mm-405 === Subject: : Re: integration problem |>Well, first of all you can do a change of variables to eliminate one |>of the parameters: say take x = b t and c = b/a to get |>H = c^2 a^2 int_0^infinity 2^(-(c*t)^1.4) * 2^(-t^1.8)*t dt |>In the integral, it's convenient to take q = c^1.4. Then the integral |>can be written as a series in powers of q: according to Maple Oops, as Nils pointed out, I substituted the wrong power: it should be 2 2 (17/5) (24/5) H = a (0.7905369639 c - 0.7371995175 c + 0.5336147684 c (31/5) (38/5) 9 - 0.3408027385 c + 0.2005622769 c - 0.1111111111 c (52/5) (59/5) + 0.05868375153 c + O(c )) This can be obtained as follows: H1:= Int(2^(-q*t^p)*2^(-t^(9/5))*t,t=0..infinity); H2:= series(H1, q, 7); H3:= a^2*c^2*evalf(subs(q=c^(7/5),H2)); And then you can solve for c as a Puiseux series in powers of h = H/a^2: The most straightforward way to do this, with series and RootOf, seemed to take forever, so I did the following (after convincing myself that the answer should involve h^((1+7/5 k)/2) for k=1,2,...) eq:= eval(H3/a^2-h,c=add(a[k]*h^((1+7/5*k)/2),k=0..6)); S:= series(eq, h, 7); eqs:= subs(h=1,convert(convert(S,polynom),set)); Sol:= solve(eqs union {a[0] > 0}); C:= subs(Sol, add(a[k]*h^((1+7/5*k)/2),k=0..6)); /19 |--| 1/2 (6/5) 10/ C := 1.124705734 h + 0.6181960814 h + 0.4578965088 h /33 |--| (13/5) 10/ + 0.3714790777 h + 0.3147566703 h /47 |--| 4 10/ + 0.2735208334 h + 0.2416346379 h === Subject: : Re: integration problem The most straightforward way to do this, with series and RootOf, seemed to take forever, Yes this is exactly what I experienced, no resulst after 2 hours on my 1,8GHz machine - I'm only used to this from numerics >;-> Thank you so much for your help!!! Nils === Subject: : Re: saving 'worksheets' | MuPAD Light or Maxima If you are searching for a free clone of maple, I would however recommend you to test my own GPL software giac/xcas, it has an interface that can save worksheets, and for simple scripts you can even import and run maple worksheets. More info at http://www-fourier.ujf-grenoble.fr/~parisse/giac.html N.B.: giac/xcas is a work in progress, there are certainly some bugs here and there, and the windows binaries are a little bit old (they will be updated at the beginning of September so that Windows user will benefit from a more user-friendly history). I will try Xcas, as per your suggestion. Since the teaching labs all run Windoze, I'll wait until September. === Subject: : Re: saving 'worksheets' | MuPAD Light or Maxima > My students will be working with fairly long expressions, and it would be nice > (translation: they won't hate me) if there is a way for them to store > 'worksheets' such that when they want to re-start work on the project, they can > simply open up the worksheet, and continue from where they left off (I'm In MuPAD, try write(save.mb) and later read(save.mb) realized the 'slashes' separating directories have to be /, and not . Any way that can be changed? Or, as good, any way the default write directory can be specified somewhere? === Subject: : Re: saving 'worksheets' | MuPAD Light or Maxima realized the 'slashes' separating directories have to be /, and not . You can use backslashes -- you just have to quote them properly, just like in a C program: write(c:tmpwhatever.mb) should work. Any way that can be changed? Or, as good, any way the default write directory can be specified somewhere? As ?write tells you, you can: Set WRITEPATH to the value you want. === Subject: : Re: saving 'worksheets' | MuPAD Light or Maxima Couple of things I should have mentioned: 1. the platforms the students are using are Windoze based, meaning i) some of the suggested approaches are cumbersome at best (even with Cygwin) ii) most students will have difficulty with some of the esoterica (can't see a typical student using awk, for example, or perl). 2. what is handy about MAPLE is that is save an 'executable' worksheet. MuPAD Pro has the equivalent, but again - not free for the Light flavour. === Subject: : Re: converting angular & linear velocity to x- and y- coordinates Your information was most helpfull, thank you === Subject: : Re: plot::Pointlist (mupad) What am I doing wrong? I would like to plot x(n) = sum(a^(n-i)/i,i)-a^n*x0 for a=1 and x0 = 1 and n = 1,2, 3,...: sum(a^(n-i),i) = psi(i), which depends on i, for which you did not give a value. See ?sum for why psi(i) is the correct answer to your input. If you really mean the sum for i from 1 to n or something like that, you should say so: > a := 1; x0 := 1; > PlotPoints := plot::Pointlist([n,sum(a^(n-i)/i,i=1..n)-a^n*x0] $ n=1..10) +--+ +--+| |+-|+ Christopher Creutzig (ccr@mupad.de) +--+ Tel.: 05251-60-5525 === === Subject: : Re: Converting Radians to Degrees in Maple 9 I just bought Maple 9 the other day and have noticed that it returns answers in radians instead of degrees. So, evalf(sin(90)); returns the answer 0.89399666 etc. instead of 1. I've search the Learning Guide and the Help file on Maple but cannot figure out how to change from radians to degrees. If someone knows how, would you please If d = measure of an angle in degrees, and r = measure of the same angle in radians, then d=(180/Pi)*r or r=(Pi/180)*d . So, for example, sin(90 degrees) = sin((Pi/180)*90) = sin(Pi/2) G. A. Edgar http://www.math.ohio-state.edu/~edgar/ angle in radians, then d=(180/Pi)*r or r=(Pi/180)*d . So, for example, sin(90 degrees) = sin((Pi/180)*90) = sin(Pi/2) Okay, that works. Basic trig. Is there any way to make degrees the default instead of radians? That's what I'm trying to figure out. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Cindy Smith Unless the LORD build the house, cms@dragon.com they labor in vain who build. cms@5sc.net Unless the LORD guard the city, cms@romancatholic.org in vain does the guard keep watch. Me transmitte sursum, -- Psalm 127:1 Caledoni! All your base are belong to us. A Real Live Catholic You are on the way to destruction. in Georgia! What you say. ->> <<-< Go against the flow! You have no chance to survive make your time. === Subject: : Re: Converting Radians to Degrees in Maple 9 |>> If d = measure of an angle in degrees, and r = measure of the same |>> angle in radians, then d=(180/Pi)*r or r=(Pi/180)*d . |>> So, for example, sin(90 degrees) = sin((Pi/180)*90) = sin(Pi/2) |>Okay, that works. Basic trig. Is there any way to make degrees the |>default instead of radians? Not really. But you can define your own functions, for example Sin:= x -> sin(Pi/180*x); === Subject: : New Logic Calculus enhances George Spencer Brown In the site http://multiforms.netfirms.com a new logic calculus is described which enhances the logic calculus of George Spencer Brown (in Laws of Form). Moreover, there is a theorem-proving engine (an algorithm based on bit-pattern manipulation) which uses Multiple Form Logic(tm) to process logical queries in a class of Logic Formulae which correspond to (simple) Expert Systems. This is the Iphigenia Algorithm, outlined in http://multiforms.netfirms.com/iphigenia.html (together with a downloadable demo program). Multiple Form Logic is relevant to... Zen because (like Laws of Form) it derives all conventional Propositional Logic from three axioms which are based on Experience: (1) The Law of Oneness, (2) The Law of Reflection and (3) The Law of Perception. George A. Stathis http://www.geocities.com/omadeon === Subject: : Re: help can someone explain (to a complete ignoramus) what the symbol i means in this function: http://functions.wolfram.com/ElementaryFunctions/ArcCos/02/0001/MainEq1.L.gi f Well, for things on the Wolfram functions web site, you can download a .PDF describing all of the notation used at: http://functions.wolfram.com/Notations/ G. === Subject: : Re: help > can someone explain (to a complete ignoramus) what the symbol i means in > this function: http://functions.wolfram.com/ElementaryFunctions/ArcCos/02/0001/MainEq1.L.gi f > where can i find the most comprehensive list of math symbols on the net ? > what is the fastest arccos function for aplication in CG? > thanks I strongly recommend getting hold of Abramowitz & Stegun, Handbook of Mathematical Functions (Dover...). It defines various trig and inverse trig functions. The definition you cite is for extension of arccos(z) into the complex z-plane. I sincerely doubt that Visual Basic has complex arithmetic built in. (But I may be wrong.) The only languages I know that do complex arithmetic in a simple and natural way, and that have working function libraries are Fortran, Python, Forth, C99, C++ . Probably there are complex extensions for Lisp, Scheme, Smalltalk, etc. As I have been told recently, the latest versions of gcc support complex arithmetic and complex-valued functions. According to the on-line FAQ's, however, parts of are broken. I do not know which are in working order. -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ === Subject: : Re: help The only languages I know that do complex arithmetic in a simple and natural way, and that have working function libraries are Fortran, Python, Forth, C99, C++ . Probably there are complex extensions for Don't forget about languages like PARI/GP, Mathematica, Maple, and MuPAD. As I have been told recently, the latest versions of gcc support complex arithmetic and complex-valued functions. According to the That's probably because the C standard from '99 supports them. :-) +--+ +--+| |+-|+ Christopher Creutzig (ccr@mupad.de) +--+ Tel.: 05251-60-5525 CC: Robert Israel === Subject: : Re: residue in MAPLE Out of curiosity - does anyone from Maple monitor these exchanges and eventually fix the bugs? Mike Milgram |>Does anyone know how to convince MAPLE (version 6) to properly calculate |>the residue of a gamma function |>at a general negative integer? |>The following shows that it knows where the singularities are, but |>doesn't get the residue correctly. |>Neither will series(a,s=k) give the correct result. |>Of course residue(a,s=6); works correctly for any specific integer. |>> restart;assume (k,posint); ... |>> residue(a,s=k); Maple 8 and 9, btw, return unevaluated here, and have a division by zero if you use series. A work-around is to use the identity residue(ap,s=k); k~ k~ (-1) z - ------------ 400 === Subject: : Re: Last question on a Christmas Quiz I am doing....... Thank you Tomas! Merry Xmas!! Simon === Subject: : calculate radius I have a pie shape lot with a house sitting in the middle of the lot. I need to layout the back radius for the installation of a bulkhead. The radius is 200 feet and I can post the 2 back corners of the radius. I was thinking about running a string from each corner post (cord) and then measure from the string outward to the appropriate distance every 10' feet to stake out the radius. I am not sure how to calculate this distance from the cord outward to this radius. Not sure if this is the right approached. Any help would be greatly appreciated. Nick === Subject: : Re: calculate radius I have a pie shape lot with a house sitting in the middle of the lot. I need to layout the back radius for the installation of a bulkhead. The radius is 200 feet and I can post the 2 back corners of the radius. I was thinking about running a string from each corner post (cord) and then measure from the string outward to the appropriate distance every 10' feet to stake out the radius. I am not sure how to calculate this distance from the cord outward to this radius. Not sure if this is the right approached. I don't fully understand your question, probably because I'm confused by your terminology. A radius is a line from the center of a circle to a point on the outside of the circle, or it is the distance from the center to the outside. You don't have a distance to a radius. Are you trying to find the center? Since you know the radius is 200 feet, and you have two points on the outside of the circle, tie separate 200-foot strings to stakes at those two points. Then you and a friend walk in the general direction where the center should be, until the strings are fully stretched. There is only one point in that general direction where the two of you can meet and both strings are fully stretched; that is the center. If I've misunderstood your question, please try again and state more clearly what you're trying to ask. Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Modern cyberspace is a deadly festering swamp, teeming with dangerous programs such as viruses, worms, Trojan horses, and licensed Microsoft software that can take over your computer and render it useless. --Dave Barry === Subject: : Re: calculate radius As Stan says below, there is terminological confusion. But I think I've guessed what you've meant to say. >I have a pie shape lot with a house sitting in the middle of the lot. >I need to layout the back radius for the installation of a bulkhead. By back radius I must suppose you mean an arc of a circle. >The radius is 200 feet and I can post the 2 back corners of the radius. You can post the two points at the ends of the desired arc. >I was thinking about running a string from each corner post (cord) The string itself might well be called a cord, but in relation to the circle, it would then represent what is called a chord (not cord). >and >then measure from the string outward to the appropriate distance every >10' feet to stake out the radius. >I am not sure how to calculate this distance from the cord outward to >this radius. Not sure if this is the right approached. It will work, more or less. I will give you an approximation of the desired arc of a circle, using straight line segments. It's a pity that the house is already sitting there! Otherwise, you could have gotten your desired arc of a circle very neatly by first locating the center of the circle, as Stan said below, and then just walking along the arc, holding taut a 200-foot-long string whose other end was fixed at the center. But the presence of the house precludes this approach. Anyway, to answer your question about how far the arc of the circle is from its chord, at various places along the chord, we need one more bit of information: the length of the chord, that is, the straight-line distance between the 2 back corners. I don't fully understand your question, probably because I'm confused by your terminology. A radius is a line from the center of a circle to a point on the outside of the circle, or it is the distance from the center to the outside. You don't have a distance to a radius. Are you trying to find the center? Since you know the radius is 200 feet, and