ON A WEIGHTED INTERPOLATION OF FUNCTIONS WITH CIRCULAR MAJORANT
|
|
- Dennis McDowell
- 6 years ago
- Views:
Transcription
1 ON A WEIGHTED INTERPOLATION OF FUNCTIONS WITH CIRCULAR MAJORANT Received: 31 July, 2008 Accepted: 06 February, 2009 Communicated by: SIMON J SMITH Department of Mathematics and Statistics La Trobe University, PO Box 199, Bendigo Victoria 3552, Australia ssmith@latrobeeduau QI Rahman 2000 AMS Sub Class: 41A05, 41A10 Key words: Abstract: Interpolation, Lagrange interpolation, Weighted interpolation, Circular majorant, Projection norm, Lebesgue constant, Chebyshev polynomial Denote by L n the projection operator obtained by applying the Lagrange interpolation method, weighted by (1 x 2 ) 1/2, at the zeros of the Chebyshev polynomial of the second kind of degree n + 1 The norm L n = max Lnf, f 1 where denotes the supremum norm on 1, 1, is known to be asymptotically the same as the minimum possible norm over all choices of interpolation nodes for unweighted Lagrange interpolation Because the projection forces the interpolating function to vanish at ±1, it is appropriate to consider a modified projection norm L n ψ = Lnf, where ψ C 1, 1 is a given max f(x) ψ(x) function (a curved majorant) that satisfies 0 ψ(x) 1 and ψ(±1) = 0 In this paper the asymptotic behaviour of the modified projection norm is studied in the case when ψ(x) is the circular majorant w(x) = (1 x 2 ) 1/2 In particular, it is shown that asymptotically L n w is smaller than L n by the quantity 2π 1 (1 log 2) Page 1 of 20
2 1 Introduction 3 2 Some Lemmas 7 3 Proof of the Theorem 17 Page 2 of 20
3 1 Introduction Suppose n 1 is an integer, and for any s, let θ s = θ s,n = (s + 1)π/(n + 2) For i = 0, 1,, n, put x i = cos θ i The x i are the zeros of the Chebyshev polynomial of the second kind of degree n + 1, defined by U n+1 (x) = sin(n + 2)θ/ sin θ where x = cos θ and 0 θ π Also let w be the weight function w(x) = 1 x 2, and denote the set of all polynomials of degree n or less by P n In the paper 5, JC Mason and GH Elliott introduced the interpolating projection L n of C 1, 1 on {wp n : p n P n } that is defined by (11) (L n f)(x) = w(x) n l i (x) f(x i) w(x i ), where l i (x) is the fundamental Lagrange polynomial n x x k U n+1 (x) (12) l i (x) = = x i x k U n+1(x i )(x x i ) k=0 k i Mason and Elliott studied the projection norm L n = max L n f, f 1 where denotes the uniform norm g = max 1 x 1 g(x), and obtained results that led to the conjecture (13) L n = 2 π log n + 2 (log 4π ) π + γ + o(1) as n, Page 3 of 20 where γ = 0577 is Euler s constant Smith 8 This result (13) was proved later by
4 As pointed out by Mason and Elliott, the projection norm for the much-studied Lagrange interpolation method based on the zeros of the Chebyshev polynomial of the first kind T n+1 (x) = cos(n + 1)θ, where x = cos θ and 0 θ π, is 2 π log n + 2 π (log 8π ) + γ + o(1) (See Luttmann and Rivlin 4 for a short proof of this result based on a conjecture that was later established by Ehlich and Zeller 3) Therefore the norm of the weighted interpolation method (11) is smaller by a quantity asymptotic to 2π 1 log 2 In addition, (13) means that L n, which is based on a simple node system, has (to within o(1) terms) the same norm as the Lagrange method of minimal norm over all possible choices of nodes and the optimal nodes for Lagrange interpolation are not known explicitly (See Brutman 2, Section 3 for further discussion and references on the optimal choice of nodes for Lagrange interpolation) Now, an immediate consequence of (11) is that for all f, (L n f)(±1) = 0 Thus L n is particularly appropriate for approximations of those f for which f(±1) = 0 This leads naturally to a study of the norm (14) L n ψ = max f(x) ψ(x) L nf, where ψ C 1, 1 is a given function (a curved majorant) that satisfies 0 ψ(x) 1 and ψ(±1) = 0 Evidently L n ψ L n In this paper we will look at the particular case when ψ(x) is the circular majorant w(x) = 1 x 2 Note that studies of this nature were initiated by P Turán in the early 1970s, in the context of obtaining Markov and Bernstein type estimates for p if p P n satisfies p(x) w(x) for x 1, 1 see Rahman 6 for a key early paper in this area Our principal result is the following theorem, the proof of which will be developed in Sections 2 and 3 Page 4 of 20
5 Theorem 11 The modified projection norm L n w, defined by (14) with w(x) = 1 x2, satisfies (15) L n w = 2 π log n + 2 (log 8π ) π + γ 1 + o(1) as n Observe that (15) shows L n w is smaller than L n by an amount that is asymptotic to 2π 1 (1 log 2) Before proving the theorem, we make a few remarks about the method to be used By (11), ( ) L n w = max 1 x 1 w(x) n l i (x) Since the x i are arranged symmetrically about 0, then w(x) n l i(x) is even, and so by (12), L n w = max 0 θ π/2 F n (θ), where (16) F n (θ) = sin(n + 2)θ n + 2 n cos θ cos θ i Figure 1 shows the graph of a typical F n (θ) if n is even, and it suggests that the local maximum values of F n (θ) are monotonic increasing as θ moves from left to right, so that the maximum of F n (θ) occurs close to π/2 For n odd, similar graphs suggest that the maximum occurs precisely at π/2 These observations help to motivate the strategy used in Sections 2 and 3 to prove the theorem the approach is akin to that used by Brutman 1 in his investigation of the Lebesgue function for Lagrange interpolation based on the zeros of Chebyshev polynomials of the first kind Page 5 of 20
