# Convergent Sequence Examples Pdf

Pointwise convergence is a very weak kind of convergence. 10 Procedure for Estimating Adjusted Net Saving 61 2. We will see two. We say that the sequence n D U converges to zero in D U if. , there is some c so that, for all k, kx kk c. The polarization identity expresses the norm of an inner product space in terms of the inner product. 1 Introduction 23 2. -Mix two forms of data in different ways. For example, 10 + 20 + 20…does not converge (it just keeps on getting bigger). It is nearly identical to existing sample sequences. Since the product of two convergent sequences is convergent the sequence fa2. The relationships between different types of convergence are summarized in Figure 4. 1 Weak convergence in normed spaces We recall that the notion of convergence on a normed space X, which we used so far, is the convergence with respect to the norm on X: namely, for a sequence (x n) n 1, we say that x n!xif kx n xk!0 as n!1. 1 For the geometric sequence with r ~ 0; i. Convergent sequences, Divergent sequences, Sequences with limit, sequences without limit, Oscillating sequences. examples below, these are differences that should make a difference in the planning and management activities of any crisis relevant groups. , λ*) corresponding to λ*. Convergent Sequences Subsequences Cauchy Sequences Examples Notice that our de nition of convergent depends not only on fp ng but also on X. 5, c) 9, -3, 1, 5. Relative to convergence, it is the behavior in the large-n limit that matters. and Xis a r. A rather complete treatment of these and related problems was. ( ) 0 0 (4. for all sequences (x n) in X. The proof can be found in a number of texts, for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl; New York: Dover, 1956). Hence, we have, which implies. A sequence is bounded if it is bounded above and below, that is to say, if there is a number, k, less than or equal to all the terms of sequence and another number, K', greater than or equal to all the terms of the sequence. Transition Kernel of a Reversible Markov. Sample Quizzes with Answers Search by content rather than week number. Now we will investigate what may happen when we add all terms of a sequence together to form what will be called an infinite series. Ratio Test. A sequence in R is a list or ordered set: (a 1, a 2, a 3,. If the limit of s k is infinite or does not exist, the series is said to diverge. A sequence { } is Cauchy if, for every ,there exists an such that ( ) for every Thus, a Cauchy sequence is one such that its elements become arbitrarily 'close together' as we move down the sequence. The Riemann Integral and the Mean Value Theorem for Integrals 4 6. A Cauchy sequence (pronounced CO-she) is an infinite sequence that converges in a particular way. De nition 0. The limit of a sequence is said to be the fundamental notion on which the whole of analysis ultimately rests. Nested intervals. ANALYSIS I 7 Monotone Sequences Example. For example, the divergent sequence of partial sums of the harmonic series (see this earlier example) does satisfy this property, but not the condition for a Cauchy sequence. Conversely, it follows from Theorem 1. In Rk, every Cauchy sequence converges. We’ll look at this one in a moment. The language of this test emphasizes an important point: the convergence or divergence of a series depends entirely upon what happens for large n. An approximation theory for sequences of this kind has recently been developed, with the aim of providing tools for computing their asymptotic singular value and eigenvalue distributions. The sequences are progressive (hierarchical): any prefix is well distributed, making them suitable for incremental rendering and adaptive sampling. For example, 1 + x+ x2 + + xn+ is a power series. A sequence has the Cauchy property if and only if it is convergent. This says that if the series eventually behaves like a convergent (divergent) geometric series, it converges (diverges). Convergence generally means coming together, while divergence generally means moving apart. It also explores particular types of sequence known as arithmetic progressions (APs) and geometric progressions (GPs), and the corresponding series. 2) The sequence can approach one of the two infinities. Nets Take a moment to verify to yourself that the use of the word \tail" in this context agrees with its use in the context of sequences, and that in the case where D= N with its usual ordering, this agrees with the usual de nition of sequence convergence. a sequence does not have to converge to a given ﬁxed point (unless a0 is already equal to the ﬁxed point). Let f: D → C be a function. (In fact, the only books. Show that (X,d) in Example 4 is a metric. We will now look at two very important terms when it comes to categorizing sequences. Calculus III: Sequences and Series Notes (Rigorous Version) Logic De nition (Proposition) A proposition is a statement which is either true or false. Also in different example, you learn to generate the Fibonacci sequence up to a certain number. Note that each x n is an irrational number (i. The ratio of successive pairs of numbers in this sequence converges on 1. A definition is given of convergence of a sequence of sets to a set, written X„ -> X, where X and the Xn are subsets of Euclidean m-space £"'. Meaning 'the sum of all terms like', sigma notation is a convenient way to show where a series begins and ends. I think we must be getting close to some calculus. For one thing, it is common for the sum to be a relatively arbitrary irrational number:. Determine if the sequence converges or diverges. The sequence xn converges to something if and only if this holds: for every >0 there. its limit doesn't exist or is plus or minus infinity) then the series is also called divergent. 2 Limit Laws The theorems below are useful when -nding the limit of a sequence. If the limit of s k is infinite or does not exist, the series is said to diverge. Stayton (2008) demonstrated that rates of convergence can be. Take a neighborhood U of x. We know that a n!q. This is a collection of lecture notes I’ve used several times in the two-semester senior/graduate-level real analysis course at the University of Louisville. Show that ($\sum \frac{\ln(n)}{n}$) diverges. Divergent Sequences. Mixed Methods Research •Characteristics of mixed methods research -Collect and analyze both quantitative and qualitative data. The sum of convergent and divergent series Kyle Miller Wednesday, 2 September 2015 Theorem 8 in section 11. For example, take any three numbers and sum them to make a fourth, then continue summing the last three numbers in the sequence to make the next. The Coupon Collector’s Problem 13 2. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. I Integral test, direct comparison and limit comparison tests,. A convergent sequence has a limit — that is, it approaches a real number. If (an)1=n ‚ 1 for all su-ciently large n, then P n an is divergent. Properties of the sample autocovariance function The sample autocovariance function: ˆγ(h) = 1 n nX−|h| t=1 (xt+|h| −x¯)(xt −x¯), for −n 0 c ﬁnite & an,bn > 0? Does. This says that if the series eventually behaves like a convergent (divergent) geometric series, it converges (diverges). In Chapter 1 we discussed the limit of sequences that were monotone; this restriction allowed some short-cuts and gave a quick introduction to the concept. A unifying approach to convergence of linear sampling type operators in Orlicz spaces Vinti, Gianluca and Zampogni, Luca, Advances in Differential Equations, 2011; On the Graph Convergence of Sequences of Functions Grande, Zbigniew, Real Analysis Exchange, 2008. Metric spaces are generalizations of the real line, in which some of the theorems that hold for R remain valid. ©v Q2G0U1T6N dKQuKtJaY rS]oBfzt]wuaTrGe] _LpLTCH. 4 Convergence 32 2. Since this test for convergence of a basic-type improper integral makes use of a limit, it's called the limit comparison test , abbreviated as LCT. Solutions to Problems in Chapter 2 2. A Convergence Test for Sequences Thm: lim n!1 ﬂ ﬂ ﬂ ﬂ an+1 an ﬂ ﬂ ﬂ ﬂ = L < 1 =) lim n!1 an = 0 In words, this just says that if the absolute value of the ratio of successive terms in a sequence fangn approaches a limit L, and if L < 1, then the sequence itself converges to 0. For example, 1 + x+ x2 + + xn+ is a power series. Introduction One of the most important parts of probability theory concerns the be-havior of sequences of random variables. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. Recall the sequence (x n) de ned inductively by x 1 = 1; x n+1 = (1=2)x n + 1;n2N:. Sequences that are not convergent are said to be divergent. • If f 0 (r) = 0, the sequence converges at least quadratically to the ﬁxed point (this is sometimes called superconvergence in the dynamical systems literature). WIJSMAN(i) 0. Absolute and conditional convergence Remarks: I Several convergence tests apply only to positive series. These results are not original, and similar results on the relation between the limits of the series and these two sequences (or related sequences) have appeared in the literature before. Thanks to all of you who support me on Patreon. Subsequences. ROC contains strip lines parallel to jω axis in s-plane. Then as n→∞, and for x∈R F Xn (x) → (0 x≤0 1 x>0. ˆ1 + i 2 , 2 + i 22. 3 Complexiﬁcation of the Integrand. Some of the earliest and best examples of convergent sequence evolution include the stomach lysozymes of langurs and cows (Stewart et al. A sequence {xn} is inﬁnitely large if for any ε > 0 only a ﬁnite number of points (n,xn) are between the two horizontal lines y = −ε, and y = +ε. p This integral converges for all p > 0, so the series converges for all p > 0. , After measuring, we choose a set of parameters i and build our. Ratio Test. Before we discuss the idea behind successive approximations, let’s ﬁrst express a ﬁrst- order IVP as an integral equation. A series which is larger than a convergent series might converge or diverge. Let (a n) be the sequence de ned by a n= 1 1 n; n 1: Evaluate limsup n!1 a nand liminf. For a Cauchy sequence, the terms get "closer together" the "farther out" you go in the sequence. This is a set of exercises and problems for a (more or less) standard beginning calculus sequence. Sequences of Functions We now explore two notions of what it means for a sequence of functions ff ng n2N to converge to a function f. Assume that lim n!1 an exists for anC1 D p 3an with a0 D2: Find lim n!1 an. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. is convergent. It's important to understand what is meant by convergence of series be fore getting to numerical analysis proper. Cauchy Sequences ⇔ Convergent Sequences A sequence of real numbers is said to be Cauchy if , limit X( ) ( ) 0 →∞ −=X nm nm. >1/-normed space. 4 and Example 3. • If f 0 (r) = 0, the sequence converges at least quadratically to the ﬁxed point (this is sometimes called superconvergence in the dynamical systems literature). The meanings of the terms “convergence” and “the limit of a sequence”. Example: A convergent sequence in a metric space is bounded; therefore the set of convergent real sequences is a subset of '1. convergence synonyms, convergence pronunciation, convergence translation, English dictionary definition of convergence. Series •Given a sequence {a 0, a 1, a2,…, a n} •The sum of the series, S n = •A series is convergent if, as n gets larger and larger, S n goes to some finite number. 2 Sequences: Convergence and Divergence In Section 2. In the world of finance and trading, convergence and divergence are terms used to describe the. The relationships between different types of convergence are summarized in Figure 4. Let be a convergent series of real nonnegative terms. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. Cauchy Sequences and Complete Metric Spaces Let's rst consider two examples of convergent sequences in R: Example 1: Let x n = 1 n p 2 for each n2N. The proof can be found in a number of texts, for example, Infinite Sequences and Series, by Konrad Knopp (translated by Frederick Bagemihl; New York: Dover, 1956). 8 Order Properties of Limits 47 2. Summary of Convergence estsT for Series estT Series Convergence or Divergence Comments n th term test (or the zero test) X a n Diverges if lim n !1 a n 6= 0 Inconclusive if lim a n = 0. Using models developed by Garcia. Let (a n) be the sequence de ned by a n= 1 1 n; n 1: Evaluate limsup n!1 a nand liminf. Convergent,Divergent & Oacillatory Sequences with examples - Lesson 2-In Hindi-{Infinite Sequences} - Duration: 53:27. Convergence of Random Variables 5. • Answer all questions. 2 Convergence Index 7. A sequence can be defined by a formula (or generator) which generates each term. We note that absolute convergence of an inﬁnite series is necessary and sufﬁcient to allow the terms of a series to be. Denition 7. Nets and lters (are better than sequences) 3. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. Alternating series and absolute convergence (Sect. 2 Tests for Convergence Let us determine the convergence or the divergence of a series by comparing it to one whose behavior is already known. Let us refer to these metrics as d 1 and d 2 respectively, and suppose that the sequence (x k) converges to in the 6. (1) is pointwise convergent over the interval x ∈ A. If this limit is one , the test is inconclusive and a different test is required. Whereas, in this case the output of the experiment is a random sequence, i. A complete normed linear space is called a Banach space. n) is convergent to 1 and the subsequence (a 2n 1) of (a n) is convergent to 1: Later, we will prove that in general, the limit supremum and the limit in mum of a bounded sequence are always the limits of some subsequences of the given sequence. 6 Boundedness Properties of Limits 39 2. sequence are increasing. Thus, fx ngconverges in R (i. striatum and C. Let {fn}∞ n=1 be a sequence of real or complex-valued functions deﬁned on a domain D. Algebraic manipulations give, since. These notes are sef-contained, but two good extra references for this chapter are Tao, Analysis I; and Dahlquist. Thus the space is not sequentially compact and by Lemma 3 it is not compact, a contradiction to our hypothesis. pdf from MATH 2 at Wuhan University of Technology. We know when a geometric series converges and what it converges to. 1 The pattern may for instance be that: there is a convergence of X. the merging of distinct technologies, industries, or devices into a unified whole n. Series of Numbers 4. Almost sure convergence, convergence in probability and asymptotic normality In the previous chapter we considered estimator of several diﬀerent parameters. The definition of convergence of a sequence was given in Section 11. Now we will investigate what may happen when we add all terms of a sequence together to form what will be called an infinite series. Some are quite easy to understand: If r = 1 the sequence converges to 1 since every term is 1, and likewise if r = 0 the sequence converges to 0. Coupling Constructions and Convergence of Markov Chains 10 2. 