Presence-Condition Simplification in Highly - uni-passaude
presence condition have been introduced by the global variabil-ity model (e g , encrypt requires keys) A desirable simplifica-tion of the presence condition is to identify encrypt and autore-sponder as the sole cause of the defect and to “hide” the parts in-troduced by the variability model Simplifying the condition and
4 BOOLEAN ALGEBRA AND LOGIC SIMPLIFICATION
LOGIC SIMPLIFICATION BOOLEAN OPERATIONS AND EXPRESSIONS Variable, complement, and literal are terms used in Boolean algebra A variable is a symbol used to represent a logical quantity Any single variable can have a 1 or a 0 value The complement is the inverse of a variable and is indicated by a bar over variable (overbar)
Simplifying finite sums - CMU
Jan 26, 2014 · Given any function f , f is another function de ned by f (x) = f (x + 1) f (x) (A thing that turns functions into other functions is called an operator is called the di erence operator ) The basic identity we rely on is this: bX 1 k=a f (k) = f (b) f (a): To simplify any nite sum whatsoever, all we need to do is nd a
An Introduction to Partial Differential Equations
we find from the differential equation (DE) that (2 20) XT t = DX xxT and dividing by XTwe find (2 21) T t DT = X xx X = −λ where λis to be determined Now because T t/DT is only a function of t and X xx/X is only a function of x we know that λ must be independent of xand trespectively, and therefore must be a constant
E12 Digital Electronics I 51 Cot 2007 E12 Digital
Complete Simplification Process 1 Construct the K map and place 1s and 0s in the squares according to the truth table 2 Group the isolated 1s which are not adjacent to any other 1s (single loops) 3 Group any pair which contains a 1 adjacent to only one other 1 (double loops) 4 Group any octet even if it contains one or more 1s that have
Karnaugh Maps (K-map) - Auburn University
• The output for a don’t care condition can be either 0 or 1 WE DON’T CARE • Don’t Care conditions denoted by: X, -, d, 2 • X is probably the most often used • Can also be used to denote inputs Example: ABC = 1X1 = AC • B can be a 0 or a 1
K-maps
Simplification of Boolean Functions Using K-maps •This is equivalent to the algebraic operation, aP + a P =P where P is a product term not containing a or a •A group of cells can be combined only if all cells in the group have the same value for some set of variables
Covered Entity Guidance - CMS
Jun 17, 2016 · Administrative Simplification: Covered Entity Guidance 2 Background The Administrative Simplification standards adopted by HHS under the Health Insurance Portability and Accountability Act of 1996 (HIPAA) apply to any entity that is: • A health care provider that conducts certain transactions in electronic form (referred to
2 Propositional Equivalences 21 Tautology/Contradiction
,::p^:q De Morgan’s Law,p^:q Double Negation Law This method is very similar to simplifying an algebraic expression You are using the basic equivalences in somewhat the same way you use algebraic rules like 2x 3x= xor (x+ 1)(x 3) x 3 = x+ 1 Exercise 2 4 1 Use the propositional equivalences in the list of important logical
Principes fondamentaux des oscilloscopes
Condition de configuration initiale (exemple) Condition de configuration optimale Niveau de déclenchement Configurer la mise à l’échelle des signaux de l’oscilloscope est un processus répétitif qui consiste à effectuer des réglages sur le panneau avant jusqu’à ce que « l’image » souhaitée soit affichée à l’écran
[PDF] fraction algébrique
[PDF] fraction algébrique exercices
[PDF] fractions algébriques conditions d'existences
[PDF] fraction algébrique théorie
[PDF] fraction algébrique definition
[PDF] fraction algébrique 3ème exercices
[PDF] domaine d'une fonction definition
[PDF] condition d'existence math
[PDF] condition de germination des graines 6eme
[PDF] pour germer une graine a besoin
[PDF] tp de germination des graines pdf
[PDF] conditions germination graines pdf
[PDF] la germination des graines cours pdf
[PDF] les étapes de la germination pdf
Simplifying nite sums
Misha Lavrov
ARML Practice 1/26/2014
Warm-up
1. Find 1 + 2 + 3 + + 100. (The story goes that Gauss was given this problem by his teacher in elementary school to keep him busy so he'd quit asking hard questions. But he gured it out in his head in about ten seconds.) 2. (2003 AIME I.) One hundred c oncentriccircles with radii1;2;3;:::;100 are drawn in a plane. The smallest circle is
colored red, the strip around it green, and from there the colors alternate. What fraction of the total area of the largest circle is colored green?Warm-up
Solutions
1.The classic a rgumentgo eslik ethis:
1 + 2 + 3 ++ 100
+ 100 + 99 + 98 ++ 1101 + 101 + 101 ++ 101IfSis the sum, 2S= 100101 = 10100, soS= 5050.