6 π/4 π/2 Figure 1: Plot of F 12 (θ) for 0 θ π/2 Page 6 of 20
7 2 Some Lemmas This section contains several lemmas that will be needed to prove the theorem The first such lemma provides alternative representations of the function F n (θ) that was defined in (16) Lemma 21 If j is an integer with 0 j n + 1, and θ j 1 θ θ j, then ( j 1 ) j sin(n + 2)θ n F n (θ) = ( 1) n + 2 cos θ i cos θ + (21) cos θ cos θ i=j i j 1 = ( 1) j 2 sin(n + 2)θ (22) sin(n + 1)θ + n + 2 cos θ i cos θ Proof The result (21) follows immediately from (16) For (22), note that the Lagrange interpolation polynomial for U n (x) based on the zeros of U n+1 (x) is simply U n (x) itself, so, with l i (x) defined by (12), U n (x) = n l i (x)u n (x i ) = U n+1(x) n + 2 n 1 x 2 i x x i (This formula appears in Rivlin 7, p 23, Exercise 132) Therefore sin(n + 1)θ = sin(n + 2)θ n + 2 n cos θ cos θ i If this expression is used to rewrite the second sum in (21), the result (22) is obtained Page 7 of 20
8 We now show that on the interval 0, π/2, the values of F n (θ) at the midpoints between consecutive θ-nodes are increasing this result is established in the next two lemmas Lemma 22 If j is an integer with 0 j n, then where n,j := (n + 2) ( F n (θ j+1/2 ) F n (θ j 1/2 ) ) = 2 sin θ j sin θ 1/2 n,j, j (23) n,j := (j n 1) + cot θ (2j+2i 3)/4 cot θ (2j 2i 1)/4 Proof By (22), + cot θ j 1/4 cot θ 1/ csc θ j 1/4 csc θ j+1/4 n,j = 2(n + 2) sin θ j sin θ 1/2 j sin 2 j 1 θ i + 2 cos θ i cos θ j+1/2 cos θ i cos θ j 1/2 From the trigonometric identity Page 8 of 20 (24) sin 2 A cos A cos B = 1 ( ) ( ) B A B + A cot 2 sin B + cot cos A cos B, 2 2
9 it follows that j Therefore sin 2 j 1 θ i cos θ i cos θ j+1/2 = 1 2 sin θ j+1/2 2j+2 i j+1 cos θ i cos θ j 1/2 cot θ (2i 3)/4 1 2 sin θ j 1/2 2j i j cos θ j (j + 1) cos θ j+1/2 + j cos θ j 1/2 cot θ (2i 3)/4 2j = cos θ j sin θ 1/2 cot θ (2i 3)/4 + (2j + 2) sin θ j sin θ 1/ sin θ j+1/2 ( cot θj 1/4 + cot θ j+1/4 ) 2j (25) n,j = (4j 2n) sin θ j sin θ 1/2 + 2 cos θ j sin θ 1/2 cot θ (2i 3)/4 Next consider j sin θ j + cos θ j 2j = cot θ (2i 3)/4 + sin θ j+1/2 ( cot θj 1/4 + cot θ j+1/4 ) j ( ) sin θj + cos θ j cot θ(2i 3)/4 + cot θ (4j 2i 1)/4 Page 9 of 20
10 (26) Also = sin θ j j 1 + cos θ j sin θ (2i 3)/4 sin θ (4j 2i 1)/4 j = sin θ j cot θ (4j 2i 1)/4 cot θ (2i 3)/4 j = sin θ j cot θ (2j+2i 3)/4 cot θ (2j 2i 1)/4 sin θ j+1/2 ( cot θj 1/4 + cot θ j+1/4 ) (27) = 2 sin θ j cos θ j sin θ j+1/2 sin θ j 1/4 sin θ j+1/4 2 cos θ j+1/4 sin θ j+1/4 + sin θ 1/2 = sin θ j sin θ j 1/4 sin θ j+1/4 2 cos θ j+1/4 = sin θ j sin θ 1/2 + csc θ j 1/4 csc θ j+1/4 sin θ j 1/4 sin θ 1/2 = sin θ j sin θ 1/ cot θj 1/4 cot θ 1/2 + csc θ j 1/4 csc θ j+1/4 The lemma is now established by substituting (26) and (27) into (25) Lemma 23 If j is an integer with 0 j (n 1)/2, then F n (θ j+1/2 ) > F n (θ j 1/2 ) Proof By Lemma 22 we need to show that n,j > 0, where n,j is defined by (23) Now, if 0 < a < π/4 and 0 < b < a, then cot(a + b) cot(a b) > cot 2 a Page 10 of 20
11 Also, x csc 2 x is decreasing on (0, π/4), so csc 2 x > π/(2x) if 0 < x < π/4 Thus and so j + j cot θ (2j+2i 3)/4 cot θ (2j 2i 1)/4 > j + j cot 2 θ (j 1)/2 = j csc 2 θ (j 1)/2 > (n + 2)j j + 1, n,j > 1 + cot θ j 1/4 cot θ 1/ csc θ j 1/4 csc θ j+1/4 n + 2 j + 1 > csc 2 θ (4j 3)/ csc θ j 1/4 csc θ j+1/4 n + 2 j + 1 Because θ (4j 3)/8 < π/4, the first term in this expression can be estimated using csc 2 x > π/(2x), while the second term can be estimated using csc x > 1/x Therefore n + 2 n,j > j + 5/4 n + 2 (n + 2) 2 + j + 1 2π 2 (j + 3/4)(j + 5/4) n + 2 > 1 (j + 1)(j + 5/4) 4 + n + 2 2π 2 This latter quantity is positive if n 3 Since 0 j (n 1)/2, the only unresolved cases are when j = 0 and n = 1, 2, and it is a trivial calculation using (23) to show that n,j > 0 in these cases as well Page 11 of 20 We next show that in any interval between successive θ-nodes, F n (θ) achieves its maximum in the right half of the interval
12 Lemma 24 If j is an integer with 0 j (n + 1)/2, and 0 < t < 1/2, then (28) F n (θ j 1/2+t ) F n (θ j 1/2 t ) Proof If j = (n + 1)/2, then θ j 1/2 = π/2, so equality holds in (28) because F n (θ) is symmetric about π/2 Thus we can assume j n/2 For convenience, write a = j 1/2 t, b = j 1/2 + t Since sin(n + 2)θ a = sin(n + 2)θ b = ( 1) j cos tπ, it follows from (21) that F n (θ b ) F n (θ a ) has the same sign as G n,j (t) := n i=j (cos θ b cos θ i )(cos θ a cos θ i ) j 1 If j = 0 this is clearly positive, and otherwise G n,j (t) > 2j 1 i=j (cos θ b cos θ i )(cos θ a cos θ i ) j 1 (cos θ i cos θ b )(cos θ i cos θ a ) (cos θ i cos θ b )(cos θ i cos θ a ) j 1 sin 2 θ 2j i 1 = (cos θ b cos θ 2j i 1 )(cos θ a cos θ 2j i 1 ) (cos θ i cos θ b )(cos θ i cos θ a ) Page 12 of 20
