mm-466 === Subject: : Re: How to integrate x*e^(x^4)> x*e^(x^4)Using Mike Henry's substitution, u = x^2,this integral reduces to (1/2)int(e^(u^2)du,which is known to have NO elementaryantiderivative. Maybe the original problemwas a definite integral, which is different,and you did not give us the limits of integration.=== === Subject: : Re: HOW TO INTEGRATE X*e^(x^4) ??> Hi guyz, just wondering if anyone can help me integrate the function> x*e^(X^4), it x times e to the power of x to the 4.> ThanxTry substituting u=x^2.=== === Subject: : Re: Graph theory by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3ALgOv15254;> I just have a question about the solution that you have posted:> in this part where you have said> If v = 2, e > 0, then G is e-1 edge connected and e > (e-1)v/2> For induction step add vertex, thus is added at least k edges.> e + k > kv/2 + k > k(v+1)/2> I don't understand how you have replaced (e-1)v/2 by kv/2 in the> inequality.>That's in the wrong place. Put it after> If v = 2, e > 0, then G is e-1 edge connected and e > (e-1)v/2>As G is e-1 edge connected, then by definition k = e-1.> Since you have been so kind I was wondering if you can help onthese> two.....I really appreciate your help, thanx again>You're taking the class, I don't know anything about graph theoryexcept>what I learn solving problems posted by students.> If a k-chromatic graph G has a coloring in which each color is> assigned to at least two vertices, show that G has a k-coloring of> this type.>I'll skip on that unless the night is slow.> Show that if G is simple and 3-regular, then k=k'?>What's k' ?k'is the minimum size of a disconnecting set === === Subject: : Graph theory>Help!> 1.Show that if G is k-edge-connected,> then e(G)bigger than or equal to kv/2My previous proof was wrong. I leave to you to find the error.Also the definition I guessed you didn't correct me on.G is k-edge-connected when k is the smallest number of edge removals that disconnects GEquivallently: some removal of k edges disconnects G not removal of k-1 edges disconnects GIf G is k-edge connected then for all vertices v, k <= degree vkv(G)/2 <= (sum_v degree v)/2 = e(G)A simple connected 3-regular graph can be disconnected by removal of threeedges. What three edges? Hard part is to show it can't be disconnectedby removing two edges which I strongly suspect to be the case, that thegraph is 3-edge connected.=== === Subject: : Abstract algebra by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3ALgOI15243;How to construct a field with p^2 elements, where p is an odd prime?=== === Subject: : Re: Abstract algebra> How to construct a field with p^2 elements, where p is an odd prime?The same way one constructs the field GF(p^2) with p^2 elements,where p is the even prime:Adjoin to the field GF(p), with p elements, a root of anirreducible quadratic of GF(p)[x], its polynomial ring.When p is odd, the irreducible quadratic can always be taken tobe (x^2 + 1). So, for example, GF(9) can be taken to be the set{0, 1, 2, x, x+1, x+2, 2x, 2x+1, 2x+2} with addition andmultiplication mod 3 and mod (x^2 + 1).=== === Subject: : Re: Abstract algebra[...]> When p is odd, the irreducible quadratic can always be taken to> be (x^2 + 1). So, for example, GF(9) can be taken to be the set> {0, 1, 2, x, x+1, x+2, 2x, 2x+1, 2x+2} with addition and> multiplication mod 3 and mod (x^2 + 1).Oops, that's garbage. The irreducible quadratic can *not* alwaysbe taken to be (x^2 + 1). What *is* true is that since themultiplicative group GF(p)* of GF(p){0} is cyclic of even orderp-1, it has a cyclic subgroup of index 2 and so half of theelements of GF(p)* are non-squares r, so (p-1)/2 of the p(p-1)/2irreducible quadratics of GF(p)[x] are of the form (x^2 - r).