10 Examples of Limits 56 2. A series ∑a n is said to converge or to be convergent when the sequence (s k) of partial sums has a finite limit Examples of convergent sequences. Definition of Convergence and Divergence in Series The n th partial sum of the series a n is given by S n = a 1 + a 2 + a 3 + + a n. One possibility is ˆ ( 1)n 1 n ˙ +1 n=1 = 1; 1 2; 1 3; 1 4;:::, which converges to 0 but is not monotonic. is convergent. 1) occur in applications ([7, 8, and 18], for example), and it can be shown that all slowly convergent sequences occurring as examples in the references of the present paper satisfy (1. , to an element of R). The Cauchy criterion for uniform convergence of a series gives a condition for the uniform convergence of the series (1) on without using the sum of the series. 11 Subsequences 78 2. Universal nets 12 4. For real inner product spaces it is (x,y) = 1 4 (kx+ yk2 −kx−yk2). Convergence and Divergence of Sequences. In this post, we will focus on examples of. 12 Adjusted Net Saving, by Region, 1995–2015 63. If a series is divergent and you erroneously believe it is convergent, then applying these tests will lead only to extreme frustration. Sequences and series of functions: uniform convergence Pointwise and uniform convergence We have said a good deal about sequences of numbers. It is also possible to prove that a convergent sequence has a unique limit, i. Can you find an example ? While we now know how to deal with convergent sequences, we still need an easy criteria that will tell us whether a sequence converges. CONVERGENCE PETE L. the merging of distinct technologies, industries, or devices into a unified whole n. Therefore, {fn} converges pointwise to the function f = 0 on R. We note that absolute convergence of an inﬁnite series is necessary and sufﬁcient to allow the terms of a series to be. Order and Rates of Convergence 1 Order of convergence 11 Suppose we have that Then the convergence of the sequence x k to ¯x is said. ter estimates as well as the Potential Scale Reduction (PSR) convergence criteria, which compares several independent MCMC sequences. 1 n, 2 3 n are examples of null sequences since lim n = 0 and lim 2 3 n = 0. Exercises 15 2. You should be able to verify that the set is actually a vector subspace of ‘1. a sequence of xed numbers. F-convergence, lters and nets The main purpose of these notes is to compare several notions that describe convergence in topological spaces. Deﬁnition: A sequence {v k} of vectors in a normed linear space V is Cauchy conver-gent if kv m − v nk → 0 as m,n → ∞. Show that weak* convergent sequences in the dual of a Banach space are bounded. Neal, WKU MATH 532 Sequences of Functions Throughout, let (X ,F, ) be a measure space and let {fn}n=1 ∞ be a sequence of real- valued functions defined on X. Fatou's lemma and the dominated convergence theorem are other theorems in this vein,. Find the radius of convergence R and the domain of convergence S for each of the following power series: X∞ n=0 xn, X∞ n=1 x n n, X∞ n=0 x nn, X∞ n=0 nnxn, X∞ n=0 x n!, X∞ n=0 (−1)n n2 x2n Hwk problem: if the series P ∞ k=0 4 na n is convergent, then P ∞ n=0 a n(−2) n is also con-vergent. In Rk, every Cauchy sequence converges. A convergent sequence has a limit — that is, it approaches a real number. For example, once we have computed from the first equation, its value is then used in the second equation to obtain the new. Thus the space is not sequentially compact and by Lemma 3 it is not compact, a contradiction to our hypothesis. 9 Monotone Convergence Criterion 52 2. Let † > 0. Before introducing almost sure convergence let us look at an example. Quadratic Convergence of Newton’s Method Michael Overton, Numerical Computing, Spring 2017 The quadratic convergence rate of Newton’s Method is not given in A&G, except as Exercise 3. Such a sequence does not exist, indeed, if it had a convergent subsequence a m k. Two examples of nets in analysis 11 3. Theorem 317 Let (a n. Similarities among protein sequences are reminiscent of homology and convergent evolution via common ancestry and/or selective pressure, respectively. , x n 2Qc) and that fx ngconverges to 0. Discuss the pointwise convergence of the sequence. There are three main results: the rst one is that uniform convergence of a sequence of continuous. For positive term series, convergence of the sequence of partial sums is simple. 9 Uniform Convergence of Sequences of Functions In this chapter we consider sequences and. weakly convergent and weak* convergent sequences are likewise bounded. Of these, 10 have two heads and three tails. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. A sequence is "converging" if its terms approach a specific value as we progress through them to infinity. an e–cient way and will lead us to criteria for the convergence of rearrangements. You da real mvps! \$1 per month helps!! :) https://www. Prove that if ff ngconverges uniformly to f on X, then f is bounded. 1 n, 2 3 n are examples of null sequences since lim n = 0 and lim 2 3 n = 0. The almost sure convergence of Zn to Z means that there is an event N such that P(N) = 0 and for every element w 2Nc, limn!¥Zn(w) = Z(w), which is almost the same as point-wise convergence for deterministic functions (Example 5. Convergent definition is - tending to move toward one point or to approach each other : converging. Convergent Sequences Subsequences Cauchy Sequences Examples Notice that our de nition of convergent depends not only on fp ng but also on X. Sequences and series of functions: uniform convergence Pointwise and uniform convergence We have said a good deal about sequences of numbers. In the Algebra world, mathematical induction is the first one you usually learn because it's just a set list of steps you work through. Hint: The dual space of c00 under the ℓ∞ norm is (c00)∗ ∼= ℓ1. Scalable Convex Multiple Sequence Alignment via Entropy-Regularized Dual Decomposition. Answer: We will use the Ratio-Test (try to use the Root-Test to see how difficult it is). Then for any integer n there is an x n in S such that |x n| > n. We have shown above that the sequence {f n} ∞ n=1 converges pointwise. Alternating sequences change the signs of its terms. The range variation of σ for which the Laplace transform converges is called region of convergence. Let us ﬁrst make precise what we mean by "linear. 5 Divergence 37 2. Thus convergent sequences do not distinguish between the compact topology of βD and the discrete topology on its underlying set. However, it has huge computational complexity, which is square of that of the. p This integral converges for all p > 0, so the series converges for all p > 0. Convergence and Divergence Lecture Notes It is not always possible to determine the sum of a series exactly. EXAMPLE 2 EXAMPLE 1 common difference arithmetic sequence, GOAL 1 Write rules for arithmetic sequences and find sums of arithmetic series. Solution First, it is easy to see the pointwise limit function is x(t) = 0 on [0;1]. Stayton (2008) demonstrated that rates of convergence can be. Proposition 2. Certainly, uniform convergence implies pointwise convergence, but the converse is false (as we have seen), so that uniform convergence is a stronger \type" of convergence than pointwise convergence. View Notes - Notes-3-2019-version2. Cases of convergent evolution — where different lineages have evolved similar traits independently — are common and have proven central to our understanding of selection. If you take any number in the sequence then subtract it by the previous one, and the result is always the same or constant then it is an arithmetic sequence. 1 Deﬁnition of limit. We have the following useful test for checking the uniform convergence of (fn) when its pointwise limit is known. Properties of the sample autocovariance function The sample autocovariance function: ˆγ(h) = 1 n nX−|h| t=1 (xt+|h| −x¯)(xt −x¯), for −n 0 c ﬁnite & an,bn > 0? Does. 2 More deﬁnitions and terms 1. Convergent sequences in topological spaces 1. f (x) = (1 1 x x. Roughly speaking, a "convergence theorem" states that integrability is preserved under taking limits. Initial values of the nchains = 4 sequences are indicated by solid squares. Pick ϵ = 1 and N1 the corresponding rank. Therefore, with the L 2-norm of Eq. Universal nets 12 4. Weaklawoflargenumbers. (Enough to quote previous homework problem. The hope is that as the sample size increases the estimator should get 'closer' to the parameter of interest. 2008 issue of Mathematics Magazine [1], the questions of convergence, density, and correspondence of rational numbers that can be written as infinitely nested radicals are explored. Application of du Bois-Reymond’s comparison of. Michael Boardman March 1999 Abstract Convergence criteria for spectral sequences are developed that apply more widely than the traditional concepts. This part of probability is often called \large sample theory" or \limit theory" or \asymptotic theory. We can break this problem down into parts and apply the theorem for convergent series to combine each part together. Two examples of nets in analysis 11 3. By this we mean that a function f from IN to some set A is given and f(n) = an ∈ A for n ∈ IN. If a complete metric space has a norm defined by an inner product (such as in a Euclidean space), it is called a Hilbert space. Each number in the sequence is the sum of the two numbers that precede it. Likewise, if the sequence of partial sums is a divergent sequence (i. Deﬁnition: A normed linear space is complete if all Cauchy convergent sequences are convergent. ( ) 0 0 (4. However, there are many diﬀerent ways of deﬁning convergence of a sequence of functions. 34 144 12 =12 Note that you may use parenthesis in the usual ways. Determine if the sequence converges or diverges. Convergence In Distribution (Law). rather than selection pressure, and because it is important to distinguish between founder effects and convergent evolution. , and all of them are de ned on the same probability space (;F;P). 1023 = 4092. Show that weakly convergent sequences in a normed space are bounded. If fn! f on E, and if there is a sequence (an) of real numbers such that an! 0 and. The class of Cauchy sequences should be viewed as minor generalization of Example 1 as the proof of the following theorem will indicate. Fibonacci sequences occur frequently in nature. Give an example to show that this statement is false if uniform convergence is replaced by pointwise convergence. Now that we have seen some more examples of sequences we can discuss how to look for patterns and figure out given a list, how to find the sequence in question. In our previous lesson, Taylor Series, we learned how to create a Taylor Polynomial (Taylor Series) using our center, which in turn, helps us to generate our radius and interval of convergence, derivatives, and factorials. In other words, if one has a sequence (f n)∞ =1 of integrable functions, and if f is some. 10) 2 1 ; 3 2 ; 4 3 ; 5 4 ::: We know this converges to 1 and can verify this using the same logic used in the proof under the de nition of convergence showing that 1 n converges to zero. k ≤ a n ≤ K'. The sum of convergent and divergent series Kyle Miller Wednesday, 2 September 2015 Theorem 8 in section 11. Definition of Convergence and Divergence in Series The n th partial sum of the series a n is given by S n = a 1 + a 2 + a 3 + + a n. Nets and subnets 7 3. A power series is an infinite series. Convergence in the space of test functions Clearly D U is a linear space of functions but it turns out to be impossible to define a norm on the space. EXAMPLES USING MATHCAD 14 Basic Operations: 22+ =4 Type the = sign to get a result. Deﬁnition: A sequence f. 3 Convergence of Subsequences of a Convergent Sequence Theorem. Convergence and (Quasi-)Compactness 13 4. convergence failure during the sample period of 2000 – 2011. The importance of the Cauchy property is to characterize a convergent sequence without using the actual value of its limit, but only the relative distance between terms. 635, and the infinite sum is around 1. Introduces the de nition of rate of convergence for sequences and applies this to xed-point root- nding iterative methods. Concludes with the development of a formula to estimate the rate of convergence for these methods when the actual root is not known. Give an example of a convergent sequence that is not a monotone sequence. For example. For one thing, it is common for the sum to be a relatively arbitrary irrational number:. so the series 0. Chapter 5 Sequences and Series of Functions In this chapter, we deﬁne and study the convergence of sequences and series of functions. If the sequence of these partial sums {S n} converges to L, then the sum of the series converges to L. So in a first countable space “sequences determine the topology. is an example of a convergent sequence since lim n n+1 = 1. n) is convergent to 1 and the subsequence (a 2n 1) of (a n) is convergent to 1: Later, we will prove that in general, the limit supremum and the limit in mum of a bounded sequence are always the limits of some subsequences of the given sequence. For example, random evolutionary change can cause species to become more similar to each other than were their ancestors. Increasing sequence IS-17 Induction terminology IS-1 Inductive step IS-1 Inﬁnite sequence see Sequence Inﬁnite series see Series Integral test for series IS-24 Limit of a sequence IS-13 sum of inﬁnite series IS-20 Logarithm, rate of growth of IS-18 Monotone sequence IS-17 Polynomial, rate of growth of IS-18 Powers sum of IS-5 Prime factorization IS-2. Proposition 2. Nair EXAMPLE 1. Get an intuitive sense of what that even means!. Application of du Bois-Reymond’s comparison of. This week, we will see that within a given range of x values the Taylor series converges to the function itself. Recall the sequence (x n) de ned inductively by x 1 = 1; x n+1 = (1=2)x n + 1;n2N:. ( ) 0 0 (4. Likewise, if the sequence of partial sums is a divergent sequence (i.