13 We will show that each term in this sum is positive Because sin θ 2j i 1 > sin θ i, this will be true if for 0 i j 1, sin θ 2j i 1 (cos θ i cos θ b )(cos θ i cos θ a ) sin θ i (cos θ b cos θ 2j i 1 )(cos θ a cos θ 2j i 1 ) > 0 By rewriting each difference of cosine terms as a product of sine terms, it follows that we require sin θ 2j i 1 sin θ (j+i 1/2+t)/2 sin θ (j+i 1/2 t)/2 sin θ i sin θ (3j i 3/2+t)/2 sin θ (3j i 3/2 t)/2 > 0 To establish this inequality, note that sin θ 2j i 1 sin θ (j+i 1/2+t)/2 sin θ (j+i 1/2 t)/2 sin θ i sin θ (3j i 3/2+t)/2 sin θ (3j i 3/2 t)/2 = 1 cos θt 1 (sin θ 2j i 1 sin θ i ) sin θ 2j i 1 cos θ j+i+1/2 + sin θ i cos θ 3j i 1/2 2 = cos θ t 1 sin θ j i 3/2 cos θ j 1/2 1 sin θj 2i 5/2 + sin θ 3j 2i 3/2 4 = cos θ j 1/2 cos θ t 1 sin θ j i 3/2 1 2 sin θ 2j 2i 2 = cos θ j 1/2 sin θ j i 3/2 cos θt 1 cos θ j i 3/2 > 0, Page 13 of 20 and so the lemma is proved The final major step in the proof of the theorem is to show that in each interval between successive θ-nodes, the maximum value of F n (θ) is achieved essentially at the midpoint of the interval
14 Lemma 25 If n, j are integers with n 2 and 0 j (n + 1)/2, then (29) max θ j 1 θ θ j F n (θ) = F n (θ j 1/2 ) + O ( (log n) 1), where the O ((log n) 1 ) term is independent of j Proof By Lemma 24, it is sufficient to show that G n,j,t := F n (θ j 1/2+t ) F n (θ j 1/2 ) is bounded above by an O ((log n) 1 ) term that is independent of j and t for 0 t 1/2 Now, by (22) we have (210) G n,j,t = 2 j 1 n + 2 cos tπ cos θ i cos θ j 1/2+t + 2 sin (n + 1)tπ 2(n + 2) sin cos θ i cos θ j 1/2 ( (2j + 1)π 2(n + 2) (n + 1)tπ 2(n + 2) Since cos tπ 1 4t 2 if 0 t 1/2, then each summation term can be estimated by cos tπ 4t 2 cos θ i cos θ j 1/2+t cos θ i cos θ j 1/2 cos θ i cos θ j 1/2+t From (2x)/π sin x x for 0 x π/2, it follows that cos θ i cos θ j 1/2+t = 2 sin θ (j+i 1/2+t)/2 sin θ (j i 5/2+t)/2 8(i + 1) 2 π 2 (j i)(j + i + 2), ) Page 14 of 20
15 and so j 1 cos tπ cos θ i cos θ j 1/2+t cos θ i cos θ j 1/2 32t2 π 2 = 32t2 π 2 j 1 (i + 1) 2 (j i)(j + i + 2) j j j+1 1 k (211) 16t2 (j + 1) (log(j + 1) 1), π2 where the final inequality follows from Also, (212) sin 2j+1 1 k 1 + log(j + 1) ( ) (n + 1)tπ (2j + 1)π 2(n + 2) sin (n + 1)tπ sin tπ 2(n + 2) 2(n + 2) 2 tπ2 (j + 1) 2(n + 2) sin (2j + 1)π 2(n + 2) We now return to the characterization (210) of G n,j,t By (212), G n,0,t π 2 /(2(n + 2)) For j 1, it follows from (211) and (212) that (213) G n,j,t 2π2 t(j + 1) 1 16t π 6 j + 1 log(j + 1), n + 2 π4 32(n + 2) log(j + 1) Page 15 of 20
16 where the latter inequality follows by maximizing the quadratic in t On the interval 1 j (n + 1)/2, the maximum of (j + 1)/ log(j + 1) occurs at an endpoint, so { } j (214) log(j + 1) max log 2, n log((n + 3)/2) The result (29) then follows from (213) and (214) Page 16 of 20
17 3 Proof of the Theorem Since L n w = max 0 θ π/2 F n (θ), it follows from Lemmas 23 and 25 that ( F π ) n 2 + O ((log n) 1 ) if n is odd, L n w = ( ) F π(n+1) n + O ((log n) 1 ) 2(n+2) if n is even To obtain the asymptotic result (15) for L n w we use a method that was introduced by Luttmann and Rivlin 4, Theorem 3, and used also by Mason and Elliott 5, Section 9 If n is odd, then by (22) with n = 2m 1, (31) ( π ) F n = 2 = m 1 2 2m + 1 m 2 2m + 1 cos θ i (k 1/2)π csc sin 2m + 1 (k 1/2)π, 2m + 1 where the second equality follows by reversing the order of summation Now, π 2m + 1 m (k 1/2)π csc 2m + 1 π m = 2m + 1 (k 1/2)π csc 2m m + 1 (k 1/2)π m 1 k 1/2 The asymptotic behaviour as m of each of the sums in this expression is given Page 17 of 20
18 by lim m π 2m + 1 m (k 1/2)π csc 2m + 1 = 2m + 1 (k 1/2)π π/2 0 csc x 1 dx x and Also, m sin m 2m 1 k 1/2 = 2 (k 1/2)π 2m + 1 = csc m 1 k 1 k = log 4 π = log(4m) + γ + o(1) π mπ 4m + 2 sin2 4m + 2 = 2m O(1) π Substituting these asymptotic results into (31) yields the desired result (15) if n is odd On the other hand, if n = 2m is even, then by (22) and (24), (32) F n ( π(n + 1) 2(n + 2) ) π = sin 4m m + 1 = 1 m + 1 m 1 ( 2m cos π 4m + 4 cos θ i cos (2m+1)π 4m+4 ) m 1 (2i 1)π cot 8m + 8 cos θ i + O(m 1 ) Page 18 of 20
19 The sum of the cotangent terms can be estimated by a similar argument to that above, using to obtain Also, π/2 2m+2 1 cot 2m + 2 m 1 1 cos θ i = m (cot x x 1 ) dx = log 2 π, (2i 1)π 8m + 8 = 2 π 1 2(m + 1) (cos = 2 π + O(m 1 ) ( log 16m ) π + γ + o(1) mπ 4m + 4 csc π 4m + 4 ) 2 If these asymptotic results are substituted into (32), the result (15) is obtained if n is even, and so the proof of Theorem 11 is completed Page 19 of 20
20 References 1 L BRUTMAN, On the Lebesgue function for polynomial interpolation, SIAM J Numer Anal, 15 (1978), L BRUTMAN, Lebesgue functions for polynomial interpolation a survey, Ann Numer Math, 4 (1997), H EHLICH AND K ZELLER, Auswertung der Normen von Interpolationsoperatoren, Math Ann, 164 (1966), FW LUTTMANN AND TJ RIVLIN, Some numerical experiments in the theory of polynomial interpolation, IBM J Res Develop, 9 (1965), JC MASON AND GH ELLIOTT, Constrained near-minimax approximation by weighted expansion and interpolation using Chebyshev polynomials of the second, third, and fourth kinds, Numer Algorithms, 9 (1995), QI RAHMAN, On a problem of Turán about polynomials with curved majorants, Trans Amer Math Soc, 163 (1972), TJ RIVLIN, Chebyshev Polynomials From Approximation Theory to Algebra and Number Theory, 2nd ed, John Wiley and Sons, New York, SJ SMITH, On the projection norm for a weighted interpolation using Chebyshev polynomials of the second kind, Math Pannon, 16 (2005), Page 20 of 20
On the Lebesgue constant of subperiodic trigonometric interpolation
On the Lebesgue constant of subperiodic trigonometric interpolation Gaspare Da Fies and Marco Vianello November 4, 202 Abstract We solve a recent conjecture, proving that the Lebesgue constant of Chebyshev-like