=== === Subject: : Check my answer please - Laplace Transform One more question about Inverse Laplace TransformActually its not a question, I just wanted to see if someonecan check my answer and see if I am right..The problem is:Find the inverse laplace transform of:F(s) = 100s / (s^2 + 4)(s^2 + 2s + 10)Factoring (s^2 + 4) I got s = 2jFactoring (s^2 +2s + 10) I got s = -1+3jFor M1 One more question about Inverse Laplace Transform> Actually its not a question, I just wanted to see if someone> can check my answer and see if I am right..> The problem is:> Find the inverse laplace transform of:> F(s) = 100s / (s^2 + 4)(s^2 + 2s + 10)> Factoring (s^2 + 4) I got s = 2j> Factoring (s^2 +2s + 10) I got s = -1+3j> For M1 For M2 My final answer is:> F(t) = 13.87 sin (2t + 56.31) + 14.62 E^-t sin (3t - 15.26)Yikes! I've no idea what you have done but the Laplace of your answerhas a cubic in the numerator.The way I would do it is with partial fractions (again)100s/((s^2 + 4)(s^2 + 2s +10)) = (As + B)/(s^2 + 4) + (Cs + D)/(s^2 + 2s + 10).Add up RHS, equate numerators and equate coefficients of like powers ofs.You'll get A= 150/13; B=200/13; C = -150/13; D = -500/13. Write s^2 + 2s + 10 = (s + 1)^2 + 3^2 and go for it.I get the inverse Laplace to be (100/13)sin(2t) + (150/13)cos(2t) - (350/39)exp(-t)sin(3t) - (150/13)exp(-t)cos(3t).=== === Subject: : Re: Check my answer please - Laplace TransformCc: plsperry@sc.rr.com> My final answer is:> F(t) = 13.87 sin (2t + 56.31) + 14.62 E^-t sin (3t - 15.26)> Yikes! I've no idea what you have done but the Laplace of your answer> has a cubic in the numerator.How do you figure that, Paul? Your answer (as you'll see) isalmost identical to the answer you are criticising (only thephase angle of the second sine term is different).> The way I would do it is with partial fractions (again)> 100s/((s^2 + 4)(s^2 + 2s +10)) = > (As + B)/(s^2 + 4) + (Cs + D)/(s^2 + 2s + 10).> Add up RHS, equate numerators and equate coefficients of like powers of> s.> You'll get A= 150/13; B=200/13; C = -150/13; D = -500/13. Write > s^2 + 2s + 10 = (s + 1)^2 + 3^2 and go for it.> I get the inverse Laplace to be (100/13)sin(2t) + (150/13)cos(2t) -> (350/39)exp(-t)sin(3t) - (150/13)exp(-t)cos(3t).Now let's take that in pieces... (150/13)cos(2t) + (100/13)sin(2t) = (sqrt(150^2 + 100^2)/13)cos(2t - arctan(100/150)) = 13.88 cos(2t - 33.69deg) = 13.88 sin(2t + 56.31deg)And the other piece is: exp(-t)[(-150/13)cos(3t) + (350/39)sin(3t)] = exp(-t)[sqrt((-150/13)^2 + (350/39)^2)cos(3t - arctan((350/39)/(-150/13))) = exp(-t)[14.62 cos(3t - 142.13deg)] = exp(-t)[14.62 sin(3t - 52.13deg)]So Paul's answer expressed in the form the original poster is using is: F(t) = 13.88 sin(2t + 56.31deg) + 14.62 exp(-t) sin(3t - 52.13deg)Both answers agree on the term containing 2t, and both answers agreeon the magnitude of the term containing the 3t. However, the answersdisagree on the 3t phase angle. Given that they both have the 14.62,I imagine the original poster made a typo or quadrant error computingthe 15.26deg phase ange.=== === Subject: : Re: Check my answer please - Laplace Transform> My final answer is:> F(t) = 13.87 sin (2t + 56.31) + 14.62 E^-t sin (3t - 15.26)> Yikes! I've no idea what you have done but the Laplace of your answer> has a cubic in the numerator.> How do you figure that, Paul? Simple enough - I really _didn't_ recognize what he had done and Maplesaid the Laplace of his answer was a rational function with cubicnumerator.