⇐
Previous Lecture Notes
|
⇑ Up |
Next Lecture Notes ⇒
|
Summary
|
|
---|---|
MENU
|
Technical Writing
|
General Writing
|
Document Type
| All professional, scientific and specialist documents, drafts, reports, letters, papers, theses | Literary and other types of writing |
Style
| Familiar, simple, clear and precise and of evaluating value | Poetic, rhetorical or elegant and carries stamp of individuality |
Skills
| Acquired through practice | Creative and innovative with an inborn talent |
Format
| Strict and well defined so that the reader can understand the organization of document | No set pattern and predefined organization |
Language
| Simple, straightforward, objective, rational and scientific | Elegant or creative. Can be poetic, literary or generic |
Words
| Technical words and their explanation | Descriptive and literary composition |
Content
| Preplanned on the basis of information collected | Spontaneous and written on-the-spur of the moment |
Profession
| Pertains to profession | Relates to society in general |
Function
| Instructs, informs and persuades | Amuses, inspires and educates |
Diction
| It is simple and effective | May use complex and long sentences, but the meaning will be clear |
---|
Bad: I present the time complexity of an algorithm in this paper. OK: We present the time complexity of an algorithm in this paper. Better: The time complexity of an algorithm is presented in this paper. [traditional style: passive voice] This paper presents the time complexity of an algorithm. [modern style: active voice]
Bad: We first sort points in the nondecreasing order of their x-coordinate values. Then, we find a closest pair of points. Good: An algorithm first sorts points in the nondecreasing order of their x-coordinate values. Then, the algorithm finds a closest pair of points.
"This paper models a problem of finding shortest paths from a single source to all other vertices in a digraph." "This paper presents an algorithm for determining whether a given positive integer is prime."
"In 1883, the puzzle called the Towers of Hanoi was invented by the French mathematician Édouard Lucas." "E. W. Dijkstra proposed a greedy algorithm for finding single-source shortest paths in 1959." "Although we have presented an O(n log n) algorithm for MPT, it remains unknown whether the algorithm has an optimum time complexity. That will be one of our future research topics."
→ Step-wise Refinement in Top-Down Design
One Paragraph ←→ One Issue (Modularity)
This paper is organized as follows. Section 2 presents necessary definitions and notations on transactions in a distributed database system. Section 3 discusses properties of a transaction that are required for fault tolerance. Based on the properties, Section 4 presents a distributed algorithm for recovery from failures of transactions in a distributed database system. Then, Section 5 analyzes the communication complexity of the algorithm and compares it with those of the previously proposed algorithms. Finally, Section 6 summaries our results and discusses future directions of our research.
Example:
It is widely known that NP-complete problems are time-consuming to solve [2,5]. . . . References [1] E. W. Dijkstra, "A note on two problems in connexion with graphs," Numerische Mathematik, vol. 1, pp.269-271, 1959. [2] M. R. Garey and D. S. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completeness, W. H. Freeman and Co., 1979. . . .
A nonempty set F of n factories fi (1 ≤ i ≤ n)
Bad Example
g(x) = ∑∞k=0 ak xk
g(x) − 3 x g(x)
= ∑∞k=0 ak xk − 3 ∑∞k=1 ak-1 xk
= a0 + ∑∞k=1 (ak − 3 ak-1) xk
g(x) − 3 x g(x) = a0 = 2
(1 − 3x) g(x) = 2
g(x) = 2/(1 − 3x) = 2 ∑∞k=0 (3k xk)
= ∑∞k=0 2 (3k xk)
= ∑∞k=0 (2⋅3k) xk
ak = 2⋅3k
f(n) = 2⋅3n
Bad
2y2 = 2x2 + 4x + 2 = 2 (x + 1)2 / 2 = (x + 1)2
= y = x + 1
Good
The following is a given equation consisting of two variables x and y. 2y2 = 2x2 + 4x + 2By dividing the both sides of the equation by 2, we have the following.y2 = x2 + 2x + 1The right-hand side of the equation can be factorized as follows.x2 + 2x + 1 = (x + 1)2Thus, the given equation is transformed as follows.y2 = (x + 1)2This quadratic equation has two distinct solutions of y in terms of x: y = x + 1 and y = −(x + 1). Note that there are an infinite number of pairs of x and y that satisfy the given equation.Therefore, solutions of the given equation are all pairs of x and y that satisfy at least one of the two linear equations y = x + 1 and y = −(x + 1).
x4 + 1 = 2x2
= x4 - 2x2 + 1 = 0
= (x2 - 1)2 = 0
= x2 - 1 = 0
= x2 = 1
= x = 1, -1
The given equation x4 + 1 = 2x2 is rewritten by moving all terms to the left-hand side as follows. x4 - 2x2 + 1 = 0By considering x2 as a varible (say, y = x2), we can apply the well-known factorization formula of a quadratic polynomial (say, y2 - 2y + 1 = (y - 1)2) for factoring the polynomial of degree 4 in the left-hand side as follows.(x2 − 1)2 = 0It implies that x2 − 1 = 0. Then, we get the equation x2 = 1. Thus, the given equation has two roots: x = 1 and x = −1.
f(n)
= 8 ( 5 n2 + 7 n5 + 2) / ( (1/2) n2 )
given a computational complexity function f(n)
= 16( 5 n2 + 7 n5 + 2) / n2
by mulpiplying 2 to both numerator and denominator
≤ 16( 5 n5 + 7 n5 + 2 n5) / n2
for all n ≥ 1
since n5 ≥ 1 and n5 ≥ n2 for all n ≥ 1
= 16( 14 n5 ) / n2
for all n ≥ 1
= 16( 14 n3 )
for all n ≥ 1
since nc / nd = n(c−d) for all n ≥ 1 when d ≠ 0
= 224 n3
for all n ≥ 1
an upper bound O(n3) derived
Conditional p → q |
|
¬ q → p | p unless q. |
Biconditional p ↔ q |
|
Universal Quantifier ∀ |
|
Existential Quantifier ∃ |
|
Definitions |
|
Assumptions |
|
Either the election is not decided and the votes have been counted or the votes have not been counted.
For every integer x, if there exists some integer y such that x times y is not equal to x, then x is not equal to 0.
A natural number p is a prime iff p is greater than 1 and there is no factor of p other than 1 and itself p.
Let x be an arbitrary integer. We prove the above assertion, using proof by cases. There are three cases to be considered.
∃ x [ ∃ y [ p(x) ∧ p(y) ∧ q(x,y) ] ]
Proof:
We prove it by constructive proof.
Let x = 2 and y = 3.
Obviously, x and y are prime, i.e., p(x) ∧ p(y) holds.
Since x + y = 5 is also prime, q(x,y) holds.
Therefore, p(x) ∧ p(y) ∧ q(x,y) is true for x = 2 and y = 3.