Metodología DMAIC

CAPÍTULO I: MARCO TEÓRICO

1.9. Six Sigma

1.9.2. Metodología DMAIC

Domain (Region): A set S of points in the Aigand plane is said to be connected set if any two of ' its points can be joined by a continuous curve, till of whose joints belong to S.

-*»An open connected set is called an open domain. If the boundary points of S are also added to an open domain, then it Ts called cloied domain,

Jordan A rc : Let x(t) and y(t) be continuous function of a real variable / in the interval a s t <,fi. Then the set of points x in the Argand plane given by die equation

is called a continuous arc if, corresponding to one value of t, there exists more then one value for f z, then z is said to be a multiple point A continuous arc with no multiple point on it is called a

Jordan curve.

$ Contour : By contour, we mean a continuous chain of a finite number of regular arcs. If the contour is closed and does not intersect itself, then it is called closed contour.

h Example: Boundaries o f triangles and rectangles.

* Sfrrtpiy and

Multiply

Connected Domains :

f/ A domain in which every closed curve can be shrunk to a point without passing out of Die region

is called a simply connected domain. If a domain is not simply connected, then it is called multiply connected domain.

path of the definite integral

I I | & wg«mi»3 Mxmounc»4H

IM COMPLEX LINE INTEGRAL

Suppose,/?# is continuous at every point of a closed curve C having a finite length, Le.; C is a rectifiable curve. Divide C into n parts by means of (it + 1) points z 9

Let a = z Q,b * z H,

We choose a point £* on each arc joining z*_j to z k.

Form the sum

Sn=t. /(•)(,~*r,l).

\ rml

Suppose maximum value of (zr - z r_,) 0 as n -* « .

Then the sum S„ tends to a fixed limit which does not depend upon the mode of subdivision and i denote this limit by

hi : .

£ f(z ) d z or I f{ z )d z

L

which is called thecomplex line integral or line integral offfz) along C. An evaluation of integral

by such method is alsocalled ab-initio method i

r * :

By the symbol f(z )d z we mean the along a boundary C intfce positive sense. In case of closed paths, the positive direction in anticlockwise. The integral along C is often called a contour integral.

F u N r tiO H iw C n w H f lf Vm b w u s | 5 5

1.62 ; Evaluate ](3x* + 4xy + 3y* ) d x +2 (x* + 3xy + 4y* )d y ] (I) a lo n g y = x* (U) a lo n g y = x .

Does the value o f the Integral depend upon the path ?

(i) Along curve, y = x 2 => d y - 2x d x and limit jc -► 0 to 1.

I * £ [(3x2 + Ax3 + 3x 4) d x +(:2x2 + 6x3 + 8x4)2xdx]

= £ (3x2 + 4x3 + 3x4 + 4x3 + 12x4 + 16xs)d x = (ii) Along curve, y * x x> dy “ dx and limit x -> 0 to 1.

I » £ (3x2 + 4x2 + 3x2)dx + (2x2 + 6x2 + %x2)d x =

Clearly, both values are the same, hepce the value of integration does not depend upon the

path o f integration. Am.

pRppfc 1.63 ; Evaluate

J[

(z)2dz, along

t, (I) the real axis from z ~ 0 to z ~ 2 and then along a line parallel to y-axisfrom z m 2 to Z m2 + L

(ii) along the line 2 y - x . IRGPV June 2905 and June 2007]

Since (z )2 = (x - iy)2 - (x2 - y 2)~ 2 ixy

(i) Along the path OMP where M is (2,0) and P is (2, 1).

i

2+i .

Qc) d t M o n .

= (x2 - y 2 -2 ixy)d z+ (x2 - y 2 -2 ix y )d z _...(0

M |

Now, along OM, y**Q => dz = dx and x varies from 0 to 2.