# Determine the standard deviation of these results

I've managed to work out the below for the sample standard deviation. Any error's that I should be aware of?

## Homework Statement

The ultimate tensile strength of a material was tested using 10 samples. The results of the tests were as follows.

711 N 〖mm〗^(-2),732 N 〖mm〗^(-2),759 N 〖mm〗^(-2),670 N 〖mm〗^(-2),701 N 〖mm〗^(-2),
765 N 〖mm〗^(-2),743 N 〖mm〗^(-2),755 N 〖mm〗^(-2),715 N 〖mm〗^(-2),713 N 〖mm〗^(-2).

[/B]
(a) Determine the mean standard deviation of these results.
(a) Express the values found in (a) in GPa.

## Homework Equations

s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )

## The Attempt at a Solution

(a) As we are referring to a 'sample' we apply the below formula for Standard Deviation:

s= √(1/(N-1) ∑_(i=1)^N▒(x_i-x ̅ )^2 )

Calculating the mean value x ̅

(711 + 732 +759+670+701 +765 +743+755 +715 +713)/10

x ̅= 726.4 〖N mm〗^(-2)
Determining the (x_i-x ̅)^2

(711-726.4 )^2= 237.16
(732-726.4 )^2= 31.36
(759-726.4 )^2= 1062.76
(670-726.4 )^2= 3180.95
(701-726.4 )^2= 625.16
(765-726.4 )^2= 1489.96
(743-726.4 )^2= 275.56
(755-726.4 )^2= 817.96
(715-726.4 )^2= 129.96
(713-726.4 )^2= 179.56

Determining the sum of the above

∑▒237.16+31.36 +1062.76 +3180.95 +625.16 +1489.96 +275.56 +817.96 +129.96 +179.56

= 8,030.29 〖N mm〗^(-2)

Divide by N-1

(1/5)∙8030.29= 1606.06 N 〖mm〗^(-2)

Determining the sample standard deviation σ

σ= √((1606.06)=40.08 N 〖mm〗^(-2)=4.008 MPa

(b)4.008 MPa=0.004008 GPa

Or in standard form, 4.008∙10^(-3) GPa.

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##\sigma=40 ## N/mm^2=40 MPa=.040 GPa if my arithmetic is correct. ## \\ ## And you don't need to keep all the extra sig figs on the standard deviation. ## \\ ## And I think they want the mean and standard deviation. You didn't give the value of the mean in GPa.

Mark44
Mentor
Determine the mean standard deviation of these results.
There's a mean and there's a standard deviation. I'm not aware of a term called mean standard deviation, but there is a standard error, where you look at the variation of a collection of samples.

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There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.

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There is a standard deviation, ##\sigma_{\bar x}##, of the sample mean ##\bar x##. It is closely related to the confidence interval of the population mean ##\mu##. $$\sigma_{\bar x} = \sigma/\sqrt M \approx s/\sqrt M$$ where ##s## is the sample standard deviation and ##M## is the sample size.
Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.

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FactChecker
FactChecker
Gold Member
Let the mean be ## Y ##, so that ## Y=\bar{X}=\frac{X_1+X_2+...+X_M}{M} ##. ## \\ ## If the random measurements are uncorrelated, ## \sigma_Y^2=\frac{\sigma_{X_1}^2+\sigma_{X_2}^2+...+\sigma_{X_M}^2}{M^2}=\frac{M \, \sigma_X^2}{M^2}=\frac{\sigma_X^2}{M} ##. ## \\ ## That is what I think @FactChecker is referring to, but I don't think it is what the problem is asking for. ## \\ ## I believe the OP @Tiberious left off the word "and" between "mean" and "standard deviation", and it has created some confusion.
They are all very standard things to ask for. But I agree that the sample mean and sample standard deviation are the first two things to calculate.

Yes - it does seem my OP was missing the word 'and' apologies for the confusion. Just to confirm;

The answer for standard deviation is 0.040 Gpa ?

So, would the mean be 0.016Gpa ?

FactChecker
Gold Member
In your original post, why are you dividing by 5 instead of 9 for the sample variance?

PS. small error: (701-726.4 )^2=645.16, not 625.16.

So, amending (701-726.4 )^2 from 625.16 to 645.16 and amending the following would lead to 805.02N mm^-2

Amendments being: -

(701-726.4 )^2 = 645.16

(1/10)∙8050.29= 805.02 N mm^-2

σ= √((805.02)=28.37 N mm^(-2)= 0.02837 Mpa

Converting to Gpa yields: -

2.837*10^-5 Gpa

Looking any better ?

FactChecker
Gold Member
(1/10)∙8050.29= 805.02 N mm^-2
Why are you dividing by 10 instead of N-1=10-1=9?

If you know the true mean, ##\mu##, you should use that in the formula for the sample standard deviation and divide by N. But if you estimate the mean with ##\bar x ##, your results of ##\sum_{i=1}^{i=N} (x_i-\bar x)^2## will be smaller and you should divide by ##N-1##.

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Ah! thanks, brief moment of confusion. I assume this would amend the overall error ?


# Prove this formula $\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$

I am trying to use prove, by just simple algebraic manipulation, to prove the equality of this formula. $$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}= 1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)$$

I have been given hints and instructions from this thread

1. Write using Euler’s identity $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$
2. Factor the denominator
3. Find the partial fractions decomposition, expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again).

This is how I have done:

1. RHS: $$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{1-2r(\dfrac{e^{ix}+e^{-ix}}{2})+r^{2}}=\dfrac{1-r(\dfrac{e^{ix}-e^{-ix}}{2})}{1-re^{ix}-re^{-ix}+r^2}$$

Then from this thread, I have learnt how to factorize the denominator (See Jack D'Aurizio answer, first answer of the thread, first line). The trick is to write $$1=e^0$$.

1. $$\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(re^{ix})(r-e^{-ix})}$$
1. The third step is to decompose the fraction:

$$\dfrac{A}{r-e^{ix}}+\dfrac{B}{r-e^{-ix}}=\dfrac{1-r(\dfrac{e^{ix}+e^{-ix}}{2})}{(r-e^{ix})(r-e^{-ix})}$$

$$A(r-e^{-ix})+B(r-e^{ix})=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I notice that let $$x=0$$, we will have

$$A(r-1)+B(r-1)=1-r(\dfrac{e^{ix}+e^{-ix}}{2})$$

I am stuck here, I don't know to find $$A$$ and $$B$$

Also, could you provide in details how to finish the part "expand the parts as geometric series, and convert from exponential functions back to trigonometric functions (Euler’s identity again)".

My symbolic manipulation skill is not very good, so a detailed answer is great!

Thanks!

• You have some typos where you write $e^{ix}$ instead of $e^{-ix}$.
– J.G.
Jan 27, 2021 at 22:30
• @ J.G. Thanks, I just copy and paste, let me correct them all Jan 27, 2021 at 22:30
• There are sign errors in the $e^{-ix}$ coefficients too.
– J.G.
Jan 27, 2021 at 22:31
• the sum spans over $-\infty , \infty$ or $1, \infty$, although $\cos$ is symmetric you might miss some piece Jan 27, 2021 at 23:00

There is possibly a better way, and I think this was discussed as solved example in tristam needham's book(*). Any who, we begin with geometric series:

$$\frac{1}{1-x} = \sum_{j=0}^{\infty} x^j$$

Sub: $$x \to re^{ i \theta}$$ and simplfy

$$\frac{1}{(1- r \cos \theta) - i \sin \theta} = \sum_{j=0}^{\infty} r^j e^{i j\theta} \tag{2}$$

For LHS, by multiplying with complex conjguate

$$\frac{1}{(1-r \cos \theta) - i r\sin \theta} = \frac{ (1-r \cos \theta) + i \sin \theta}{ 2 - 2 r \cos \theta+r^2} \tag{1}$$

Equating real part in (1) to real part in (2),

$$\frac{1 - r \cos \theta}{ 2 - 2r \cos \theta + r^2} = \sum_{j=0}^{\infty} r^j \cos j \theta$$

Done!