More informationVectors in Function Spaces
Jim Lambers MAT 66 Spring Semester 15-16 Lecture 18 Notes These notes correspond to Section 6.3 in the text. Vectors in Function Spaces We begin with some necessary terminology. A vector space V, also
More information1 + lim. n n+1. f(x) = x + 1, x 1. and we check that f is increasing, instead. Using the quotient rule, we easily find that. 1 (x + 1) 1 x (x + 1) 2 =
Chapter 5 Sequences and series 5. Sequences Definition 5. (Sequence). A sequence is a function which is defined on the set N of natural numbers. Since such a function is uniquely determined by its values
More information1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.
Math120 - Precalculus. Final Review. Fall, 2011 Prepared by Dr. P. Babaali 1 Algebra 1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.
More informationNumerical Analysis: Interpolation Part 1
Numerical Analysis: Interpolation Part 1 Computer Science, Ben-Gurion University (slides based mostly on Prof. Ben-Shahar s notes) 2018/2019, Fall Semester BGU CS Interpolation (ver. 1.00) AY 2018/2019,
More information12.0 Properties of orthogonal polynomials
12.0 Properties of orthogonal polynomials In this section we study orthogonal polynomials to use them for the construction of quadrature formulas investigate projections on polynomial spaces and their
More informationApproximation theory
Approximation theory Xiaojing Ye, Math & Stat, Georgia State University Spring 2019 Numerical Analysis II Xiaojing Ye, Math & Stat, Georgia State University 1 1 1.3 6 8.8 2 3.5 7 10.1 Least 3squares 4.2
More informationUNIFORM BOUNDS FOR BESSEL FUNCTIONS
Journal of Applied Analysis Vol. 1, No. 1 (006), pp. 83 91 UNIFORM BOUNDS FOR BESSEL FUNCTIONS I. KRASIKOV Received October 8, 001 and, in revised form, July 6, 004 Abstract. For ν > 1/ and x real we shall
More informationA numerical study of the Xu polynomial interpolation formula in two variables
A numerical study of the Xu polynomial interpolation formula in two variables Len Bos Dept. of Mathematics and Statistics, University of Calgary (Canada) Marco Caliari, Stefano De Marchi Dept. of Computer
More informationATHS FC Math Department Al Ain Revision worksheet
ATHS FC Math Department Al Ain Revision worksheet Section Name ID Date Lesson Marks 3.3, 13.1 to 13.6, 5.1, 5.2, 5.3 Lesson 3.3 (Solving Systems of Inequalities by Graphing) Question: 1 Solve each system
More informationAP Calculus (BC) Chapter 9 Test No Calculator Section Name: Date: Period:
WORKSHEET: Series, Taylor Series AP Calculus (BC) Chapter 9 Test No Calculator Section Name: Date: Period: 1 Part I. Multiple-Choice Questions (5 points each; please circle the correct answer.) 1. The
More informationi x i y i
Department of Mathematics MTL107: Numerical Methods and Computations Exercise Set 8: Approximation-Linear Least Squares Polynomial approximation, Chebyshev Polynomial approximation. 1. Compute the linear
More informationScientific Computing
2301678 Scientific Computing Chapter 2 Interpolation and Approximation Paisan Nakmahachalasint Paisan.N@chula.ac.th Chapter 2 Interpolation and Approximation p. 1/66 Contents 1. Polynomial interpolation
More informationTAYLOR AND MACLAURIN SERIES
TAYLOR AND MACLAURIN SERIES. Introduction Last time, we were able to represent a certain restricted class of functions as power series. This leads us to the question: can we represent more general functions
More informationMathematics 324 Riemann Zeta Function August 5, 2005
Mathematics 324 Riemann Zeta Function August 5, 25 In this note we give an introduction to the Riemann zeta function, which connects the ideas of real analysis with the arithmetic of the integers. Define
More informationOn the Lebesgue constant of barycentric rational interpolation at equidistant nodes
On the Lebesgue constant of barycentric rational interpolation at equidistant nodes by Len Bos, Stefano De Marchi, Kai Hormann and Georges Klein Report No. 0- May 0 Université de Fribourg (Suisse Département
More informationINTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES
INTRODUCTION TO REAL ANALYSIS II MATH 433 BLECHER NOTES. As in earlier classnotes. As in earlier classnotes (Fourier series) 3. Fourier series (continued) (NOTE: UNDERGRADS IN THE CLASS ARE NOT RESPONSIBLE
More informationFourier series
11.1-11.2. Fourier series Yurii Lyubarskii, NTNU September 5, 2016 Periodic functions Function f defined on the whole real axis has period p if Properties f (t) = f (t + p) for all t R If f and g have
More informationLimits at Infinity. Horizontal Asymptotes. Definition (Limits at Infinity) Horizontal Asymptotes
Limits at Infinity If a function f has a domain that is unbounded, that is, one of the endpoints of its domain is ±, we can determine the long term behavior of the function using a it at infinity. Definition
More information12) y = -2 sin 1 2 x - 2
Review -Test 1 - Unit 1 and - Math 41 Name SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Find and simplify the difference quotient f(x + h) - f(x),
More informationChapter 4: Interpolation and Approximation. October 28, 2005
Chapter 4: Interpolation and Approximation October 28, 2005 Outline 1 2.4 Linear Interpolation 2 4.1 Lagrange Interpolation 3 4.2 Newton Interpolation and Divided Differences 4 4.3 Interpolation Error
