Homework - Algorithms - Summer 2008 | CMSC 351, Assignments of Algorithms and Programming

Material Type: Assignment; Professor: Kruskal; Class: Algorithms; Subject: Computer Science; University: University of Maryland; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 02/13/2009

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Summer 2008 CMSC 351: Homework 0 Clyde Kruskal
Problem 1. Use mathematical induction to show that
(a)
n
X
i=1
i(i+ 1) = n(n+ 1)(n+ 2)
3(b)
n
X
i=0
2i= 2n+1 1
Problem 2. See bottom of page 53 of CLRS (bottom of page 34 in CLR) and/or the
bottom of this sheet.
(a) Assume bx=a. What is x(in terms of aand b)?
(b) Using only part (a), show that logc(ab) = logca+ logcb.
(c) Show that alogbn=nlogba
Problem 3. Differentiate the following functions:
(a) ln(x2+ 5)
(b) lg(x2+ 5)
(c) 1
ln(x2+5)
Problem 4. Integrate the following functions:
(a) 1
x
(b) 1
3x+7
(c) ln x[HINT: Use integration by parts.]
(d) xln x[HINT: Use integration by parts.]
(e) xlg x
lg n= log2n
ln n= logen
lgkn= (lg n)k
lg lg n= lg(lg n)
For all real a > 0, b > 0, c > 0, and n,
a=blogba
logc(ab) = logca+ logcb
logban=nlogba
logba=logca
logcb
logb(1/a) = logba
logba=1
logab
alogbn=nlogba

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Summer 2008 CMSC 351: Homework 0 Clyde Kruskal

Problem 1. Use mathematical induction to show that

(a)

∑^ n i=

i(i + 1) = n(n^ + 1)( 3 n^ + 2) (b)

∑^ n i=

2 i^ = 2n+1^ − 1

Problem 2. See bottom of page 53 of CLRS (bottom of page 34 in CLR) and/or the bottom of this sheet. (a) Assume bx^ = a. What is x (in terms of a and b)? (b) Using only part (a), show that logc(ab) = logc a + logc b. (c) Show that alogb^ n^ = nlogb^ a

Problem 3. Differentiate the following functions:

(a) ln(x^2 + 5) (b) lg(x^2 + 5) (c) (^) ln(x^12 +5)

Problem 4. Integrate the following functions:

(a) (^1) x (b) (^3) x^1 + (c) ln x [HINT: Use integration by parts.] (d) x ln x [HINT: Use integration by parts.] (e) x lg x

lg n = log 2 n ln n = loge n lgk^ n = (lg n)k lg lg n = lg(lg n) For all real a > 0, b > 0, c > 0, and n, a = blogb^ a logc(ab) = logc a + logc b logb an^ = n logb a logb a = logc^ a logc b logb(1/a) = − logb a logb a =

loga b alogb^ n^ = nlogb^ a