x) = 0 means that for every >0 there existsa2R(suciently negative) such that x < aimplies thatjf(x)j< ; and lim x!1f(x) = 0 means that for every >0 there existsb2R (suciently positive) such that x > bimplies thatjf(x)j< : (b) Iff0 is identically zero, then the result follows immediately. If not, choosec2Rsuch thatf(c)6= 0. Taking=jf(c)j>0 in the limit denitions, we nd that there exista;b2Rsuch that jf(x)j
b;(1) whereacb(sincef(x)6=f(c) ifx < aorx > b). Sincefis continuous on the compact interval [a;b] it is bounded on [a;b] and attains its maximum and minimum values on [a;b]. It follows from (1) thatfis bounded onR. Moreover, iff(c)>0, then maxff(x) :x2[a;b]g f(c) sofattains its global maximum onRat some point in [a;b]. Similarly, iff(c)<0, then minff(x) :x2[a;b]g f(c) sofattains its global minimum onRat some point in [a;b]. (c) The function,f:R!Rdened by f(x) =11 +x2 attains its maximum value,f(0) = 1, but not its inmum 0 onR. The functiong:R!Rdened by
g(x) =x1 +x2: attains both its maximum value,g(1) = 1=2, and minimum value, g(1) =1=2, onR.quotesdbs_dbs16.pdfusesText_22
Introduction to the R Language - Functions