More informationExamination paper for TMA4130 Matematikk 4N: SOLUTION
Department of Mathematical Sciences Examination paper for TMA4 Matematikk 4N: SOLUTION Academic contact during examination: Morten Nome Phone: 9 84 97 8 Examination date: December 7 Examination time (from
More informationNEWMAN S INEQUALITY FOR INCREASING EXPONENTIAL SUMS
NEWMAN S INEQUALITY FOR INCREASING EXPONENTIAL SUMS Tamás Erdélyi Dedicated to the memory of George G Lorentz Abstract Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers The collection of all linear
More information8.5 Taylor Polynomials and Taylor Series
8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES 50 8.5 Taylor Polynomials and Taylor Series Motivating Questions In this section, we strive to understand the ideas generated by the following important questions:
More informationFourier Series. 1. Review of Linear Algebra
Fourier Series In this section we give a short introduction to Fourier Analysis. If you are interested in Fourier analysis and would like to know more detail, I highly recommend the following book: Fourier
More informationAdvanced Computational Fluid Dynamics AA215A Lecture 2 Approximation Theory. Antony Jameson
Advanced Computational Fluid Dynamics AA5A Lecture Approximation Theory Antony Jameson Winter Quarter, 6, Stanford, CA Last revised on January 7, 6 Contents Approximation Theory. Least Squares Approximation
More informationEvaluating Limits Analytically. By Tuesday J. Johnson
Evaluating Limits Analytically By Tuesday J. Johnson Suggested Review Topics Algebra skills reviews suggested: Evaluating functions Rationalizing numerators and/or denominators Trigonometric skills reviews
More informationTRIGONOMETRIC FUNCTIONS. Copyright Cengage Learning. All rights reserved.
12 TRIGONOMETRIC FUNCTIONS Copyright Cengage Learning. All rights reserved. 12.2 The Trigonometric Functions Copyright Cengage Learning. All rights reserved. The Trigonometric Functions and Their Graphs
More information6.5 Trigonometric Equations
6. Trigonometric Equations In this section, we discuss conditional trigonometric equations, that is, equations involving trigonometric functions that are satisfied only by some values of the variable (or
More informationNext, we ll use all of the tools we ve covered in our study of trigonometry to solve some equations.
Section 6.3 - Solving Trigonometric Equations Next, we ll use all of the tools we ve covered in our study of trigonometry to solve some equations. These are equations from algebra: Linear Equation: Solve:
More informationNeville s Method. MATH 375 Numerical Analysis. J. Robert Buchanan. Fall Department of Mathematics. J. Robert Buchanan Neville s Method
Neville s Method MATH 375 Numerical Analysis J. Robert Buchanan Department of Mathematics Fall 2013 Motivation We have learned how to approximate a function using Lagrange polynomials and how to estimate
More informationCK- 12 Algebra II with Trigonometry Concepts 1
14.1 Graphing Sine and Cosine 1. A.,1 B. (, 1) C. 3,0 D. 11 1, 6 E. (, 1) F. G. H. 11, 4 7, 1 11, 3. 3. 5 9,,,,,,, 4 4 4 4 3 5 3, and, 3 3 CK- 1 Algebra II with Trigonometry Concepts 1 4.ans-1401-01 5.
More informationTrigonometric Functions. Section 1.6
Trigonometric Functions Section 1.6 Quick Review Radian Measure The radian measure of the angle ACB at the center of the unit circle equals the length of the arc that ACB cuts from the unit circle. Radian
More informationBivariate Lagrange interpolation at the Padua points: The generating curve approach
Journal of Approximation Theory 43 (6) 5 5 www.elsevier.com/locate/jat Bivariate Lagrange interpolation at the Padua points: The generating curve approach Len Bos a, Marco Caliari b, Stefano De Marchi
More information256 Summary. D n f(x j ) = f j+n f j n 2n x. j n=1. α m n = 2( 1) n (m!) 2 (m n)!(m + n)!. PPW = 2π k x 2 N + 1. i=0?d i,j. N/2} N + 1-dim.
56 Summary High order FD Finite-order finite differences: Points per Wavelength: Number of passes: D n f(x j ) = f j+n f j n n x df xj = m α m dx n D n f j j n= α m n = ( ) n (m!) (m n)!(m + n)!. PPW =
More informationFourier and Partial Differential Equations
Chapter 5 Fourier and Partial Differential Equations 5.1 Fourier MATH 294 SPRING 1982 FINAL # 5 5.1.1 Consider the function 2x, 0 x 1. a) Sketch the odd extension of this function on 1 x 1. b) Expand the
More informationINSTRUCTOR SAMPLE E. Check that your exam contains 25 questions numbered sequentially. Answer Questions 1-25 on your scantron.
MATH 41 FINAL EXAM NAME SECTION NUMBER INSTRUCTOR SAMPLE E On your scantron, write and bubble your PSU ID, Section Number, and Test Version. Failure to correctly code these items may result in a loss of
More informationReconstruction of sparse Legendre and Gegenbauer expansions
Reconstruction of sparse Legendre and Gegenbauer expansions Daniel Potts Manfred Tasche We present a new deterministic algorithm for the reconstruction of sparse Legendre expansions from a small number
More informationError Estimates for Trigonometric Interpolation. of Periodic Functions in Lip 1. Knut Petras
Error Estimates for Trigonometric Interpolation of Periodic Functions in Lip Knut Petras Abstract. The Peano kernel method is used in order to calculate the best possible constants c, independent of M,
More information1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents.
