  EGM4313 INTERMEDIATE ENGINEERING ANALYSIS Instructor-U.H.Kurzweg

"The winds and waves are always on the side of the ablest navigator"-Edward Gibbon (1737-1794)

"Eternity is really long, especially near the end"-Woody Allen (1935- ) This is one of several WEB pages which I have constructed over the past decade for courses in mechanics and applied mathematics here at the University of Florida . The present page is intended as a suplement the the four credit course EGM 3413 dealing with the topics of vector fields, solution of ODEs by matrix methods, partial differential equations, and functions of a complex variable.
The fractal shown above is generated by a MATLAB program using the iteration Z[n+1]=Z[n]^6 - 1.12 . It's amazing that such a simple iteration will generate such an intricate six-fold symmetric figure.

The book we are using is Advanced Engineering Mathematics by E. Kreyszig 8th Ed. John Wiley&Sons, Inc. The book has a few flaws such as using Ao instead of Ao/2 in the standard Fourier series, using c^2 instead of alpha as the thermal diffusivity, and giving a poor discusion of the characteristic variables in PDEs, but then it is still the best of the comprehensive intermediate engineering math texts available. An OUTLINE for the course is found by clicking HERE.You will find this material to be an essential part of the modern engineer's repertoire.

You can reach me anytime at e-mail      kurzweg@ufl.edu .

Your course grade will based on  three hour exams(30% each) plus weekly homeworks (10% total). There will be no final exam. To see the topics of this week's lectures scroll down the page until you encounter the knight on horeseback. Click on the underlined stuff below to obtain additional information on the topics being discussed. Here is a little cartoon to start things off- <>
FIRST WEEK: Today will be our first lecture in EGM 4313. We will devote the time to a review of vectors, talking about dot and cross products , the scalar and vector triple. What we mean by scalar and vector fields, and introduce the directional derivative and the gradient. (Chapter 8)

HERON'S FORMULA: A famous formula due to Heron of Alexandria is that for the area of any triangle when the length of its three sides are given. This formula states that the square of the area of any triangle is given by A^2=s(s-a)(s-b)(s-c), where a, b and c are the lengths of the sides of the triangle and s= (a+b+c)/2 is the half-perimeter. I first ran into this formula during my high school geometry class and remember asking my teacher how this formula is derived. He did not know and I found out later that the geometric proof, first given by Heron * in about 100AD, is quite complicated and based on inscribing a circle in a triangle. I'll give you here a quick reverse proof using algebra and vector concepts we have talked about in the first lecture. Assume the formula is correct and rewrite it as A^2=[(a+b)^2-c^2][c^2-(a-b)^2]/16 . Now place the three corners of your triangle at A with coordinates (0,0), at B with coordinates (bx,by) and at C with coordinates (cx,0). The corresponding side lengths are then a=sqrt[(bx-cx)^2+by^2], b=cx, and c=sqrt[bx^2+by^2]. Substituting these quantities into the above A^2 formula, we find after a bit of manipulation that things reduce to the very simple form A=[by][cx]/2 , which is the known area of the triangle via vector calculus when taking half of the absolute value of the cross product between the triangle edge vectors ibx+jby and icx. This vector product result is much simpler than (but equivalent to) the Heron formula and can be used to calculate the area of any polygon by simply adding up the areas of the triangles making up the polygon. Such formulas are often used for determining the acreage of land within a polygonal boundary. For a right triangle one has , via the Pythagorean Theorem**, that a^2+b^2=c^2 and hence that the triangle area becomes A=a*b/2.

*-Heron of Alexandria(65-125AD) was an expert in geometry and mechanics working at the greek school in Alexandria, Egypt . In addition to the Heron formula, he is also known for the invention of the aeolipile, a steam filled hollow sphere which rotates about a fixed axis by the action of steam jets. He can thus also be considered the grandfather of the steam turbine and the jet engine.

**-Pythagoras was born in 569BC on the island of Samos, Greece and is considered by many the first pure mathematician . He  developed mathematical concepts including a proof for the Pythagoran Theorem. The actual result that a^2+b^2=c^2 for a right triangle was already known earlier(about 1900BC) by the Babylonians and Pythagoras may have become aware of it during his travels to Syria and Egypt. Click HEREto see the best kown ancient proof of the Pythagorian Theorem given by Euclid(325-265BC) in his book Elements. A simple extention of a geometrical Pythagorean Theorem proof can be used to derive the familiar Law of Cosines shown HERE.

LAWS FOR TRIANGLES: In much of our subsequent discussions it will be assumed that the students are thoroughly familiar with the basic laws for oblique triangles. Go HERE to review these.

A SUPPLEMENTAL SOURCE FOR VECTOR CALCULUS DEFINITIONS-There are numerous links on vectors and vector fields found on the WEB. Here is one you might want to look at-

http://www-solar.mcs.st-and.ac.uk/~alan/MT3601/Fundamentals/Fundamentals.html

TANGENT PLANE TO A SURFACE: In class today we defined the gradient to a surface and showed how a tangent plane to the surface has its normal point in the same direction as the gradient. I give HEREanother example of such a calculation for a point xo,yo,zo on a paraboloid z=1-x^2-y^2. A plane tangent to this surface is given by a(x-xo)+b(y-yo)+c(z-zo)=0, where an b and c are constants to be found. Now the gradient of the paraboloid is grad(u)=2xoi+2yoj+k and the normal to the plane is n=ai+bj+ck. Since the vectors n and grad(u) must be parallel, we conclude that a=2xo, b=2yo, and c=1. Hence the tangent plane becomes 2xo(x-xo)+2yo(y-yo)+(z-zo)=0. More on the gradient and the directional derivative. Length of a Space Curve. Also concept of the divergence of a vector and the curl of a vector. Some vector identities involving these three operations.

DIVERGENCE OF A VECTOR FIELD: A second important operation in vector analysis is the divergence of a vector field V. The simplest way to envision the divergence of  V(x,y,z) is to think of V as the velocity field for a fluid flow passing through an incremental cube as shown  HERE. The net outflow of fluid through this volume divided  by the volume then represents the divergence of this vector flow field V and is designated as div(V). Note that div(V) is a scalar quantity while V is a vector quantity.

CURL OF A VECTOR FIELD: This is the third important vector manipulation in vector calculus and is equal to the circulation  of a vector field V about a closed curve divided by the area within the curve. Mathematically one has curl(V)=det[i, j, k; d/dx, d/dy, d/dz; u, v, w] , which is a new vector field. For example curl(iy-jx)=det[i, j, k; d/dx, d/dy, d/dz; y, -x, 0]= -2k. Note that the curl[grad(F)] is always zero as is div(curl(V)).

SECOND WEEK: Expression of grad,div and curl in other orthogonal coordinate systems.Vector manipulations used to formulate equations of electromagnetic wave propagation, heat flow, and 2D incompressible and inviscid fluid flow behaviour. Line integrals and independence of path. (End of Chapter 8 and beginning of Chapter 9)

INTEGRATION OF SURFACE AND VOLUME INTEGRALS USING ORTHOGONAL COORDINATE SYSTEMS: I have noticed from some of the questions you have been asking me, that many of you have difficulty in evaluating multiple integrals in coordinate systems other than cartesian. The procedure is really very much straight forward. If you take any orthogonal system u,v,w, then a volume increment there is dV= huhvhw dudvdw and a surface increment is dS=huhvdudv, where hu, h v and hw are the scale factors. These scale factors depend on the particular orthogonal coordinate system being used and can be determined by the invariance of an increment of length squared ds^2 in space. For a cylindrical system we have (h r, htheta , hz )=(1, r, 1) while for a spherical system one has(1, r, r*sin(theta)). Consider now the volume of a sphere of radius r=a. It is given by Integral[dV]=Integral[r^2*sin(theta)*dr*d(theta)*d(phi)], when expressed in spherical coordinates. The range of integration is 0< r<a, 0<theta<Pi for the polar angle theta, and 0<phi<2*Pi for the azimuthal angle phi. This very simple triple integration yields V=(4/3)*Pi*a^3 and is a much easier integration than if done in a cartesian system(x,y,z). Consider next the total surface area S of a cylinder of radius b and height H. Here one has (choosing cylindrical coordinates ) that S==2*Pi*b^2+Integral[r*d(theta)*dz], with range 0<theta<2*Pi , 0<z<H and r=b. Thus S= 2*Pi*[r^2+b*H]. It is also easy to show that the cylinder volume is V=Pi*b^2*H . The famous mathematician Archimedes * first showed that the ratio of the volume of the largest sphere of radius b which can be put into a cylinder of the same radius and of height H=2b is[(4/3)*Pi*b^3]/[2b*Pi*b^2]=2/3. He was so proud of this result that he had a picture of a sphere in a cylinder engraved on his tombstone.

