We can now see that k-th term is (−1)k 1/k, and that there are 100 terms, so we would write the sum in sigma notation as X100 k=1 (−1)k 1 k Key Point To write a sum in sigma notation, try to find a formula involving a variable k where the first term can be obtained by setting k = 1, the second term by k = 2, and so on Exercises 3
Sigma notation is a way of writing a sum of many terms, in a concise 1 k + 1 Here, the index k takes the values 0, 1, 2, and 3 We’ll plug those each into 1 k+1
For each natural number k, we de ne p k(n) = Xn j=1 jk; for n = 1;2;::: Then p k(n) is a polynomial in n of degree k + 1, and its leading coe cient is 1 k + 1 In other words: Xn j=1 jk = nk+1 k + 1 + nkc k + n k 1c k 1 + :::nc 1 + c 0; where c 0;c 1;:::c k are some coe cients
Sigma Notation Questions Question 4 - Jan 2011 9 A sequence a 1, a 2, a 3, , is de ned by a 1 = k; a n+1 = 5a n + 3; n 1 where k is a positive integer (a) Write down an expression for a 2 in terms of k [1] (b) Show that a 3 = 25k + 18 [2] (c) i Find P4 r=1 a r in terms of k, in its simplest form ii Show that P4 r=1 a r is divisible by
Remark 0 1 It follows from the de nition that a countable intersection of sets in Ais also in A De nition 0 2 Let fA ng1 n=1 belong to a sigma algebra A We de ne limsupfA ng= \1 k=1 " [1 n=k A n #; and liminffA ng= [1 k=1 " \1 n=k A n #: Remark 0 2 (1) limsupfA ngis the set of points that are in in nitely many of the A n, and liminffA
Se lee: la sumatoria de los elementos a k desde k = n hasta k = m Nota: 1 Los límites superior e inferior de la sumatoria son constantes respecto del índice de la sumatoria 2 El límite inferior es cualquier número entero menor o igual que el límite superior; es decir, n m Ejemplo: Calcular: 5 k 3 (2k 1) Solución: 1(3)-1 k=3 2(4)-1 k=4
Title: MacroExercicesSupCorrige dvi Created Date: 1/24/2016 4:54:59 PM
• Sigma algebras can be generated from arbitrary sets This will be useful in developing the probability space Theorem: For some set X, the intersection of all σ-algebras, Ai, containing X −that is, x ∈X x ∈Ai for all i− is itself a σ-algebra, denoted σ(X) This is called the σ-algebra generatedby X Sigma-algebra Sample Space, Ω
time delta sigma modulators (DTDSMs) are recommended to see another tutorial paper [8] 2 Fundamentals of Delta-Sigma Modulators 2 1 Basic Characteristics Figure 1 shows the basic concept of the delta-sigma modula-tor The modulator is a kind of filter system for quantization noise by using a feedback loop
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Sigma Algebras and Borel Sets.
A.{Algebras.
Denition 0.1A collectionAof subsets of a setXis a-algebraprovided that (1); 2 A, (2) ifA2 Athen its complement is inA, and (3) a countable union of sets inAis also in A. Remark 0.1It follows from the denition that a countable intersection of sets inAis also inA. Denition 0.2LetfAng1n=1belong to a sigma algebraA. We dene limsupfAng=1 k=1" 1[ n=kA n# and liminffAng=1 k=1" 1\ n=kA n# Remark 0.2(1) limsupfAngis the set of points that are in innitely many of theAn, and liminffAngis the set of points that fail to be in at most nitely many of theAn, in other wordsx2liminffAngif and only if there is an indexksuch thatx2Anfor allnk. (2) Recall that iffxngis a bounded sequence of real numbers, then limsup n!1fxng= limn!1sup nkxn= infnsup nkxn because the sequenceyk= supnkxnis nonincreasing and bounded below. Also liminf n!1fxng= limn!1infnkxn= sup ninfnkxn because the sequenceyk= infnkxnis nondecreasing and bounded above. If we partially order the sets in the-algebraAby inclusion, then for any sequencefAng of sets, supfAng=1[ n=1A n; andinffAng=1\ n=1A n:
With this notation,
limsupfAng= inf [supfAng1n=k] and liminffAng= sup[inffAng1n=k] in analogy with the denition for sequences of real numbers. (3) In further analogy with the situation for sequences of real numbers, we have the following propositions. 1 Proposition 0.1Letfxngbe a sequence of real numbers and letAn= (1;xn). Then limsupfAng= (1;x)where x= limsupfxng and liminffAng= (1;x)where x= liminffxng:
Proposition 0.2liminffAng liminffAng.
B. Borel Sets.
Denition 0.3A setERis anFset provided that it is the countable union of closed sets and is aGset if it is the countable intersection of open sets. The collection ofBorel sets, denotedB, is the smallest-algebra containing the open sets. Remark 0.3(1) EveryGset is a Borel set. Since the complement of aGset is anFset, everyFset is a Borel set. (2) Every interval of the form [a;b) is both aGset and anFset and hence is a Borel set. In fact, the Borel sets can be characterized as the smallest-algebra containing intervals of the form [a;b) for real numbersaandb.
C. Example: Problem 44, Section 1.5.
Claim:Letpbe a natural number,p >1, andx2[0;1]. Then there is a sequence of integersfangwhere 0an< pand such that x=1X n=1a np n: This expansion is unique except whenx=q=pnfor some natural numberqin which case there are exactly two such expansions. Proof:What follows in an outline of the proof of the Claim. For 0k < pdene I
1;k= [k=p;(k+1)=p]. Clearly the intervalsI1;kare essentially disjoint in the sense that they
overlap in at most one point, and their union is [0;1]. Denea1to be a number such that x2I1;a1. The choice ofa1is unique except whenx=k=pfor some 1k < p. If we choose a
1=kand the sequence required by the Claim isfk;0;0;:::g. If we choosea1=k1 then
the sequence required by the Claim isfk1;p1;p1;:::g. (Note that for any natural numberk,1 X n=kp1p n=1p k1. This explains why the second sequence works.) In the case where there is no ambiguity in the choice ofa1, deney=xa1=pand the intervalsI2;k= [k=p2;(k+ 1)=p2] for 0k < p. Choosea2so thaty2I2;a2. Ambiguity will occur only ify=k=p2for some 1k < pand in this case the sequences required are either fa1;k;0;0;:::gorfa1;k1;p1;p1;:::g. 2 In general we proceed as follows. Assuming that there was no ambiguity in the choice of a
1; a2; :::; an1, deney=xPn1k=1ak=pk, and the intervalsIn;k= [k=pn;(k+ 1)=pn] for
0k < p. Chooseanso thaty2In;an. Ambiguity occurs only wheny=k=pnfor some
1k < p, and in this case the sequences required are eitherfa1;:::;an1;k;0;0;:::gor
fa1;:::;an1;k1;p1;p1;:::g. It remains to show that in factx=P1k=1ak=pk, and that the converse of the claim is true. Remark 0.4(1) Ifp= 10 then the expansion given above is the familiar decimal expansion of a number. Ifp= 2 the expansion is called the binary expansion, and ifp= 3 the ternary expansion. (2) Later on we will construct theCantor setC, a set with unusual and interesting properties. By comparing the construction ofCwith the above problem, it can be seen thatCconsists of all points in [0;1] that have a ternary expansion such thatan6= 1 for alln. 3quotesdbs_dbs2.pdfusesText_2
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