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The fractions obtained from a continuedfraction by omitting all its elements from a certain one on are called the convergents.The value assigned to an infinite continued fraction is the limit of its convergents.Show that:a) Every rational numberner as a continued fraction:mn,where m, n ∈ N can be expanded in a unique man-m= q1 +nq2 +1q3 +,1...1qn−1 + q1nassuming that qn = 1 for n > 1.Hint: The numbers q, .
. . , qn , called the incomplete quotients or elements, can beobtained from the Euclidean algorithmm = n · q1 + r1 ,n = r1 · q2 + r2 ,r1 = r2 · q3 + r3 ,...by writing it in the formm11= q1 += q1 +nn/r1q2 +b) The convergents R1 = q1 , R2 = q1 +R1 < R3 < · · · < R2k−1 <1q2 , . . .....satisfy the inequalitiesm< R2k < R2k−2 < · · · < R2 .nc) The numerators Pk and denominators Qk of the convergents are formed according to the following rule:Pk = Pk−1 qk + Pk−2 ,P2 = q1 q2 ,Qk = Qk−1 qk + Qk−2 ,Q2 = q2 ,P1 = q1 ,Q1 = 1.3.1 The Limit of a Sequence105d) The difference of successive convergents can be computed from the formulaRk − Rk−1 =(−1)kQk Qk−1(k > 1).e) Every infinite continued fraction has a determinate value.f) The value of an infinite continued fraction is irrational.g)√1+ 51=1+.121+1+...h) The Fibonacci numbers 1, 1, 2, 3, 5, 8, . .
. (that is, un = un−1 + un−2 andu1 = u2 = 1), which are obtained as the denominators of the convergents in g),are given by the formula√ √ %$1+ 5 n11− 5 n.un = √−225i) The convergents Rk =PkQk√in g) are such that | 1+25this result with the assertions of Exercise 11 in Sect. 2.2.Pk−Q|>k1√.Q2k 5Compare5.
Show thata) the equality1+111111+ + ··· + +=3−− ··· −1! 2!n! n!n1 · 2 · 2!(n − 1) · n · n!holds for n ≥ 2;b) e = 3 − ∞n=01(n+1)(n+2)(n+2)! ;c) for computing the number e approximately the formula e ≈ 1 + 1!1 + 2!1 +111+ n!nis much better than the original formula e ≈ 1 + 1!1 + 2!1 + · · · + n!.· · · + n!(Estimate the errors, and compare the result with the value of e given on p. 103.)6.
If a and b are positive numbers and p an arbitrary nonzero real number, then themean of order p of the numbers a and b is the quantitySp (a, b) =a p + bp21p.In particular for p = 1 we obtain the arithmetic mean of a and b, for p = 2 theirsquare-mean, and for p = −1 their harmonic mean.a) Show that the mean Sp (a, b) of any order lies between the numbers a and b.1063b) Find the limits of the sequencesSn (a, b) ,LimitsS−n (a, b) .7.
Show that if a > 0, the sequence xn+1 = 12 (xn + xan ) converges to the square rootof a for any x1 > 0.Estimatethe rate of convergence, that is, the magnitude of the absolute error√|xn − a| = |Δn | as a function of n.8. Show thata)S0 (n) = 10 + · · · + n0 = n,n(n + 1) 1 2 1= n + n,222n(n+1)(2n+1)111S2 (n) = 12 + · · · + n2 == n3 + n2 + n,6326S1 (n) = 11 + · · · + n1 =S3 (n) =n2 (n + 1)2 1 4 1 3 1 2= n + n + n ,4424and in general thatSk (n) = ak+1 nk+1 + · · · + a1 n + a0is a polynomial in n of degree k + 1.(n)1b) limn→∞ Snkk+1= k+1.3.2 The Limit of a Function3.2.1 Definitions and ExamplesLet E be a subset of R and a a limit point of E. Let f : E → R be a real-valuedfunction defined on E.We wish to write out what it means to say that the value f (x) of the functionf approaches some number A as the point x ∈ E approaches a.