Note that the scale factor triple product huhvhw is equivalent to the absolute value of the Jacobian of the transformation equations relating the cartesian system (x,y,z) to the particular orthogonal system(u,v,w) of interest. A good discussion of scale factors and vector operations in various orthogonal coordinate systems (including cylindrical, spherical, elliptic, parabolic, toroidal, and bipolar) can be found in the book"Mathematical Methods for Physicists" by G. Arfken(3rd ed. Academic Press 1985).

* Archimedes of Syracuse(287BC-212BC)-Greatest mathematician and mechanical genius of ancient times. Born and died in Sicily but spent a good part of his time at the greek school(library) in Alexandria, Egypt. He discovered the Archimedes Principle of Buoyancy , showed that Pi =3.14159...has a value less than 3+1/7 but greater than 3+10/71, invented the Archimedes screw for pumping water, and developed defense machines including a solar concentrator for burning the sails of ships. He was killed by a Roman soldier during the Second Punic War. On a recent(2005) visit to Syracuse in Sicily, I asked our tour guide about the location of Archimedes's grave. It seems nobody knows although there are three to four sites throughout the city which claim that distinction. Sounds to me a lot like "Washigton Slept Here"stories. Anyway HERE is a picture of the Ear of Dionysus, an extant archeological site in Syracuse dating to the time of Archimedes.

CURVE LENGTH USING THE POSITION VECTOR: We can describe a space curve in terms of its position vector R(t)=ix(t)+jy(t)+kz(t). Taking the time derivative dR/dt one gets a vector increment along the curve itself, so that the curve length becomes L=Int[sqrt(R'(t).R'(t)),t=t1..t2]. this last dot product term is equivalent to ds=sqrt(dx^2+dy^2+dz^2) as you encountered earlier in calculus. Click HEREto see the method applied for determining the length of the cardioid
curve r=(1-cos(t). Line Integrals and Surface Integrals. Derivation and application of the Divergence(Gauss)
Theorem , and the Stokes Theorem. (Chapter 9)

WHO WAS GAUSS? -Karl Friedrich Gauss(1777-1855) was a mathematician and astronomer who spent most of his professional life at the University of Goettingen as director of their observatory. He is considered the "Prince of Mathematicians" on a footing equal to Archimedes and Newton. His works include investigations on the fundamental theorem of algebra, the prime number theorem, the least squares method, and non-Euclidian geometry. He also carried out geodesic surveys, invented the heliotrope and calculated the orbit of the asteroid Ceres. He showed how one can construct a regular seventeen sided polygon (heptadecagon) using only a straight edge and compass. Furthermore he showed that it is always possible to construct a regular polygon this way as long as the number of sides equals a Fermat Prime (ie. (2^2^n) +1=3,17,257,etc ). The divergence theorem is named after him and also the hypergeometric series . You can find out more about him by going to Gauss. An image of KFG wearing a cap and a rather dour expression is found HERE.

WHO WAS GEORGE GREEN? George Green(1793-1841) was the son of a baker in Sneinton, Nottingham, England . He was a mathematical genius, although his formal education had stopped with grade shool. His occupation was that of miller and his hobby was mathematics, which he largely learned on his own. In 1828 he wrote the first of his ten or so remarkable scientific papers. This first paper, entitled "An Essay on the Application of Mathematical Analysis to Theories of Electricity and Magnetism" , brought him to the attention of the scientific establishment and he was invited and then began to attend Cambridge University as an undergraduate at the ripe old age of forty. Unfortunately his health deteriorated and he died a few years later at the age of 48. His name is associated with Green's theorem and the various Green's formulas and he is also credited with invention of the Green's function. He was never married but had seven children with the daughter of his mill foreman.

3RD WEEK: (more from Chapter 9) Green's Theorem as a special case of Stokes Theorem. Green's Theorem used to determine the areas bounded by various curves including the cardiod. Also the derivation of Green's first and second formulas from the divergence theorem. Evaluation of Surface Integrals.Evaluation of some problems from the book.

AREA DETERMINATION USING GREEN'S THEOREM: In class last time we derived Green's Theorem from the Stokes theorem and showed that the area integral of the partial of Q(x,y) with respect to x minus the partial of P(x,y) with respect to y is equal to the line integral of Pdx+Qdy around the curve C bounding the area. By setting Q=x and P=0 one finds that the area  for any simply connected region equals the line integral of xdy about the bounding curve. If one uses polar coordinates the area becomes 0.5*lineintegral[r^2 d(theta)]. To demonstrate this result we consider the area contained within the Rhodonea curve r=cos(2*theta) . Doing the line integration using MAPLE we find the following

EVALUATION OF A SURFACE INTEGRAL USING THE DIVERGENCE THEOREM: Suppose that we wanted to find the surface integral of x^2  taken over the hemisphere x^2+y^2+z^2=1, z>0. One way to quickly do this is to consider the divergence theorem for the vector field V=ix over this hemisphere. Here the theorem  says the  hemisphere volume, which is 2*Pi/3, just equals the surface integral of V doted into the surface normal n. Across the bottom boundary of the volume, where n=-k, there is no contribution since V.n=0 there. However, over the hemisphere n=ix+jy+kz so that V.n becomes x^2. Thus the surface integral of x^2 taken over the hemisphere just equals 2p/3=2.094395.

VOLUME OF A TETRAHEDRON VIA THE DIVERGENCE THEOREM: Consider the volume of a tetrahedron formed by the slanted plane x+y+z=1 and the planes x=0, y=0 and z=0. We can obtain this volume V quickly by applying the divergence theorem for the vector field F=ix whose divF=1. One has V=Surface Integral[ix.n dS]. This surface integral vanishes along  the x=0, y=0, and z=0 surfaces,  leaving one with only V=Surface Integral[ix.(i+j+k)/sqrt(3)dS]. Projecting dS into the x,y plane via the substitution dS=dxdy/(n.k)=sqrt(3)dxdy, one finds V=DoubleIntegral[xdxdy, over 0<y<(1-x), 0<x<1]=1/6. That is, the volume of the tetrahedron formed by cutting a unit cube by a slanted plane passing through three of its corners at (1,0,0), (0,1,0), and(0,0,1) is just one-sixth of the cube volume. Introduction to matrices and matric multiplication. Transpose and inverse of a matrix. Gauss-Jordan elimination method. Solution of simultaneous algebraic and differential equations by matrix methods. (Chapter 6)

A QUICK TUTORIAL FOR NUMERICAL MATRIX MANIPULATIONS USING MATLAB: One of the easiest to use canned programs available for matrix calculations isMATLAB . Many of you already own a student edition of MATLAB and the extended version is also available in the Department's Computer Lab. Here is a brief summary of what you can do-

1. First define a Matrix, say: A=[3,4,5; 2,5,0;2,1,8] . Here element a2,3 in the second row and third column is 0.

2. The determinant is found by typing: det(A) and for this matrix equals 16.

3. The inverse of A is gotten by the operation: B=inv(A) and here yields [2.5,-1.6875,-1.5625;-1,0.875,0.625;-0.5,0.3125,0.4375].

4. Matrix multiplication is defined by: A*B and, as expected, here yields the identity(or unit) matrix I =[1,0,0;0,1,0;0,0,1].