3 Jo* ~ y2 ~ 2“ 0f) cfe =

Also* along MP, x ~ 2 ,d z ~ id y and>- varies from 0 to 1

Imp “ y 2 " 2ixy^d z = I (4 “ y2 ~ Aiy ^ y 4 iy ~ i^ Y + 2 y7 ~ 4 i i + 2 = 2 + i i f 3 3 Hence 0 ) becomes, r2+t r r * 8 _ I I . 14 11. / = 1 ( z ) d z = — + 2 + —1= — + —i. * 3 3 3 3

Example 1.64: Evaluate f * (2x+/;y + t)d z along the two p a th s: (i) x - t + 1, y - I t 2- - I

(ii) The straight line joining (1 - i) and (2 + i) ' '

Ans.

Solution* Let £** (2x+iy+ l)dz

fftG PV Dec. 2009 (N)f

.-(1) (i) Alongpath: x = / + 1 andy - 2t2 - 1

=> dx ~ d t and dy = At dt.

Then dz = dx + idy.

Also from path x ~ t + 1 :

when x = 0, then t ~ 0 when x = 2, then t = 1 i.e.. limit = / -> 0 to I. Hence (1) becomes : i / = j[2(t +1) + 1(2*8 -1) + l](<fc + i4tdt) o j = j(2t + 2 + 2it2 - i + l)(\ + 4it)dt o . 25. = 4 + —-i. 3

(ii) Along path is straight line joining (1 - i) and (2 + f)'i.e., point (I, - 1) to (2, 1)

1 + 1 Equation of path; .y + 1 = — ~(r -1 ) => ^ +1 = 2jc - 2 2 ^1 => y+ 1 = 2 x - 2 => y = 2x - 3 => dy ~ 2dx

Fuwcttowi <ibCm i» ¥ m m u8 [1 7 Also dz - dx + idy and limit x -> I to 2.

Hence (I) becomes :

/ = J2[2x + i(2x - 3)+1 )(d z+ i2dx) = j^[2x +1 + i(2x - 3](2i + l>ijc

» 4 + 8 i. Ans.

mxample 1.65 : Evaluate £ (** + 3z+ 2)dzt where C is the arc o f the cycloid x - a ( 0 + sin 0),

■ y = a ( l- c o s 6) between the points (0, 0) to (no, 2a).

'Solution. Let us consider the path of integration along a curve C consisting o f :

I (i) the part ofthe real axis from the point (0,0) to the point (*0,0). On this line: z = x , d z = dx and x goes from 0 to na.

(ii) followed by a line parallel to the imaginary axis from the point (m , 0) to the point

(na, 2a). On this line:

z ~ m + iy, d z =■ idy an^y goes from 0 to 2a.

Let, / = £ (z 1 + lz + 2)dz '

“ F + 3x + 2)c£x + £ ° {(>ro + iy)2 + 3 (fro+iy) + 2]idy

I = f"i x + -x2 + 2x1 + i|"-(?ra + iy)s + ^(>ra + iy)2 + 21yl*

L4 2 J 0 [_3 2 J0

/ = | j ( W + |( ^ a ) 2 +2*aJ

+ i[i(^ a + i2 o )s +^(>ra+i2a)2 + 4 o - i ( x a ) 8

-^(/ra)2l

|_3 2 3 2 J

1

I - 2 ;ra + - ( * a + i2a)4 + —(tra+i2a) + 4ia . Ans.

3 2

Example 1.66: Integrate z2 along the straight line OA and also along the path OBA consisting o f two straight line segments OB and BA where O is the origin, B is the point z = 3 and A the

point z * 3 + L fRGPV Dec, 2004]

Solution. Since, z = x + i y => d z - d x + i d y . Along Die curve C, we have *

£ z 2d z = j c (x + iy)2(dx+idy)= £ (x2 - y 2 +2ixy)(dx+idy) ...(I) (O'ThOpoint Afe z ** 3 + /, i.e:% A fa (3,1).

The equation of the line OA i s :

5 S | FmrnOTi^Kdi rattATjca-MI

Hence, x*=3y=>dx « 3dy and .y varies from 0 to 1.

