Up until now, you've been told that you can't take the square root of a negative number. That's because you had no numbers which were negative after you'd squared them (so you couldn't "go backwards" by taking the square root). Every number was positive after you squared it. So you couldn't very well square-root a negative and expect to come up with anything sensible.

Now, however, you can take the square root of a negative number, but it involves using a new number to do it. This new number was invented (discovered?) around the time of the Reformation. At that time, nobody believed that any "real world" use would be found for this new number, other than easing the computations involved in solving certain equations, so the new number was viewed as being a pretend number invented for convenience sake.

(But then, when you think about it, aren't all numbers inventions? It's not like numbers grow on trees! They live in our heads. We made them all up! Why not invent a new one, as long as it works okay with what we already have?)

Anyway, this new number was called "i", standing for "imaginary", because "everybody knew" that iwasn't "real". (That's why you couldn't take the square root of a negative number before: you only had "real" numbers; that is, numbers without the "i" in them.) The imaginary is defined to be:

Then:   Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved

Now, you may think you can do this:

But this doesn't make any sense! You already have two numbers that square to 1; namely –1 and +1. And i already squares to –1. So it's not reasonable that i would also square to 1. This points out an important detail: When dealing with imaginaries, you gain something (the ability to deal with negatives inside square roots), but you also lose something (some of the flexibility and convenient rules you used to have when dealing with square roots). In particular, YOU MUST ALWAYS DO THE i-PART FIRST!

Simplify sqrt(–9).

(Warning: The step that goes through the third "equals" sign is "  ", not " ". The i isoutside the radical.)

Simplify sqrt(–25).

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Simplify sqrt(–18).

Simplify –sqrt(–6).

In your computations, you will deal with i just as you would with x, except for the fact that x2 is just x2, but i2 is –1:

Simplify 2i + 3i.

2i + 3i = (2 + 3)i = 5i

Simplify 16i – 5i.

16i – 5i = (16 – 5)i = 11i

Multiply and simplify (3i)(4i).

(3i)(4i) = (3·4)(i·i) = (12)(i2) = (12)(–1) = –12

Multiply and simplify (i)(2i)(–3i).

(i)(2i)(–3i) = (2 · –3)(i · i · i) = (–6)(i2 · i)

=(–6)(–1 · i) = (–6)(–i) = 6i

Note this last problem. Within it, you can see that  , because i2 = –1. Continuing, we get:

This pattern of powers, signs, 1's, and i's is a cycle:

In other words, to calculate any high power of i, you can convert it to a lower power by taking the closest multiple of 4 that's no bigger than the exponent and subtracting this multiple from the exponent. For example, a common trick question on tests is something along the lines of "Simplify i99", the idea being that you'll try to multiply i ninety-nine times and you'll run out of time, and the teachers will get a good giggle at your expense in the faculty lounge. Here's how the shortcut works:

i99 = i96+3 = i(4×24)+3 = i3 = –i

That is, i99 = i3, because you can just lop off the i96. (Ninety-six is a multiple of four, so i96 is just 1, which you can ignore.) In other words, you can divide the exponent by 4 (using long division), discard the answer, and use only the remainder. This will give you the part of the exponent that you care about. Here are a few more examples:

Simplify i17.

i17 = i16 + 1 = i4 · 4 + 1 = i1 = i

Simplify i120.

i120 = i4 · 30 = i4· 30 + 0 = i0 = 1

Simplify i64,002.

i64,002 = i64,000 + 2 = i4 · 16,000 + 2 = i2 = –1

Now you've seen how imaginaries work; it's time to move on to complex numbers. "Complex" numbers have two parts, a "real" part (being any "real" number that you're used to dealing with) and an "imaginary" part (being any number with an "i" in it). The "standard" format for complex numbers is "a + bi"; that is, real-part first and i-part last.

Top  |  1 | 2 | 3  |  Return to Index  Next >>

Cite this article as:

Stapel, Elizabeth. "Complex Numbers: Introduction." Purplemath. Available from    http://www.purplemath.com/modules/complex.htm. Accessed 22 May 2012

MATH 221A Multivariate Calculus Spring 2004Notes on surface integralsVersion 1.2 (April 01, 2005)