5. The eigenvalues of matrix A are given by: eig(A) and here are found to be 0.2829, 10.1396, and 5.5775.

6. The corresponding eigenvectors are obtained by: [V,D]=eig(A) This prints out the three vectors corresponding to the three different eigenvalues. Here we find the three vectors are the transpose of [-0.9048,0.3836,0.1848], [0.6457,0.2513,0.7210],and [-0.2352,-0.8145,0.5304]. Note that one can always make one of the elements of an eigenvector unity by multiplying all elements in the vector by the same number.

USE OF MATRIX METHODS TO DEFINE A PLANE IN SPACE PASSING THROUGH THREE POINTS: A very nice application of matric methods is to find the equation for the plane Ax+By+Cz=1 containing three specified points P(x1,y1,z1),P(x2,y2,z2), and P(x3,y3,z3). Go HERE to see our method for obtaining  the values of A, B, and C using the linalg  portion of the canned math program MAPLE.

MATRIX MANIPULATOR AVAILABLE ON THE WEB:  To quickly calculate the inverse, eigenvalues and determinent values of matrices go to   http://wims.unice.fr/wims/wims.cgi

4TH WEEK:(Chapters 3and part of 7) Eigenvalues, eigenvectors, and the fundamental matrix. Incorporation of initial values and non-homogneous terms into the matrix solution. A brief , easy to grasp, one-page tutorial on basic matrix operations is found at - http://www.math.hmc.edu/calculus/tutorials/matrixalgebra/
SAMPLE CALCULATION FOR THE SOLUTION OF AN ODE BY MATRIX METHODS: We have shown in class how to reduce an nth order ode to a set of first order equations expressible in matric form as X'=MX. The eigenvalues k  needed in the solution follow from det[M-Ik]=0, where I is the unit matrix I=[1,0,0;0,1,0;0,0,1]. Once the eigenvalue have been determined one next constructs the eigenvectors X from the recipe x1/cof(a11)=x2/cof(a12+x3/cof(a13+..where the cofactor cof(a)= (-1)^(i+j)*minor(aij) and i is the row number, j the column number and the minor is that part of the determinant left after one strikes out the row and column containing the element aij. The final solution will then be X=c1*X1*exp(k1*t)+c2*X2*exp(k2*t) with c1 and c2 being arbitrary constants and k1 and k2 the two eigenvalues when dealing with a second order ODE. Click HEREto see this solution procedure demonstrated for a special case.

MULTIPLE MASS-SPRING PROBLEM SOLVED BY MATRIX METHODS: An interesting application of the matrix methods we have talked about in the last few lectures is the determination of the eigenfrequencies of a multiple mass spring system. One such problem is that shown HERE.For this problem the two equations of motion are d2x1/dt2=-(k1/m1)x1+(k2/m1)(x2-x1) and d2x2/dt2 = -(k3/m2)x2+(k2/m2)(x1-x2). These can be rewritten as four simultaneous first order ODEs whose matrix form is as shown above. The eigenvalues and corresponding eigenfunctions for this 4x4 coefficient matrix are readily determined via the above matrix evaluation link.  We have carried out such an evaluation for m1=m2=1 and k1=k2=k3=1 and find the eigenvalues to be ± i*sqrt(3) and ± i. The corresponding column eigenvectors read x1=[1,-1,sqrt(3),-sqrt(3)] Tand x2=[1,1,0,0]T. Thus one sees that the higher frequency mode consists of the  masses moving in opposite directions while the lower frequency oscillation is characterized by the masses moving in the same direction. More on the solution of dX/dt=M X , eigenvalues and eigenvectors. Application of initial conditions. Solution of non-homogeneous matrix equations by variation of parameters. Phase plane techniques in connection with two simultaneous first order differential equations.
5TH WEEK Review for the First Hour In-Class Exam coming up later in the week.

FIRST HOUR IN CLASS EXAM. Exam will cover material on vector fields and matrix solution methods. Will be tested only on those topics of Chapters 3, 6, 7, 8 and 9 discussed in class. Chapter 3 deals with converting higher order ODEs to matrix form , which you should be able to do. The exam is closed book except you can bring your calulator and one 3"x5" card containing whatever you want to write on it. You can answer any three out of four questions for a maximum of 10 points each. Partial credit will be given .

6TH WEEK: Introduction to Fourier Series( Chapter 10 ). Representation of any bounded periodic function by an infinite series involving sine and cosine terms. The Fourier coefficients a n and b n. Several examples of Fourier series.

WHO WAS JOSEPH FOURIER?- Jean Babtiste Joseph Fourier was a mathematical physicist, teacher, revolutionary, politician , friend of Napoleon, and prefect of Grenoble. He was born in Auxerre , France in 1768, the son of a tailor, and died in Paris in 1830. He is today best known for the Fourier Series and the Fourier Integral. He accompanied Napoleon on his Egyptian campaign as scientific advisor, was for a short time the governor of lower Egypt, and earlier almost lost his head during the later stages of the French revolution when he was accused of being a supporter of Robespierre , the most radical of the revolutionists. When Fourier first submitted his paper on heat conduction and Fourier series to the French Academy of Sciences in 1807, a panel consisting of Laplace, Lagrange , Monge, Biot and others rejected the paper as not being sufficiently rigorous. It was not until 1822 that this groundbraking work was published by Fourier in bookform under the title "Theorie Analytique de la Chaleur". The law of heat conduction( i.e. heat flow in a solid conductor is directly proportional to the temperature gradient ) is also named after him. The curly haired fellow shown is J.Fourier.

SAMPLES OF FOURIER SERIES FOR DIFFERENT PERIODIC FUNCTIONS: If you go HERE you will find applet animations for five different periodic functions and how the Fourier series approaches these as the number of terms increases toward infinity. Note the Gibbs phenomenon at those points where the functions have discontinuities. More on Fourier Series. Identities involving p. Expansion for even and odd functions. Expansion for period p=2L. Parseval Identity.

GENERATION OF FOURIER SERIES USING MATHCAD: HERE and HERE are some examples of even Fourier series generated by Mathcad including the one we spent a good portion of the period on during the last lecture.

GENERATION OF FOURIER SERIES USING MAPLE: We have shown in class that that any function of period 2L can be expanded in a Fourier series f(x)=(1/2)a 0+Sum[ancos(npx/L)+b n sin(npx/L), n=1..Infinity). Here an =(1/L )*Int[f(x)*cos(npx/L),x=- L... L] and bn=(1/ L)*Int[f(x)*sin(npx/L), x=- L.. L]. One can easily automate this procedure, as we have done in the attached jpg, by using the canned program MAPLE. I show HEREthe plot of the Fourier series for  f(x)=[sin(x)] 5 for period 2p. If you tried this by hand it would require considerable time. Also, you will notice that, the series requires only 5 terms n=1,...,5 to represent f(x) exactly. A second example consider the odd function of period 4 whose value is zero for 0<x<1 and (x-1) for 1<x<2. It has only nonvanishing bn coefficients and a Fourier series of 100 terms produces the results shown HERE.As a third example consider the use of MAPLE to obtain the Fourier series for an even more complicated triangular periodic function F(x)=0 for 0<x<2, =(x-2) for 2<x<3, and =(4-x) for 3<x<4  where L=2. Here the function has no symmetry and hence both the an and bn terms are non-vanishing. The results of the MAPLE calculation are shown HERE. Again to carry out such a calculation by hand would be rather tedious.

REPRESENTING A PARABOLA BY A FOURIER COSINE SERIES: If you want to represent a non-peridic function F(x) by a Fourier series, you can always do so by defining the range -L<x<+L over which you want to represent the function. I do this HERE by looking at F(x)=x2 in -p<x<p. A simple evaluation yields the even Fourier series representation
F(x)=p2/3+4*Sum[(-1)n*cos(nx)/n^2, n=1..infinity] for this case. As the graph shows, this peridic function nicely represents the parabola F(x)=x2 inside -L<x<+L but clearly does not do so outside. An interesting equality which follows from this result (when one sets x=0) is that  p2/12=1-1/4+1/9-1/25+... =0.8224670...Also one could think of the periodic parabolic structure shown in red as the shape of a reflecting surface capable of collecting light, incoming parallel to the y axis, and concentrating the radiation at the points (x, y)=(2pn, 1/4).