Another identity could be made by equating imaginary parts

*: Indeed it was! See page-78, 79 of Visual Complex Analysis to see how this is simply the fourier series corresponding to the geometric series. Simply two ways to view a function: As an infinite trignometric sum or as a infinite polynomial series. Pretty neat.

• in geo series sum is starting from $0$ Jan 27, 2021 at 23:02

Similar to Buraian's answer, a method I enjoy using is using the equivalent series for $$\sin$$, and then adding $$C+iS$$, as follows:

Let \begin{align} C&=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\ S&=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\ \end{align} Then \begin{align} C+iS&=1+r(\cos x+i\sin x)+r^2(\cos 2x+i\sin2x)+r^3(\cos3x+i\sin3x)+\cdots\\ &=1+re^{ix}+(re^{ix})^2+(re^{ix})^3+\cdots\\ &=\frac{1}{1-re^{ix}}=\frac{(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{1-r\cos x+ri\sin x}{1-2r\cos x+r^2} \end{align} Hence, equating real and imaginary parts we obtain \begin{align} C&=\frac{1-r\cos x}{1-2r\cos x+r^2}\\ S&=\frac{r\sin x}{1-2r\cos x+r^2}\\ \end{align} I hope that was helpful and gives you a new and interesting method for attacking these sorts of problems.

• I like your answer most, but I also vote for Buraian since he post this method first. Your solution has a "Eulerish" feeling to it. haha :) Cheers! Jan 28, 2021 at 15:42
• @JamesWarthington :)) I'm so glad I helped you! This method is pretty powerful: for other examples of me using it in other answers/questions of mine see here: math.stackexchange.com/questions/3886649/… and math.stackexchange.com/questions/3988168/… Jan 28, 2021 at 17:18
• Could you help me here as well: math.stackexchange.com/questions/4003509/… Jan 28, 2021 at 17:41
• @JamesWarthington I will have a go, I suspect the proof lies in using trig identities. Jan 28, 2021 at 17:50
• thank you, I wish my algebraic manipulation skill can be as good as you Jan 28, 2021 at 17:53

$$\cos(nx)=2\cos(x)\cos((n-1)x)-\cos((n-2)x)$$, then $$\sum_{n=1}^\infty r^n\cos(nx)=2\cos(x)\sum_{n=1}^\infty r^n\cos((n-1)x)-\sum_{n=1}^\infty r^n \cos((n-2)x)$$ $$=2r\cos(x)\sum_{n=0}^\infty r^n\cos(nx)-r^2\sum_{n=0}^\infty r^n \cos(nx)$$ Then, after some algebra, $$\sum_{n=1}^\infty r^n\cos(nx)=\frac{r\cos(x)-r^2}{1-2r\cos(x)+r^2}$$

• Can you expand your answer? They are so compacted that I don't know how did you derive these from "some algebra". Jan 28, 2021 at 3:55
• @JamesWarthington set $$f=\sum_{n\ge0}r^n\cos(nx).$$ Then $$\sum_{n\ge1}r^n\cos(nx)=f-1$$ and $$f-1=2r\cos(x)f-r^2f.$$ Solve for $f$. Jan 28, 2021 at 6:46

I might derive the series from $$\frac{1-r \cos (x)}{r^2-2 r \cos (x)+1}$$ by setting $$\cos x = (e^{ix}+e^{-ix})/2 = (z+z^{-1})/2$$ with $$z=e^{ix}$$. Then \eqalign{ \frac{1-r \cos (x)}{r^2-2 r \cos (x)+1} &= \frac{r z^2+r-2 z}{2 (r-z) (r z-1)} = \frac{1}{2}\left(1+\frac{r/z}{(1-r/z)}+\frac{1}{(1-r z)}\right) \cr &= \frac{1}{2}\left(2+\frac{r/z}{(1-r/z)}+\frac{r/z}{(1-r z)}\right) = \frac{1}{2}\left(2+\sum_{n=1}^{\infty}{(r/z)^n}+\sum_{n=1}^{\infty}{(r z)^n}\right) \cr &= 1+\sum_{n=1}^{\infty}{z^n+1/z^n\over2} {r^n} = 1+\sum_{n=1}^\infty r^n \cos nx \ \ \text{or}\ \sum_{n=0}^\infty r^n \cos nx \,. \cr }

• @Micheal E2: Man, your answer is quite difficult. I think you write with great brevity. I am trying to retrace every step you post here but they are so hard. Jan 28, 2021 at 15:39

Since\begin{align}\frac{1-r(e^{ix}+e^{-ix})/2}{(r-e^{ix})(r-e^{-ix})}&=\frac{A}{r-e^{ix}}+\frac{B}{r-e^{-ix}}\\\implies 1-r(e^{ix}+e^{-ix})/2&=A(r-e^{-ix})+B(r-e^{ix}),\end{align}the next step is to consider the $$r^0$$ and $$r^1$$ terms separately, giving$$1=-e^{-ix}A-e^{ix}B,\,-(e^{ix}+e^{-ix})/2=A+B.$$Now solve simultaneous equations. You should find$$A=-\frac12e^{ix},\,B=-\frac12e^{-ix}.$$

• @ J.G. what do you by $r^{0}$ and $r^{1}$ terms? By setting $r=r^{0}$ and $r=r^{1}$. This is not right, I think my interpretation is wrong. Jan 27, 2021 at 23:32
• @.J.G could you edit your reply and show me how to solve this system of equations? I have tried for 2 hours and only obtain $A=\frac{e^{ix}+3e^{ix}}{2(1-e^{-2ix})}$, increadily more complicated than your result. Jan 28, 2021 at 3:16
• @.J.G how do you know this system of equations only has 1 roots for A and for B? Jan 28, 2021 at 3:41
• @.J.G Here are my efforts to solve your system of equations: math.stackexchange.com/questions/4002655/… Jan 28, 2021 at 4:00

With $$1 - 2r\cos x + r^2 = (1-re^{i x})(1-re^{-ix})$$

\begin{align} \frac{1-r\cos x}{1-2r\cos x+r^2} = Re \frac{1}{1-re^{ix}} = Re\sum_{n=0}^\infty(re^{ix})^n =1+\sum_{n=1}^{\infty}r^{n}\cos nx \end{align}


# How to prove this delta-epsilon proof involving $x^2$?

Unlike my last question I want to try something where $x$ can be any real value in $f(x)$ so it's not just $x \geq 0$. I want to fix an $\epsilon$ and find the largest $\delta$ I can get away with using to make the necessary inequalities hold.

$$\lim_{x \rightarrow 2} x^2= 4$$

Say I pick some $\epsilon = 0.5$ which means I need to find some $\delta > 0$ such that for all $x$ satisfying $0 < |x-2| < \delta$, the inequality $|f(x) - L| = |x^2 - 4| < 0.5$ holds.

1. Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?

2. How do I pick the right $\delta$ for something like this?

Trying to simplify:

$|x^2 - 4| < 0.5$

$|x+2||x-2| < 0.5$

$|x-2| < \frac{0.5}{|x+2|}$

Now I'm stuck. Is there a better way to approach these problems? So far I've been trying to manipulate the epsilon inequality so it looks more like the delta one and then try to set the delta and epsilon expressions equal to each other, but maybe there is a more reliable way to prove these relationships?

Update:

Trying another way:

$|x^2 - 4| < 0.5$ simplifies to

$\sqrt{3.5} < x < \sqrt{4.5}$

This gives me two $x$-values away from $a=2$, either $2 - \sqrt{3.5} = .1291...$ or $\sqrt{4.5} - 2 = .1213...$

So if I pick the smallest of the two, $\delta = \sqrt{4.5} - 2$ which satisfies the epsilon condition?

• Choose $\delta=\epsilon/5$ for $\epsilon<1$ – Fakemistake Jan 23 '18 at 20:46
• "Am I stating this correctly so far / understanding the delta-epsilon relationship and goals?" For the most part. But there are two considerations you are eliding. i)$\epsilon$ doesn't have to be $.5$ but could be $.05$ or $.000000005$ of $5\times 10^{534}$. It can be any value. ii) And $\delta$ ... well it's best not to think of delta as a constant, but as a number whose value is determined by the value of $\epsilon$. – fleablood Jan 23 '18 at 20:58
• If $\epsilon = .5$ then just let $\delta$ be something really really small. If $\delta = .01$ then $|x - 2| < \delta$ means $1.99 < x < 2.01$ so $3.9601< x^2 < 4.0401$ so $-.0399< x^2 - 4 <.0401$ so $|x^2 - 4| < .05 < .5$. So.... that doesn't tell you much about how to do it in general. Does it? – fleablood Jan 23 '18 at 21:04
• @fleablood I tried using a graph instead and edited my post. Does this approach make sense? – Aruka J Jan 23 '18 at 21:05
• I understand $\epsilon$ can be any positive value, I'm just picking one arbitrarily, fixing some $\epsilon$, and then finding the largest $\delta$ that makes $|f(x)-L| < \epsilon$ hold. We could pick something very small and it would work but that doesn't show how large we could get away with going. – Aruka J Jan 23 '18 at 21:05

We need to show that $\forall \epsilon>0$ $\exists\delta>0$ such that

$$\forall x\neq2 \quad |x-2|<\delta \implies\left|f\left(x\right)-l\right|<\varepsilon$$

that is

$$|x^2-4|<\epsilon\iff-\epsilon<x^2-4<\epsilon\iff4-\epsilon<x^2<4+\epsilon\iff \sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}\iff \sqrt{4-\epsilon}-2<x-2<\sqrt{4+\epsilon}-2\\\iff |x-2|<min\{\sqrt{4+\epsilon}-2,2-\sqrt{4-\epsilon}\}=\sqrt{4+\epsilon}-2=\delta \quad \square$$

• I think your result is the same as mine correct? $\delta = \sqrt{4.5} - 2$? (I picked $\epsilon = 0.5$) – Aruka J Jan 23 '18 at 21:07
• @ArukaJ Yes but you need to prove it in general for genrec $\epsilon$ and $\delta$ values. – gimusi Jan 23 '18 at 21:19
• For the third time. !!!STOP!!! picking specific values of $\epsilon$. It will NOT help you. – fleablood Jan 23 '18 at 21:32
• @gimusi Oh, yeah. Absolutely. Looks good. i haven't gone through it with a fine tooth comb but, yeah, that's exactly how do it. So barring any arithmetic errors, It's good. My comment was for Aruka who seems hell bent on finding specific deltas for specific elements and not actually trying to find a general formula for deltas for all possible epsilons. – fleablood Jan 23 '18 at 21:39
• Well, it's fine to get an insight but there comes a point where you will have to say, insight is done, now we must do a proof. And you will NEVER be allowed to say "It worked for all examples I tried therefore it must be true" as a proof. – fleablood Jan 23 '18 at 21:45

Hint $$|x^{ 2 }-4|=\left| x-2 \right| \left| x-2+4 \right| <{ \left| x-2 \right| }^{ 2 }+4\left| x-2 \right|$$

If $|x- 2| < \delta$ then

$- \delta < x -2 < \delta$

$2 - \delta < x < 2 + \delta$. Let's assume for the moment that $\delta < 2$.

