Many Cheerful Facts

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Many Cheerful Facts for EE 20N/120

GSI: James Fung

March 8, 2011

1 Introduction

Like many math classes, EE 20N and 120 assume students know of a number of denitions, concepts,formulas, and operations. This is necessary to provide a solid foundation upon which to build theconcepts introduced in these courses. This document attempts to list, summarize, and highlight anumber of these useful facts as a study aid. In no way is this a replacement for attending lecture,discussion, or o ce hours.

Structure. This document is broken into a number of short blocks. Each block contains up to 4sections:

Statement : where the item is dened, expressed as an equation, etc.

Meaning : a prose description to put concepts into words, which will facilitate thinking about

the item Usage : while not an exhaustive list, situations where the item is likely to be applicable

Derivation : an outline of how the result can be derived, which may help students see howdi erent concepts are related or be used as an a practice exercise

2 Complex Numbers

2.1 Common complex numbers

Statement. These will probably be the most common complex numbers youll deal with becausethey are the most fundamental:

1 =1 ei0

i =1 ei

2

1 =1 ei

i =1 ei3 2

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Because this course likes to use nice numbers, these will also be helpful:

p 32

+ i12

=1 ei

6

p 22

+ ip 22

=1 ei

4

12

+ i p 32

=1 ei

3

As noted in Section 2.2, all of these phases are to within 2k, k 2Z .Usage. You should just know these, as they will make converting between rectangular and polarcoordinates much easier. If you have trouble, the easiest way to memorize them is to picture themaround the unit circle.

2.2 Magnitude and phase

Statement. Let z = a + bi = Re i, R 0. Its magnitude is:

R = |z | = p a2 + b2Its phase is:

= \ z = arctanba+ 2k, k 2

Z

Note: arctan on a calculator will return a value in 2 ,

2

.

Meaning. Because ei2k = 1, we can add 2 k to the phase and not change the number. Thus ei

and e i are both the same number, namely -1.

Note that since magnitude is always non-negative, phase between 2 and

2 corresponds to thepositive real half of the complex plane (i.e. quadrants I and IV, see Figure 2.2). Instead of magnitude going negative, keep the magnitude positive and shift the phase by .

Usage. Deriving magnitude and phase response from frequency response.

2.3 Multiplication and division

Statement. Let x = Rx eix and y = Ryeiy . Then

x y = Rx Ryei(x + y )

Note that the magnitudes multiply while the phases add:

|x y| = Rx Ry = |x | |y|\ (x y) = x + y = \ x + \ y

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Figure 1: Since b/a = b/ a, be careful when taking arctan. Choose the appropriate phase basedon the real and imaginary components.

For division:

xy

=RxRy

ei (x y )

Meaning. Note:

1x

=1

Rxe ix

See Figure 2.3

Usage. Multiplication and division of complex numbers is easier when they are in polar form. Incontrast, addition and subtraction are easiest in cartesian form.

Useful for nding the magnitude and phase of some frequency responses, especially rational transfer

functions. In particular, it is often easier to understand and plot a complicated frequency responseby factoring its numerator and denominator into 1st-order components and then combining theircontributions.

2.4 Complex conjugate

Statement. Let z = a + bi = Re i. Its conjugate is z= a bi = Re i .

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Figure 2: Reciprocal and conjugate of a complex number.

Meaning. The complex conjugate ips the sign of the imaginary component of a complex numberwhile leaving the real component unchanged. Graphically, in the complex plane, z is a reectionof z with respect to the real (horizontal) axis (see Figure 2.3).

Usage. Useful for relating complex and real numbers. For instance, using the example above,z + z= 2 a = 2Re {z} and z z= R2 = a2 + b2. See Figure 2.4.

Also, note that if a polynomial with real coe cients has a complex root, the complex roots come inpairs. Thus, if an polynomial in z has complex root p, then p is also a root. Thus, if p = + i =

ei , then(z p)(z p) =0

z2 ( p + p)z + p p=0z2 2z + 2 =0

Thus we can determine (its real component) and (its magnitude) directly from the polynomial.Furthermore, since 2 = 2 + 2, we can easily determine its complex component. This is exactlywhat your familiar quadratic equation from algebra does.