DIFFERENTIATION AND INTEGRATION OF FOURIER SERIES: It is possible to both differentiate and integrate the two sides of a Fourier series and in most cases recover a correct form for the resultant function. We demonstrate this HERE for the case of the discontinous function f(x)=+1 for 0<x<1 and f(x)=-1 for -1<x<0 which has a simple Fourier sine series representation. Note that an integration yields a sawtooth function g(x) and a differentiation yields a bunch of delta functions h(x). Mathematicians tend to be very uncomfortable with differentiating functions with discontinuities but as seen the results aren't bad in this case and have the form of what is expected by looking at the slope of the original function f(x).

WHO WAS PARSEVAL?-Antoine Parseval(1755-1836) was a French royalist jailed during the French Revolution and almost beheaded. Later got in trouble for publishing an  anti-Napoleonic treatise and  had to flee the country. Published just five mathematical papers during his lifetime, but these contained important new results including the Parseval inequality and the Parseval theorem.

AN APPLICATION USING THE PARSEVAL THEOREM: We have shown in class that when squaring the Fourier series for a function f(x) by itself and then integrating the result over the range -L<x<+L, one arrives at the Parseval result
(1/L)*Int[f(x)2, x=-L..L)=(1/2)ao^2+Sum[an2+bn2, n=0..infinity]]
, where an and bn are the Fourier coefficients. This result can be used, when f(x)=x and L=1, to show that the sum of the reciprocals of the square of all integers equals p2/6=1.644934066848.. ( a result already known to Euler prior to the invention of Fourier series). The Parseval relation is one way to relate certain definite integrals to an infinite series. Other techniques exist including a very powerful approach using the geometric series and Laplace transforms.

7TH WEEK: Complex Form of the Fourier Series, Development of the Fourier Transform. The Dirac delta function and the Heaviside step function and their Fourier transforms.

WORKING WITH THE COMPLEX VERSION OF THE FOURIER SERIES: The standard Fourier series can readily be converted to its complex form by using the substitutions cos(x)=[exp(ix)+exp(-ix)]/2 and sin(x)=[exp(ix)-exp(-ix)]/2i. This leads to f(x)=Sum[cn exp(inpx/L), n=-infinity..infinity] with cn=[1/(2L)]*Int[f(x)*exp(-inpx/L, x=-L..L]. Here L is the function half period and n represents all integers. Applying this result to the repetative rectangular pulse of period 2L=4 with f(x)=+1 for -1<x<+1 and f(x)=0 for -2<x<-1 and 1<x<2, we find that f(x)=Sum[1/(n*p)*sin(p/2)*exp(inpx/2),n=-infinity..+infinity]. This is equivalent to f(x)=0.5+2*Sum[sin(np/2)*cos(npx/2)/(np),n=1..infinity]. Click HEREto see an approximation of this last result using a 200 term approximation. Note that the rectangular pulse is nicely reproduced except for the unavoidable Gibbs ear phenomenon at its discontinuities.

RELATION BETWEEN THE FOURIER SERIES AND THE FOURIER TRANSFORM: A question which arises when studying Fourier series of period 2L, is what happens when L is allowed to go to infinity but the actual function f(x) has appreciable value only near x=0. In this limit the summation in the Fourier series goes over to an integral and one ends up with two integrals , one giving the Fourier transform g(w) of the function f(x) and the second the inverse which returns one to the original function f(x). Such transforms are of tremendous importance in numerous modern research areas and methods, such as the FFT(Fast Fourier Transform ) , which aid in the speeding up the Fourier transform process, are very much in voque. The standard Fourier Transform of f(x) is given by g(w)=1/sqrt(2Pi)*Int[f(x)exp+iwx, x=-infinity..+infinity] and its inverse is f(x)=1/sqrt(2*Pi)*Int[g(w)exp(-iwx), w=-infinity..+infinity]. Note  many engineers use a less symmetric version of the Fourier Transform Pair [f(x),g(w)] in which the integral for f(x) is multiplied by one and the integral for g(w)by 1/(2Pi) and the signs in the exponents may be switched. It makes no difference as to which of these definitions is used as long as one is consistent. To get some feel for the Fourier integral, we look at the Fourier integral for the rectangular pulse f(x)=1 for -1<x<1 , namely, (1/p)*Int[cos(kx)*sin(w)/w, w=-N..N) with N->infinity. A numerical approximation obtained via MAPLE using N=60 yields the result shown HERE. A comparison with a standard Fourier series expansion for the same function when the period is 2L=5 is also shown and , as expected, the two results almost coincide. Completion of Chapter 10. More on some manipulations with Fourier Integrals. Also working out some problems from the book involving both Fourier Series and Fourier Transforms.

FOURIER SINE TRANSFORM FOR A TRIANGLE: Consider the triangle function f(x)=x in -L<x<L. Its Fourier Sine Series is readilly shown to be f(x)=Sum[b nsin(n p x/L), n=1..infinity], where b n=(2/L)*Int[f(x)*sin(n p x/L), x=0..L]. If we now let w=np /L so that dw=( p/L)dn and allow L to go to infinity, we can replace the Sum[( )dn by an integral Int[( )L/ p)dw. This results in a double integral with the integral limits on both the variable w and x extending from 0 to infinity. We thus have obtained the Fourier Sine Transform where the transform of f(x) is defined as g(w)=Int[f(x)*sin(wx),x=0..infinity]  and its inverse is f(x)=(2/ p)*Int[g(w)*sin(wx), w=0..infinity]. Specifically , for the triangle function one finds that g(w)=-cos(w)/w+sin(w)/w^2. Upon inverting,  this yields the result f(x)=(2/ p)*Int[(sin(w)/w^2-cos(w)/w)*sin(wx), w=0..infinity]. We have plotted an approximation to this integral HERE. Note that I have approximated w=0 by 0.001 to avoid the singularity at w=0 and have replaced the upper limit of infinity by w=200 to give a reasonable computer run time when using MAPLE.

USING MAPLE OR MATLAB TO FIND  FOURIER TRANSFORMS: Existing canned programs such as MAPLE  or MATLAB can very quickly calculate for you the Fourier transform g(w) of a function f(x). We  demonstrate this HERE by showing some graphs of f(x) and the corresponding transform.

USE OF FOURIER TRANSFORMS IN RETURNING FILTERED SIGNALS: If one treats the x as time t and the w as angular frequency w , then the Fourier transform g(w) can be thought of as a frequency spectrum (decomposition) of the original signal f(x). One can introduce some filtering into this g(w) by multiplying it by a window function H(x-a)-H(x-b), where H is the Heavyside step function. The result, on inverting the product, produces an approximation to the original f(x) which is missing those frequency components outside the window. I demonstrate this HEREby looking at the inverse of the Fourier transform of the square pulse f(x)=H(x-2)-H(x-3), when filtered by the windows  -5<w<5  and  by -50<w<50. You will note that a lot of information about the detailed structure of f(x) is lost when inverse Fourier transforming the filtered transform. On the other hand, when a signal is received containing a lot of noise , use of filtering can help bring about a clearer view of the original signal. In connection with such filtered decompositions, I also want to make you aware of the wavelet approach to signal analysis. This has become a very active  area of applied mathematics research in the last few years and is a serious competitor to Fast Fourier Transform approaches for signal compression and de-noising.

THE DIRAC DELTA FUNCTION AND ITS FOURIER TRANSFORM: An important function encountered in various applications including mechanical vibrations, quantum mechanics, and control theory is the Dirac delta function defined as d(x-a)=infinity at x=a and zero for all other x. It has the further property that Int[d(x-a)*f(x), x=-infinity..+infinity)=f(a). Note that the delta function also equals the derivative of the Heaviside step function H(x)=0 for x<a and H(x)=1 for x>a. The Fourier transform of d(x-a) equals g(w)=Int[d(x-a)*exp(-i*w*x), x=-infinity..+infinity]=exp(-i*w*a). Taking the inverse Fourier transform of this result, we find another representaion for the Dirac delta function , namely, d(x-a)=(1/p)*Int[cos(w*(a-x)), w=0...+infinity]=sin(w*(x-a))/(p*(x-a)) . As an approximation for the value of this last integral, we show you HEREa plot of this integral for a=0 when the upper limit on w is taken as w=10.