Definition of surface integralWe are given a vector field ~F in space and a surface S in the domain of ~F. Thegeneral idea of surface integral issurface integral of ~F over surface S=the limit of a sum of terms each having the form(component of ~F normal to a piece S)(area of that piece of S).Here’s how we make the idea more precise. Break the surface S into pieces andlabel these pieces ¢Sij . We use two indices because the surface is a two-dimensionalthing. Think of the index i as running from 1 to m and the index j as running from1 to n for a total of mn pieces. (See Figure 1 at the end.) Define ¢~Aij as the vectorwith direction normal to the piece ¢Sij and magnitude equal to the area of the piece¢Sij . In each piece, pick a point Pij . At each of the points, compute the vector fieldoutput ~F(Pij). Recall that the dot product ~F(Pij) · ¢~Aij can be written~F(Pij) · ¢~Aij = k ~F(Pij)kk¢~Aijk cos µ = ³k ~F(Pij)k cos µ´k¢~Aijk.The last expression shows that this dot product gives the component of ~F normal to¢Sij times the area of ¢Sij . This is what we want to add up. We define the surfaceintegral of ~F for the surface S as the limit of such a sum:ZZS

~F · d~A = limm,n!1n

Xi=jn

Xi=1

F~ (Pij) · ¢A~ij

You can think of d~A as an “infinitesimal” version of ¢~Aij . The direction of d~A isnormal to the surface at each point.In order to compute a surface integral, it is useful to think of d~A in the followingway. Consider the “grid” on the surface S (as shown in Figure 1 at the end) asconsisting of two families of curves, with each family consisting of those curves thatare locally parallel, but not globally parallel in general. Here, locally parallel meansthat the family of curves is parallel if we zoom in on any point. An important thinghere is that within each family, no two curves intersect. The curves in the first familydo not have to be perpendicular to the curves in the second family. At a point P onthe surface, take C1 to be the curve from one family that goes through P and C2 tobe the curve from the other family that goes through P. Let d~R1 be the infinitesimaldisplacement vector tangent to C1 at P and let d~R2 be the infinitesimal displacementvector tangent to C2 at P. Consider the cross product d~R1×d~R2. Recall the geometricdefinition of cross product: (1) the direction of the cross product is perpendicular toboth vectors in the product (as given by the right hand rule); and (2) the magnitudeof the cross product is the area of the parallelogram formed by the two vectors in theproduct. The cross product d~R1 × d~R2 is thus normal to the surface at P (since d~R1

and d~R2 are both tangent to the surface) and has magnitude equal to the area of thesurface piece with edges d~R1 and d~R2. Thusd~A = d~R1 × d~R2.Computing surface integralsIn computing line integrals, the general plan is to express everything in terms oftwo variables. This is a reasonable thing to do because a surface is a two-dimensionalobject. The essential things are to determine the form of d~A for the surface S andthe outputs ~F(P) along the surface S, all in terms of two variables. How to proceed

depends on how we describe the curve. In general, we have two choices: a relationamong the coordinates or a parametric description. The solution to the followingexample illustrates how to work with the first of these.Example: Compute the line integral of ~F(x, y, z) = xˆı + y ˆ| + z ˆk for the surface Sthat is the piece of the plane 12x − 6y + 3z = 24 with x ¸ 0, y · 0, and z ¸ 0.Note: To get started, you should draw a picture showing the surface and a few of thevector field outputs along the surface.Solution : From the equation of the plane, we compute12 dx − 6 dy + 3 dz = 0.This relates small displacements dx, dy, and dz along the plane. See Figure 2.To generate one family of curves on the surface, we can use x = constant. Thisgives us dx = 0. Using this in the previous relation and solving for dz gives dz = 2dy.We can now use these to getd~R1 = dxˆı + dy ˆ| + dz ˆk = 0ˆı + dy ˆ| + 2dy ˆk = ¡ˆ| + 2 ˆk¢dy.To generate the other family of curves on the surface, we can use y = constant.This gives us dy = 0. Using this in the above relation and solving for dz givesdz = −4dx. We can now use these to getd~R2 = dxˆı + dy ˆ| + dz ˆk = dxˆı + 0 ˆ| − 4dx ˆk = ¡ˆı − 4 ˆk¢dx.With d~R1 and d~R2 in hand, we can compute d~A asd~A = d~R1 × d~R2 = ¡ˆ| + 2 ˆk¢ × ¡ˆı − 4 ˆk¢dxdy = ¡−4ˆı + 2 ˆ| − ˆk¢dxdy.You should think about the direction these vectors point. We have made choices thatresult in d~A pointing in a certain direction. A different set of choices could result ind~A pointing in the opposite direction.2We now want to express the vector field outputs along the surface S in terms ofthe same two variables (x and y in this case) that we have used for d~A. We will usethe equation of the plane to solve for z givingz = 8 − 4x + 2y.Thus, on the surface, the vector field has outputs~F = xˆı + y ˆ| + (8 − 4x + 2y) ˆk.We now compute~F · d~A = ¡xˆı + y ˆ| + (8 − 4x + 2y) ˆk¢ · ¡−4ˆı + 2 ˆ| − ˆk¢dxdy= ¡−4x + 2y − (8 − 4x + 2y)¢dxdy

