mm-1159 === Subject: Matrix algebra 1 - Consider the application alpha : R2[x]-E R3[x] defined by alpha (P(x)) = x^2P'(x),where P' it's the derivative of P a) Show that the operation alpha is linear application b)Determinate the matrix of alpha in relation to the base 1,2x,3x^2 from R2[x],and for the base 1,1+x,1+x+x^2,1+x+x^2+x^3 from R3[x] c)Ddeterminate the nucleus and the image of alpha. === Subject: Re: Matrix algebra A linear function (your linear application) must satisfy alpha(bP1(x)+ cP2(x))= balpha(P1(x))+ c alpha(P2(x)). Here, alpha(bP1)(x)+ cP2(x))= x^2(bP1(x)+ cP2(x))'= x^2(bP1'(x)+ cP2'(x))= b x^2P1'(x)+ cx^2P2'(x)= b alpha(P1(x))+ c alpha(P2(x)). Simplest way to do that is to apply alpha to each of the basis vectors in R2[x] and write the result in terms of the basis for R3[x}. Those become \ the columns of the matrix. alph(1)= x^2(1)'= 0= 0(1)+ 0(1+x)+ 0(1+x+x^2)+ 0(1+x+x^2+x^3). The first column consists of all 0's. alpha(2x)= x^2(2x)'= 2x^2. To write that in terms of the basis for R2[x] means to find A, B, C so that 2x^2= A(1)+ B(1+x)+ C(1+ x+ x^2)= (A+ B+ C)+ (B+C)x+ Cx^2, for all x. Equating coefficients of like powers, obviously C= 2, then B+ C= B+ 2= 0 so B= -2 and then A+ B+ C= A- 2+ 2= A= 0. 2x^2= -2(1+x)+ 2(1+ x+ x^2). The second column consists of 0, -2, 2, 0. alpha(3x^2)= x^2(3x^2)'= 6x^3. Now we must find A, B, C, D so that 6x^3= A(1)+ B(1+x)+ C(1+x+x^2)+ D(1+x+x^2+x^3)= (A+B+C+D)+(B+C+D)x+ (C+D)x^2+ \ Dx^3. Again equating coefficients of like powers D= 6, C+ D= C+ 6= 0 so C= -6, B+ C+ D= B- 6+ 6= B= 0, A+ B+ C+ D= A+ 0- 6+ 6= A= 0. The third column \ consists of 0, 0, -6, 6. The matrix is [0 0 0] [0 -2 0] [0 2 -6] [0 0 6] The usual English term is kernel which certainly could be interpreted as nucleus. It consists of all vectors in R2[x] that this linear function maps to 0. You could do that by looking at the definition of the function: alpha(f)= x^2f'= 0 for all x. Then f'= 0 so f is a constant: f= c(1). The kernel consists of the subspace of R2[x] spanned by 1: all constant polynomials. You could also do this using the matrix form: find all solutions to [0 0 0][x] [0] [0 -2 0][y]=[0] [0 2 -6][z] [0] [0 0 6] [0] (The spaces won't show properly on that!) It's easy to see that y= z= 0 while x can be anything: again, polyomials of the form c(1). The image is the subspace of R3[x] that is the result of something in R2[x]. It is easy to see that alpha(a+ 2bx+ 3cx^2)= 2bx^2+ 6cx^3. Again setting that equal to A(1)+ B(1+x)+ C(1+x+x^2)+ D(1+ x+ x^2+ x^3)= \ (A+B+C+D)+ (B+C+D)x+ (C+D)x^2+ Dx^2 and setting the coefficients of like power equal, \ we get A+ B+ C+ D= 0, B+ C+ D= 0, C+ D= 2b, and D= 6c. Then C+ D= C+ 6c= 2b so C= 2b- 6c, B+ C+ D= B+ 2b- 6c+ 6c= B+ 2b= 0 so B= -2b and A+ B+ C+ D= A- \ 2b+ 2b- 6c+ 6c= A= 0. The image is the subspace spanned by (-2b)(1+x)+ (2b-6c)(1+x+x^2)+ 6c(1+ x+ x^2+ x^3). That is two dimensional since it depends on the two numbers b and c. A basis for the image can be gotten by taking b= 1, c= 0: -2(1+x)+ 2(1+ x+ x^2)= 2x^2 and then tking b= 0, c= 1: -6(1+x+x^2)+ 6(1+x+x^2+x^3)= 6x^3: the image is the subspace of R3[x] \ spanned by 2x^2 and 6x^3. === Subject: Re: Matrix algebra c) Ker(alpha)= 1 a) alpha (x+y) = x^2 p'(x) + y^2 p'(x) alpha (p'+p'x+p'x^2)+(p'+p'x+p'x^2) = (p'+p',p'x+p'x,p'x^2+p'x^2) = (p'+p'x+p'x^2)+(p'+p'x+p'x^2) = Alpha (x+y) Alpha(delta*x) = delta x^2 p'(x) = alpha p'x^2(x) Results from well known properties of derivatives,that alpha (P(P(X)) it's a linear application THIS IS WHAT I HAVE. === Subject: Re: maxima / minima problem : 2 variables Certainly. What partial derivatives do you have, and how did you get them? -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Re: maxima / minima problem : 2 variables So I redid my partials and i got the right answer, which is q* = 89.44 and s* = 22.36 . So the second part of the question asks, for each stationary point, is C(q*, s*) a max or a min? local or global? I set up the Hessian matrix, and took the determinant (which I got (96000 - 48s^2)/q^4 ), and from that I determined that it is a local maximum, because at some point the determinant becomes negative, but for all the other s,q combos it's positive. === Subject: Re: maxima / minima problem : 2 variables For f(q, s) = 6000/q + 3s^2/q + (q-s)^2/q, the q-partial is (q^2 - 4s^2 - 6000)/(q^2) and the s-partial is (8s - 2q)/q. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Number of solutions If I am not mistaken, you can't distinguish between those cases from the determinant alone. Roughly, if the coefficients of one row are all the same multiple of the coefficients of another row, then either (a) the constant terms are related by the same multiple, in which case those two rows are redundant and there are an infinite number of solutions, or (b) they are not, in which case those two rows are contradictory and there are no solutions. This is a lot easier to do through Gauss-Jordan elimination, IMHO. http://aspire.cs.uah.edu/textbook/gauss.html -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Re: Number of solutions Thats right... all the determinant tells you is if you have a unique solution or not. The OP needs to put the system into Echelon form to tell what type of solution the system has === Subject: Akhila Raman posted: Fourier curio... i noticed F.T. has its share of curiosity: http://mathworld.wolfram.com/FourierTransform1.html let us do this without using duality property. we want the Fourier transform of f(x)=1: if we wish to argue F(k)= delta(k), then by definition of delta function, for k # 0, F(k)=0. [ http://mathworld.wolfram.com/DeltaFunction.html ] that means for k # 0, is not quite right by the definition of sinusoidal function. it should remain oscillating between [-1,1] as -Akhila Raman === Subject: Re: Akhila Raman posted: Fourier curio... If you want to post the same question in multiple newsgroups, then please crosspost, do not multipost. You have gotten answers to this question in sci.math === Subject: River problem I hope this question is appropriate for this forum: A river flows east at 3km/h. A swimmer, starting from the south shore can swim at 2 km/h. In which direction should the swimmer head in order to travel the least resultant distance? I have used a simple vector navigational diagram starting with the river vector, then head to tail with the swimmer's vector somewhat upstream so that it meets head-to-head the resultant velocity vector at 90 degrees. This gives me a nice right triangle with a resultant velocity of 2.24km/h and the swimmer's heading North 42degrees West. My question: Is it possible to solve this as a calculus problem, such as a rate problem, to get the maximum resultant angle without assuming a right angle? Robecall === Subject: Re: River problem Yes. One must, in this process, take the time of crossing into account. The minimal time occurs when the swimmer heads directly across and ignores the current, but then the distance covered is not necessarily minimal, as the current sweeps the swimmer along a longer than necessary path. So one must balance the direction of crossing to minimize the effect of the current. Let the river run from left to right with river current given by the vector Vr = [3,0] and the swimmer's velocity vector be given by Vs = [2*cos(theta), 2*sin(theta)] with theta = 90.bc for swimming directly across and betwen 90.bc and 180.bc for swimming against the current. If the width of the river is W, then the time of crossing will be T = W/(2*sin(theta)), as 2*sin(theta) is the component of swimming velocity perpendicular to the river. The distance covered in crossing vector will be T*(Vc + VC). The square of the length of that crossing vector will have zero derivative with respect to theta when the distance is minimal. I leave the other details to you. === Subject: Re: River problem can a You have made an old man very happy :-) Your suggestion to use the SQUARE of D made all the difference , leading to a simple cosine quadratic. I had been working just with D, and a consequent nightmare trig equation. Robert Callcott === Subject: Re: River problem I don't see any reasonable way to do so nor do I see any reason to want to. This simply isn't as hard as a calculus problem! There is no rate of change involved. === Subject: Re: Integral operator? Maybe it is a noncense,but you have to prove it.You can calculate this eliptic integral of the first kind then you post the result. I did a debuging to this operation and I understood that it can be compared with I... 1/sqrt (1-x),I also understood that i don't have this \ operation.If limt is involved it's the convergence of the integral we're trying to \ prove. It's a convergent one. But I'm not too concerned about it as you seem to be. I don't belong to an organization where integral operations are a reflex that shows primal interest in mathematics. Successions ,perhaps... I don't plan to extract my sorrows from an hanging rope. === Subject: Re: Integral operator? You started by asking for help. You were given help and complained that you