A TABLE OF FOURIER TRANSFORM PAIRS: By going  HEREyou can find a table of some of the better known Fourier transforms and their inverses. We see, for example, that the gaussian f(x)=ex[(-x2) has the transform g(k)=sqrt(p)*exp(-k2/4) and that the function f(x)=Heaviside(x-0)*exp(-x) has the transform g(k)=1/(1+I*k). From this last result at x=1 one can infer (after a bit of manipulation which I leave as a challenge to the reader) that  int(w*sin(w)/(1+w^2), k=0..infinity)=p/(2*exp(1))=0.57786..Note that this last value is close but not equal to the famous Euler-Mascheroni constant g=0.577215...=lim n-> infinity[1+1/2+1/3+..+1/n-ln(n) ].

FOURIER TRANSFORM OF CONVOLUTION INTEGRALS: We showed in class that the Fourier transform of the convolution integral f*g=Int[f(z)g(x-z), z=-infinity..+infinity] equals sqrt(2Pi)*F(f(x))F(g(x)), where F indicates the Fourier transform of the separate functions. One can also invert this result to state that (f*g)=F -1[ F(f(x))F(g(x))]. This last form can help in the evaluation of some complicated integrals. We demonstrate this HERE for the case of the rectangular pulse f(x)=+1 for -1<x<=1 and zero everywhere else and the function g(x)=1/(1+x2 ). After some manipulations( which I leave for the reader to carry out), one finds in this case, that-

(1/2) arctan(2/x2) =  Int[sin(w)cos(wx)exp(-w)/w, x=0..infinity]

from which also follows that Pi/4=Int[(1/w)sin(w)exp(-x), x=0..infinity]

8TH WEEK: Introduction to Partial Differential Equations.( Chapter 11) . Formulation and Solution of the 1D Wave Equation. Separation of Variables Solution Method and d'Alembert's Solution.

THE VIBRATING STRING PROBLEM:- One of the best known solutions of a partial differential equation is that for a  vibrating string of length L tied down at x=0 and x=L. Here the PDE reads ytt =c2yxx , where the subscripts indicate partial derivatives of the transverse string displacement y(x,t) and c =sqrt(T/ r) is the constant speed of signal propagation depending on the tension T and the string's linear density r. The two boundary conditions are y(0,t)=y(L,t) and the initial conditions are y(x,0)= f(x) and yt (x,0)= y(x). This equation can be solved by a separation of variables approach y(x,t)=F(x)*G(t) and use of the Fourier sine series. We show you HERE the solution obtained when L=1, c=1, f (x) = sin(3 p x), and y (x)=0. For this case one finds the standing wave form y(x,t)=cos(3 p t)*sin(3 px) which has the fixed angular frequency of w=3p. It is the uniqueness of the vibration frequency for a string vibrating in a given oscillation mode which makes stringed musical instruments possible. Another interesting solution of the vibrating string corresponds to f(x)=x/Pi for 0<x<Pi/2 and f(x)=(1-x/Pi) for Pi/2<x>Pi , with y(x)=0. In this case the initial triangular displacement yields the following pattern(HERE.) at later times. More on the Wave Equation and solution in Higher Dimensions. Vibrating Membrane. Sound Waves in a Sphere. Waveguide problem.

D'ALEMBERT'S SOLUTION OF THE 1D WAVE EQUATION:-We have shown in class that substitution of the characteristic variables h =x-ct and x=x+ct converts the one  dimensional wave equation to y hx =0 , which on simple integration yields the general solution y(x,t)=f(x-ct)+g(x+ct). If one now considers a string of infinite length and uniform density, one need only apply the initial conditions y(x,0)= f (x) and yt(x,0)= y(x) to find the unique values of f and g. This is what D'Alembert did (and also what we showed in class). The result is-
y(x,t)=[f(x-ct)+f(c+ct)]/2 + [1/(2c)]*Int[ y ( m), m =x-ct ...x+ct]
which is known as D'Alembert's Solution. We show you HEREa picture of D'Alembert(1717-1783) and a MATLAB created graph of the solution when initial conditions are f=exp(-x^2) and y=0. Note that the initial gaussian breaks up into two gaussians of half the original height and that these travel in opposite directions at speed +c and -c along the string. Another view of wave development for the same initial Gaussian displacement is found by going HERE.

A VIBRATING RECTANGULAR MEMBRANE:- A good demonstration of the solution of the wave equation in 2D is that associated with a vibrating rectangular membrane. Here the governing equation for the membrane displacement is
Ztt=c 2(Z xx+Zyy) and one tries the separation of variables substitution Z=T(t)*F(x,y)=T(t)sin(n p x/a)sin(n py/b), assuming that the membrane is clamped at its edges at x=0, x=a, y=0, and y=b. This leads to the double Fourier sine series solution running over the integers n and m from 0 to infinity. The imposed initial conditions  F(x,y,0)= f(x,y) and Ft (x,y,0)= y (t)  determine the values of the constants Anm and B nm  appearing in  T(t)=A nm sin( wt)+B nm cos( wt), where w =c p *sqrt([(n/a)^2+m/b)^2]. The spatial part of this solution, for a given n and m ,is referred to as an eigenmode F(x,y) and the corresponding angular frequency w is the eigenfrequency. We show you HEREa MATLAB drawn 3D surface representing Z(x,y,0) for n=4 and m=5 when a=b=1.

HOW TO USE MATLAB TO GRAPH SOLUTION SURFACES U(X,Y): Some of you , who have available computer programs such as MATHCAD, MAPLE, MATHEMATICA, or MATLAB , have been asking me how one goes about plotting some of the solution surfaces U(x,y) we have been encountering in solving PDEs. I'll show you here one example based on the solution of Uxx=Uxy subject to U(x,0)=x^2 and U(0,y)=sin(y). By the method of characteristics we know the general solution to this PDE is U=F(y+x)+G(y) where the functions F and G can be evaluated by use of the specified boundary comditions. We find U(x,y)=sin(y)+2yx+x^2. A plot of this finction is given HERE.

9TH WEEK: Derivation and Solution of the 1D  Heat Conduction Equation. Time-dependent development of temperature in a bar with zero end temperatures but x dependent initial condition. Treatment of heat conduction problems in bar with finite end temperatures. 1D Diffusion problems.

TEMPERATURE IN A BAR: The temperature T(x,t) in a bar of  length L , and maintained at zero end temperature, is governed by the 1D Heat Conduction Equation T t =aTxx subject to the IC of T(x,0)= f(x) plus two BCs of T(0,t)=T((L,t)=0. As shown in class, this equation can readily be solved by a separation of variables approach by setting T(x,t)=G(t)*F(x). This leads to the result T(x,t)=Sum[C n *sin(n px/L)*exp(- a (n p /L)^2*t), n=1..infinity] , where the coefficient Cn is given by Cn=(2/L)*Int[f(x)*sin(n px/L), x=0..L]. We show you HERE the time development of this temperature profile when the bar has an initial temperature of T(x,0)= f(x)=1 and length L=1. As expected, the temperature will be very close to the end temperature by the time the non-dimensional parameter a t/L 2 >1.

TIME-DEPENDENT TEMPERATURE IN A SLAB: The time dependent temperature in a slab (where 0<x<a, 0<y<b) represents a good example of heat conduction in 2D. The governing equation is Tt= a (Txx+Tyy) and this has the very simple separation of variables solution T(x,y,t)=DoubleSum[(1-(-1)^n)*(1-(-1)^m)*exp[- a*p^2*t*((n/a)^2+(m/b)^2)]*sin(n*p *x/a)*sin(m* p*y/b)/(n*m), n=0..infinity, m=0..infinity]
whenver the temperature vanishes at the edge of the slab and the initial temperature is T(x,y,0)=1. We show your HEREa 3D color plot of the temperature in the slab at at/a^2=0.05 when a=b=1. Conduction in bars of infinite length using the Fourier Integral approach. The Error Function and its properties. Use of Laplace transforms to solve the 1D heat conduction equation. Heat conduction in a cylindrical geometry.