Math120 - Precalculus. Final Review Prepared by Dr. P. Babaali 1 Algebra 1. Use the properties of exponents to simplify the following expression, writing your answer with only positive exponents. (a) 5
More information1.2 A List of Commonly Occurring Functions
Arkansas Tech University MATH 2914: Calculus I Dr. Marcel B. Finan 1.2 A List of Commonly Occurring Functions In this section, we discuss the most common functions occurring in calculus. Linear Functions
More informationarxiv: v1 [math.ca] 23 Oct 2018
A REMARK ON THE ARCSINE DISTRIBUTION AND THE HILBERT TRANSFORM arxiv:80.08v [math.ca] 3 Oct 08 RONALD R. COIFMAN AND STEFAN STEINERBERGER Abstract. We prove that if fx) ) /4 L,) and its Hilbert transform
More informationfunction independent dependent domain range graph of the function The Vertical Line Test
Functions A quantity y is a function of another quantity x if there is some rule (an algebraic equation, a graph, a table, or as an English description) by which a unique value is assigned to y by a corresponding
More informationPre-Calculus Chapter 0. Solving Equations and Inequalities 0.1 Solving Equations with Absolute Value 0.2 Solving Quadratic Equations
Pre-Calculus Chapter 0. Solving Equations and Inequalities 0.1 Solving Equations with Absolute Value 0.1.1 Solve Simple Equations Involving Absolute Value 0.2 Solving Quadratic Equations 0.2.1 Use the
More informationApproximation of Circular Arcs by Parametric Polynomial Curves
Approximation of Circular Arcs by Parametric Polynomial Curves Gašper Jaklič Jernej Kozak Marjeta Krajnc Emil Žagar September 19, 005 Abstract In this paper the approximation of circular arcs by parametric
More informationRemarks on the Rademacher-Menshov Theorem
Remarks on the Rademacher-Menshov Theorem Christopher Meaney Abstract We describe Salem s proof of the Rademacher-Menshov Theorem, which shows that one constant works for all orthogonal expansions in all
More informationMAT137 Calculus! Lecture 6
MAT137 Calculus! Lecture 6 Today: 3.2 Differentiation Rules; 3.3 Derivatives of higher order. 3.4 Related rates 3.5 Chain Rule 3.6 Derivative of Trig. Functions Next: 3.7 Implicit Differentiation 4.10
More informationNYS Algebra II and Trigonometry Suggested Sequence of Units (P.I's within each unit are NOT in any suggested order)
1 of 6 UNIT P.I. 1 - INTEGERS 1 A2.A.1 Solve absolute value equations and inequalities involving linear expressions in one variable 1 A2.A.4 * Solve quadratic inequalities in one and two variables, algebraically
More informationExamples of the Fourier Theorem (Sect. 10.3). The Fourier Theorem: Continuous case.
s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem. The Fourier Theorem:
More informationPre-Calculus Exam 2009 University of Houston Math Contest. Name: School: There is no penalty for guessing.
Pre-Calculus Exam 009 University of Houston Math Contest Name: School: Please read the questions carefully and give a clear indication of your answer on each question. There is no penalty for guessing.
More informationAlgebra 2 Khan Academy Video Correlations By SpringBoard Activity
SB Activity Activity 1 Creating Equations 1-1 Learning Targets: Create an equation in one variable from a real-world context. Solve an equation in one variable. 1-2 Learning Targets: Create equations in
More informationAlgebra 2 Khan Academy Video Correlations By SpringBoard Activity
SB Activity Activity 1 Creating Equations 1-1 Learning Targets: Create an equation in one variable from a real-world context. Solve an equation in one variable. 1-2 Learning Targets: Create equations in
More informationChapter 5 Analytic Trigonometry
Chapter 5 Analytic Trigonometry Section 1 Section 2 Section 3 Section 4 Section 5 Using Fundamental Identities Verifying Trigonometric Identities Solving Trigonometric Equations Sum and Difference Formulas
More information2.2 The derivative as a Function
2.2 The derivative as a Function Recall: The derivative of a function f at a fixed number a: f a f a+h f(a) = lim h 0 h Definition (Derivative of f) For any number x, the derivative of f is f x f x+h f(x)
More informationMu Alpha Theta National Convention 2013
Practice Round Alpha School Bowl P1. What is the common difference of the arithmetic sequence 10, 23,? P2. Find the sum of the digits of the base ten representation of 2 15. P3. Find the smaller value
More informationCHAPTERS 5-7 TRIG. FORMULAS PACKET
CHAPTERS 5-7 TRIG. FORMULAS PACKET PRE-CALCULUS SECTION 5-2 IDENTITIES Reciprocal Identities sin x = ( 1 / csc x ) csc x = ( 1 / sin x ) cos x = ( 1 / sec x ) sec x = ( 1 / cos x ) tan x = ( 1 / cot x
More informationOn the Chebyshev quadrature rule of finite part integrals
Journal of Computational and Applied Mathematics 18 25) 147 159 www.elsevier.com/locate/cam On the Chebyshev quadrature rule of finite part integrals Philsu Kim a,,1, Soyoung Ahn b, U. Jin Choi b a Major
More informationIncreasing and decreasing intervals *
OpenStax-CNX module: m15474 1 Increasing and decreasing intervals * Sunil Kumar Singh This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 2.0 A function is
More informationSome approximation theorems in Math 522. I. Approximations of continuous functions by smooth functions
Some approximation theorems in Math 522 I. Approximations of continuous functions by smooth functions The goal is to approximate continuous functions vanishing at ± by smooth functions. In a previous homewor
More informationPre-Calculus MATH 119 Fall Section 1.1. Section objectives. Section 1.3. Section objectives. Section A.10. Section objectives
Pre-Calculus MATH 119 Fall 2013 Learning Objectives Section 1.1 1. Use the Distance Formula 2. Use the Midpoint Formula 4. Graph Equations Using a Graphing Utility 5. Use a Graphing Utility to Create Tables
More information= 10 such triples. If it is 5, there is = 1 such triple. Therefore, there are a total of = 46 such triples.