did not want to do the problem that way. Now you are just being insulting. === Subject: Re: Integral operator? c) It's in the form Integral [from a to b) dx/[(b-x)^k] = Limit[x-E b-] of integral [from a to x] dx/[(b-x)^k] In this way, Lim [alpha -E 1-] (1/(1/2))(1-x^4)^(1/2),with the superior limit of integration equal to alpha and the lower equal to 1/2. = -1/2 Lim [x-E 1-] (1-x^4)^(1/2) - (1/2)(1/2)^(1/2) = -1/2(0-sqr2/4) = Square root of two over eight = 1/2 === Subject: Re: Integral operator? This is not the form of the integral you asked about. What you have written is nonsense. You said the problem is to evaluate the integral Int[1/2..1] dx / sqrt(1 - x^4) This integral can be done, and limits are not involved. If the problem you have been asked to solve is something different, then you need to say exactly what that original problem is. === Subject: Re: Integral operator? This is part (c) from your original post. What do you mean by calculate operationally and integrally transformed? Please give a clear example. Do you want the integral solved in terms of the usual mathematical functions that are involved for integrals of this type? If not, then what are you really asking for? If this is homework for a class, what is the name and subject matter of the course? === Subject: Re: Succession series? b) fn = { - n^2/(n^2+1) { n^2/(n^2+1) In this way it's divergent with no limit. It can't be written the serie of modules. or, Lim fn = 1 By Leibniz it's divergent. === Subject: Theory of Equations - Solutions for Quartic Equations i am revising my degree math from a book called Further Engineering Mathematics by K A Stroud. I have come to this problem: x^4-4x^3+8x+3 This should be reduced by x=y+1 Still I have not solved it. Can anyone help out. Anton (engineer, 51 from Malta EU) acutajar@euroweb.net.mt === Subject: Re: Theory of Equations - Solutions for Quartic Equations If you substitute y + 1 for x and do all the algebra you will get y^4 - 6y^2 + 8 = (y^2 - 4)(y^2 - 2). However, I wonder about the word reduced. The usual way to work this problem is to see if any of the divisors of 3, 1, -1, 3 and -3 are roots. It turns out -1 is a root so x - (-1) = x + 1 is a factor. Using ordinary long division, (x^4 - 4x^3 + 8x + 3)/(x + 1) = x^3 - 5x^2 + 5x + 3. Again checking 1, -1, 3 and -3 gives 3 as a root. Divide again (x^3 - 5x^2 + 5x + 3)/(x - 3) = x^2 - 2x -1. Now use the Quadratic Formula on x^2 - 2x -1. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Theory of Equations - Solutions for Quartic Equations Substitute y+1 for x and then expand the equation, collect terms. You should end up with an equation in y that you can solve in terms of y^2. This will lead you to some roots of the original: Let z = y^2: z^2 - 6z + 8 roots are: z = 2, z = 4 So y = +/- sqrt(2) and y = +/- 2, and thus x = sqrt(2) + 1 x = -sqrt(2) + 1 x = -1 x = 3 === Subject: Re: Theory of Equations - Solutions for Quartic Equations You are right- the substitution x= y+ 1 reduce it to the bi-quadratic equation y^4+ 6y^2+ 11= 0. Let z= y^2 to reduce to the quadratic z^2+ 6z+ 11= 0. You can solve that with the quadratic formula (solutions are complex), then solve y^2= z to get four values for y and then used y= x+ 1 \ to find x. === Subject: linear equation How would I go about solving the following? An item costs $900, has a scrap value of $50, and a useful life of five years. The linear equation relating book value and number of years is a. BV = -50x + 850 b. BV = -50x + 900 c. BV = -170x + 850 d. BV = -170x + 900 === Subject: Re: linear equation if there is a choice between several linear functions for the solution, you can just choose the one with BV(x=0) = 900 BV(x=5) = 50 because a line is uniquely determined by two given points. If the linear functions were not given explicitly, just set BV(x) = a*x + b and determine a and b by the conditions Best wishes Torsten. === Subject: Inverse of matrix Is my answer correct? Find the inverse of 1 -1 0 0 1 1 1 0 -1 My answer is 1/2 1/2 1/2 -1/2 1/2 1/2 1/2 1/2 -1/2 === Subject: Re: Inverse of matrix days. My association with the Department is that of an alumnus. Multiply them out and see! -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes by Bill Watterson) Arturo Magidin magidin-at-member-ams-org === Subject: k12.ed.math? I posted there maybe 7 years ago. It used to be very active. It appears to be non-existent now. Has that group been replaced?? Basically, I'm looking for a math newsgroup with high school emphasis Mel Mel === Subject: Re: what