[...]=== === Subject: : Re: Check my answer please - Laplace TransformX-RFC2646: OriginalI found one place where I made a mistake. I didnt put the negative sign on -123.69When calculating M2 /_ Theta2 I did the following:100s / s^2 + 4 | where s = -1+3jThen I got this:100(-1+3j) / (-1+3j)^2 +4=(316.23 /_108.43) / (7.211 /_ -123.69)=43.85 /_ 232.12I am guessing that 232.12 deg would be the same as 52.13 deg?232.13 deg - 180 deg = 52.12 deg (slight rounding diff.)Is that correct?Steve, I-Net+Free PC Tech Support - http://www.webzila.comDLL Files - http://www.dllplanet.comLive Support - http://www.webzila.com/livesupport> My final answer is:> F(t) = 13.87 sin (2t + 56.31) + 14.62 E^-t sin (3t - 15.26)> Yikes! I've no idea what you have done but the Laplace of your answer> has a cubic in the numerator.> How do you figure that, Paul? Your answer (as you'll see) is> almost identical to the answer you are criticising (only the> phase angle of the second sine term is different).> The way I would do it is with partial fractions (again)> 100s/((s^2 + 4)(s^2 + 2s +10)) => (As + B)/(s^2 + 4) + (Cs + D)/(s^2 + 2s + 10).> Add up RHS, equate numerators and equate coefficients of like powers of> s.> You'll get A= 150/13; B=200/13; C = -150/13; D = -500/13. Write> s^2 + 2s + 10 = (s + 1)^2 + 3^2 and go for it.> I get the inverse Laplace to be (100/13)sin(2t) + (150/13)cos(2t) -> (350/39)exp(-t)sin(3t) - (150/13)exp(-t)cos(3t).> Now let's take that in pieces...> (150/13)cos(2t) + (100/13)sin(2t) => (sqrt(150^2 + 100^2)/13)cos(2t - arctan(100/150)) => 13.88 cos(2t - 33.69deg) => 13.88 sin(2t + 56.31deg)> And the other piece is:> exp(-t)[(-150/13)cos(3t) + (350/39)sin(3t)] => exp(-t)[sqrt((-150/13)^2 + (350/39)^2)cos(3t - > arctan((350/39)/(-150/13))) => exp(-t)[14.62 cos(3t - 142.13deg)] => exp(-t)[14.62 sin(3t - 52.13deg)]> So Paul's answer expressed in the form the original poster is using is:> F(t) = 13.88 sin(2t + 56.31deg) + 14.62 exp(-t) sin(3t - 52.13deg)> Both answers agree on the term containing 2t, and both answers agree> on the magnitude of the term containing the 3t. However, the answers> disagree on the 3t phase angle. Given that they both have the 14.62,> I imagine the original poster made a typo or quadrant error computing> the 15.26deg phase ange.> -- > Rich Carreiro rlcarr@animato.arlington.ma.us === === Subject: : Re: Vector Subspace test on polynomials by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3B1oDN11177;> Let p1(x) = 6 + 4x - 2x^2 - 8x^3 (so (6 + 4 - 2 - 8 = 0))> and I let p2(x) = 1 - x^3 (so (1 + 0 + 0 - 1 = 0))> Adding p1(x) + p2(x) gives me: 7 + 4x - 2x^2 - 9x^3> so my new a0,a1,a2,a3 for the result is 7,4,-2,-9>Expanding my understanding of polynomial addition (unless I've been>wrong all these years):>p1(x) + p2(x) = p3(x) = 7 + 4x - 2x^2 - 9x^3 >where p3(x)'s a=7, p3(x)'s b = 4, p3(x)'s c = -2 and p3(x)'s d = -9>and my conclusion was based on the notion that p3(x) is not in the>subspace that p1(x) and p2(x) is due to the sum of the coefficiantsof>p3(x) not equaling 0.>Yet if the answer sheet is saying that p3(x) *IS* in the subspace(where>the sum of the coefficiants of p3(x) = 0), then I apparently need a>crash course in polynomial addition. >Why the answer sheet is treating this as a function space rather thana>polynomial space is beyond my comprehension. Your math is excellent. Like any true mathematician, you'rearithmetic stinks!=== === Subject: : Re: Vector Subspace test on polynomials> Well 7 + 4 - 2 - 9 is not = 0.