2.5 Eulers Formula

Statement.

ei= cos() + i sin()

That is

Reneio=cos( ), andImneio=sin( )

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Figure 3: This xkcd comic means youll never forget what a complex conjugate does.

Meaning. As ei traces a path around the unit circle, both the real and imaginary parts of thispath are sinusoids.

Usage. Complex exponentials are everywhere in this course. This formula is often used to converta complex exponential into real and imaginary parts. Also, using [2.4], we can derive two very

important formulas:

cos() =ei+ e i

2

sin() =ei e i

2i

Tangentially related to these formulas, it is often useful to exploit symmetry to simplify expressions.For example:

e i+ e 3i=

ei+ e i

e 2i

=2 cos e 2i

Thus we have separated the expression into magnitude and phase components, though note thatmagnitude must always be non-negative, so magnitude is |2cos| and phase must be shifted by when 2 cos< 0.

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As a further digression, did you know:

e3i=(cos + i sin)3

=cos 3+ 3 i cos2 sin 3cossin2 i sin3

=

)Re

ne3i

o=cos3= cos 3 3cossin2

Imne3io=sin3 = 3 cos 2sin sin32.6 Complex N th roots

Statement. There are N distinct N th roots of complex number z = Re i:

z1/N = R1/N ei(+2 k

N ) , k 2 hN iMeaning. The N roots are scattered uniformly in a circle of radius R1/N . One N th root has phaseN , and the rest can be placed in relation to it.

Usage. For solving an equation such as zN = p, or zN p = 0.

Derivation. Instead of thinking of the problem as nding a root, invert the problem and ask whatnumber, when raised to the N th power, produces the original number. Let y = Ryeiy be a root.Note that z = Re i(+2 k) for k 2Z . From [2.3]:

yN = RN y eiN y

Thus we need match up magnitude and phase: RN y = R and N y = + 2k.

3 Miscellaneous Math

3.1 Periodicity

Statement. A continuous time function, f : R ! C , is periodic if there exists a real number p suchthat f (t) = f (t + p), 8t 2R .A discrete time function, f : Z ! C , is periodic if there exists an integer p such that f (n) =f (n + p), 8n 2Z .The smallest such positive p is called the fundamental period.

Meaning. The function repeats every p. This must be an integer for discrete time signals becausef (n) is undened over non-integers.

Usage. While the statement If f has a fundamental period p then its fundamental frequency is! 0 = 2/p , is true, its converse is not. Just because a discrete time function has the form ei! 0 n

does not mean it is periodic with period p = 2/ ! 0 since p must be an integer. We cannot evenuse the formula p = 2/ ! 0 because this assumes the signal is periodic. Instead, one should arguefrom the denition of periodicity:

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From the denition of periodicity, f (n) = ei! 0 n is periodic if there exists an integer p such thatf (n) = f (n + p), 8n 2Z .

f (n) = f (n + p)

ei ! 0 n = ei! 0 (n + p)

1 = ei! 0 p

=) ! 0 p =2k, k 2ZSince p must also be an integer, ! 0 must be a rational multiple of 2 for f to be periodic, and thefundamental period is the smallest positive integer p that allows that to happen.

3.2 Odd and even functions

Statement. An even continuous-time function satises xe(t) = xe( t). An odd continuous-timefunction satises xo(t) = xo( t). The discrete-time counterparts would be xe(n) = xe( n) andxo(n) = xo( n).

Meaning. These are di erent forms of symmetry. Even symmetry is a mirror symmetry withrespect to a vertical line running through the origin. Odd symmetry is a rotational symmetry withrespect to the origin.

Usage. The most common even function students will probably encounter is cos( ), and the mostcommon odd function sin( ).

Often in math, symmetry can be exploited to simplify or evaluate expressions. For instance:

Z T

T xe(t)dt = 2 Z

T

0xe(t)dt

That is, since both the positive and negative sides are the same, we can e

ectively integrate oneside and double the result. For odd functions:

Z T

T xo(t)dt = 0

That is, the integration over one side cancels out the other side.