ERROR FUNCTION: In solving the 1D heat conduction equation over the infinite range -infinity<x<infinity one finds that T(x,t)=1/(2sqrt( pa t))*Int[T( x,0)*exp((x- x)^2/(4 at)), x=-infinity...infinity] . For cases where the initial condition T(x,0) is a constant over part of the x range and zero everywhere else, this integral can be converted via the substitution u=( x-x)/(2sqrt( at)) to an integral of the form erf(x)=[2/sqrt( p )]*Int[exp(-u^2),u=0..x)] . This last integral is referred to as the error function and has the property that erf(0)=0 and erf(infinity)=1. It is a tabulated function and I show you its graph HERE  as obtained via MATLAB. The error function arises all the time in both diffusion and conduction problems and so is one you should be familiar with.

FOURIER TRANSFORM SOLUTION FOR THE TEMPERATURE IN A BAR OF INFINITE LENGTH: The heat conduction equation in 1D when x extends from minus to plus infinity can be conveniently solved by Fourier transform methods. Applying the standard Fourier transform to Tt=aTxx yields df(k,t)/dt=-ak^2 f(k,t) where f(k,t) is the Forier transform of the unknown temperature T(x,t). Solving we get f(k,t)=C(k)exp(-ak^2t), where C(k) is the Fourier transform of the initial condition T(x,0).
Inverting f(k,t) leads to the solution T(x,t). With a little manipulation this solution can be written as-
T(x,t)=1/(2*sqrt(apt)*Int[T(z,0)*exp(-(z-x)^2/(4at)), z=-ininity...infinity] so that the temperature will be known at all later time once T(x,0) is specified. We show you HERE the solution for the case of an initial temperature condition represented by the  double pulse T(x,0)=H(x+2)-H(x+1)+H(x-1)-H(x-2). The solution is expressible exactly in terms of error functions or in terms of the integral shown in the figure.

AGE OF THE EARTH: An interesting heat conduction problem concerns the cooling of a sphere of radius r=a and an intial constant temperature T o. This is a problem first looked at by Lord Kelvin in the 19th century to determine the age of the earth. Casting the problem into spherical coordinates one needs to solve Tt= a [Trr +(2/r)Tr]  subject to T(0,t) finite, T(a,t)=0, and T(r,0)=T o. Using the substitution T=R(r)/r]exp(- al 2t), this leads to R"+l 2R=0. From this follows the closed form solution T(r,t)=Sum[(2T oa/npr)(-1)n+1 sin(npr/a) exp( a(np/a)2t), n=1..infinity]. We have plotted this result on the accompanying graph for at=0.1, 0.5, and 1. The approximate e-folding time for the original temperature is seen from this result to be t*=(a/ p )2/a. Although Kelvin estimated from his solution (based on the temperature rise in deep mines) that the earth was only some 24 million years old and this value is clearly in error as pointed out by numerous sources at the time(Huxley etc), the value of t*obtained for the earth ( assuming it to be made essentially of iron where a=0.205cm 2/sec) is actually t*=(6.378x10 8cm/ p ) 2/0.205=2.01x10 17sec=6.38 billion years, which is in the right ballpark for the earth's current estimated age of about 4 billion years and a bit higher than this value because of the neglect of known convection which speeds up the heat transfer process . Perhaps Kelvin's shorter cooling time estimate was partially influenced by the Victorian belief that , according to Bishop Usher(1581-1656) , the earth was created precisely in 4004 BC. Also  deep mine temperatures are partially influenced by heating due to radioactive decay and thus throw off his calculations.

10TH WEEK Derivation of the Laplace Equation. Dirichlet Solution in the Rectangle. Application to Inviscid 2D Flow and Electrostatics.

WHO WAS LAPLACE?- Pierre Simon de Laplace(1749-1827) was a French Astronomer and Mathematician with many scientific credits to his name. He proposed the nebular theory of the evolution of the solar system, introduced the concept of a potential , was involved in setting up the metric system, did work on probability, wrote a monumental multi-volume work on planetary mechanics entitled "Traite du Mecanique Celeste"and proved (the obvious) that the solar system is stable. He was also very adept at always allying himself with the "in" political powers throughout his lifetime. He was a member of the French Academy of Sciences , taught at the Ecole Normal in Paris, and was made Count of the Empire and Chancellor of the Senate by Napoleon(although he was removed from this last post after only six weeks because of his tendency to micro-manage things). Laplace became a marquis after restoration of the Bourbons in 1817. He was also a friend (and later enemy)of Benjamin Thompson(alias Count Rumford), the famous American Tory who fled to England during the American revolution and later became well known in scientific circles for his mechanical equivalent of heat measurements while observing the boring of cannons for the elector of Bavaria. A picture of Laplace is found HERE.

SOLUTION OF THE LAPLACE EQUATION IN A SQUARE FOR DIRICHLET BOUNDARY CONDTIONS- One can solve the 2D Lapace equation for a rectangular cartesian geometry and for Dirichlet bcs by a simple separation of variables and superposition approach. As an example of this consider the solution when V(x,y) satisfies the boundary conditions V(0,y)=V(1,y)=1 and V(x,0)=V(x,1)=0. Here one superimposes the solution corresponding to the bc's V(0,y)=V(x,0)=V(x,1)=0 and V(1,y)=1 onto the solution satisfying the bcs V(1,y)=V(x,0)=V(x,1)=0 and V(0,y)=1. This leads to-
V(x,y)=(2/p)*Sum{(1-(-1)^n)/(n*sinh(n* p))*[sinh(n*p*x)+sinh(n* p*(1-x))]*sin(n*p*y), n=1..infinity}
A graph for the contours V(x,y)=Const. predicted by this solution is found by going HERE.

SOLUTION OF THE LAPLACE EQUATION INSIDE A CIRCLE: Another interesting problem concerns the value of the electric potential V inside a circle of radius r=a when the boundary condition V(a,q)=f(q) is specified. Here the appropriate PDE is (Vrr+(1/r)Vr+(1/r^2)Vqq=0 and a separation of variables solution yields
V(r,q)= Sum[(r/a)^n (Ansin(nq)+Bn cos(nq), n=0..infinity] where An and Bn are constants to be evaluated from the bc at r=a. We show you HERE a graph of the  solution V(r,q)=1/2-1/2(r/a)2cos(2q) found when f(q)= sin(q)2=(1/2)*(1-cos(2q). Note that V equals its mean value of 0.5 along the two diagonal lines where cos(2q)=0.

OCCURRENCE OF BESSEL FUNCTIONS: When dealing with the solutions of the wave or heat conduction equation in  polar coordinates or Laplace's equation in cylindrical coordinates with a z  variation , one finds that the radial part of the solution involves Bessel functions of the first kind in the form Jm(umn r/a) , where umn is the nth zero of the mth order Bessel function and r=a is the radius of the circular boundary. For example, the wave equation for a vibrating circular membrane of radius r=a has the solution U(r,theta,t)=G(t)*F(r,theta), where G(t)=Asin(wt)+Bcos(wt) and F(r,theta)=Jm(umnr/a)*[Csin(m theta)+Dcos(m theta)]. Here A,B,C,and D are constants which can be evaluated by the given initial values and the boundary condition at r=a , w=umnc/a is the angular frequency, and c the propagation speed. The product G*F is double summed over the integer values of n and m. We show you HERE a graph of a typical non-axisymetric eigenfunction Fmn=F1,2 encountered in such a solution.
By going HERE you will find the important orthogonality property for Bessel functions which allows one to expand any bounded functions in terms of a Fourier-Bessel series. REVIEW FOR SECOND HOUR EXAM , including going over some exam questions from previous years. Topics to be covered on the exam will be Fourier series and transforms , and the solution of PDE's , including the wave, heat conduction, and Laplace equations.

