Dear all, I've been given the following procedure:

> int_part:=proc(f,h,n::integer) local k,u,v,s;
     > u:=f;
     > v:=h;
     > s:=0;

     > for k from 1 to n do;
     > u:=int(u,t);
     > s:=s-(-1)^(k)*u*v;
     > v:=diff(v,t);

> od;
> s; end: 

To apply on the following function: f:=int((exp(-x*t))/sqrt(t(t+1)),t=1..infinity);

And f is a function of x. F(x).

It should give the asymptotic expansion up to O(x^-6). But probably I do something wrong. Can somebody help me please?

Best wishes, Math

I have a trigonometric equation that outputs with a solution in terms of _B1 which I want to remove.

restart: solve({7*cos(2*t)=7*cos(t)^2-5, t>=0, t<=2*Pi}, t, allsolutions, explicit);

output:

{t = arccos((1/7)*sqrt(14))},

{t = 2*Pi-arccos((1/7)*sqrt(14))},

{t = 2*arccos((1/7)*sqrt(14))*_B1-2*_B1*Pi+2*Pi*_Z1-arccos((1/7)*sqrt(14))+Pi}

Is there anyway to get rid of the _B1, or somehow evaluate it by a substitution?

Even numerically the answer still retains the _B1.

{t = 1.006853685}, {t = 5.276331623}, {t = -4.269477938*_B1+6.283185308*_Z1+2.134738969}

Also it would be nice to remove the _Z1 subscript too, as the domain of the equation is [0, 2pi].

I tried removing the 'AllSolutions' command , but then I am missing two solutions:

solve({7*cos(2*t)=7*cos(t)^2-5., t>=0 and t<=2*Pi}, t, Explicit);

 {t = 1.006853685}, {t = 2.134738969}

There should be 4 solutions in the domain [0, 2pi].

If the headline is not the appropriate means, then what is the command line based means of generating a new math container in a worksheet, suppose one has a button with the label "Create New Container inside Selected Section" if that makes my intent any more clear.

At first look the above movement seems to be impossible in real life.

The question is sort of convoluted, but I hope somebody can help.

I have a data set I pulled from an Excel Sheet of the temperature for every day for a year.

Currently, it's in an Array 365x2, where the first column is what day of the year it is and the second column is the temperature. I want to create a plot that on the x-axis shows the day, but on the y-axis shows the amount of a certain chemical in the trees. I have a function of temperature vs chemicals, but I want my graph to show the day of the year vs amount of chemical. How would I go about making such a plot? I don't know how to connect all three. 

Thanks!

Hi guys,

how do I rearrange the expression eq16, in order to get eq17 by using maple commands?

thank you!

Hello,

Can someone explain to me why the following lists do not have the same content?  And suggest how I might be able to fix the problem?

restart;

a := (x^2+y^2)^3 = y^2:

b := (x^2+y^2)^(3/2) = y:

listA := sort([seq(evalf(solve({y >= 0, subs(x = i, a)}, {y})), i = 0.5e-1 .. .5, 0.5e-1)]);

listB := sort([seq(evalf(solve({y >= 0, subs(x = i, b)}, {y})), i = 0.5e-1 .. .5, 0.5e-1)]);

evalb(listA = listB);

If I have a position vector r(t) i polar coordinates and want to calculate velocity I can't get maple to differentiate the unit vectors w.r.t time. How do I do that?

hi

I want to equating coefficients of like powers of  in expr732vc and expr732wc, but

this error is appeared.please help me.

Error, unable to compute coeff
3.mw
 

Download 3.mw

Download maplehelp_limit_inferior_superior.mw

Hello, i was just curious about why when i plot the function it indicates the values for both the limits are actually the limit sups, and if anyone has a good reference for me as far as computing these limits by hand and verifying that the precise limit does not actually exist, but can only be indicated by the range (inf(L),sup(L)) as the output suggests.

Thankyou in advance!