The product of odd and even functions is also odd or even (see Table 1). Applied to the Fouriertransform equations, we see some interesting properties (Table 2). Students should be able to derivethese properties easily and use them to see other results, such as x(t) even (or odd) ) X (! ) even(or odd), x(t) real ) X (! ) = X ( ! ), and x(t) imaginary ) X (! ) = X ( ! ).Any function can be decomposed into odd and even components as follows:

x(t) = xe(t) + xo(t)

xe(t) =x(t) + x( t)

2

xo(t) =x(t) x( t)

2

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x y xyodd odd evenodd even oddeven odd oddeven even even

Table 1: Multiplication of odd and even functions

x(t) X (! )real even real evenreal odd imaginary odd

imaginary even imaginary evenimaginary odd real odd

Table 2: Dependence of CTFT on real/imaginary and even/odd

3.3 Sum of geometric series

Statement.

N

Xk= M k =

M

N +1

1

Furthermore

1

Xk= M k =

M

1 , || < 1

Usage. Summation (in discrete-time or over discrete frequencies) of complex exponentials.

Derivation. Let

S =N

Xk= M kExpand S S .

3.4 Summation of complex exponential

Statement. Let k (n) = eik ! 0 n be a p-periodic discrete-time signal [3.1]. A su cient condition is if ! 0 = 2/p and k 2Z . Summing over one period:

Xn = h pi eik ! 0 n = ( p, k = Np, N 2Z0, otherwise

Meaning. Over one period, unless the complex exponential is constant (the frequency k! 0 is aninteger multiple of 2 rad/dot), the signal will cancel itself out.

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Usage. Discrete-time complex exponentials.

Derivation. Use symmetry (such as [3.2]) or [3.3]

3.5 Integration of complex exponential

Statement. Let k (t) = eik ! 0 t be a p-periodic continuous-time signal [3.1]. A su cient condition isif ! 0 = 2/p and k 2Z . Integrating over one period:

Z t= h pi eik ! 0 t dt = ( p, k = 00, otherwiseMeaning. Over one period, unless the complex exponential is constant (the k! 0 = 0 rad/timefrequency), the signal will cancel itself out.

Usage. Continuous-time complex exponentials.

Derivation. Use symmetry (such as [3.2]) or calculus. Recall

Z eax dx = eax

a, a 2C , a 6= 0

3.6 Di erence of squares

Statement.

a2 b2 =( a + b)(a b)

a2 + b2 =( a + ib)(a ib)

Usage. Occasionally useful for factoring, such as nding poles and zeros in a rational transferfunction. Then [2.3] can be used.

4 Inner Products

4.1 For vectors

Statement. For this class, we dene the inner product of vectors x and y as follows. For discrete-time:

hx, yi = xT y= Xk xkyk

For continuous-time:

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4.2 Orthogonality

Statement. Two elements of a vector space, x and y, are orthogonal if hx, yi = 0. A set,{. . . , x k , . . .}, is mutually orthogonal if hxk , x li = 0 , 8k 6= l.Meaning. In 2 or 3 dimensions, a simple conceptual image is that the two orthogonal vectors areperpendicular. Another conceptual interpretation is that the projection of one element onto theother has zero length (also known as zero norm).

Usage. It is often useful to represent things in terms of an orthogonal basis. For example, if {. . . , x k , . . .} are mutually orthogonal and we can express y as a linear sum of xk :

y = Xk Y kxkFurthermore, the mutual orthagonality makes it easy to nd the coe cients Y k :

hy, x li =

XkhY kxk , x li

= Y l hx l , x li

Y l = hy, x lihx l , x li

E ectively, we have isolated one particular component of y.