= −8 dxdyThe last things we need in order to carry out the integration are the relevantbounds on the variables x and y. The projection of the surface into the xy-plane isthe triangular region shown in Figure 2. We can use0 · x · 2 and 2x − 4 · y · 0to describe this region.Putting all of this together, we haveZZS

~F · d~A = Z 20 Z 04−2x¡−8¢dxdy = −8(area of triangle in xy-plane) = −16.The sign here is a result of the choices we made in computing d~ A. A different setof choices could result in the value +16. You should think about the two possibledirections for d~ A.If you have corrections or suggestions for improvements to these notes, please contact MartinJackson, Department of Mathematics and Computer Science, University of Puget Sound,Tacoma, WA 98416, martinj@ups.edu.3Figure 1. The elements used in the definition of surface integral. Top left: The surfaceS broken into pieces ¢Sij . Top right: The points Pij . Bottom left: Thearea vectors ¢~Aik. Bottom right: The vector field outputs ~F(Pij). Notethat vectors are displayed without arrow heads to reduce clutter. The baseof each vector is on the surface.x yz2-48-1 1 2-4-3-2-11

Figure 2. The piece of the plane that is the surface for the example (right) and theprojection of this plane into the xy -plane (left).4

************************************************************************************************

In this section we are now going to introduce a new kind of integral.  However, before we do that

it is important to note that you will need to remember how to parameterize equations, or put another way, you will need to be able to write down a set of parametric equations for a given curve.  You should have seen some of this in your Calculus II course.  If you need some review you should go back and review some of the basics of parametric equations and curves.

Here are some of the more basic curves that we’ll need to know how to do as well as limits on the parameter if they are required.

Curve Parametric Equations

(Ellipse)

          Counter-Clockwise                 Clockwise

 

                

(Circle)

           Counter-Clockwise                 Clockwise

 

                

Line Segment From

 

to 

With the final one we gave both the vector form of the equation as well as the parametric form and if we need the two-dimensional version then we just drop the z components.  In fact, we will be using the two-dimensional version of this in this section.

For the ellipse and the circle we’ve given two parameterizations, one tracing out the curve clockwise and the other counter-clockwise.  As we’ll eventually see the direction that the curve is traced out can, on occasion, change the answer.  Also, both of these “start” on the positive x-

axis at  .

Now let’s move on to line integrals.  In Calculus I we integrated  , a function of

a single variable, over an interval  .  In this case we were thinking of x as taking all the values in this interval starting at a and ending at b.  With line integrals we will start with

integrating the function  , a function of two variables, and the values

of x and y that we’re going to use will be the points,  , that lie on a curve C.  Note that this is different from the double integrals that we were working with in the previous chapter where the points came out of some two-dimensional region.

Let’s start with the curve C that the points come from.  We will assume that the curve is smooth (defined shortly) and is given by the parametric equations,

We will often want to write the parameterization of the curve as a vector function.  In this case the curve is given by,

The curve is called smooth if   is continuous and   for all t.

The line integral of   along C is denoted by,

We use a ds here to acknowledge the fact that we are moving along the curve, C, instead of the x-axis (denoted by dx) or the y-axis (denoted by dy).  Because of the ds this is sometimes called the line integral of f with respect to arc length.

We’ve seen the notation ds before.  If you recall from Calculus II when we looked at the arc length of a curve given by parametric equations we found it to be,

It is no coincidence that we use ds for both of these problems.  The ds is the same for both the arc length integral and the notation for the line integral.

So, to compute a line integral we will convert everything over to the parametric equations.  The line integral is then,

Don’t forget to plug the parametric equations into the function as well.

If we use the vector form of the parameterization we can simplify the notation up somewhat by noticing that,

where   is the magnitude or norm of  .  Using this notation the line integral becomes,

Note that as long as the parameterization of the curve C is traced out exactly once as t increases from a to b the value of the line integral will be independent of the parameterization of the curve.

Let’s take a look at an example of a line integral.

Example 1  Evaluate   where C is the right half of the

circle,  . rotated in the counter clockwise direction.SolutionWe first need a parameterization of the circle.  This is given by,

We now need a range of t’s that will give the right half of the circle.  The following range of t’s will do this.

                                                               

Now, we need the derivatives of the parametric equations and let’s compute ds.