> 7 + 4 - 2 - 9 does equal 0. The book's answer is correct.I think it must have been those idiot pills (I mistook for vitamins)that I took this morning. I am very sorry about this. === === Subject: : Re: Vector subspace test on polynomials> Well 7 + 4 - 2 - 9 is not = 0.Since when?=== === Subject: : Calculus Problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3B1oG711233;do you guys can set up this problem ? I've already drawn the picture.I labeled Length of shodow by h and the distance of the man from thebase of the light x. A light ball is hanging from the floor 15 ft. I man 6 ft tall walksaway from the base of the light ball with a speed of 5 ft / sec. Whenthe man is 10 ft away from the base of the light ball ?a) Find speed of his shadow.b) what is the rate of change for the lengh of the shadow ?=== === Subject: : Re: Calculus Problem> I labeled length of shadow by s and the distance of the man from> the base of the light by x.> A light ball is hanging from the floor 15 ft. A man 6 ft tall walks> away from the base of the light ball with a speed of 5 ft / sec.> When the man is 10 ft away from the base of the light ball: | * | * 15 | |* | 6 | * | | * |------------ |-----------------* x sBy similar triangles, (x+s)/15 = s/6. 6x+6s = 15s 6x = 9s 2x = 3s 2(dx/dt) = 3(ds/dt)b) what is the rate of change for the length of the shadow ? 2(dx/dt) = 3(ds/dt) 2(5) = 3(ds/dt) ds/dt = 10/3 ft/secFind the speed of the end of his shadow.(d/dt)(x+s) = dx/dt+ds/dt = (5 +10/3)ft/sec.= (25/3) ft/sec.=== === Subject: : Sierpinski TriangleCan anyone help me with fractals? I have never heard of fractals untilI took this math course. I understand the square, cube, and snowflakefractal. I can't figure out a formual for the area of the SierpinskiTriangle. We're supposed to figure out a formula for perimeter too. Iknow that the perimeter of a triangle is just the 3 sides addedtogether. How can I calculate the N and R?=== === Subject: : Re: Sierpinski Triangle by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i3BHqWP18458;>Can anyone help me with fractals? I have never heard of fractalsuntil>I took this math course. I understand the square, cube, andsnowflake>fractal. I can't figure out a formual for the area of the Sierpinski>Triangle. We're supposed to figure out a formula for perimeter too. I>know that the perimeter of a triangle is just the 3 sides added>together. How can I calculate the N and R? To construct the Sierpinski Triangle, you start with one (solid)equilateral triangle, divide it into 4 equilateral triangles, bydividing each side into two equal parts, then throw away the centerone. Repeat with each of the three triangles, then repeat the wholeprocess infinitely. Notice that, after the first step, you have 3triangles. After the second step, since each of those triangles isdivided into three, you have a total of 9 triangles. In each step youdivide each triangle into 3 so you have 3 times the previous number:after n steps you have 3^n triangles. That's your N. To subdivideeach triangle you divide each side in two so the length of a side of atriangle is divided by 2- after n steps, each triangle has side length(1/2)^n times that of the original triangle. I think that's your R. IF we measured area in the normal way- proportional to a lengthsquared (since a triangle is a 2 dimensional figure), The area of eachtriangle would be k(1/2)^(2n)= k(1/4)^n for some k and there would be3^n of them: total area k(3/4)^n which goes to 0. Instead, we thinkof the figure as having dimension d so that area is proportional tolength to the d power: the area of each triangle, after n steps, isk(1/(2^d))^n and the total area is k(3/2^d)^n. That will be finite ifand only if 3/2^d= 1 or 3= 2^d. That requires that log(3)= dlog2 ord= log(3)/log(2). The fractal dimension (also called the Hausdorffdimension) of Sierpinski's triangle is log(3)/log(2) (between 1 and2). The Sierpinski triangle is so chopped up that it no longer hasdimension 2. Oddly enough, the Sierpinski triangle's boundary has the samedimension as the area- it is so convoluted that it's fractal dimensionis also log(3)/log(2), between 1 and 2.=== === Subject: : Re: Sierpinski Triangle> Can anyone help me with fractals? I have never heard of fractals until> I took this math course. I understand the square, cube, and snowflake> fractal. I can't figure out a formual for the area of the Sierpinski> Triangle. We're supposed to figure out a formula for perimeter too. I> know that the perimeter of a triangle is just the 3 sides added> together. How can I calculate the N and R?At the first stage you have a single triangle with perimeterP, say. You chop it up into four smaller triangles, each ofwhich is similar to the original triangle. In fact, eachsmaller triangle has sides that are exactly half as long asthe original sides, so each of these smaller triangles has aperimeter of P/2. You throw away the middle triangle,keeping the other three, so the total perimeter that remainsis 3*(P/2), or (3/2)*P.At the next stage you do the same thing to each of thesethree smaller triangles: now you have 9 triangles, and eachhas a perimeter that is half that of the little triangles atthe previous stage, i.e., P/4. The total perimeter istherefore 9*(P/4) = (9/4)*P = (3/2)^2 * P.In general, at each stage you multiply the number oftriangles by 3, but each new triangle has only half theperimeter of the old ones, so the net effect is to multiplythe total perimeter by 3/2.Nr. of triangles after N steps: 3^NPerim. of small triangles after N steps: (1/2)^N * PTotal perim. after N steps: 3^N * (1/2)^N * P = (3/2)^N * PYou want to reason similarly for the area: when you performone step, each new triangle is 1/4 of an old one, so it has1/4 the area of the triangles at the previous step.=== === Subject: : Re: Sierpinski Trianglehttp://www.efg2.com/Lab/FractalsAndChaos/ SierpinskiTriangle.htmIt seems the area is finite and its perimiter infinite.I hope that this helps.=== === Subject: : Re: Linear Algebra again!!> 1.(Simultaneous Diagonalization) Suppose that T and U are two normal> operators over the complex field on a finite dimensional vector space V.> If TU=UT. Prove that there exists an orthonormal basis for V consisting> of eigenvectors of both T and U.Hint: If |t> is an eigenvector of T, then so is U|t>, with the sameeigenvalue.> 2.Is there any alternative proof of the 'orthogonal diagonalizability' of> complex normal or real symmetric linear operators on a finite dimensional> vector space, other than the proof involving the use of Schur's lemma? In> fact, can we prove directly that complex normal or real symmetric linear> operators are diagonalizable at the first place?Choose an eigenvector |t> of normal T and extend to a basis ofV. In this basis all matrix elements T{21}, ..., T{n1} exceptthe first, T{11}, of the first column of T are zero, so allelements T*{12}, ..., T*{1n}, except the first, T*{11} = T{11}*,of the first row of T* are zero, while T*{21} = T{12}*, etc.Since TT* = T*T, |T{11}|^2 = |T{11}|^2 + |T{12}|^2 + ... +|T{1n}|^2, so T{12}, ..., T{1n} are all zero and T is blockdiagonal 1x1 + (n-1)x(n-1). Proceed by induction.