It is natural to callsuch a number A the limit of the values of the function f , or the limit of f as xtends to a.Definition 1 We shall say (following Cauchy) that the function f : E → R tends toA as x tends to a, or that A is the limit of f as x tends to a, if for every ε > 0 thereexists δ > 0 such that |f (x) − A| < ε for every x ∈ E such that 0 < |x − a| < δ.In logical symbolism these conditions are written as∀ε > 0 ∃δ > 0 ∀x ∈ E 0 < |x − a| < δ ⇒ f (x) − A < ε .3.2 The Limit of a Function107If A is the limit of f (x) as x tends to a in the set E, we write f (x) → Aas x → a, x ∈ E, or limx→a,x∈E f (x) = A. Instead of the expression x → a,x ∈ E, we shall as a rule use the shorter notation E x → a, and instead oflimx→a,x∈E f (x) we shall write limEx→a f (x).Example 1 Let E = R\0, and f (x) = x sin x1 .
We shall verify thatlim x sinEx→01= 0.xIndeed, for a given ε > 0 we choose δ = ε. Then for 0 < |x| < δ = ε, takingaccount of the inequality |x sin x1 | ≤ |x|, we shall have |x sin x1 | < ε.Incidentally, one can see from this example that a function f : E → R may havea limit as E x → a without even being defined at the point a itself. This is exactlythe situation that most often arises when limits must be computed; and, if you werepaying attention, you may have noticed that this circumstance is taken into accountin our definition of limit, where we wrote the strict inequality 0 < |x − a|.We recall that a neighborhood of a point a ∈ R is any open interval containingthe point.Definition 2 A deleted neighborhood of a point is a neighborhood of the point fromwhich the point itself has been removed.If U (a) denotes a neighborhood of a, we shall denote the corresponding deletedneighborhood by Ů (a).The setsUE (a) := E ∩ U (a)Ů E (a) := E ∩ Ů (a)will be called respectively a neighborhood of a in E and a deleted neighborhood ofa in E.If a is a limit point of E, then Ů E (a) = ∅ for every neighborhood U (a).δIf we temporarily adopt the cumbersome symbols Ů E (a) and VRε (A) to denotethe deleted δ-neighborhood of a in E and the ε-neighborhood of A in R, thenCauchy’s so-called “ε–δ-definition” of the limit of a function can be rewritten as δδlim f (x) = A := ∀VRε (A) ∃Ů E (a) f Ů E (a) ⊂ VRε (A) .Ex→aThis expression says that A is the limit of the function f : E → R as x tendsto a in the set E if for every ε-neighborhood VRε (A) of A there exists a deletedδδneighborhood Ů E (a) of a in E whose image f (Ů E (a)) under the mapping f :E → R is entirely contained in VRε (A).1083LimitsTaking into account that every neighborhood of a point on the real line containsa symmetric neighborhood (a δ-neighborhood) of the same point, we arrive at thefollowing expression for the definition of a limit, which we shall take as our maindefinition:Definition 3 lim f (x) = A := ∀VR (A) ∃Ů E (a) f Ů E (a) ⊂ VR (A) .Ex→aThus the number A is called the limit of the function f : E → R as x tends to awhile remaining in the set E (a must be a limit point of E) if for every neighborhoodof A there is a deleted neighborhood of a in E whose image under the mappingf : E → R is contained in the given neighborhood of A.We have given several statements of the definition of the limit of a function.
Fornumerical functions, when a and A belong to R, as we have seen, these statementsare equivalent. In this connection, we note that one or another of these statementsmay be more convenient in different situations. For example, the original form isconvenient in numerical computations, since it shows the allowable magnitude ofthe deviation of x from a needed to ensure that the deviation of f (x) from A willnot exceed a specified value. But from the point of view of extending the concept ofa limit to more general functions the last statement the definition is most convenient.It shows that we can define the concept of a limit of a mapping f : X → Y providedwe have been told what is meant by a neighborhood of a point in X and Y , that is,as we say, a topology is given on X and Y .Let us consider a few more examples that are illustrative of the main definition.Example 2 The function⎧if x > 0,⎨1if x = 0,sgn x = 0⎩−1 if x < 0(read “signum x”8 ) is defined on the whole real line.