Now from (1), we have

JL *2<3k = £ ( ^ 2 - ^ 2 +

2i3y.y).{3dy

+ icfr) = £

(&+6i)0+i)y2dy

= 0 8 + 2 6 0 ^ j = i(l8 + 2 6 .-)= 6 + j ^ . ...(2 )

00 Again, z 'd z = z 'd z = J ^ ^ + *2<fc

(x2 - y 2 +2wy)(dx + M f y ) + ( x 2 - y 2 + 2ixy)(dx+ idy) ...(3) On the line OB, y = 0=>dy = 0 and x varies from 0 to 3.

On the line BA, x = 3=>dx = 0 and >> varies from 0 to 1. Hence, (3) becomes:

z 2dz *

x 2dx+

£ ( 9 - y 2 + 6iy).idy

*

= 9 + i r 9 - I + 3 i j = 6 + i ^ . Thus equations (2) and (4X show th a t:

L

z 2dz = I

JOBA

z 2dz.

x 3 3 + i 9^--^-+ 3i,y2 3 ₀ 3 ...(4) Am.

U l CAUCHY’S INTEGRAL THEOREM OR CAUCHY’S THEOREM

Statement:

i

I f f ® & m anotytic function and f ( x ) k continuous at each point wtthim an# on a simple closed curve C, then

F u N C T fO N S ^ rC o tffte x Va r ia b les |* W

fc* • 6)

.<afqrr

Let R be the region bounded by the curve C.

Let f( z ) - u ( x , y ) + iv(x,y) = u + iv and z - x + iy

=> d z = dx+idy, then

£ f ( z ) d z = (u + iv)(dx + idy)

= £ ( u d x - v d y ) + i £ (vdx+udy)

du du dv dv

Since f ( z ) is continuous, the partial derivatives are also continuous in region R. So using Green's Theorem M dx

+

Ndy »

JJ

1) becomes:

I f( z ) d z = JJ / / ...(2)

Since f(z) is analytic so using Cauchy-Rientann equations, we get

d u _ d v d u __ du 8 x ~ By and S y ~ 3x

Thus, (2) becomes : jj, /(*)<& = 0.

wit •

r n *

Extortion o f Cauchy's Theorem: Ifffz) is analytic in the region R between

two simple closed curve C\ and C* then /(*)<** = k f ( z ) d z .

CAUCHY'S INTEGRAL FORMULA

Statement:

Iff® is anotytic within and an a etaaed curve C and *a * is any point within C, then

^ " 5 5 1:

Je -a )dX' °T

i

o e C

[RGPVDec. 2007, June 08/

f( z )

Proof. Consider the function--- > which is analytic at every point within C except at z - cl

■ 2 - 0

Draw a circle C\ with a as centre and radius r such that C\ lies entirely inside C. ie., C\ : \ z - a \ = r.

- »

f Thus function ——- is analytic in the region between C and Ci.

z - a

By using extension of Cauchy's theorem, we have

f f J V L d z m

Jc ( z - a ) Jc, ( z - a ) *(1) Since,

C,

: | * - a | = r or * - a = rew or z = a + new

=> d z ss i r e ^ d S and 0 -► o to 2 n .

Hence (1) becomes:

r j w _ <fe= [ J ^ - d z k (Z '-a ) k ( z - a )

= f ^ ’^ - )'.irei9d& * i f t a + r e ^ )d 0 ...(2) In Die limiting form, as the circle C\ shrinks to the point a, i.e., r -* 0, then (2) becomes:

Jc <** * * f / W * - * » r 60 a ^ < a >

Hence /( ° ) = ^ < J e ...(3) froved.

Deduction :

Differentiating both sides of (3) w.r.t. a, we get

/ ’(a) = — — f f ..ftfcK <fe (4|

' 1 2 x i d a * ( z - a) 2« (z - a f ~ (4)

Similarly, r ( a ) = ^ l J ^ f dz' -(5)

In general, /<' >(<,) = ^ t ( I - a p * * 1 -< 6)

Results (4) to (6), are Cauchy's integral formula fo r the derivative qff(z).

