engineering mathematics for computer science pdf
Mathematics for Computer Science pot
... simplification Therefore, P (n) is true for all natural n by induction, and the theorem is proved Induction was helpful for proving the correctness of this summation formula, but not helpful for discovering ... core truth of mathematics 2.3 Using Induction Induction is by far the most important proof technique in computer science Generally, induction is used to prove that some statement holds for all natural ... Eric Lehman and Tom Leighton Mathematics for Computer Science 2004 Contents What is a Proof? 15 1.1 Propositions ...
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Mathematics for Computer Science pptx
... of numbers of the form rn, where r is a positive real number and n N Well ordering commonly comes up in Computer Science as a method for proving that computations won’t run forever The idea is ... proposition for each possible set of truth values for the variables For example, the truth table for the proposition “P AND Q” has four lines, since there are four settings of truth values for the ... before we start into mathematics, we need to investigate the problem of how to talk about mathematics To get around the ambiguity of English, mathematicians have devised a special language for...
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concrete mathematics a foundation for computer science phần 1 pdf
... “Discrete Mathematics! ’ Therefore the subject needs a distinctive name, and “Concrete Mathematics has proved to be as suitable as any other The original textbook for Stanford’s course on concrete mathematics ... foundation for computer science / Ronald L Graham, Donald E Knuth, Oren Patashnik xiii,625 p 24 cm Bibliography: p 578 Includes index ISBN o-201-14236-8 Mathematics 19612 Electronic data processing Mathematics ... expression for the quantity of interest For the Tower of Hanoi, this is the recurrence (1.1) that allows us, given the inclination, to compute T,, for any n Find and prove a closed form for our...
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Vic broquard c++ for computer science and engineering
... C++ for Computer Science and Engineering Vic Broquard Copyright 2000, 2002, 2003, 2006 by Vic Broquard All rights reserved No part of this book may be reproduced or transmitted in any form without ... substantially longer to perform.) Thus, it is the ability of the modern computer to perform reliably and to perform at great speed that has made it so powerful Introduction to Programming Computers have ... was formally standardized by ISO (International Standards Organization) This means that now your C++ program can be written and compiled on any computer platform (PCs, minicomputers, mainframe computers)...
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Faculty of Computer Science and Engineering Department of Computer Science Part 2 pdf
... Faculty of Computer Science and Engineering Department of Computer Science a b c d e f–k f *k f\ 10 f\ x f* f2 Page 2/10 Faculty of Computer Science and Engineering Department of Computer Science ... x) } } 31 Page 5/10 Faculty of Computer Science and Engineering Department of Computer Science Appendix Formal parameters and actual parameters Simply speaking, formal parameters are those that ... global function Page 7/10 Faculty of Computer Science and Engineering Department of Computer Science In Example 2, we are in the situation of developing a method for the class List, thus we can assume...
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concrete mathematics a foundation for computer science phần 2 pptx
... closed form for f(n) , when n L Provethatf(n)=n-l+f([n/2~)+f(~n/Z])foralln~l 35 Simplify the formula \(n + )‘n! e] mod n 36 Assuming that n is a nonnegative integer, find a closed form for the ... arbitrary positive integer n in the form n = 2m + 1, where < < 2” Give explicit formulas for and m as functions of n, using floor and/or ceiling brackets What is a formula for the nearest integer to a ... v(x+l) Au(x) (2.54) This formula can be put into a convenient form using the shij?! operator E, defined by Ef(x) = f(x+l) Substituting this for v(x+l) yields a compact rule for the difference of...
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concrete mathematics a foundation for computer science phần 3 ppsx
... 4) QED for the case m = 12 So far we’ve proved our congruence for prime m, for m = 4, and for m = 12 Now let’s try to prove it for prime powers For concreteness we may suppose that m = p3 for some ... cl(P) = -1; p(pk) = for k > Therefore by (4.52), we have the general formula ifm=pjpz p,; if m is divisible by some p2 (4.57) That’s F If we regard (4.54) as a recurrence for the function q(m), ... write for gcd( m, n), a for m’, and b for -n’.) The Farey series gives us another proof of (4.32), because we can let b/a be the fraction that precedes m/n in 3,, Thus (4.5) is just (4.31) again For...