11TH WEEK: SECOND HOUR EXAM. Closed book except you can bring one 3"x5" card and your hand calculator. Introduction to Complex Variables. Cartesian and polar representations of a complex number and its Argand diagram representation. Addition, subtraction , multiplication and division. DeMoivre Formula for finding multiple roots of complex numbers.

ARGAND DIAGRAM- This is a convenient way to plot a complex number z=x+iy within the z plane. The x axis represents the real part of z and the y axis the imaginary part. In polar form one has z=r exp(i q ), so that r=sqrt(x^2+y^2)=amp(z) and q =arctan(y/x)=arg(z) is the angle measured in the counterclockwise sense. Click HEREto see the location of the complex number z=exp(i q) at intervals of d q = p/4 starting with q=0.

WHO WAS DE MOIVRE? -Abraham de Moivre(1667-1754) was French born protestant who emigrated to England after the revocation of the Edict of Nantes and during the expulsion of all Huguenots from France. He was unable to obtain a university position in England as a foreigner and had to maintain himself as a tutor of mathematics. He wrote several books in the area of analytic geometry, probability, and statistics and was a member of the Royal Society and a friend of Newton. He discovered the Stirling formula before Stirling and is today mainly remembered for his formula for obtaining the roots of a complex number.

USE OF COMPLEX VARIABLE METHODS TO QUICKLY OBTAIN TRIGNOMETRIC IDENTITIES: We can use complex variable methods to obtain some of the trignometric identities you encountered in some of your earlier math classes. The key
for doing so is the Euler Identity exp(iz)=cos(z)+isin(z). Go HERE to see some of these identities derived.

PROPERTIES OF COMPLEX HYPERBOLIC FUNCTIONS: In discussing functions of complex variables one often runs into the hyperbolic functions sinh(z), cosh(z) etc. The properties of such functions are easily established by replacing z by x+iy. Thus , for example, sinh(x+iy)={exp(x)[cos(y)+isin(y)]-exp(-x)[cos(y)-isin(y)]}/2=cos(y)sinh(x)+isin(y)cosh(x) , so that sinh(1+i)=0.63496..+i 1.2984.. . In case you are a little rusty on hyperbolic functions go HERE to refresh your memory.

THE COMPLEX NUMBER Z=(1+I)^n AND THE BERNOULLI SPIRAL: Sometimes powers of complex numbers lead to very interesting trajectories in the Argand plane as the power is varied continiously. Take, for example, the number Z=(1+i)^n and let n vary from n=0 to infinity. It is clear that Z=2i when n=2 and Z=-4 when n=4. Casting the number into polar form we find that Abs(z)=r=2^(2 q/p) and Arg(z)= q=n p /4. One can eliminate the n from these last two results to find r=exp[2ln(2) q/p ] which is recognized to be a standard Bernoulli logarithmic spiral . Click HEREto see its form. So what is the value of (1+i)^16 ?  J.Bernoulli was so proud of his discovery of the logarithmic spiral r=a*exp(b*q )that he had it inscribed on his tombstone. The engraver got the figure slightly wrong and it rather looks like a bunch of concentric circles. Click HERE to see me pointing to the engraving of the spiral on a recent visit to Basel, Switzerland.

MNEMONIC FOR THE NUMBER EXP(1): As you know there are an infinite number of irrational numbers arising in mathematics. The most famous of these are e, p and sqrt(2). Approximations to these numbers can be readily retained by the use of mnemonics. Usually the mnemonics are generated by counting the number of letters in a word. Thus for example p="How I like a drink, alcoholic of course, after the heavy lectures involving quantum mechanics"=3.14259265358979. A shorter mnemonic for p is"May(3) I(1) have(4) a(1) small(5) container(9) of(2) coffee(6)".You can also construct a mnemonic using other knowledge involving dates etc . I show you HEREa way to do this to 30 place accuracy for my favorite irrational number e.

AN INTERESTING ITERATION PATTERN GENERATED BY A COMPLEX NUMBER: Another interesting pattern involving a complex number in the Argand Plane( and one I have not seen before) is that obtained by  the iteration a[n+1]=i^a[n]starting with a=0 . We get a=i^0=1, a=i^1=i, a=i^i=exp(- p/2),etc. We have automated this iteration procedure with a one line MAPLE program and have plotted the results HERE for the first 40 iterations. Notice the interesting three arm spiral pattern generated with the large n limit given by a[n]=i^a[n] whose solution can be expressed in terms of the Lambert function and reads a[infinity]=(2i/ p )*Lambert( p/2i)=0.43828..+i*0.36059..

ROOTS OF ALGEBRAIC EQUATIONS: It is known that an nth order algebraic equation has n roots some of which may be real and others complex. Thus for example the second order equation ax2+bx+c=0 can be written as (ax2+bx+b2/4a)=b2/4a-c or x=[-b +- sqrt(b2-4ac)]/(2a). Thus if a, b, and c are real numbers, then the two roots are real if b2>4ac and complex conjugates if b2<4ac. For the cubic equation ax3+bx2+cx+d=0 things are a bit more complicated. We show you HERE its analytic solution based on some algebraic manipulations. It is also possible to analytically solve the quartic algebraic equation in closed form but not the quintic and beyond. For the roots of higher order algebraic equations it is best to simply used a canned numerical program . For example, using MAPLE, the  four roots of x4-x2+x-1=0 are the two real values x= 1 and x= -1.465571232,  and the two complex conjugate roots x=0.2327856159-.7925519930 I and  x= 0.2327856159+.7925519930 I . According to the Descartes Rule an algebraic equation has its number of positive roots equal to the number of sign changes or less by an even integer. In the last example we have three sign changes so that the number of positive real roots are either 3 or 1. As seen the number is one root in this case.

12TH WEEK Defining the function of a complex variable f(z)=u+iv. Cauchy-Riemann conditions for analytic functions. Some identities involving complex variables. Orthogonality of the curves u=const. and v=const. Complex velocity potential and some simple 2D inviscid flows described by it.

DEMONSTRATION OF THE ORTHOGONALITY OF THE u=Const and v=const CURVES: We have shown in class that for any  analytic function f(z)=u+iv,  the curves u=const and v=const form an orthogonal set of curves. Lets demonstrate this fact for the function f(z)=z*sin(z)=[xsin(x)cosh(y)-ycos(x)sinh(y)]+i[xcos(x)sinh(y)+ysin(x)cosh(y)]. The contour map obtained via MAPLE in -4<x<4,-4<y<4 can be found by clicking HERE. More on functions of a complex variable. Complex electrostatic and velocity potentials. Line integrals in the complex z plane. Cauchy's Theorem.

ELECTROSTATIC POTENTIAL USING COMPLEX VARIABLE METHODS: We know that the 2D electrostatic potential V(x,y) in a vaccuum is given mathematically by solving the Laplace equation . Since  both u and v of any analytic function f=u+i*v also satisfy the Laplace equation, one can define a complex electrostatic potential as F(z)=V(x,y)+iE(x,y) and consider the real part of any analytic function f(z) to be a possible solution for V(x,y) in a 2D electrostatic problem. We demonstrate this HERE for the potential between two parallel wires maintained at different constant potentials using the complex potential function F(z)=ln(z-1)-ln(z+1). There are an infinite number of other complex harmonic functions F(z), some of which can also be used directly to describe electrostatic problems for specified boundary conditions.

STREAMFUNCTION AND VELOCITY POTENTIAL USING COMPLEX VARIABLE METHODS: A 2D inviscid flow is characterized by having a velocity field (U,V) having both zero divergence and zero curl. This allows the definition of a complex velocity potential F(z)=j(x,y)+iy(x,y), where y is the streamfunction satisfying U=yy and V=-yx and j is the velocity potenetial satisfying U=jx and V=jy. Thus any analytic function f(z) can be considered to represent a possible 2D flow field. Certain forms have particular utility such as F(z)=±(Q/2p)ln(z), F(z)=±i(G/2p)ln(z), and F(z)=k/z which represent a source or sink, a clockwise or counterclockwise rotating vortex, and a doublet, in that order. We show you HERE the flow field created by superimposing a rectlinear flow F(z)=z and a doublet F(z)=1/z. The streamfunctions(in green) here are given by
y= Im(z+1/z)= (r-1/r)sin(q)  and the velocity potentials(in red) as j=Re(z+1/z)=(r+1/r)cos(q). The resultant pattern corresponds to inviscid flow about a cylinder.