$(2- \delta)^2 < x^2 < (2+ \delta)^2$

$4 - 4\delta + \delta^2 < x^2 < 4 + 4\delta + \delta^2$

$-4\delta + \delta^2 < x^2 - 4 < 4\delta + \delta^2$.

$-4 \delta - \delta^2 < x^2 - 4 < 4\delta + \delta^2$

$|x^2 - 4| < |4\delta + \delta^2| = 4\delta + \delta^2$ (because $\delta$ is positive)

So we want $\epsilon \ge 4\delta + \delta^2$. Given that we know what $\epsilon$ is, can we find a way of figuring out $\delta$ in terms of $\epsilon$ so that that would be true?

If we assume $\delta \le 1$ then $\delta^2 \le \delta$ so $5\delta \ge 4\delta + \delta^2$.

So if we choose any $\delta$ so that i) $\delta < 2$ and ii) $\delta \le 1$ and iii) $5\delta < \epsilon$ that will do.

So for any $\delta < \min (\frac \epsilon 5, 1)$ that will do.

So to do the proof:

For any $\epsilon > 0$, let $\delta = \min (\frac \epsilon 5, 1)$

Then if $|x - 2| < \delta$ implies by all the work we did above that

$-5\delta \le -4\delta -\delta^2 < -4\delta + \delta^2 < x^2 -4 < 4\delta + \delta^2 \le 5\delta$ so

$|x^2 - 4| < 5\delta \le \epsilon$.

And that's the proof.

It is my honest opinion that your question lies in the realm of questions of the type: "what is the best approach for riding a bicycle?" - Most answers you will receive will send you snapshots of happy riders; but you don't really want these answers, do you? I suggest you get on the bike, and keep falling until you don't.

• Are delta-epsilon proofs more art than science, is that sort of what you are saying? There is no standard methodical way to get the answer? – Aruka J Jan 23 '18 at 20:39
• Is bicycle riding an art? No, it is a skill. delta-epsilon proofs are a skill. There are many skills that you can only pick up by actually doing. Swimming, bike-riding are two out of three immediate examples that come to mind. – uniquesolution Jan 23 '18 at 20:41
• Proofs in general are like this. There are some approaches that work more often than others, but it's more about the intuition of what works than following a particular step-by-step approach. In this case, the best advice you can get is to poke around with the triangle inequality and when you get something that works, try to understand what made it work. – AlexanderJ93 Jan 23 '18 at 20:43


# $\sum_{n=1}^{\infty}\frac{1}{n^22^n}$ by integration or differentiation

There is an infinite sum given: $$\sum_{n=1}^{\infty}\frac{1}{n^22^n}$$ It should be solved using integration, derivation or both. I think using power series can help but I don't know how to finish the calculation. Any help will be appreciated!

• $Li_{2} \left( \frac{1}{2} \right) = \frac{\pi^2}{12}-\frac{ (\ln 2)^2}{2}$. – Donald Splutterwit Apr 23 '17 at 19:26
• Thank you! But how did you get that? @DonaldSplutterwit – Hendrra Apr 23 '17 at 19:32
• I copied it from here ... en.wikipedia.org/wiki/Spence%27s_function ... I am scribbling in my note book now ... I will get back to you when I have managed to derive it ... – Donald Splutterwit Apr 23 '17 at 19:34
• Thank you very much :) I'm really looking forward to the solution! – Hendrra Apr 23 '17 at 19:35
• Have a look at math.stackexchange.com/a/1056111/44121 – Jack D'Aurizio Apr 23 '17 at 20:10

Note that we have $\int_0^x t^{n-1}\,dt=\frac{x^n}{n}$. Then, we can write

\begin{align} \sum_{n=1}^\infty \frac{x^{2n}}{n^2}&=\sum_{n=1}^\infty \int_0^x t^{n-1}\,dt\int_0^x s^{n-1}\,ds\\\\ &=\int_0^x\int_0^x \frac{1}{1-st}\,ds\,dt\\\\ &=-\int_0^{x} \frac{\log(1-sx)}{s}\,ds\\\\ &=-\int_0^{x^2} \frac{\log(1-s)}{s}\,ds\\\\ &=\text{Li}_2(x^2) \end{align}

Evaluating at $x=1/\sqrt{2}$ yields

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty \frac{1}{n^2\,2^n}=\text{Li}_2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)}$$

And we are done!

To evaluate $\text{Li}_2(1/2)$, we exploit the relationship

$$\text{Li}_2(1-x)=-\text{Li}_2\left(1-\frac1x\right)-\frac12\log^2(x)$$

Letting $x=1/2$ yields

$$\text{Li}_2(1/2)=-\text{Li}_2\left(-1\right)-\frac12\log^2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)$$

where we used

\begin{align} \text{Li}_2(-1)&=-\int_0^{-1}\frac{\log(1-x)}{x}\,dx\\\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}\\\\ &=\frac{\pi^2}{12} \end{align}

• That's a nice solution! Thanks. However I must ask about $\sum_{n=1}^{\infty}\frac{x^{2n}}{n^2}$. Why are we considering such a sum? – Hendrra Apr 23 '17 at 19:43
• You're welcome. My pleasure. The value of that sum when $x=1/\sqrt 2$ is $\sum_{n=1}^\infty \frac{1}{n^2\,2^n}$. – Mark Viola Apr 23 '17 at 19:45
• The value truly is $\sum_{n=1}^{\infty}\frac{1}{n^22^n}$! Thus I have the next question. Why did you decided to take an $x = \frac{1}{\sqrt{2}}$? Just because it works? – Hendrra Apr 23 '17 at 19:48
• Well, yes. We took $x=1/\sqrt 2$ in order that we would have the sum of interest. – Mark Viola Apr 23 '17 at 19:51
• @Tyberius The first integral yields $$\int_0^x \int_0^x \frac{1}{1-st}\,dt\,ds=-\int_0^x \frac{\log(1-sx)}{s}\,ds=-\int_0^{x^2}\frac{\log(1-s)}{s}\,ds=\text{Li}_2(x^2)$$ – Mark Viola Apr 23 '17 at 20:20

Another way to calculate it: consider $$f(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n^2}.$$ Differentiating w.r.t. $x$ gives $$f'(x)=\sum_{n=1}^{+\infty}\frac{x^{n-1}}{n}=\frac{1}{x}\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad xf'(x)=\sum_{n=1}^{+\infty}\frac{x^n}{n}\quad\Rightarrow\quad (xf'(x))'=\sum_{n=1}^{+\infty}x^{n-1}=\frac{1}{1-x}.$$ Now integrating with $f(0)=0$ $$xf'(x)=-\ln(1-x)\quad\Rightarrow\quad f(x)=-\int_0^x\frac{\ln(1-t)}{t}\,dt.$$ Motivation for termwise differentiation for power series is straightforward.

My answer to a duplicate question says:

$$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n}$$ Integrating from 0 to t we get $$\int_{0}^{t}\frac{1}{(1-x)}dx=\sum_{n=0}^{\infty}\int_{0}^{t} x^{n}dx$$$$-\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^{n}}{n}$$ Dividing by t and integrating $$\int_{0}^{0.5}-\frac{\ln(1-t)}{t}dt=\sum_{n=1}^{\infty}\int_{0}^{0.5} \frac{t^{n-1}}{n}dt=\sum_{n=1}^{\infty} \frac{1}{n^{2}2^{n}}$$ This on calculating is $$\dfrac{\pi^2}{12}-\dfrac{ln^2(2)}{2},$$

• This answer is identical to your answer to this question. If the questions are the same, this one should be flagged as a duplicate. If not, then it would be more efficient and less noisy to cite or quote (with attribution) from the other answer. The link between the answers would also serve as a link between two closely related questions. In this case, the questions were duplicates. – robjohn May 26 at 1:31
• @robjohn So should I change anything ??. I posted it here because the other one was marked as duplicate. Should I remove it from here?? – DivMit May 26 at 3:12
• Even though the other question was closed, your answer there is still active and getting votes. One should be removed, but you might edit one to reference the other; you might just say something like, "my answer to a duplicate question says..." – robjohn May 26 at 5:28
• Okay , I will link my answers – DivMit May 26 at 9:55


How do limits work with floor/ceiling?

I'm interested in the below equation:

$$\frac{n}{\operatorname{floor}(\frac{x}{n})}$$

Plotting with $$n = 1..100$$ shows the graph being slightly more aliased as $$n$$ increases and a discontinuity forms from $$0..n$$. The domain from what I can tell is $$(-\infty, 0) \cup [n, \infty)$$. I wanted to investigate the limits at each $$n$$ but not entirely certain how floor and ceiling factor into algebra.

Take for example $$n = 2$$,

$$f(x) = \frac{2}{\operatorname{floor}(\frac{x}{2})}$$

How could I find $$\lim_{x\to0} f(x)$$? Would the limit even exist considering the discontinuity between $$[0, 2)$$?

I know you can break down the function and help find the limit using the rules of limits:

$$\lim_{x\to b} \frac{p}{q} = \frac{\lim_{x\to b}p}{\lim_{x\to b}q}$$

So more precisely I'm looking for $$\frac{2}{\lim_{x\to0}\operatorname{floor}(\frac{x}{2})}$$.

For what it's worth, the plot looks like:

I can surmise $$\lim_{x\to0^-} f(x) = -2$$, am I right in assuming $$\lim_{x\to0^+} f(x)$$ doesn't exist?

• When dealing with a domain that is not the whole real line, you need to be careful with limits on the border of the domain. In this case, the limit is the same as $\lim_{x\to 0^{-}} f(x)$ because the small open neighborhoods of $0$ in the domain of $f$ are only negative. This means the limit will be $-n$ since $\lfloor x/n\rfloor=-1$ is constant for $x$ negative near $0.$ – Thomas Andrews Apr 8 at 16:36
• Your function not only has a discontinuity but fails to be defined at all for $x\in[0,n)$. – Henning Makholm Apr 8 at 16:36
• On the other hand it is clear that $f(x)=-n$ for every $x\in[-n,0)$, and since these numbers are the only ones in the domain that are close to $0$, the function very trivially goes to $-n$ for $x\to 0$. There's no algebra involved in this. – Henning Makholm Apr 8 at 16:38

It helps to carefully state definitions. A workable definition of a limit (of a real function) is something like the following:

Definition: Let $$f$$ be a function defined on some domain $$D\subseteq \mathbb{R}$$ and let $$a \in \mathbb{R}$$. Further suppose that there is some $$L\in\mathbb{R}$$ such that for every $$\varepsilon > 0$$ there exists some $$\delta > 0$$ such that if $$x \in D$$ and $$0 < |x-a|<\delta$$, then $$|f(x) - L| < \varepsilon.$$ $$L$$ is said to be the limit of $$f(x)$$ as $$x$$ approaches $$a$$, denoted $$\lim_{x\to a} f(x) = L.$$

I claim that $$\lim_{x\to 0} f(x) = \lim_{x\to 0} \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} = -2.$$ To prove this claim, I have to show that if $$\varepsilon$$ is any positive number, then I can find a value $$\delta$$ such that $$f(x)$$ is within $$\varepsilon$$ of $$-2$$ whenever $$x$$ is in the domain of $$f$$ and within $$\delta$$ of zero.