is the average digit of decimal expansion of pi, 7? reply-type=response I agree that if pi is normal (which we have assumed), then there is probability 1 that the string 000000 appears in the expansion. On average, it will appear in (say) position million in the decimal expansion of pi, as it is about a one-in-a-million sequence. Lets say the running average up to that point is 4.5. Then the new average after the string is (4.5 * 1,000,000)/1,000,006 = 4.499973 ... You can also find a string of 10 9 characters, but it will probably appear somewhere of the order of 10^10 places into pi. The effect on the running total will be a factor of (4.5 * 10^10 + 90)/(4.5* 10^10 + 10) - a string \ of 10 9 characters at the end of a list of 10^10 characters isn't going to change the total or average very much at all. So the distance you have to go to find the string grows exponentially with the length of the string, but the effect on the total grows linearly. You are on a hiding to nothing. I looked at the first 26 digits of pi, and from the 16th position all \ values of the running average are between 4.0 and 5.0 Based upon this, I can make prediction which I am pretty confident about. The minimim value for the average digits in the sequence 1, 4, 1, 5, 9, ... is 1, which occurs at the first position. The largest value is 5.28, which occurs at position 14. The running average is always between 4.0 and 5.0, excepting for n = 1,2,3,4,5,6,14,16. For any epsilon, there are only finite n where the average of the first n digits lies out 0.45 plus or minus epsilon. (Note the similarity of this problem to the asymmetric 1D random walk problem). If the long sequence was a sequence on 9s, then the But m will be very much greater than n. If you want to find a string of n 0 characters, you will typically need m to be of the order of 10^n. With that approximation, the average remains b. Adding 6 zeroes to the end of a sequence of digits of length 10^6 isn't going to change the average by much. === Subject: Re: what is the average digit of decimal expansion of pi, 7? m will *usually* be very much greater than n. However, it will not *always* be greater than n and if you search long enough you should be able to find sequences of bs which are very much longer than the 'head' of the string. Cliff -- Have you ever noticed that if something is advertised as 'amusing' or 'hilarious', it usually isn't? === Subject: Re: what is the average digit of decimal expansion of pi, 7? reply-type=response Why do you think that is true? I doubt it very much. Whilst I can't *prove that the first googol places which may have average 4.49999999999..., aren't immediately followed by a googol 0s in a row dragging the average down to 0.25, it would strike me as most odd to see a googol zeroes in row as part of a random string of length 2 * googol. So it is possible but extremely unlikely that there are more than (say) 20 positions where the rolling average isn't between 0.4 and 0.5, and in fact \ I think I have found them all, as well as the highest average that is ever achieved (5.28). I am pretty confident you cannot prove your assertion, because it appears clearly wrong. === Subject: Re: what is the average digit of decimal expansion of pi, 7? Why do you think that it isn't? I can't think of any reason why m is constrained to be greater than n. A googol is a very small number in terms of the lengths of the sequences that I am talking about. Vanishingly small, in fact. At some stage the decimal expansion of pi will repeat the whole of the preceding sequence of numbers up to that point. That is also 'odd'. It will also at some stage repeat the sequence of digits up to that point more than once. You can think of all sorts of 'odd' occurrences and will be sure of coming across them, if you search long enough. You cannot have found them all. The sequence is infinite, so unless you can prove that there are only a finite number of times that the average isn't between 0.4 and 0.5, then it is extremely likely that there are other occasions when the average will be outside those limits. I believe it to be true. Cliff -- Have you ever noticed that if something is advertised as 'amusing' or 'hilarious', it usually isn't? === Subject: Re: what is the average digit of decimal expansion of pi, 7? reply-type=response I gave you a reason, based upon the probability of a string of low numbers decreases exponentially, but the number of opportunities increases only linearly. But that is hardly relevant. Your proof requires n<