We shall show that it has nolimit as x tends to 0. The nonexistence of this limit is expressed by∀A ∈ R ∃V (A) ∀Ů (0) ∃x ∈ Ů (0) f (x) ∈/ V (A) ,that is, no matter what A we take (claiming to be the limit of sgn x as x → 0), thereis a neighborhood V (A) of A such that no matter how small a deleted neighborhoodŮ (0) of 0 we take, that deleted neighborhood contains at least one point x at whichthe value of the function does not lie in V (A).8 TheLatin word for sign.3.2 The Limit of a Function109Since sgn x assumes only the values −1, 0, and 1, it is clear that no numberdistinct from them can be the limit of the function.
For any such number has aneighborhood that does not contain any of these three numbers.But if A ∈ {−1, 0, 1} we choose as V (A) the ε-neighborhood of A with ε = 12 .The points −1 and 1 certainly cannot both lie in this neighborhood. But, no matterwhat deleted neighborhood Ů (0) of 0 we may take, that neighborhood contains bothpositive and negative numbers, that is, points x where f (x) = 1 and points wheref (x) = −1.Hence there is a point x ∈ Ů (0) such that f (x) ∈/ V (A).If the function f : E → R is defined on a whole deleted neighborhood of a pointa ∈ R, that is, when Ů E (a) = Ů R (a) = Ů (a), we shall agree to write more brieflyx → a instead of E x → a.Example 3 Let us show that limx→0 | sgn x| = 1.Indeed, for x ∈ R\0 we have | sgn x| = 1, that is, the function is constant andequal to 1 in any deleted neighborhood Ů (0) of 0.
Hence for any neighborhoodV (1) we obtain f (Ů (0)) = 1 ∈ V (1).Note carefully that although the function | sgn x| is defined at the point 0 itselfand | sgn 0| = 0, this value has no influence on the value of the limit in question.Thus one must not confuse the value f (a) of the function at the point a with thelimit limx→a f (x) that the function has as x → a.Let R− and R+ be the sets of negative and positive numbers respectively.Example 4 We saw in Example 2 that the limit limRx→0 sgn x does not exist.
Remarking, however, that the restriction sgn |R− of sgn to R− is a constant functionequal to −1 and sgn |R+ is a constant function equal to 1, we can show, as in Example 3, thatlimR− x→0sgn x = −1,andlimR+ x→0sgn x = 1,that is, the restrictions of the same function to different sets may have different limitsat the same point, or even fail to have a limit, as shown in Example 2.Example 5 Developing the idea of Example 2, one can show similarly that sin x1 hasno limit as x → 0.Indeed, in any deleted neighborhood Ů (0) of 0 there are always points of the11form −π/2+2πnand π/2+2πn, where n ∈ N.
At these points the function assumesthe values −1 and 1 respectively. But these two numbers cannot both lie in the εneighborhood V (A) of a point A ∈ R if ε < 1. Hence no number A ∈ R can be thelimit of this function as x → 0.Example 6 IfE− = x ∈ R | x =1,n ∈ N−π/2 + 2πn110and3E+ = x ∈ R | x =Limits1,n ∈ N ,π/2 + 2πnthen, as shown in Example 4, we find thatlimE− x→0sin1= −1 andxlimE+ x→0sin1= 1.xThere is a close connection between the concept of the limit of a sequence studied in the preceding section and the limit of an arbitrary numerical-valued functionintroduced in the present section, expressed by the following proposition.Proposition 1 9 The relation limEx→a f (x) = A holds if and only if for everysequence {xn } of points xn ∈ E\a converging to a, the sequence {f (xn )} convergesto A.Proof The fact that (limEx→a f (x) = A) ⇒ (limn→∞ f (xn ) = A) follows immediately from the definitions.
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