4.3 Orthogonality of complex exponentials (discrete)

Statement. Let k be drawn from the set of p-periodic discrete-time complex exponentials. Inparticular, let k(n) = eik ! 0 n , where ! 0 = 2/p . If we dene the inner product over one period,n 2 h pi, then k and l are orthogonal [4.2] if they are di erent ( k l 6= Np, N 2Z ).Usage. [4.2] applied to complex exponentials, which are everywhere in this course.

h k , li = Xn = h pi eik ! 0 n (eil ! 0 n )= Xn = h pi ei(k l)! 0 n

h k , li = ( p, k l = Np, N 2Z0, otherwiseThat is, their inner product equals p when k = l or they are separated by p, 2 p, . . . (but recall,for any integer M , if k l = Mp, k = l = eik ! 0 n ), so the inner product is only non-zero whenthe two complex exponentials are the same.

Derivation. From denition of inner product [4.1] and summation of complex exponentials [3.4].

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4.4 Orthogonality of complex exponentials (continuous)

Statement. Let k be drawn from the set of p-periodic continuous-time complex exponentials. Inparticular, let k (t) = eik ! 0 t , where ! 0 = 2/p . If we dene the inner product over one period,t 2 h pi, then k and l are orthogonal [4.2] if k 6= l.Usage. [4.2] applied to complex exponentials, which are everywhere in this course.

h k , li = Z h pi eik ! 0 t (eil ! 0 t )dt= Z h pi ei(k l)! 0 t dt

h k , li = ( p, k = l0, otherwiseUnlike the discrete case, where k (n) = k+ p(n), there is no corresponding occurance in thecontinuous case. Thus, the inner product is only non-zero when k = l. I.e. when the two complexexponentials are the same, which also applies in the discrete case.

Derivation. From denition of inner product [4.1] and integration of complex exponentials [3.5].

5 LTI Systems

5.1 Linearity

Statement. A system, H , is linear if it satises the scaling and superposition properties.Scaling: For any input-output pair x H ! y

ax H ! ayis also an input-output pair for all a 2C .Superposition: For any input-output pairs x1

H

! y1 and x2H

! y2x1 + x2

H

! y1 + y2is also an input-output pair.

Meaning. If inputs are scaled and/or added together, the corresponding outputs are scaled and/oradded together in the same way.

Usage. Linearity can used to create new input-output pairs from known input-output pairs. Notethat both scaling and superposition are satised if, for any input-output pairs x1

H

! y1 andx2

H

! y2ax 1 + bx2

H

! ay1 + by2

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is also an input-output pair 8a, b 2C . By extension, if xkH

! yk are input-output pairs, then

Xk akxkH

! Xk akykAlso, a linear system must satisfy the zero-input, zero-output (ZIZO) property. That is, if theinput signal x(t) = 0 , 8t , then the corresponding output must be y(t) = 0 , 8t . This is seen bysetting a = 0 in the scaling property above.

5.2 Time-invariance

Statement. A system, H , is time-invariant if, for any input-output pair x(t) H ! y(t)x(t ) H ! y(t )

is also an input-output pair for all (for a continuous-time system,

8

2R ; for a discrete-time

system, 8 2Z ).Meaning. If inputs are time-shifted, the corresponding outputs are time-shifted by the same amount.

Usage. Time invariance can used to create new input-output pairs from known input-output pairs.

5.3 Impulse response

Statement. For a discrete-time LTI system, H , the impulse response , h(n), is dened as theresponse to the system when the input is a Kronecker delta: (n) H ! h(n). For a general inputsignal, x(n), the response is

x(n) H ! (xh)(n) =1

Xk= 1 x(k)h(n k)=

1

Xk= 1 x(n k)h(k)Similarly, for a continuous-time system, we dene the impulse response as the response to a Diracdelta, (t) H ! h(t), and the response to input x(t) is

x(t) H

!(x

h)( t) =

Z 1

1

x( )h(t )d

= Z 1

1x(t )h( )d

These summations and integrals are known as convolution operations.

Meaning. One important result that follows from the above is that the impulse response, h(),completely describes the behavior of an LTI system, and any other information is either not neededor can be derived from the impulse response.

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Usage. If the impulse response is known, this is one method of nding the output of the system giventhe input signal. As it requires a convolution operation, this method is best when the expressionsfor x() and h() are relatively simple. One alternative is to go via the frequency domain, such asusing Y (! ) = X (! )H (! ).