                                             

The line integral is then,

                                            

Next we need to talk about line integrals over piecewise smooth curves.  A piecewise smooth

curve is any curve that can be written as the union of a finite number of smooth curves,  ,

…,   where the end point of   is the starting point of  .  Below is an illustration of a piecewise smooth curve.

Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do.  All we do is evaluate the line integral over each of the pieces and then add them up.  The line integral for some function over the above piecewise curve would be,

Let’s see an example of this.

Example 2  Evaluate   where C is the curve shown below.

SolutionSo, first we need to parameterize each of the curves.

                                             

Now let’s do the line integral over each of these curves.                            

                                

                                    

Finally, the line integral that we were asked to compute is,

                                         

Notice that we put direction arrows on the curve in the above example.  The direction of motion along a curve may change the value of the line integral as we will see in the next section.  Also

note that the curve can be thought of a curve that takes us from the point 

 to the point  .  Let’s first see what happens to the line integral if we change the path between these two points.

Example 3  Evaluate   where C is the line segment from 

 to  .

SolutionFrom the parameterization formulas at the start of this section we know that the line segment

starting at   and ending at   is given by,

                                                 

for  .  This means that the individual parametric equations are,

                                              

Using this path the line integral is,

                                              

When doing these integrals don’t forget simple Calc I substitutions to avoid having to do things like cubing out a term.  Cubing it out is not that difficult, but it is more work than a simple substitution.

So, the previous two examples seem to suggest that if we change the path between two points then the value of the line integral (with respect to arc length) will change.  While this will happen fairly regularly we can’t assume that it will always happen.  In a later section we will investigate this idea in more detail.

Next, let’s see what happens if we change the direction of a path.

Example 4  Evaluate   where C is the line segment from 

 to  .

SolutionThis one isn’t much different, work wise, from the previous example.  Here is the parameterization of the curve.

                                                 

for  .  Remember that we are switching the direction of the curve and this will also change the parameterization so we can make sure that we start/end at the proper point.

Here is the line integral.

                                               

So, it looks like when we switch the direction of the curve the line integral (with respect to arc length) will not change.  This will always be true for these kinds of line integrals.  However, there are other kinds of line integrals in which this won’t be the case.  We will see more examples of this in the next couple of sections so don’t get it into your head that changing the

direction will never change the value of the line integral.

Before working another example let’s formalize this idea up somewhat.  Let’s suppose that the

curve C has the parameterization  ,  .  Let’s also suppose that the initial point on the curve is A and the final point on the curve is B.

The parameterization  ,   will then determine an orientation for the curve where the positive direction is the direction that is traced

out as t increases.  Finally, let   be the curve with the same points as C, however in this case the curve has B as the initial point and A as the final point, again t is increasing as we

traverse this curve.  In other words, given a curve C, the curve   is the same curve as C except the direction has been reversed.

We then have the following fact about line integrals with respect to arc length.

Fact

So, for a line integral with respect to arc length we can change the direction of the curve and not change the value of the integral.  This is a useful fact to remember as some line integrals will be easier in one direction than the other.

Now, let’s work another example

Example 5  Evaluate   for each of the following curves.

(a)     [Solution]

(b)  : The line segment from   to  .   [Solution]

(c)  : The line segment from   to  .   [Solution]

SolutionBefore working any of these line integrals let’s notice that all of these curves are paths that

connect the points   and  .  Also notice that 

 and so by the fact above these two should give the same answer.

Here is a sketch of the three curves and note that the curves illustrating   and   have been separated a little to show that they are separate curves in some way even though they are the same line.

(a) 

Here is a parameterization for this curve.

                                                   

Here is the line integral.

                                      

[Return to Problems]

(b)  : The line segment from   to  .

There are two parameterizations that we could use here for this curve.  The first is to use the formula we used in the previous couple of examples.  That parameterization is,

                                                

for  .

Sometimes we have no choice but to use this parameterization.  However, in this case there is a second (probably) easier parameterization.  The second one uses the fact that we are really

just graphing a portion of the line  .  Using this the parameterization is,

                                                    

This will be a much easier parameterization to use so we will use this.  Here is the line integral for this curve.

                                              

Note that this time, unlike the line integral we worked with in Examples 2, 3, and 4 we got the same value for the integral despite the fact that the path is different.  This will happen on occasion.  We should also not expect this integral to be the same for all paths between these two points.  At this point all we know is that for these two paths the line integral will have the same value.  It is completely possible that there is another path between these two points that will give a different value for the line integral.[Return to Problems]

(c)  : The line segment from   to  .