So, let $$\varepsilon > 0$$ be arbitrary and take $$\delta = 1$$. Observe that $$f$$ is only defined on the set $$D = \mathbb{R} \setminus [0,2).$$ If $$x \in D$$ and $$|x| < \delta = 1$$, then $$x \in (-1,0)$$, and so $$\lfloor x/2 \rfloor = -1$$. But then $$|f(x) - (-2)| = \left| \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} + 2 \right| = \left| \frac{2}{-1} + 2 \right| = 0.$$ Hence whenever $$|x-0| < \delta$$ and $$x\in D$$, we have $$|f(x)-(-2)| = 0 < \varepsilon$$. Therefore $$\lim_{x\to 0} \frac{2}{\left\lfloor \frac{x}{2} \right\rfloor} = -2,$$ as claimed.

The important point here is that the definition of a limit only cares about what is happening in the domain of the function. Points where the function is undefined are irrelevant. As long as we are working with a real-valued function of a real variable, the function of interest is simply not defined on the interval $$[0,2)$$, and so we don't have to worry about those points.

• My understanding in finding a limit was the limit for $x\to b$ only exists if the limit exists for both $x\to b^-$ and $x\to b^+$ and they are equal. If one doesn't exist, the limit doesn't exist; or, if they are not equivalent, the limit doesn't exist. I understand the leftside limit is $-2$ but the rightside limit is not defined. Doesn't then, the limit not exist? I'll disclaim I'm in Calculus I. – gator Apr 8 at 16:49
• As I said in the preamble, you have to carefully state your definitions. The definition of a limit which I gave is pretty standard. If you want to discuss left- and right-hand limits, you need to first carefully define what those mean. I would claim that $\lim_{x\to 0^+} f(x) = -2$ vacuously---indeed, I would claim that if $L$ is any real number, then $\lim_{x\to 0^+} f(x) = L$, since if I make $\delta$ small enough, I cannot find any values of $x$ such that $0 < x < \delta$. Thus all of the conditions are met. Again, state the definition very carefully, then see what happens. – Xander Henderson Apr 8 at 16:53
• You need to exclude the case where $x=a\in D$ or you'd have $\lim_{x\to a} f(x)$ would never be defined if $a$ was a point of discontinuity, even if it could be made continuous. You also want that for $\delta>0$ that there is at least one $x\in D\setminus\{a\}$ so that $|x-a|<\delta.$ Otherwise, all limit values would be vacuously true. – Thomas Andrews Apr 8 at 16:55
• @gator Using that definition, then you are at a loss here. You'd just have to leave it undefined. But in later math, that definition proves inadequate in many ways. – Thomas Andrews Apr 8 at 16:57
• @ThomasAndrews Regarding $|x-a| > 0$, I added that---omitting it was an oversight on my part. Regarding vacuous limits, I have no problem allowing the degenerate case in which the limit is anything by vacuity. But, again, everything comes down to definitions. If one wants to omit vacuous limits (which, in reality, one probably does), then the definition should be written slightly differently. – Xander Henderson Apr 8 at 19:21


# What is this pattern called?

## Back-Story

I became interested in the patterns in multiplication tables for different base number systems a while ago. Specifically, the pattern made by the last digit of each number in the multiplication table. So, base 10 would look like this:

1|2|3|4|5|6|7|8|9|0
2|4|6|8|0|2|4|6|8|0
3|6|9|2|5|8|1|4|7|0
4|8|2|6|0|4|8|2|6|0
5|0|5|0|5|0|5|0|5|0
6|2|8|4|0|6|2|8|4|0
7|4|1|8|5|2|9|6|3|0
8|6|4|2|0|8|6|4|2|0
0|0|0|0|0|0|0|0|0|0

I thought it was interesting that when you move to number systems with different bases, the patterns don't follow the number. They follow it's relative position in the number system. E.g., in base 12, the pattern is 6,0,6,0,6......

## Images

I then realized, I could see the pattern better if I just assigned each number a color. I started with using 10 greyscale colors, with 0 being black and 9 being white. So now base 10 looks like this:

Then, I figured that I really could use as many colors as I wanted to, and see if a larger pattern forms. Using all 256 greyscale colors, I came up with this image representing a base 256 multiplication table:

Or I could go from black to white to black, and smooth out the image:

## Animation

I decided to animate the pattern to better see what was going on. To do this, I defined my 1-n color scale as [w,w,w,w,b,b,b,b,b,b,b,b....]. Where w is white and b is black. I would create a frame, shift my colors down one [b,w,w,w,w,b,b,b,b,b,b,b....], and create the next frame. I repeated this until they colors fully cycled and got this animated image.

Here's a site where you can modify the settings.

## What is this pattern called?

My question is, what is this pattern called? I'm having a hard time finding anything about it. It seems to be a bunch of hyperbolic curves imposed on each other. There is a bunch of "stars" at the corners of where you would divide the image into 4ths, 9ths, etc.

Any insight into this would be appreciated.

• This pattern is called "cool". – Lee Mosher Apr 25 '15 at 16:42
• I'm not an expert, but look at moire patterns. – Bob Krueger Apr 25 '15 at 16:46
• @Bob1123 It does seem similar to a moire pattern. But if it is, the question becomes what patterns are making up this pattern! – Alex McKenzie Apr 25 '15 at 16:50
• Should the question title be changed to "What is the visual pattern in the multiplication table of modular arithmetic?" The OP didn't mention modular arithmetic but it does make this question easier to search for – man and laptop Apr 25 '15 at 17:08
• I love that you posted this! I stumbled across this same pattern years ago but didn't think to ask about it online. (Then again, that might have been before I knew about Math.SE.) So, thanks for doing this! :) – El'endia Starman Apr 26 '15 at 0:28

What you've discovered is essentially modular arithmetic. By looking at only the last digits of a product (in whatever base you're looking at at the moment), you're in effect saying 'I don't care about things that differ by multiples of $n$; I want to consider them as the same digit'. For instance, in base $7$, $5\times 2=10_{10}=13$ has the same last digit as $4\times 6=24_{10}=33$; we put both of these numbers into a bucket labeled '$[3]$', along with $3$, $23=17_{10}$, $43=31_{10}$, etc. In mathematics, when we talk about $31 \bmod 7$ we sometimes just mean the number $3$ itself (that is, the 'label' on this bucket that's between $0$ and $6$, but it's often convenient to think of it as representing the whole bucket: whatever number we pick out of the $[3]$ bucket, when we add it to a number in the $[2]$ bucket, we know that our result will be in the $[5]$ bucket, and when we multiply a number in the $[3]$ bucket by a number in the $[4]$ bucket, we know that our result will be in the $[5]$ bucket; etc. "Last digits" are just a convenient way of talking about these buckets (though things get a little sketchier when you talk about negative numbers - note that according to these rules, $-3$ goes into the $[4]$ bucket!).

Meanwhile, the bands in your pattern are actually (pieces of) hyperbolas. Since $a\times (n-b)\equiv -(a\times b)\pmod n$ (the statement '$x=y\pmod n$' is a mathematical way of phrasing '$x$ and $y$ are in the same bucket in base $n$'; here, the difference between $a\times (n-b)$ and $-(a\times b)$ is $a\times n$), the far right hand side is essentially a reflection of the left, and similarly the bottom is a reflection of the top. If you rearrange the four quarters of your square so that the center of symmetry is (what was previously) the top left corner — i.e., take $A\ B\atop C\ D$ to $D\ C\atop B\ A$ — and then put the origin at the center, then the bands will exactly be (scaled versions) of the hyperbolae $xy=C$ (which are the hyperbolae $y^2-x^2=2C$ rotated by $45^\circ$). This happens because each 'cycle' of black-to-white or black-to-white-to-black will be separated by one multiple of $n$; e.g., the first transition between cycles occurs along the hyperbola $xy=n$; the second along the hyperbola $xy=2n$; etc.

(As for the moiré patterns, they're related to the usual way that such patterns are generated, and in particular they're somewhat related to aliasing near the Nyquist limit when the frequency between hyperbolic bands starts coming close to the frequency of the 'pixels' you're sampling with, but that's another story altogether...)

• So could you say our base 10 number systyem (or any base number system) can be described with modular arithmetic? And is this pattern called anything specific? E.g., modulus pattern? – Alex McKenzie Apr 25 '15 at 17:10
• @AlexMcKenzie The last-digits multiplication that you're talking about is exactly modular arithmetic (specifically, it's the multiplication table mod $n$). A cute example: look at the table of last digits for the base-7 multiplication table. Notice that every row has each non-zero number exactly once, and (by symmetry) so does every column? This isn't a coincidence! This will happen whenever your base is a prime; the numbers mod $p$ form what's called a group under multiplication. – Steven Stadnicki Apr 25 '15 at 17:14
• @AlexMcKenzie In fact, let me flesh out my answer a little bit to explain what I mean by that first sentence... – Steven Stadnicki Apr 25 '15 at 17:22
• Correct me if I'm wrong, but essentially this is a graph of multiple instances of $xy = C$ separated by a multiple of $n$. this is what forms the pattern in the corners. Then, since I'm sampling across a grid of finite pixels, a moire pattern forms. So if I had an infinitely large image, I would only see the $xy = C$ pattern, and not the artifacts. – Alex McKenzie Apr 25 '15 at 17:22
• @AlexMcKenzie Also, congratulations on finding this! This sort of pattern-hunting is an excellent way of experimenting in mathematics and learning about all sorts of facets of its vast world. – Steven Stadnicki Apr 25 '15 at 17:38

You can model your graphics as computing $f_n(x,y)=n\left[\frac{xy}{n}\right]$, where here the square brackets are ad-hoc notation to mean taking the fractional part (or "reduce modulo 1"). This takes $z=xy$ and cuts it at a bunch of horizontal hyperbolae, collapsing the graph like a Fresnel lens. Then, what you are doing is sampling $f_n$ on integer points $\{(i,j)\in\mathbb{Z}^2:1\leq i,j\leq n\}$, but $f_n$ oscillates faster than your sample grid, leading to a Moire pattern.

These would be the powers in a discrete Fourier transform matrix: http://en.wikipedia.org/wiki/Discrete_Fourier_transform_(general)


# What's the inverse operation of exponents?

You know, like addition is the inverse operation of subtraction, vice versa, multiplication is the inverse of division, vice versa , square is the inverse of square root, vice versa.

What's the inverse operation of exponents (exponents: 3^5)

Addition and multiplication are commutative, so there is just one inverse function.

Exponents are not commutative; $2^8 \not= 8^2$. So we need two different inverse functions.

Given $b^e = r$, we have the "$n$th root" operation, $b = \sqrt[e] r$. It turns out that this can actually be written as an exponent itself: $\sqrt[e] r = r^{1/e}$.

Again, given $b^e = r$, we have $e = \log_b r$, the "base-$b$ logarithm of $r$".