Derivation. This perhaps most easily seen in the discrete-time case. First, note that x(n) can beexpressed as:

x(n) =1

Xk= 1 x(k) (n k)The right-hand side is a sum of scaled and shifted impulses. Since the system is LTI, from [5.1] and[5.2], the system response would therefore be a summation of scaled and shifted impulse responses,a summation equal to ( xh)(n). The details are left to the student to ll in.

5.4 Eigenfunction for discrete-time systems

Statement. Let H be a discrete-time LTI system. If its input is a complex exponential withfrequency ! = rad/dot, x(n) = ei n , then its output is also a complex exponential with frequency

:

ei n H ! H ( )ei n

where

H ( ) =1

Xk= 1 h(k)e i kMeaning. If the input to the system is a complex exponential, the output is a complex exponential,possibly shifted and scaled, but of the same frequency as the original. Thus, the LTI system cannotgenerate new frequencies.

Furthermore the frequency response H ( ), how much the complex exponential is shifted andscaled, varies solely on the frequency of the input, . Since ei n = ei( +2 )n , 8 2R , n 2Z , H is2-periodic.Usage. Note that the frequency response at , H ( ), only applies to that frequency: ei n H !H ( )ei n . Therefore all the following statements, which have been seen on student papers, are ingeneral not true for LTI systems:

x(n) H !H (! )x(n)cos(! 0n)

H

!H (! 0) cos(! 0n)ei! 0 t H !H (! )ei! 0 t

Derivation. From [5.3], we know the output is ( xh)(n). Substitute x(n) = ei n into the convolution

summation and then factor out the ei n . This is a good exercise for students.

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5.5 Eigenfunction for continuous-time systems

Statement. Let H be a continuous-time LTI system. If its input is a complex exponential withfrequency ! = radians per unit time, x(t) = ei t , then its output is also a complex exponentialwith frequency :

ei t H ! H ( )ei t

where frequency response H ( ) is

H ( ) = Z 1

1h( )e i d

Meaning. If the input to the system is a complex exponential, the output is a complex exponential,possibly shifted and scaled, but of the same frequency as the original. Thus, the LTI system cannotgenerate new frequencies.

Unlike discrete-time LTI systems, the frequency response for continuous-time systems are not nec-essarily periodic since ei t 6= ei( +2 )t , 8t 2R .Usage. The same warnings in [5.4] apply. The frequency response at frequency , H ( ), onlyapplies to that frequency. ei t H ! H ( )ei t .Derivation. Similar to [5.4], except with a continuous-time convolution.

5.6 Impulse response (revisited)

Statement. Observe that in both [5.4] and [5.5], the frequency response H ( ) is the Fourier trans-form (DTFT and CTFT, respectively) of the impulse response. This is not a coincidence. Notethat for discrete-time:

(n) =1

2Z ! 2h2i 1 ei ! n d!And for continuous-time:

(t) =1

2Z 1

11 ei! t d!

That is, input impulse can be thought of as hitting the system with a superposition of unit-strengthcomplex exponentials, and the resulting impulse response is the superposition of responses to thosecomplex exponentials:

h(n) =1

2Z ! 2h2i H (! )ei ! n d!h(t) =

12Z

1

1H (! )ei! t d!

So rather than describe system behavior at a single frequency, as in [5.4] and [5.5], the impulseresponse (or frequency response) captures system behavior at all frequencies.

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6 Partial Fractions

Usage. It is often useful to express a rational polynomial as a sum of lower order terms, in particularwhen taking the inverse Z- or Laplace transforms. We are interested in rational polynomials becausethat is the form transform functions from linear constant-coe cient di erence/di erential equations

(LCCDEs) take.Procedure. 1) The rational polynomial must be proper for the following partial fractions expansionto work. If the expression is not already proper, use factorization or long division to convert it toone. For example

F (z) =z3 + 4 z2 9z

z3 2z2 z + 2, |z | > 2

is not proper. However:

F (z)z

=z2 + 4 z 9

z3 2z2 z + 2

=1

z 2 +2

z 1 +2

z + 1

=) F (z) =z

z 2+

2zz 1

+2z

z + 1which is easy to take the inverse Z-transform of:

f (n) = [2n + 2 1n 2 ( 1)n ]u(n)

Sometimes the expression cannot be factored, and so long division comes in handy:

G(s) =s3 s2 + 3 s 7s3 2s2 s + 2

, Re {s} > 2

=1 +s2 + 4 s 9

s3 2s2 s + 2=1 +

1s 2

+2

s 1+

2s + 1

=) g(t) = (t) +e2t + 2 et 2e t

u(t)

2) Factor the denominator to nd the location and degree of all the poles.