 This post is similar to an earlier post of mine about an artist drawing the letter A as a hole in line paper.  Replicating that in Maple I have put in the 'too hard' basket for now.  I decided to draw the capital letters of the alphabet.  I started with the letter A, not just because it is the first letter, but that that I thought it was of intermediate difficulty.   (...as well as having lady friends with names starting  with B!:-))  Easier letters would be I and L.   One can draw these using the polygon command.  'A' is harder as it has inner regions which must be catered for viz the triangle and trapezium base.  There are about ten letters of the alphabet with curves:  {B, D, P, R, S} and  { J. G, C  O }.   They all have curved portions, but the former set have curves which "bulge out to the right", (with the possible exception of letter S.)  I decided to dive in and tackle the letter B.  I'd heard of Bezier curves and tried to learn about these.  (More later).  Owing to the bulging B, I decided I'd try to liken that part to an ellipse.  I was not happy with the curvature, so tried modifying the equation, involving powers of 4.  After quite a lot of time I managed to get the upper part of the letter  B.  There were difficulties in coloring the letter and I am indebted to Carl (Love aka DeVore) who I recalled mentioned about the order of displaying the various sections of the letter: that saved me hours of anguish!  At this stage the resulting output was that of the letter D, and I was getting tired.  I did a copy and paste of this code into the program below, to compare the output with my previous letter A.  I was disappointed  as the output for the D was almost non-existent.  Initially I thought there may be differences in the values used for A and D and changed x0, y0 for x0D, y0D - but to no effect.  The cut and paste section of D is between the # # # # .

   The output of the rotated letter A, has 'worked' but it looks unusual - I'm not sure if I suffer from slight astigmatism, or there's a bug in my code!  Comments on this please.  I'm not expecting a close inspection of the code - that may be like finding a needle in a haystack. Any ideas as to why there is little output for D?.  

Bezier Curves

         Restarting from my previous comment on letters C. J etc these have curves which are orientated something akin to y=x^2 - or in maths lingo, their base curves can be represented by functions.  I found sparse mention of Bezier curves on the Maple site, but there are many sites now describing how to construct these.  I'll just constrict talk to quadratic Bezier curves.  A Bez quad curve can be obtained by multiplying PMT - (nothing to do with women's problems!) - where (I think) P is a 1x3 row vector ( P0 P1 P2), M is a 3x3 matrix <1  -2   1,  -2  2  0,  1  0  0> and T is a 3x1 column vector <t^2, t, 1>  and 0<t<1.

  On multiplying this out it comes to something like:

C(t) =(P0-2P1+P2)t^2+ (-2P0+2P1)t +P0

   This is the equation of a quadratic, or parabola, but I'm not sure how to apply it:-(  I believe the P are related to coords of  "control points" , and when t=0 the parabolashould pass through one's begining point, and similarly when t=1 it should correspond to the end point of one's desired curve.

  I'd appreciate some feedback on whether my ideas are correct.  What are P0,P1,P2 ?  Numbers, vectors?  How are they related to the control points?  ...and the coords of initial & last points of one's curve.  It seems to me there is still a lot of trial and error work to do, but I suppose in a GUI environment the user is using a device to view the curve and can quickly choose the appropriate one.

     Thanks for reading this long missive, and again any feedback is much appreciated.

David    

restart:

# # # # # # # # # # # #  # # # # # # # #

# Second attempt at the letter A: earlier working version was

# letterA_Art_trick.mws

# Functional operators.  Use of ->

# to find coords of rotated points.

#To Do: 

#Rotate Trig through phi

#   "   trapezL

# Orientation looks wrong

#  Check - then put in a proc

# non-numeric vertex in trigArL

# # # # # # # # # # # #  # # # # # # # #

with(plots):

with(plottools):

# Following are parameters for A : to later put in proc(...)

# phi rotates the A through that angle about the origin/

l:=12:w:=2:thet:=Pi/2.7:topl:=2:x0:=2:y0:=5:phi:=3*Pi/4:

 h := (x, y) -> [x*cos(phi)-y*sin(phi), x*sin(phi)+y*cos(phi)]:

#Alist is list of coords for letter A

Alist:=[[x0,y0],[x0+l*cos(thet),y0+l*sin(thet)],[x0+l*cos(thet)+topl,y0+l*sin(thet)],[x0+2*l*cos(thet)+topl,y0], [x0+2*l*cos(thet)+topl-w,y0],[x0+5*l*cos(thet)/3+topl-w,y0+l*sin(thet)/3],[x0+l*cos(thet)/3+w,y0+l*sin(thet)/3], [x0+w,y0]]:

#outA:=polygon([[x0,y0],[x0+l*cos(thet),y0+l*sin(thet)],[x0+l*cos(thet)+topl,y0+l*sin(thet)],[x0+2*l*cos(thet)+topl,y0], [x0+2*l*cos(thet)+topl-w,y0],[x0+5*l*cos(thet)/3+topl-w,y0+l*sin(thet)/3],[x0+l*cos(thet)/3+w,y0+l*sin(thet)/3], [x0+w,y0]]):

#trigA is small inner triangle of letter A - not rotated

eps:=3*w/4:

corr:=.3*w:

#trigA:=polygon([[x0+(2*l*cos(thet)+topl)/2, #y0+l*sin(thet)-eps],[x0+w+(l/3+corr*w)*cos(thet), #y0+(l/3+corr*w)*sin(thet)],[x0+2*l*cos(thet)+topl-w-(l/3+corr*w)*cos(thet)#, y0+(l/3+corr*w)*sin(thet)]], color=white):

trigA_L:=[[x0+(2*l*cos(thet)+topl)/2, y0+l*sin(thet)-eps],[x0+w+(l/3+corr*w)*cos(thet), y0+(l/3+corr*w)*sin(thet)],[x0+2*l*cos(thet)+topl-w-(l/3+corr*w)*cos(thet), y0+(l/3+corr*w)*sin(thet)]]:

trigArL:=[]:

for i from 1 to 3 do

  #Get coords from trigA_L

x:=trigA_L[i][1]:

y:=trigA_L[i][2]:

# Add coords to new rotated list trigAr

#L := [op(L),x[5]];

trigArL:=[op(trigArL), h(x,y)]:

end do:

#trigArL;

trigAr := polygon(trigArL, color=white, linestyle=1, thickness=1):

#plots[display](trigAr);

#Read coords of Alist into another list after first transforming the #coords using func op h.  Angle of rot is phi.  thet is inclination of #letter A

#Main Outline of letter - but NOT the convex hull

Alistr:=[]:

#  h := (x, y) -> [x*cos(phi)-y*sin(phi), x*sin(phi)+y*cos(phi)]:

for i from 1 to 8 do

  #Get coords from Alist

x:=Alist[i][1]:

y:=Alist[i][2]:

# Add coords to new rotated list Alistr

#L := [op(L),x[5]];

Alistr:=[op(Alistr), h(x,y)]:

end do:

l2 := polygon(Alistr, color=wheat, linestyle=1, thickness=1):

# # # # # # # # # # # # # # # # #

#plots[display](trigAr,l2);

#plots[display](trigAr,trapezL, l2);

#trigAr,

#thet:=Pi/3:

# Rotation by thet  anti-clockwise about the origin

#h := (x, y) -> [x*cos(thet)-y*sin(thet), x*sin(thet)+y*cos(thet)]:

#h(sqrt(3), 1);

# To find a polygon (trapezium EFGH??) which is the shape of he base of A.

# This is necessary as this portion is colored same as rest of A.

#trapezr:=polygon([[(x0+2*l*cos(thet)+topl-w)*cos(phi)-y0*sin(phi),(x0+2*l*#cos(thet)+topl-w)*sin(phi)+y0*cos(phi)],[(x0+5*l*cos(thet)/3+topl-w)*cos(p#hi)-(y0+l*sin(thet)/3)*sin(phi),(x0+5*l*cos(thet)/3+topl-w)*sin(phi)+(y0+l#*sin(thet)/3)*cos(phi)],[(x0+l*cos(thet)/3+w)*cos(phi)-(y0+l*sin(thet)/3)*#sin(phi),(x0+l*cos(thet)/3+w)*sin(phi)+(y0+l*sin(thet)/3)*cos(phi)],[(x0+w#)*cos(phi)-y0*sin(phi),(x0+w)*sin(phi)+y0*cos(phi)]],color=white,filled=tr#ue,  linestyle=DOT):

#trapezL is a list of coords of the base trapezium

#THESE coords have not been rotated.  Repeat previous procedure to obtain #rotated coords

trapezr:=[]:

for i from 5 to 8 do

  #Get coords from Alist

x:=Alist[i][1]:

y:=Alist[i][2]:

# Add coords to rotated list trapezr

trapezr:=[op(trapezr), h(x,y)]:

end do:

#trap2 := polygon(trapezr, color=wheat, linestyle=1, thickness=1):

#trapezL:=polygon([[x0+2*l*cos(thet)+topl-w,y0],[x0+5*l*cos(thet)/3+topl-w,y0+l*sin(thet)/3],[x0+l*cos(thet)/3+w,y0+l*sin(thet)/3], [x0+w,y0]], color=red):

trapezL:=polygon(trapezr, color=white):

# lbase is an unseen line at base of the trapezium of A.  Colored white # as needed to be hidden

lbase:=line([(x0+2*l*cos(thet)+topl-w)*cos(phi)-y0*sin(phi),(x0+2*l*cos(thet)+topl-w)*sin(phi)+y0*cos(phi)], [(x0+w)*cos(phi)-y0*sin(phi),(x0+w)*sin(phi)+y0*cos(phi)], color=white):

# # # # # # # # # # # # # # # # # # # # # # # # # # # # # #

# Letter D  (or top part of B!)

a := .6: b := .5: x0D := 1: y0D:=7:

c1:=15:n:=4:

pl1:=plot([x, y0D+b^2*sqrt(c1-(x-x0D)^n/a^2), x=1..2.6], scaling=constrained, color=wheat,filled=true):

pl2:=plot([x, y0D-b^2*sqrt(c1-(x-x0D)^n/a^2), x=1..2.6], scaling=constrained, color=red, filled=true):  #was white

pl3:=plot([x, y0D+b^2*sqrt(c1/2-(x-x0D)^n/a^2), x=1.3..3], scaling=constrained, color=blue,filled=true):  #was white

pl4:=plot([x, y0D-b^2*sqrt(c1/2-(x-x0D)^n/a^2), x=1.3..3], scaling=constrained,color=wheat, filled=true):

x:=1.3:

#Create & Draw inner up line of D

ylow:=eval(y0D-b^2*sqrt(c1/2-(x-x0D)^n/a^2)):

yhigh:=eval(y0D+b^2*sqrt(c1/2-(x-x0D)^n/a^2)):

l_up_in:=line([x,ylow], [x,yhigh], color=wheat):

#Create & Draw outer up line of D

x:=1:

ylow:=eval(y0D-b^2*sqrt(c1-(x-x0D)^n/a^2)):

yhigh:=eval(y0D+b^2*sqrt(c1-(x-x0D)^n/a^2)):

l_up_out:=line([x,ylow], [x,yhigh], color=white):

plots[display]( pl2,pl4,pl3,l_up_in,l_up_out,pl1, scaling=constrained, view=[1..2.6, 6..8.1]);

# removed l_up_in, before l_up_out - no difference

# # # # # # # # # # # # # # # # # # # # # # # # - End of letter D - # # # #

# Display letter A

plots[display](trigAr,lbase,trapezL, l2);

Below I successfully employed the is command to compare Q1 & Q2.  However, attempting to compare T1 & T2 which is essentially Q1 & Q2 but defined as a functions dependent on m does not work so well.  Can I do this comparison?  If so what is the proper syntax to do this?

Download is_command.mw

Hello,

I am a beginner with Maple. I had some basic tutorials. In fact I want to solve a first order differential equation with non-constant coefficient:

Can you really tell me what are the steps to follow in Maple 18 in order to solve this equation. I didn't find any tutorial or work that deals with variable coefficients.

Thank you in advnace.

Hi
sry guys i have a problem with my maple when i use my fsolve sometimes it gave me no solution but i khow we have a solution ill use simplify to gave it a simplified input but also no awenser is there any other command yo make my problem easier for maple and also is there anyway that i can force my cpu and memory to focus more on solving problems

and at last what is default method for fsolve ( i mean is it newton method or something else and can i change it )

tnQ for your attentions
thats my maple file problem.mw

Using the learning sequence as an alternative to learn problems related to "balance of a body" is shown in this video; thanks to the kindness that Maple offers us in its fundamental programming syntax.

Balance_of_a_body_with_learning_sequence.mw

Lenin Araujo

Ambassador Of Maple