• Does division have just one inverse function? 8/4 ≠ 4/8 after all. Jul 31 '19 at 20:59
• 8/4 and 4/8 have a simple numerical relationship. (One is 2/1, the other is 1/2.) Whereas 2^8 and 8^2 have no particularly simple relationship (256 and 64, respectively.) Admittedly I didn't explain that very well... Aug 5 '19 at 8:02
• Good stuff, you can't express "nth root" in Excel but you can express x^(1/n) just fine. This gets doubly interesting if the exponent is e.g. 1.26 Sep 14 '21 at 18:21

These functions are the logarithms, and they are fundamentally important. For $$a = b^c$$ (where $$b > 0$$) we write: $$c = \log_b a,$$ which we can take to be the definition of $$\log_b$$. We read the operation as "logarithm, base $$b$$," or "base $$b$$ logarithm".

In particular, we have $$\log_a (a^b) = b \qquad\text{and}\qquad a^{\log_a b} = b.$$ Of special interest is the natural logarithm, denoted by $$\ln$$ or $$\log$$, the logarithm of base $$e$$. (NB that sometimes $$\log$$ can also denote base $$10$$, or base $$2$$, depending on context.)

Logarithmic identities correspond to exponential identities. From example, from the definition we can conclude that $$\log_b (pq) = \log_b p + \log_b q$$ (for $$p, q > 0$$), which corresponds to the identity $$b^{p + q} = b^p b^q$$.

Perhaps counterintuitively, sometimes it is convenient to define the natural logarithm first and then define the exponential function $$x \mapsto e^x$$ to be its inverse, which leads to the slightly antiquated name antilog for an exponential function $$x \mapsto b^x$$.

Edit Some of the other answers here pointed out quite rightly that one can also ask about the inverse of functions where the variable is in the base, i.e., functions $$x \mapsto x^a$$, and inverses of these functions$$^*$$ (at least when $$a > 0$$) are just $$x \mapsto x^{1/a}$$, which we often write as $$x \mapsto \sqrt[a]{x}$$. These functions are called power functions (note that the inverse of a power function is again a power function), and we reserve the name exponential function for functions $$x \mapsto b^x$$ where the variable is in the exponent, i.e., those to which the logarithms are inverses.

$$^*$$For some $$a$$ (in particular, even integers), we need to restrict the map $$x \mapsto x^a$$ to $$[0, \infty)$$ in order to take an inverse.

• The other answers were good, but your answer explained it best to me. Oct 3 '14 at 12:02
• What about the example in the post? What is the inverse function then? 3^(1/5) or 'base 3 logarithm of 243'? How can I know just by seeing the function which one is the variable here? Oct 24 '19 at 13:05
• That's really the point of the edit: Is the question asking about the inverse of $x \mapsto 3^x$ or the inverse of $x \mapsto x^5$? From just the expression $3^5$ alone there's no way to tell, much like asking about a function whose evaluation yields the expression $1 + 2$ does not indicate whether the question is about the function $x \mapsto x + 2$ or $x \mapsto 1 + x$. As you can see from the question, I initially understood OP to be asking about $x \mapsto 3^x$, since this is an exponential function, and OP asked about "inverse operation of exponents". Oct 24 '19 at 21:21

There are two inverse operations of exponentiation.

## Logarithm

$$\log _{b} a$$

It's read "base-$b$ logarithm of $a$". And it means "the exponent which $b$ must be raised to, so that the result is $a$".

## Root

$$\sqrt[b] a$$

It's read "$b$-th root of $a$". And it means "the number which, when raised to $b$, produces $a$".

It depends on what you see as the function and what the variable in $3^5$.

Generalising your "square is the inverse of square root" leads to reciprocal exponents being the inverse of exponents, so $3^5 = 243$ corresponds to $3 = 243^{1/5}$.

Alternatively $3^5 = 243$ corresponds to $5 =\log _{3} 243 = \frac{\log _{10} 243}{\log _{10} 3}= \frac{\log _{e} 243}{\log _{e} 3}$ using logarithms.

Logarithms: $$10^x = 100 \iff x=\log _{10} 100 = 2$$

If you take $x=3^5$, to "get the 5 back" you do $log_3(x)$ and, to "get the 3 back", you do $\sqrt[5]{x}$.
The interesting thing here is that there are 2 ways to reverse the operation, while other operations had just one: If you take $x=2+7$, to "get the 2 back" you did $x-7$ and to "get the 7 back", $x-2$. This happens because 2+7 = 7+2. The sum is "symmetrical" (the right term is commutative). If you want to "get the 2 back" from $x=2+7$, just subtract 7. If you want to "get the 2 back" from $y=7+2$, just subtract 7 again (because, after all, $x=y$).
But $x=3^5$ is not the same as $y=5^3$. So you cant expect to use the same operation to "get the 3 back from x" and "get the 3 back from y"


## PAT(A) 1126. Eulerian Path (25)

### 1126. Eulerian Path (25)

In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similarly, an Eulerian circuit is an Eulerian path which starts and ends on the same vertex. They were first discussed by Leonhard Euler while solving the famous Seven Bridges of Konigsberg problem in 1736. It has been proven that connected graphs with all vertices of even degree have an Eulerian circuit, and such graphs are called Eulerian. If there are exactly two vertices of odd degree, all Eulerian paths start at one of them and end at the other. A graph that has an Eulerian path but not an Eulerian circuit is called semi-Eulerian. (Cited from https://en.wikipedia.org/wiki/Eulerian_path) Given an undirected graph, you are supposed to tell if it is Eulerian, semi-Eulerian, or non-Eulerian.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 2 numbers N (<= 500), and M, which are the total number of vertices, and the number of edges, respectively. Then M lines follow, each describes an edge by giving the two ends of the edge (the vertices are numbered from 1 to N).

Output Specification:

For each test case, first print in a line the degrees of the vertices in ascending order of their indices. Then in the next line print your conclusion about the graph -- either "Eulerian", "Semi-Eulerian", or "Non-Eulerian". Note that all the numbers in the first line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.

Sample Input 1:
7 12
5 7
1 2
1 3
2 3
2 4
3 4
5 2
7 6
6 3
4 5
6 4
5 6

Sample Output 1:
2 4 4 4 4 4 2
Eulerian

Sample Input 2:
6 10
1 2
1 3
2 3
2 4
3 4
5 2
6 3
4 5
6 4
5 6

Sample Output 2:
2 4 4 4 3 3
Semi-Eulerian

Sample Input 3:
5 8
1 2
2 5
5 4
4 1
1 3
3 2
3 4
5 3

Sample Output 3:
3 3 4 3 3
Non-Eulerian

### 代码

/*
* Problem: 1126. Eulerian Path (25)
* Author: HQ
* Time: 2018-03-12
* State: Done
* Memo: 图，dfs
*/
#include "iostream"
using namespace std;

int N,M;
int degree[500 + 5];
int G[500 + 5][500 + 5];
int visit[500 + 5];
int even = 0;
int cnt = 0;

void dfs(int x) {
visit[x] = 1;
cnt++;
for (int i = 1; i <= N; i++) {
if (!visit[i] && G[x][i])
dfs(i);
}
}

int main() {
cin >> N >>M;
int x, y;
fill(degree, degree + 500 + 5, 0);
fill(visit, visit + 500 + 5, 0);
for (int i = 0; i < M; i++) {
cin >> x >> y;
G[x][y] = 1;
G[y][x] = true;
degree[x] ++;
degree[y] ++;
}
bool first = true;
for (int i = 1; i <= N; i++) {
if (first) {
cout << degree[i];
first = false;
}
else
cout << " " << degree[i];
if (degree[i] % 2 == 0)
even++;
}
cout << endl;
dfs(1);
if(cnt != N)
cout << "Non-Eulerian" << endl;
else if (even == N)
cout << "Eulerian" << endl;
else if(even == N-2)
cout << "Semi-Eulerian" << endl;
else
cout << "Non-Eulerian" << endl;

system("pause");
}

还没有人评论...


# Understanding Lagrange's Theorem (Group Theory)

I am beginning with Abstract Algebra and I'm trying to understand Lagrange's Theorem. The theorem reads

For any finite group $G$, the order of every subgroup $H$ of $G$ should divide the order of $G$.

It seems simple and I've used it to solve some exercices but I believe I'm missing the essence of it. Is there an example that can help me understand it better, or visualize it? Is there a geometric interpretation?

-
There is a local-representation theoretical version: if $H$ is a subgroup of $G$, then every projective module of $H$ induces a projective module of $G$. Hope you like it. :D –  awllower Mar 21 '13 at 12:01
If you like to think visually, this books.google.com/… might pique your interest (not only about Lagrange's theorem, but in general)! –  Stahl Mar 21 '13 at 12:06
The proof works by considering the cosets of $H$ in $G$, what we see is that the cosets partition $G$, so then $G$ is in some sense covered by sets of size $H$ which do not overlap. In terms of visualisation, draw a big circle, call it $G$, then you can partition your circle into equal sections of size $|H|$, for any subgroup $H$. –  user27182 Mar 21 '13 at 12:08
What would qualify as "the essence" of this? It is a very straightforward theorem. It has no hidden details or mysterious manipulations. –  Pedro Tamaroff Mar 21 '13 at 13:03
@Stahl This book is amazing, thank you! It turns out the author has also developed an open source application called GroupExplorer that helps visualize groups, homomorphisms, subgroup lattices, and more, which I find it incredibly useful in my case. –  gpo Mar 23 '13 at 21:12

The left-multiplication map $x\mapsto ax$ is bijective on $G$; its injectivity follows from the cancellative property of the group's operation, $ah=ag\iff h=g$, and overall bijectivity is a consequence of the fact that it has an inverse map, $x\mapsto a^{-1}x$. I like to view a subgroup $H\le G$ as a "puck" and the overgroup $G$ as a "air-hockey table" on which $H$ resides, and to move $H$ around we apply left multiplication by various elements. If you left-multiply by an element $a\in H$, you have not moved the puck at all since $a\in H\iff H=aH$.

Every element $g\in G$ is in some coset, or left translate, of $H$ - in particular, $g=ge\in gH$ since we know that $e\in H$. Thus, the collection of all translates (possible places for the puck to be positioned in) of $H$ "cover" the entire air-hockey table. It remains, then, to investigate the nature of the overlaps between positions, i.e. the intersections of distinct cosets. Here is the proof that cosets that overlap nontrivially must in fact be identical, put into visual form:

$\hskip 0.6in$

This means the cosets of $H$ partition the group $G$. As left multiplication is bijective, every coset is the same size, so each "looks" the same from the viewpoint of cardinality. Continuing with the idea of an air hockey table, this tells us the puck positions tile it, so we have something like:

$\hskip 1.2in$

The most fundamentally basic meaning imputed to multiplication of natural numbers is the following: if Alice has $n$ bags each containing $m$ apples, then she has $n\times m$ apples total. Similarly, our group $G$ is covered by some number $[G:H]$ of disjoint cosets, each containing $|H|$ elements, so $|G|=[G:H]\times|H|$. Note that this is even true on the level of arbitrary infinite cardinals. Thus, $|H|$ is a divisor of the order $|G|$: Lagrange's theorem.

The converse is not globally true: not every divisor $d$ of $n=|G|$ corresponds to a subgroup $H\le G$ of size $|H|=d$. Sylow theory however yields a local version of a converse: for every prime power $q=p^r$ that is a divisor $q\mid n$, there is a $p$-subgroup $H$ of size $|H|=q$.