3) The goal of partial fractions is to expand the expression into a sum simple term based on itspoles. For example:

x2 + 4 x 9(x 2)(x 1)(x + 1)

=A

x 2+

Bx 1

+C

x + 1

and the problem has been reduced to nding the value of coe cients A, B , and C . Note: thereason the rational polynomial must be proper is that this expansion cannot generate numerator of equal or higher degree than the denominator, but it is general enough to generate any polynomialof lesser degree.

One (relatively) simple method of nding the coe cients, called the Heaviside cover-up method,can be seen if we multiply both sides by the denominator:

x2 + 4 x 9 = A(x 1)(x + 1) + B (x 2)(x + 1) + C (x 2)(x 1)

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For the equality to be satised, the coe cents must be chosen such that the equality holds for allx. Given that there is a such unique A, B , and C where this is satised, we can nd their valuesby examining the equation at carefully chosen values of x. For example, say we wish to nd A.Setting x = 2:

22 + 4 2 9 = A(2 1)(2 + 1) + B 0 + C 03 =3 A =) A = 1

What we have done is isolate the term dependent on A through careful selection of x. Similarly,to solve for B , set x = 1:

12 + 4 1 9 = A 0 + B (1 2)(1 + 1) + C 04 = 2B =) B = 2

Similarly, we can nd that C = 2. Note that there are other values that make calculation easier.For instance, say we had A and B , but C was di cult to solve for. If x = 0, the expression readilysimplies to

9 = A( 1) + B ( 2) + C (2)9 = 1 2 2 + 2 C =) C = 2

from which we can solve for C . Or, as x approaches innity, the highest order terms dominate, sowe need:

x2 = Ax2 + Bx 2 + Cx 2

1 = A + B + C 1 =1 + 2 + C =) C = 2

These convenient points can also be used as a sanity check on exams and assignments.4) Higher roots complicate calculations, but the idea is still the same. For a pole of order N , wemust include terms for poles of order 1 through N . For example:

P (x)(x + 2)( x + 1) 2

=A

x + 2+

B(x + 1) 2

is not general enough to solve for all 2nd-order polynomials P (x). However

P (x)(x + 2)( x + 1) 2

=A

x + 2+

B(x + 1) 2

+C

x + 1

is general enough. To illustrate the procedure, let P (x) = x2

+ 2 x + 2. As above, we multiply bothsides by the denominator:

x2 + 2 x + 2 = A(x + 1) 2 + B (x + 2) + C (x + 2)( x + 1)

A and B are relatively easy to nd. Setting x = 2:

( 2)2 + 2 2 + 2 = A( 2 + 1) 2 + B 0 + C 02 = A

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Setting x = 1:

( 1)2 + 2 1 + 2 = A 0 + B ( 1 + 2) + C 01 = B

However, for C , we cannot set x = 1 as we did earlier. There are numerous ways to solve for C ,however. For example, if we again focus only the x2 terms:

x2 = Ax2 + 0 x2 + Cx 2

=) 1 = A + C =) C = 1Or setting x = 0:

2 = A(1)2 + B (2) + C (2) =) C = 1If we wish to make use of the Heaviside cover-up with x = 1, we can do some algebra:

(x2 + 2 x + 2) B (x + 2) = A(x + 1) 2 + C (x + 2)( x + 1)

x2 + x =( x + 1) [ A(x + 1) + C (x + 2)]x = A(x + 1) + C (x + 2)

At x = 1:

1 = A 0 + C ( 1 + 2) =) C = 1

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Many Cheerful Facts