WHO WAS CAUCHY? -Baron Augustin Louis Cauchy(1789-1857) was one of the most prolific mathematicians ever having written a total of 789 mathematical papers during his lifetime. His contributions spread over many branches of mathematics and he is especially known for development of the theory of complex variables. The Cauchy-Riemann conditions and the Cauchy integral theorem are named after him. He taught at the Ecole Polytechnique and the Sorbonne plus spent some time teaching in Turin and Prague. He won the Grand Prix of the French Academy of Sciences in 1816, helped Napoleon with the invasion plans for England(1810), was not well liked by his colleagues, and remained an ardent royalist for most of his life.(Lucky for him that he was only four years old during the reign of terror. The great French 51 year old chemist and discoverer of oxygen, Lavoisier, was not as fortunate). A picture of Cauchy is found HERE.

PROOF OF CAUCHY'S THEOREM: A complex function f(z)=u+iv is said to be analytic and hence have a unique derivative if it satisfies the Cauchy-Riemann conditions that u x =vy and v x =-u y . Let us now use these properties to prove Cauchy's famous theorem that the line integral around any closed curve in the z plane of f(z) is zero provided this function is analytic within C. The proof is based on Green's Theorem. We know from it that the double integral of (v x + uy) equals the line integral of (udx -vdy) around curve C. Likewise the double integral of ( u x -v y )corresponds to the line integral of (vdx +udy) around C. But since both of these double integrals vanish by the Cauchy-Riemann conditions , the line integral around C of f(z)dz=(u+iv)(dx+idy)=(udx-vdy)+i(vdx+udy) must be zero. QED.

13TH WEEK Cauchy Integral formula. Taylor series expansions. Residues.

EVALUATION OF A CLOSED LINE INTEGRAL FOR A FUNCTION WITH HIGHER ORDER POLES: We have shown in class how to evaluate closed line integrals involving functions with simple poles by means of the residue theorem. This residue approach continues to work for functions with higher order poles but in this latter case the evaluation of the residue involves a more complicated formula not worth committing to memory. Rather, it is better to make use of the extended Cauchy integral for such a higher order pole case. Let us demonstrate. Take the closed line integral K=Int[exp(z)/(z-1)3]. In this case the function f(z)=exp(z)/(z-1)3 has a third order pole at z=1, so treating the integral  by the Cauchy integral formula we have K=(2pi/2!)[exp(z)]" with the second derivative evaluated at z=1. We thus find K=i*pi*e. More on Taylor expansions. Laurent series and the Residue Formula.

HOW TO FIND A RESIDUE: In our discussions of solving closed line integrals in the complex plane where f(z)=g(z)/h(z) , with g(z) being analytic but h(z) having an n'th order zero at z=zo , we expanded the top and bottom term in a Taylor series about z=zo. This leads to f(z)=g(zo)+g(zo)'(z-zo)+../h(zo)'(z-zo)+[h(zo)"/2!](z-zo)^2+ with all terms in the h(z) expansion vanishing before the nth derivative term. We know that the only term in this quotient which yields a non-zero value when integated around closed curve about z=zo, is the one which goes as A/(x-xo) and that this term integrates to 2*Pi*i*A. Thus by definition, the residue Res will be A and we can conclude that for the special case of a first order pole (n=1) the residue is just Res=g(zo)/h(zo)' . For f(z)s with second and higher order poles, more complicated expressions for the residue exist. It is , however, more convenient in those instances to use the Cauchy Integral Formula for nth order poles directly.

14TH WEEK Evaluation of real integrals using complex variable methods. Filling out of teacher and course evaluation forms.

EVALUATION OF A REAL INTEGRALS BY RESIDUE METHODS: Click HEREandHERE. Also consider the integral J=Int[sin(t)4, t=0..2p]. If we convert it via the substitution z=exp(it) so that dz=izdt, we get the equivalent closed line integral Int[(z2-1)4/(16iz5)] around the unit radius circle about the origin. This can be evaluated via the Cauchy integral formula to yield (2p)/(4!16)[d4((z2-1)4)/dz4] at z=0. Thus J=3p/4, a well known result. Review for the Third Hour Exam . 15TH WEEK: THIRD HOUR EXAM covering everything in the complex variable area including those parts of Chapters 12 through 16 discussed in class. The exam will be closed book but you can bring one 3x5"card and your calculator.

Your final course grade , composed of 90% for the three tests plus 10% for the homeworks, will  be normalized to a maximum score of 100. The grade breakup will be approximately as follows:
90<A<100, 88<B+<90, 80<B<88, 78<C+<80, 67<C<78, 62<D+<67, 51<D<62.

PHOTO OF OUR FALL 1998  EGM4313 CLASS: Picture was taken with our old sub-megapixel digital camera and hence the poor resolution . Class is standing by the Stonehenge Sculpture behind the New Engineering Building. Links to our other Web Pages:

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To keep your mind sharp you might want to prove some of the following mathematical identities based on your knowledge of elementary mathematics:

(1)-  Sum[n^6*exp(-nx), n=1..inf.]={sinh(x)*[cosh(x)^2 + 28*cosh(x) +61 ]}/{2[cosh(x)-1]^4}

(2)-
Pi=48*arctan(1/38)+80*arctan(1/57)+28*arctan(1/239)+96*arctan(1/268)

(3)-  Iteration a n+1 =exp( ia n )  starting with a1=1 converges to a=x+iy= 0.576412727399..
+i 0.374699027157.. as n goes to infinity. Here x follows from
[x/cos)x)]^2=exp-2[x*tan(x)] and y=x*tan(x)

(4)- Int[t^2*sin(t)/(exp(Pi*t)-1),t=0..inf.] = 1+coth(1)-coth(1)^3

(5)- Pi= 12*arcsin[sqrt(C)/2]=(6sqrt(C)*Sum[{(2n)!*(C/16)^n}/{(n!)^2*(2n+1)}, n=0..inf.] ,  where C=2-sqrt(3)

(6)- Pi=24 arctan(B), where B=2*sqrt[2+sqrt(3)]-[2+sqrt(3)] which can be written as the continued
fraction- Pi=24sqrt(B)/{1+B/(3-B+9B/(5-3B)+25B/(7-5B)+..} and yields Pi good to eleven
places by taking terms up to 81B/(11-9B) into account.

(7)- Sum[(n^2-1)/(n^2+1)^2, n=1..inf]=[1-Pi^2*csch(Pi)^2]/2=0.443959960..

(8)- csch(x)^2=1/x^2-2*Sum[((n*Pi)^2-x^2)/((n*Pi)^2+x^2)^2,n=1..inf]

(9)- Sum[1/(n^2+a^2)^2, n=-inf..+inf]=[Pi^2/(2*a^2)]*[csch(Pi*a)^2+coth(Pi*a)^2/(Pi*a)]

(10)-Prove that tan(54°)=(1/5)*{sqrt[10+2*sqrt(5)]+sqrt[5+2*sqrt(5)]}. Hint-work with a regular pentagon  with sides of length one.

(11)-The area of a regular decahedron , each of whose ten sides equal unity,  is A=2.5*sqrt[5+2*sqrt(5)]. Also one has that
(f^2-1/4) < A/Pi < f^2, where f=(sqrt(5)+1)/2=1.608339.. is the golden ratio.

(12)-The odd number  2^(2n+1) +1 will be composite(ie.-non-prime) for all positive integer values of n.

(13)-The value of Pi is given exactly by Pi=4*int[1/sqrt((1-t^2)(2-t^2), t=0..1]*int[sqrt(1-t^2)/sqrt(2-t^2), t=0..1)] .

Your PC  can readily confirm that  these results are correct, but to come up with a derivation  or proof is a bit more challenging.
Finally, look at the following graph. Notice anything peculiar about it. Think number theory. Latest update:
July 28, 2016  