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concrete mathematics a foundation for computer science phần 4 ppsx
... summation formula of the form al, a,, z kbk = CF h, b, 1) AI, AM, '5, , BN (5.127) k’ then we also have al, a,, bl, bn k+l ’ for any integer There’s a general formula for shifting ... ignore convergence if we are using z simply as a formal symbol It is not difficult to verify that formal infinite sums of the form tk3,, (Xkzk form a field, if the coefficients ak lie in a field ... on such formal sums without worrying about convergence; any identities we derive will still be formally true For example, the hypergeometric F( “i ,’ /z) = tkZO k! zk doesn’t converge for any...
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concrete mathematics a foundation for computer science phần 5 pps
... a closed formula for them We haven’t found closed forms for Stirling numbers, Eulerian numbers, or Bernoulli numbers either; but we were able to discover the closed form H, = [“:‘]/n! for harmonic ... -F; = (-l).", for n > (6.103) When n = 6, for example, Cassini’s identity correctly claims that 3.5-tS2 = A polynomial formula that involves Fibonacci numbers of the form F,,+k for small values ... a few things For one, kim’ isn’t a polynomial if j = 0; so we will need to split off that term and handle it separately For another, we’re missing the term k = from the formula for nth difference;...
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concrete mathematics a foundation for computer science phần 6 doc
... exist for negative n Two kinds of “closed forms” come up when we work with generating functions We might have a closed form for G(z), expressed in terms of z; or we might have a closed form for ... that R(0) # 00 can be expressed in the form R(z) = S(z) t T(z), (7.28) where S(z) has the form (7.26) and T(z) is a polynomial Therefore there is a closed form for the coefficients [z”] R(z) Finding ... (7.10) and get a closed form for the coefficients, but it’s bett,er to save that for later in the chapter after we’ve gotten more experience So let’s divest ourselves of dominoes for the moment and...
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concrete mathematics a foundation for computer science phần 7 pot
... element of n has the form (T + HT)"HH for some n 0, and each term of T has the form (0 + E)n Therefore by (7.4) we have s = (I-T-HT)-'HH, and the probability generatin.g function for our problem is ... PROBABILITY For example, suppose ,the keys are names, and suppose that there are m = lists based on the first letter of a name: 1, for ,4-F; 2, for G-L; h(name) = 3, for M-R; 4, for !3-Z We start ... form for P~,~ Generalizing your a.nswer to part (a), find a closed form for the probability that exactly k matches are in the other box when an empty one is first th.rown away Find a closed form...
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concrete mathematics a foundation for computer science phần 8 doc
... that R stands for “‘real part.” valid for !.Xz > 9%~ The asymptotic formula for Bernoulli numbers B, in Table 438 illustrates this principle On the other hand, the asymptotic formulas for H,, n!, ... is negative for < x < i and positive for i < x < Therefore its integral, Bdk+~ (x)/(4k+2), decreases for < x < and increases for i < x < Moreover, we have bk+l(l - X) = -&+I (X) , for < x < 1, ... therefore a closed form for gn seems out of the question, and asymptotic information is probably the best we can hope to derive Our first handle on this problem is the observation that < gn for...
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concrete mathematics a foundation for computer science phần 9 pps
... variables Lij for all i # j Setting kii = li’ -Lji for :< i < j < n and using the constraints tifi (lij -iii) = for all i < n allows us to carry out the sums on li, for j < n and then on iii for < i ... hint, we get andasimilarformulafor&,(z) Thustheformulas (ztB;‘(z)‘B[(z)+l)Bt(z)r and (ztE;‘(z)&:(z) + l)&,(z)’ give the respective right-hand sides of (5.61) We must therefore prove that (zwwJm ... of F( a2 + j, a3, , a,,,; b2, , b,; z) for j d, thereby eliminating an upper parameter and a lower parameter Thus, for example, we get closed forms for F( a, b; a - 1; z), F( a, b; a - 2; z),...