2. The a reaof a strip b etweenthe circle of radius rand the circle of radiusr+1 is(r+1)2r2= (2r+1), which we can rewrite as (r+ 1)+r. Then x=1002992+ 982972++ 221210000100+ 99+ 98+ 97++ 2+10000
505010000= 0:505:
Basic summations
1.Arithmetic series:
n X k=1k= 1 + 2 ++n=n(n+ 1)2 =n+ 1 2 In general, given an arithmetic progression that starts ata, ends atz, and hasnterms, its sum isna+z2 2.Geometric se ries:fo rr6= 1,
n1X k=0r k= 1 +r+r2++rn1=rn1r1:As a special case,
Pn1 k=02k= 2n1.Exchanging double sums
Consider the sumS=Pn1
k=0k2k. We will evaluate this sum as follows:n1X k=0k2k=n1X k=0k1X `=02 k=n1X `=0n1X k=`+12 k:Having reordered the two sums, we rst evaluate the inner one: n1X k=`+12 k=n1X k=02 k`X k=02 k= (2n1)(2`+11) = 2n2`+1:Now the outer sum is also easy: n1X `=0(2 n2`+1) =n2n2n1X `=02 `= (n2)2n+ 2:Exchanging double sums
Consider the sumS=Pn1
k=0k2k. We will evaluate this sum as follows:n1X k=0k2k=n1X k=0k1X `=02 k=n1X `=0n1X k=`+12 k:Having reordered the two sums, we rst evaluate the inner one: n1X k=`+12 k=n1X k=02 k`X k=02 k= (2n1)(2`+11) = 2n2`+1:Now the outer sum is also easy: n1X `=0(2 n2`+1) =n2n2n1X `=02 `= (n2)2n+ 2:Exchanging double sums
Consider the sumS=Pn1
k=0k2k. We will evaluate this sum as follows:n1X k=0k2k=n1X k=0k1X `=02 k=n1X `=0n1X k=`+12 k:Having reordered the two sums, we rst evaluate the inner one: n1X k=`+12 k=n1X k=02 k`X k=02 k= (2n1)(2`+11) = 2n2`+1:Now the outer sum is also easy: n1X `=0(2 n2`+1) =n2n2n1X `=02 `= (n2)2n+ 2:Practice with exchanging double sums
1.W edene the n-th harmonic numberHnby
H n=nX k=11k =11 +12 ++1nExpress the sum
Pn k=1Hkin terms ofHn. 2.Things will get trickier when y oudo t histo
Pn k=1k2. (Recall that Pn k=1k=n(n+1)2Exchanging double sums
Solutions
1. nX k=1H k= (n+ 1)Hnn: 2.Let n=Pn
k=0k2. When we expand this out into two sums, switch the sums, and simplify, we get back n=nX `=1 n+ 1 2 2 =2n3+ 3n2+n4 12 n X `=1` 2: We don't yet know how to simplify the last sum, but since it is just 12 n, we can solve the equation fornto get n=n(n+ 1)(2n+ 1)6Review of binomial coecients
Recall that
n r=n!r!(nr)!=n(n1)()(nr+1)r!. These show up inPascal's triangle:K0
0O1K1 0O1K1 1O1K2 0O1K2 1O2K2 2O1K3 0O1K3 1O3K3 2O3K3 3O1K4 0O1K4 1O4K4 2O6K4 3O4K44O1The key point today is the identity
n r=n1 r+n1 r1, orn r=n+1 r+1n r+1. This lets us make many sums telescope: e.g., n1X k=0k=n1X k=0 k 1 =n1X k=0 k+ 1 2 k 2 =n 2Sums of binomial coecients
In general, we have:
n1X k=0 k r =n1X k=0 k+ 1 r+ 1 k r+ 1 =n r+ 1 0 r+ 1 We can use this to evaluate sums of arbitrary polynomials. 1.W ritey ourdegree- dpolynomial inkas a sum ofk
0;k1;:::;k
d. For example,k2= 2k 2+k 1: 2.No ww ecan evaluate the sum:
n X k=1k2= 2nX
k=1 k 2 +nX k=1 k 1 = 2n+ 1 3 +n+ 1 2 3.Then simplify: 2
n+1 3+n+1 2=16 (2n3+ 3n2+n).Practice with binomial coecients
1.Use this technique to nd a fo rmulafo r
Pn k=1k3. 2. Here is a systematic metho dof writing a p olynomialP(x) as a 0x 0+a1x1++adx
d: 2.1I fthe p olynomialis P(x) and has degreed, write
P(0);P(1);:::;P(d) in a row.
2.2 On the next line, write do wnthe dierences: b etweenand below two numbersaandbwriteba. 2.3 Rep eatstep 2 to the ro wof dierences, and k eepgoing until a row with only one number is left. 2.4T hec oecientsof
x0;:::;x
dare the rst entries in each row.Use this method to nd
Pn k=1k4. 3.Why do esthe metho dab ovew ork?
Practice with binomial coecients
Solutions
1.W ritingk3as 6k
3+ 6k 2+k1, we get
n X k=1k3= 6n+ 1
4 + 6n+ 1 3 +n+ 1 2 =n2(n+ 1)24 2.The ro wsof dierences w eget a re:
0 1 16 81 256
1 15 65 175
14 50 110
36 6024
Sok4= 24k
4+ 36k
3+ 14k
2+k1, and
n X k=1k4= 24n+ 1
5 + 36n+ 1 4 + 14n+ 1 3 +n+ 1 2The dierence operator
Given any functionf, fis another function dened by f(x) =f(x+ 1)f(x). (A thing that turns functions into other functions is called an operator. is called thedierence operator.)The basic identity we rely on is this:
b1X k=af(k) =f(b)f(a): To simplify any nite sum whatsoever, all we need to do is nd a functionfsuch that fis the function we're summing. This is what we did in the previous section: we discovered that xx r=x r1, which let us solve any sum with binomial coecients in it.Dierence operator problems
1.Compute (2
x), (x2x), and (x22x). 2. Use #1, and the basic rule th at( f+g) = f+g, to nd n1X k=02 k;n1X k=0k2k;andn1X k=0k 22k:3. Let Fndenote then-th Fibonacci number, dened byF0= 0, F