Remarks:

(i) If point a

©/<

then

fc

(ii) Converse of CMefcy'i Theorem (Morera's Theorem):

Let f(z) be continuous in a simply connected region R and let for every simple closed curve C, such that

£ f ( z ) d z = 0. Theft fl(z) is analytic, (iii) Cauchy Inequality :

Iff( z) is analytic within a circle C given by |s - a |s = jR a n d if l/( * } |£ M on C, then

\ f f(z)dz

Proof. Since / W = Jc ^ , q y»*i P*y Cauchy integral formula]

abv.

FuNciidJferWGbiAfcEX Variables | (fcl

|/» (o )l» J l L f J W ± , ' ' 2xi Jc (z - a )*+l2in. Jc (z — nY**

[Since, z ~ a = Re * =5 dz - iRa'8d O a x i & \ d z \ - \ i R e wd6\= R d o \

n\ e !/(*)! \dz\ [vU I-1] 2jtjc K ^ -o)**1 I n! M r2* 2* f l B+l fix

I

* * J £ - 2 * r 2n R n+i Mn\ Proved. (iv) Liouvllle's Theorem :

0 -y If a ftinctionjfc) is analytic for all finite values of z and is bounded, then f(z) is a constant

e**

1.67: Using Cauchy's integral formula, evaluate

r |* |- 3 .

jUiwftm. By derivative of Cauchy's integral formula:

Here a = - I, n = 3, and f( z ) ~ e 2*

JL

r * + i / d z, where C is the circle

[RGPV, June 2011 J

-.(1)

,2 i

= 1

2>riJc (z + 1)4

Since at pole z = a = - t , so that | z |=| - 1 1= 1 < 3 i . e point z = -1 He in circle C. /(*) = e2z => A * ) * 2e2x => f \ z ) = 22e2*

=> f m(z) = 23e2* => / '( - I ) = 8«“2. ‘• i ) ‘ _2 3 f e2* , Hence (2) becomes : 8e ~ ~ r L 7— TT®* M ** (Z + l) *♦•(2)

62 1 M il M ill—IIIiTlB III

Example 1.68 : Using Cauchy's integral formula, evaluate _I

d x

* (s+ * i)* where C is the circle

Solution.

|* + 3 I |= 1 .

For poles (or singular points), put z ( z + m ) = 0 => z = 0, « * -» * At pole z = 0

|* + 3 i|= |0 + 3 i 1=3*1, te ., pole a tz ~ 0 out side of C. At pole z - -tri :

|z + 3 i|= |-* * + 3« M ( 3 - ;r ) i |= 3 - ; r <1, /.e., pole at z = -jd, insideofC.

■ L

11-]* *

" & z ( z + m) m * I z z + n J = ± f ± _ ± f L - d z

m k z m *c (z + » )

[Using partial fractions]

= -~ [0 3 -A l2 * ./ ( - » ) ] sa m m —~ .2 * X l m 1 _{Because f} _{° e C} ■ Jc { z - a ) \ 0 , a « C [v /(z ) = 1 /(- m ) = 1) Abs.

Example 1.69 ; I f /(S)® £ fMiww C Iv the circle x? +y* =4, fin d the values of

Solution.

= -2 .

3 ^ +7»+l z - \ f ( 3 ) , f ' ( l - l ) and f X l + i ) .

The given circle C is x 2 + y 2 * 4 or C :|« |= 2.

The point z » 3 lies outside the circle \ z \ - 2 while z = 1 - / and z - 1 + / lie inside the circle. [B ecause!*M 31*34:2, \ z | * | l - i |- V 2 < 2 , \ z \ - \ \ + i \ - j 2 <2] ( 0 / ( 3 ) = X

3z2 + 72 +1 j j 3 * 2 +72 + 1 . j ^

dz a n 4 ---- -—-— in analytic within and on C.

* ~ 3

By Cauchy's integral theorem,

2 - 3

ic

3z2 +72 + 1 d z = 0,