Now, according to our fact above we really don’t need to do anything here since we know

that  .  The fact tells us that this line integral should be the same as the second part (i.e. zero).  However, let’s verify that, plus there is a point we need to make here about the parameterization.

Here is the parameterization for this curve.

                                                

for  .

Note that this time we can’t use the second parameterization that we used in part (b) since we need to move from right to left as the parameter increases and the second parameterization

used in the previous part will move in the opposite direction.

Here is the line integral for this curve.

                                      

Sure enough we got the same answer as the second part.[Return to Problems]

To this point in this section we’ve only looked at line integrals over a two-dimensional curve.  However, there is no reason to restrict ourselves like that.  We can do line integrals over three-dimensional curves as well.

Let’s suppose that the three-dimensional curve C is given by the parameterization,

then the line integral is given by,

Note that often when dealing with three-dimensional space the parameterization will be given as a vector function.

Notice that we changed up the notation for the parameterization a little.  Since we rarely use the

function names we simply kept the x, y, and z and added on the   part to denote that they may be functions of the parameter.

Also notice that, as with two-dimensional curves, we have,

and the line integral can again be written as,

So, outside of the addition of a third parametric equation line integrals in three-dimensional space work the same as those in two-dimensional space.  Let’s work a quick example.

Example 6  Evaluate   where C is the helix given

by, 

,  .

SolutionNote that we first saw the vector equation for a helix back in the Vector Functions section.  Here is a quick sketch of the helix.

Here is the line integral.

                                 

You were able to do that integral right?  It required integration by parts.

So, as we can see there really isn’t too much difference between two- and three-dimensional line integrals.

Beta function:

Beta functionFrom Wikipedia, the free encyclopedia

This article is about Euler beta function. For other uses, see Beta function (disambiguation).

In mathematics, the beta function, also called the Euler integral of the first kind, is a special function defined

by

for 

The beta function was studied by Euler and Legendre and was given its name by Jacques Binet; its

symbol Β is a Greek capital βrather than the similar Latin capital B.

Contents

  [hide]

1 Properties

2 Relationship between gamma function and beta function

3 Derivatives

4 Integrals

5 Approximation

6 Incomplete beta function

o 6.1 Properties

7 Calculation

8 See also

9 References

10 External links

[edit]Properties

The beta function is symmetric, meaning that

When x and y are positive integers, it follows trivially from the definition of the gamma function   that:

It has many other forms, including:

where   is a truncated power function and the star

denotes convolution. The second identity shows in

particular  . Some of these identities, e.g. the

trigonometric formula, can be applied to deriving the volume of an

n-ball inCartesian coordinates.

Euler's integral for the beta function may be converted into an

integral over the Pochhammer contour C as

This Pochhammer contour integral converges for all values

of α and β and so gives the analytic continuation of the beta

function.

Just as the gamma function for integers describes factorials,

the beta function can define a binomial coefficient after

adjusting indices:

Moreover, for integer n,   can be integrated to give a

closed form, an interpolation function for continuous

values of k:

The beta function was the first known scattering

amplitude in string theory, first conjectured

by Gabriele Veneziano. It also occurs in the theory

of the preferential attachment process, a type of

stochastic urn process.

[edit]Relationship between gamma function and beta function

To derive the integral representation of the beta

function, write the product of two factorials as

Changing variables by putting u=zt, v=z(1-t)

shows that this is

Hence

The stated identity may be seen as a

particular case of the identity for

the integral of a convolution. Taking

 and  , one has:

.

[edit]Derivatives

We have

where   is

the digamma function.

[edit]Integrals

The Nörlund–Rice

integral is a contour

integral involving the

beta function.

[

edit]Approximation

Stirling's

approximation gives the

asymptotic formula

for large x and

large y. If on the

other hand x is

large and y is fixed,

then

[

edit]Incomplete beta function

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Even if unavailable

directly, the complete

and incomplete Beta

function values can

be calculated using

functions commonly

included

in

mputer algebra

systems

With

example, using

the

(

distribution

we have:

Complete Beta Value = Exp(GammaLn(a) + GammaLn(b) - GammaLn(a + b))

and,

Incomplete Beta Value = BetaDist(x, a, b) * Exp(GammaLn(a) + GammaLn(b) - GammaLn(a + b)).

These result from rearranging

the formulae for the

distribution

incomplete beta and complete

beta functions, which can also

be defined as the ratio of the

logs

Similarly,

in

Octave

beta function) computes

the

beta function

fact, the Cumulative Beta

distribution - and so, to get the

actual incomplete beta

function, one must multiply the

result of

returned by the

corresponding