. Two externally tangent unit circles are constructed inside square ABCD, one tangent to AB and AD, the other to BC and CD. Compute the length of AB. Answer: + Solution: Observe that the diagonal of the
More informationAP Calculus Testbank (Chapter 9) (Mr. Surowski)
AP Calculus Testbank (Chapter 9) (Mr. Surowski) Part I. Multiple-Choice Questions n 1 1. The series will converge, provided that n 1+p + n + 1 (A) p > 1 (B) p > 2 (C) p >.5 (D) p 0 2. The series
More informationLast/Family Name First/Given Name Seat #
Math 2, Fall 27 Schaeffer/Kemeny Final Exam (December th, 27) Last/Family Name First/Given Name Seat # Failure to follow the instructions below will constitute a breach of the Stanford Honor Code: You
More informationCounting on Chebyshev Polynomials
DRAFT VOL. 8, NO., APRIL 009 1 Counting on Chebyshev Polynomials Arthur T. Benjamin Harvey Mudd College Claremont, CA 91711 benjamin@hmc.edu Daniel Walton UCLA Los Angeles, CA 90095555 waltond@ucla.edu
More informationFebruary 21 Math 1190 sec. 63 Spring 2017
February 21 Math 1190 sec. 63 Spring 2017 Chapter 2: Derivatives Let s recall the efinitions an erivative rules we have so far: Let s assume that y = f (x) is a function with c in it s omain. The erivative
More information3. Use absolute value notation to write an inequality that represents the statement: x is within 3 units of 2 on the real line.
PreCalculus Review Review Questions 1 The following transformations are applied in the given order) to the graph of y = x I Vertical Stretch by a factor of II Horizontal shift to the right by units III
More informationLecture 4. Section 2.5 The Pinching Theorem Section 2.6 Two Basic Properties of Continuity. Jiwen He. Department of Mathematics, University of Houston
Review Pinching Theorem Two Basic Properties Lecture 4 Section 2.5 The Pinching Theorem Section 2.6 Two Basic Properties of Continuity Jiwen He Department of Mathematics, University of Houston jiwenhe@math.uh.edu
More information2. Algebraic functions, power functions, exponential functions, trig functions
Math, Prep: Familiar Functions (.,.,.5, Appendix D) Name: Names of collaborators: Main Points to Review:. Functions, models, graphs, tables, domain and range. Algebraic functions, power functions, exponential
More informationChapter 2: Functions, Limits and Continuity
Chapter 2: Functions, Limits and Continuity Functions Limits Continuity Chapter 2: Functions, Limits and Continuity 1 Functions Functions are the major tools for describing the real world in mathematical
More informationMath 370 Semester Review Name
Math 370 Semester Review Name These problems will give you an idea of what may be included on the final exam. Don't worry! The final exam will not be this long! 1) State the following theorems: (a) Remainder
More informationAPPROXIMATION AND GENERALIZED LOWER ORDER OF ENTIRE FUNCTIONS OF SEVERAL COMPLEX VARIABLES. Susheel Kumar and G. S. Srivastava (Received June, 2013)
NEW ZEALAND JOURNAL OF MATHEMATICS Volume 45 (2015), 11 17 APPROXIMATION AND GENERALIZED LOWER ORDER OF ENTIRE FUNCTIONS OF SEVERAL COMPLEX VARIABLES Susheel Kumar and G. S. Srivastava (Received June,
More information5-3 Solving Trigonometric Equations
Solve each equation for all values of x. 1. 5 sin x + 2 = sin x The period of sine is 2π, so you only need to find solutions on the interval. The solutions on this interval are and. Solutions on the interval
More informationON A SINE POLYNOMIAL OF TURÁN
ROCKY MOUNTAIN JOURNAL OF MATHEMATICS Volume 48, Number 1, 18 ON A SINE POLYNOMIAL OF TURÁN HORST ALZER AND MAN KAM KWONG ABSTRACT. In 1935, Turán proved that ( n + a j ) S n,a() = sin(j) >, n j n, a N,
More information1 Chapter 2 Perform arithmetic operations with polynomial expressions containing rational coefficients 2-2, 2-3, 2-4
NYS Performance Indicators Chapter Learning Objectives Text Sections Days A.N. Perform arithmetic operations with polynomial expressions containing rational coefficients. -, -5 A.A. Solve absolute value
More informationMath Assignment 14
Math 2280 - Assignment 14 Dylan Zwick Spring 2014 Section 9.5-1, 3, 5, 7, 9 Section 9.6-1, 3, 5, 7, 14 Section 9.7-1, 2, 3, 4 1 Section 9.5 - Heat Conduction and Separation of Variables 9.5.1 - Solve the
More informationA Library of Functions
LibraryofFunctions.nb 1 A Library of Functions Any study of calculus must start with the study of functions. Functions are fundamental to mathematics. In its everyday use the word function conveys to us
More informationMath 115 ( ) Yum-Tong Siu 1. Derivation of the Poisson Kernel by Fourier Series and Convolution
Math 5 (006-007 Yum-Tong Siu. Derivation of the Poisson Kernel by Fourier Series and Convolution We are going to give a second derivation of the Poisson kernel by using Fourier series and convolution.
More informationAlgebra II B Review 5
Algebra II B Review 5 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Find the measure of the angle below. y x 40 ο a. 135º b. 50º c. 310º d. 270º Sketch
More informationMath 489AB A Very Brief Intro to Fourier Series Fall 2008
Math 489AB A Very Brief Intro to Fourier Series Fall 8 Contents Fourier Series. The coefficients........................................ Convergence......................................... 4.3 Convergence
More informationMATH 32 FALL 2013 FINAL EXAM SOLUTIONS. 1 cos( 2. is in the first quadrant, so its sine is positive. Finally, csc( π 8 ) = 2 2.