-
Thank you so much for the time and energy you put in writing such a comprehensive and detailed answer! It really helped me a lot and makes things a lot clearer. I actually understand the proof our professor gave now! –  gpo Mar 23 '13 at 19:46

You already have several great explanations, but you asked also for a way to visualise this. One way to visualise Lagrange's Theorem is to draw the Cayley table of (smallish) groups with colour highlighting.

Here is the Cayley table of a dicyclic group of order $16$ with the cosets of its centre of order $2$ highlighted. The subgroup itself consists of the elements $\{ e, a \}$ (where $e$ is the identity), and is shown in red. The other cosets appear with different colours. Because the subgroup, in this case, is normal, the table is highly regular.

# The Maple code to produce this is:
> with( GroupTheory ):
> G := DicyclicGroup( 4 ):
> H := Centre( G  ):
> DrawCayleyTable( G, cosets = H );

Here is another example, in which the subgroup is not normal. It is the Sylow $2$-subgroup of the symmetric group $S_{4}$. Since the order of $S_{4}$ is equal to $24$, the Sylow $2$-subgroup has order $8$ (and index $3$), and each coset has the same size $8$ as the subgroup. These facts are clear from the picture.

Nevertheless, the block structure of the cosets is still quite visible.

# Maple code for the second example:
> G := Symm( 4 ):
> DrawCayleyTable( G, cosets = SylowSubgroup( 2, G ) );

In each case, it is visually clear that the cosets of the subgroup form a "regular" partition of the group elements. The sizes of the blocks forming each coset are all identical, so they form a regular tiling of the Cayley table for the entire group.

Hope this helps!

-
Those are some very nice visualisations, thank you. I had no idea you could do this in Maple! –  gpo Mar 23 '13 at 20:04

I think the key idea is that groups admit "translations", that is transformation of the form $\left\{ \begin{array}{ccc} G & \to & G \\ g & \mapsto & h \cdot g \end{array} \right.$ with $h \in G$. Moreover, if $H$ is a subgroup of $G$, then the images of $H$ by two such translations are either equal or disjoint; therefore, you can cover $G$ by translating the subgroup $H$ to get a partition of $G$ into $n$ disjoint copies of $H$. You deduce that $|G|=n|H|$.

-
Any equivalence relation on a set $\,S\,$ induces a partition on $\,S\,$ (and vice versa). This is a special case where the $\,n\,$ equivalence classes have the same size $\,k,\,$ so $\,|S| = nk,\:$ so $\,k\,$ divides $|S|.\,$ The result depends crucially on this key fact that the classes have the same size (which you do not mention). –  Math Gems Mar 21 '13 at 14:38
@MathGems: I didn't give a proof but a way to understand Lagrange's theorem, and intuitively, a translation does not change the size. Moreover, implicitely a copy of $H$ has the same size than $H$. –  Seirios Mar 21 '13 at 14:45
@Serios The word "translation" is used in many ways in mathematics, not all of which are set-theoretical bijections. In any case, from a pedagogical standpoint, it is always beneficial to bring to the fore those facts which play key roles. Hence my prior comment. –  Math Gems Mar 21 '13 at 14:49

The main thing is that there is a natural one-to-one correspondence between any two (say, left) cosets of the subgroup $H$, namely the mapping $$s\mapsto yx^{-1}s$$ takes the coset $xH=\{xh\,\mid h\in H\}$ to $yH$, and its inverse is $s\mapsto xy^{-1}s$ (because $xy^{-1}yx^{-1}s=s$ for all $s$).

So, the size of each coset is the same ( $=|H|$), so, if $|G|$ is finite, $|H|$ must divide it. Stahl commented a nice link for visualizing this fact..

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For example If order of G is 12 then we can find only subgroups of order 1,2,3,4,6 and 12 which are divisiors of 12

It means that we can not find a subgroup of other order except above orders

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Home > Taylor Series > Taylor Polynomial Error

# Taylor Polynomial Error

## Contents

Sign in Share More Report Need to report the video? But, we know that the 4th derivative of is , and this has a maximum value of on the interval . We could have been a little clever here, taking advantage of the fact that a lot of the terms in the Taylor expansion of cosine at $0$ are already zero. And so when you evaluate it at a, all the terms with an x minus a disappear, because you have an a minus a on them. http://accessdtv.com/taylor-series/taylor-series-polynomial-error.html

So the error at a is equal to f of a minus P of a. Can we bound this and if we are able to bound this, if we're able to figure out an upper bound on its magnitude-- So actually, what we want to do A Taylor polynomial takes more into consideration. But what I wanna do in this video is think about if we can bound how good it's fitting this function as we move away from a. look at this site

## Taylor Series Approximation Error

Where this is an Nth degree polynomial centered at a. How well (meaning ‘within what tolerance’) does $1-x^2/2+x^4/24-x^6/720$ approximate $\cos x$ on the interval $[{ -\pi \over 2 },{ \pi \over 2 }]$? We already know that P prime of a is equal to f prime of a.

This really comes straight out of the definition of the Taylor polynomials. http://mathinsight.org/determining_tolerance_error_taylor_polynomials_refresher Keywords: ordinary derivative, Taylor polynomial Send us a message about “Determining tolerance/error in Taylor polynomials.” Name: Email address: Comment: If you enter anything in this field your comment will be I'll write two factorial. Lagrange Error Bound Calculator The system returned: (22) Invalid argument The remote host or network may be down.

Loading... Taylor Series Remainder Calculator Thus, we have a bound given as a function of . The derivation is located in the textbook just prior to Theorem 10.1. But if you took a derivative here, this term right here will disappear, it'll go to zero.

Rating is available when the video has been rented. Lagrange Error Bound Formula Loading... And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x So this thing right here, this is an N plus oneth derivative of an Nth degree polynomial.

## Taylor Series Remainder Calculator

Rather, there were two approaches taken by us to estimate how well it approximates cosine.

We might ask ‘Within what tolerance does this polynomial approximate $\cos x$ on that interval?’ To answer this, we first recall that the error term we have after those first (oh-so-familiar) Taylor Series Approximation Error Sign in 6 Loading... Taylor Polynomial Approximation Calculator Basic Examples Find the error bound for the rd Taylor polynomial of centered at on .

That is, we're looking at Since all of the derivatives of satisfy , we know that . Check This Out Professor Leonard 99,296 views 3:01:45 Taylor Polynomials - Duration: 18:06. Theorem 10.1 Lagrange Error Bound  Let be a function such that it and all of its derivatives are continuous. Because the polynomial and the function are the same there. Taylor Series Error Estimation Calculator

So let me write that. A More Interesting Example Problem: Show that the Taylor series for is actually equal to for all real numbers . If is the th Taylor polynomial for centered at , then the error is bounded by where is some value satisfying on the interval between and . http://accessdtv.com/taylor-series/taylor-polynomial-error-function.html Thread navigation Calculus Refresher Previous: Prototypes: More serious questions about Taylor polynomials Next: How large an interval with given tolerance for a Taylor polynomial?

The main idea is this: You did linear approximations in first semester calculus. Taylor Remainder Theorem Proof Especially as we go further and further from where we are centered. >From where are approximation is centered. And what I wanna do is I wanna approximate f of x with a Taylor polynomial centered around x is equal to a.

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Close Yeah, keep it Undo Close This video is unavailable. So, I'll call it P of x. Show more Language: English Content location: United States Restricted Mode: Off History Help Loading... Error Bound Formula Statistics So it might look something like this.

Of course not. Skip to main contentSubjectsMath by subjectEarly mathArithmeticAlgebraGeometryTrigonometryStatistics & probabilityCalculusDifferential equationsLinear algebraMath for fun and gloryMath by gradeK–2nd3rd4th5th6th7th8thHigh schoolScience & engineeringPhysicsChemistryOrganic chemistryBiologyHealth & medicineElectrical engineeringCosmology & astronomyComputingComputer programmingComputer scienceHour of CodeComputer animationArts patrickJMT 128,850 views 10:48 Calculus 2 Lecture 9.9: Approximation of Functions by Taylor Polynomials - Duration: 1:34:10. have a peek here And that polynomial evaluated at a should also be equal to that function evaluated at a.


## anonymous one year ago Simplify 3 square root of 5 end root minus 2 square root of 7 end root plus square root of 45 end root minus square root of 28.

• This Question is Open
1. anonymous

2. jim_thompson5910

The expression is this right? $\large 3\sqrt{5}-2\sqrt{7}+\sqrt{45}-\sqrt{28}$

3. anonymous

do you know how to solve it?

4. jim_thompson5910

first we need to simplify $$\Large \sqrt{45}$$

5. jim_thompson5910

$\large \sqrt{45}=\sqrt{9*5}$ $\large \sqrt{45}=\sqrt{9}*\sqrt{5}$ $\large \sqrt{45}=3\sqrt{5}$

6. jim_thompson5910

Notice how I factored it into 9*5 one of the factors is the largest perfect square factor possible

7. jim_thompson5910

make sense?

8. anonymous

yes

9. jim_thompson5910

how would you simplify $$\Large \sqrt{28}$$ ?

10. anonymous

2 square root of 7

11. jim_thompson5910

good

12. jim_thompson5910

So $\large 3\sqrt{5}-2\sqrt{7}+\sqrt{45}-\sqrt{28}$ turns into $\large 3\sqrt{5}-2\sqrt{7}+3\sqrt{5}-2\sqrt{7}$

13. jim_thompson5910

from here combine like terms

14. anonymous

ok

15. anonymous

i dont really know how to do this part its confusing

16. jim_thompson5910

if I gave you something like 3x+7y+2x+10y, would you be able to combine like terms?

17. anonymous

yes

18. anonymous

it would be 5x + 17y right?

19. jim_thompson5910

yes that's correct

20. jim_thompson5910

you'll use the same idea

21. jim_thompson5910

let x = sqrt(5) and y = sqrt(7) $\large 3\color{red}{\sqrt{5}}-2\color{blue}{\sqrt{7}}+3\color{red}{\sqrt{5}}-2\color{blue}{\sqrt{7}}$ $\large 3\color{red}{x}-2\color{blue}{y}+3\color{red}{x}-2\color{blue}{y}$

22. anonymous

ok

23. jim_thompson5910

simplify 3x-2y+3x-2y to get ???

24. anonymous

6x-y?

25. jim_thompson5910

6x is correct -y is not

26. anonymous

ok

27. jim_thompson5910

try again

28. anonymous

29. jim_thompson5910

-2y-2y is -4y, agreed?

30. jim_thompson5910

$\large 3\color{red}{x}-2\color{blue}{y}+3\color{red}{x}-2\color{blue}{y}$ $\large 6\color{red}{x}-4\color{blue}{y}$ $\large 6\color{red}{\sqrt{5}}-4\color{blue}{\sqrt{7}}$

31. jim_thompson5910

So, $\large 3\sqrt{5}-2\sqrt{7}+\sqrt{45}-\sqrt{28} = 6\sqrt{5}-4\sqrt{7}$

32. anonymous

ok thank a lot

33. anonymous

can you help me with this one please? if you want to. Which statement is true about the difference 2 square root of 7 − square root of 28?

34. jim_thompson5910

$2\sqrt{7} - \color{red}{\sqrt{28}} = 2\sqrt{7} - \color{red}{2\sqrt{7}} = 0\sqrt{7} = 0$