you have two points on the outside of the circle, tie separate 200-foot strings to stakes at those two points. Then you and a friend walk in the general direction where the center should be, until the strings are fully stretched. There is only one point in that general direction where the two of you can meet and both strings are fully stretched; that is the center. If I've misunderstood your question, please try again and state more clearly what you're trying to ask. === Subject: : Re: calculate radius Sorry about the wrong terminology. The survey shows an arc length of 174.65' and a cord length of 169.15' === Subject: : Re: calculate radius Sorry about the wrong terminology. That's OK. The survey shows an arc length of 174.65' and a cord length of 169.15' Excellent. Then here's an answer for you. You were wanting distances from the chord to the arc, to be measured perpendicular to the chord at various places along it. Although I think that you had intended those places to be at multiples of 10' from an end of the chord, it will actually be slightly easier (and more symmetrical) if you first locate the center of the chord. Letting x be the distance along the chord from its center, the perpendicular distance to the arc there is, in feet, Sqrt( 200^2 - x^2 ) - 181.24 where Sqrt means square root and x^2 means x squared. As examples: the arc is, in feet, 200 - 181.24 = 18.76 . perpendicular distance to the arc is, in feet, Sqrt( 200^2 - 10^2 ) - 181.24 = 18.51 . perpendicular distance to the arc is, in feet, Sqrt( 200^2 - 80^2 ) - 181.24 = 2.06 . I hope you get the idea. === Subject: : Re: calculate radius David, Wow, after looking at your solution, it seems so simple. One stupid question, where did the 181.24 come from? Thank you! >Sorry about the wrong terminology. That's OK. >The survey shows an arc length of 174.65' and a cord length of 169.15' Excellent. Then here's an answer for you. You were wanting distances from the chord to the arc, to be measured perpendicular to the chord at various places along it. Although I think that you had intended those places to be at multiples of 10' from an end of the chord, it will actually be slightly easier (and more symmetrical) if you first locate the center of the chord. Letting x be the distance along the chord from its center, the perpendicular distance to the arc there is, in feet, Sqrt( 200^2 - x^2 ) - 181.24 where Sqrt means square root and x^2 means x squared. As examples: the arc is, in feet, 200 - 181.24 = 18.76 . perpendicular distance to the arc is, in feet, Sqrt( 200^2 - 10^2 ) - 181.24 = 18.51 . perpendicular distance to the arc is, in feet, Sqrt( 200^2 - 80^2 ) - 181.24 = 2.06 . I hope you get the idea. David === Subject: : Re: calculate radius David, Wow, after looking at your solution, it seems so simple. One stupid question, where did the 181.24 come from? Thank you! You're welcome. Sorry I forgot to explain more. To answer your question above: If we think of the center of the circle as being at the origin, then we get We then need to find y at the ends of the arc (which is the same as y all along the chord) so that we can subtract it from values of y on the circle, giving then the heights from the chord up to the circle. Replacing x by half of the length of the chord, we get y = Sqrt( 200^2 - (169.15/2)^2 ) = 181.24 all along the chord. Thus, from the chord up to a general point on the circle, the height is Sqrt( 200^2 - x^2 ) - 181.24 . >>Sorry about the wrong terminology. > That's OK. >>The survey shows an arc length of 174.65' and a cord length of 169.15' > Excellent. Then here's an answer for you. > You were wanting distances from the chord to the arc, to be measured > perpendicular to the chord at various places along it. Although I think > that you had intended those places to be at multiples of 10' from an > end of the chord, it will actually be slightly easier (and more > symmetrical) if you first locate the center of the chord. Letting x be > the distance along the chord from its center, the perpendicular > distance to the arc there is, in feet, > Sqrt( 200^2 - x^2 ) - 181.24 > where Sqrt means square root and x^2 means x squared. > As examples: > to the arc is, in feet, 200 - 181.24 = 18.76 . > perpendicular distance to the arc is, in feet, > Sqrt( 200^2 - 10^2 ) - 181.24 = 18.51 . > perpendicular distance to the arc is, in feet, > Sqrt( 200^2 - 80^2 ) - 181.24 = 2.06 . > I hope you get the idea. > David === Subject: : Re: calculate radius David, thanks once again. You must be a math instructor with the nice concise explainations. Nick === Subject: : Re: Discrete FT please could someone confirm with which equation the 1/N coefficient lies, the DFT (t -> w) or the inverse DFT (w -> k) ? many thanks I don't think it matters too much as long as you are consistent. In my texts it is in the DFT (t -> w). Richard