MATH FALL 01 FINAL EXAM SOLUTIONS (1) (1 points) Evalute the following (a) tan(0) Solution: tan(0) = 0. (b) csc( π 8 ) Solution: csc( π 8 ) = 1 sin( π 8 ) To find sin( π 8 ), we ll use the half angle formula:
More informationA Proof of Markov s Theorem for Polynomials on Banach spaces
A Proof of Markov s Theorem for Polynomials on Banach spaces Lawrence A. Harris Department of Mathematics, University of Kentucky Lexington, Kentucky 40506-007 larry@ms.uky.edu Dedicated to my teachers
More informationStudent: Date: Instructor: kumnit nong Course: MATH 105 by Nong https://xlitemprodpearsoncmgcom/api/v1/print/math Assignment: CH test review 1 Find the transformation form of the quadratic function graphed
More informationNAME DATE PERIOD. Trigonometric Identities. Review Vocabulary Complete each identity. (Lesson 4-1) 1 csc θ = 1. 1 tan θ = cos θ sin θ = 1
5-1 Trigonometric Identities What You ll Learn Scan the text under the Now heading. List two things that you will learn in the lesson. 1. 2. Lesson 5-1 Active Vocabulary Review Vocabulary Complete each
More information2. Determine the domain of the function. Verify your result with a graph. f(x) = 25 x 2
29 April PreCalculus Final Review 1. Find the slope and y-intercept (if possible) of the equation of the line. Sketch the line: y = 3x + 13 2. Determine the domain of the function. Verify your result with
More informationNORMAL NUMBERS AND UNIFORM DISTRIBUTION (WEEKS 1-3) OPEN PROBLEMS IN NUMBER THEORY SPRING 2018, TEL AVIV UNIVERSITY
ORMAL UMBERS AD UIFORM DISTRIBUTIO WEEKS -3 OPE PROBLEMS I UMBER THEORY SPRIG 28, TEL AVIV UIVERSITY Contents.. ormal numbers.2. Un-natural examples 2.3. ormality and uniform distribution 2.4. Weyl s criterion
More informationsin cos 1 1 tan sec 1 cot csc Pre-Calculus Mathematics Trigonometric Identities and Equations
Pre-Calculus Mathematics 12 6.1 Trigonometric Identities and Equations Goal: 1. Identify the Fundamental Trigonometric Identities 2. Simplify a Trigonometric Expression 3. Determine the restrictions on
More informationExercise Set 6.2: Double-Angle and Half-Angle Formulas
Exercise Set : Double-Angle and Half-Angle Formulas Answer the following π 1 (a Evaluate sin π (b Evaluate π π (c Is sin = (d Graph f ( x = sin ( x and g ( x = sin ( x on the same set of axes (e Is sin
More informationFurther Mathematical Methods (Linear Algebra) 2002
Further Mathematical Methods (Linear Algebra) Solutions For Problem Sheet 9 In this problem sheet, we derived a new result about orthogonal projections and used them to find least squares approximations
More informationA. Incorrect! For a point to lie on the unit circle, the sum of the squares of its coordinates must be equal to 1.
Algebra - Problem Drill 19: Basic Trigonometry - Right Triangle No. 1 of 10 1. Which of the following points lies on the unit circle? (A) 1, 1 (B) 1, (C) (D) (E), 3, 3, For a point to lie on the unit circle,
More informationCourse Notes for Calculus , Spring 2015
Course Notes for Calculus 110.109, Spring 2015 Nishanth Gudapati In the previous course (Calculus 110.108) we introduced the notion of integration and a few basic techniques of integration like substitution
More informationNORMS OF PRODUCTS OF SINES. A trigonometric polynomial of degree n is an expression of the form. c k e ikt, c k C.
L NORMS OF PRODUCTS OF SINES JORDAN BELL Abstract Product of sines Summary of paper and motivate it Partly a survey of L p norms of trigonometric polynomials and exponential sums There are no introductory
More informationAnalytic Trigonometry. Copyright Cengage Learning. All rights reserved.
Analytic Trigonometry Copyright Cengage Learning. All rights reserved. 7.4 Basic Trigonometric Equations Copyright Cengage Learning. All rights reserved. Objectives Basic Trigonometric Equations Solving
More informationInterpolation and the Lagrange Polynomial
Interpolation and the Lagrange Polynomial MATH 375 J. Robert Buchanan Department of Mathematics Fall 2013 Introduction We often choose polynomials to approximate other classes of functions. Theorem (Weierstrass
More informationb = 2, c = 3, we get x = 0.3 for the positive root. Ans. (D) x 2-2x - 8 < 0, or (x - 4)(x + 2) < 0, Therefore -2 < x < 4 Ans. (C)
SAT II - Math Level 2 Test #02 Solution 1. The positive zero of y = x 2 + 2x is, to the nearest tenth, equal to (A) 0.8 (B) 0.7 + 1.1i (C) 0.7 (D) 0.3 (E) 2.2 ± Using Quadratic formula, x =, with a = 1,
More informationON THE CHEBYSHEV POLYNOMIALS. Contents. 2. A Result on Linear Functionals on P n 4 Acknowledgments 7 References 7
ON THE CHEBYSHEV POLYNOMIALS JOSEPH DICAPUA Abstract. This paper is a short exposition of several magnificent properties of the Chebyshev polynomials. The author illustrates how the Chebyshev polynomials
More informationPrecalculus Review. Functions to KNOW! 1. Polynomial Functions. Types: General form Generic Graph and unique properties. Constants. Linear.
Precalculus Review Functions to KNOW! 1. Polynomial Functions Types: General form Generic Graph and unique properties Constants Linear Quadratic Cubic Generalizations for Polynomial Functions - The domain
More informationA. Incorrect! This equality is true for all values of x. Therefore, this is an identity and not a conditional equation.
CLEP-Precalculus - Problem Drill : Trigonometric Identities No. of 0 Instructions: () Read the problem and answer choices carefully () Work the problems on paper as. Which of the following equalities is
More informationNUMERICAL METHODS. x n+1 = 2x n x 2 n. In particular: which of them gives faster convergence, and why? [Work to four decimal places.
NUMERICAL METHODS 1. Rearranging the equation x 3 =.5 gives the iterative formula x n+1 = g(x n ), where g(x) = (2x 2 ) 1. (a) Starting with x = 1, compute the x n up to n = 6, and describe what is happening.
More informationCALCULUS ASSESSMENT REVIEW
CALCULUS ASSESSMENT REVIEW DEPARTMENT OF MATHEMATICS CHRISTOPHER NEWPORT UNIVERSITY 1. Introduction and Topics The purpose of these notes is to give an idea of what to expect on the Calculus Readiness
More informationSection Integration of Nonnegative Measurable Functions
18.2. Integration of Nonnegative Measurable Functions 1 Section 18.2. Integration of Nonnegative Measurable Functions Note. We now define integrals of measurable functions on measure spaces. Though similar
More information