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concrete mathematics a foundation for computer science phần 10 docx
... constants Q, B, no, C such that la(n)/ Blf(n)[ for n mc and [b(n)1 Clg(n)l for n no Therefore the left-handfunctionisatmostmax(B,C)(lf(n)l+Ig(n)l),forn3max(~,no), so it is a member of the right ... 525-526 602 131 Ronald L Graham, Donald E Knuth, and Oren Patashnik, Concrete 102 Mathematics: A Foundation for Computer Science Addison-Wesley, 1989 (The first printing had a different Iversonian ... 196-204, 283-285, 306-366 for Bernoulli numbers, 271, 337, 351 for convolutions, 339-350, 355, 407 Dirichlet, 356-357, 359, 418, 437 for Eulerian numbers, 337 exponential, 350-355 for Fibonacci numbers,...
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Challenges faced by information technology students in reading english for computer science
... learned for years, others for years and the rest has learnt for years In the department of information technology at our college, they engage in ESP course, it means English for Computer Science, ... challenges in learning reading ESP for computer science, the questionnaires were designed for students Then data were analyzed and the information was displayed in the form of tables III.4 Summary ... examine the areas of challenges in reading comprehension of English for Computer Science for second-year students in the department of information technology at Nghe An JTTC To be more specific, the...
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Tài liệu Discrete Math for Computer Science Students doc
... Our algorithm performs a certain set of multiplications For any given i, the set of multiplications performed in lines through can be divided into the set S1 of multiplications performed when j ... multiplications performed when j = 2, and, in general, the set Sj of multiplications performed for any given j value Each set Sj consists of those multiplications the inner loop carries out for a particular ... using the formula for n is is straightforward to show that n n−1 = n (n − 2) However this proof just uses blind substitution and simplification Find a more conceptual explanation of why this formula...
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Faculty of Computer Science and Engineering Department of Computer Science LAB SESSION 1 pptx
... Faculty of Computer Science and Engineering Department of Computer Science pTemp->data = 3; pTemp->next = pHead; pHead = pTemp; // the ... pTemp = pTemp->next; delete pHead; pHead = pTemp; Page 2/5 Faculty of Computer Science and Engineering Department of Computer Science } } } Listing Having the List class implemented, the main function ... (num>0) pList.addFirst(num); } else valid = 0; } Page 3/5 Faculty of Computer Science and Engineering Department of Computer Science return pList; } a Rewrite the main function in Exercise 3.1...
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Faculty of Computer Science and Engineering Department of Computer Science - LAB SESSION 2 ppt
... Faculty of Computer Science and Engineering Department of Computer Science count++; } void List::display() { Node* pTemp = pHead; while ... pTemp->next;; pTemp->data += nConst; return; } Listing Page 2/4 Faculty of Computer Science and Engineering Department of Computer Science EXERCISES In this work, you are provided seven files: List.h, ... main function to test your implemented methods Page 3/4 Faculty of Computer Science and Engineering Department of Computer Science 4.8 Develop the method getIntersection of class List that find...
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Faculty of Computer Science and Engineering Department of Computer Science - LAB SESSION 3 RECURSION pot
... Faculty of Computer Science and Engineering Department of Computer Science } // Tree::~Tree() { destroy(root); ... getSizeFrom(pNode->right) + 1; return nResult; } Listing 2/3 Faculty of Computer Science and Engineering Department of Computer Science Listing gives a scenario in which we try to develop a method ... getSizeFrom(root); } int Tree::getSizeFrom(Node *pNode) { int nResult; //stop condition: what we should for the simplest case – an empty string if (pNode = NULL) nResult = 0; //recursive case: assume...
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