Maple 7 Questions and Posts

These are Posts and Questions associated with the product, Maple 7

Fivebyfive_crossword.mwsFivebyfive_crossword.txt

  I am trying to solve a 5x5 crossword  puzzle, where one must find 5 across clues, and 5 down clues - all in a 5x5 square grid.  I have done a dictionary search for all the possibilities of the ten clues.  (For brevity this is not in the attached program.)  These possible solutions have been put as lists.  The attached Maple 7 program attempts to solve the crossword by a series of nested for loops, checking if the intersection of across and down clue letters match.  In this the use of nops is made, as the number of possible words for each clue varies.  I appreciate why the program fails, but am at a loss for a "work around", and fear a complete restart may be needed. 

   I attach two files of the program.  The Fivebyfive_crossword.mws Maple 7 file includes a graphic grid of the crossword, with asterisks (*)  for unknown letters.  The corresponding .txt file probably will not have that.  Sadly, Maple doesn't allow the upload of .rtf files.  Further documentation is in the program.

   Any comments, help would be most appreciated.

The program below is a high school problem, related to the area a horse can graze, given it is tethered to a rectangular barn.  The level of difficulty is related to the length of rope.  

   I wanted to display some graphics of the field, barn and tethered horse - this latter being the most difficult.  I experimented with a .png picture of a plain silouhette of a horse, imported this into Photoshop, then saved it as a .pdf file, importing this into Maple,  I managed to import this into the worksheet, but I wanted a scaled down version of the horse in the program plots[display] section.  I was unsuccessful in this.  Undeterred, I decided to try and draw a version of a horse using the plots/plottools packages.  The resulting "horse" looks more like a cat, warthog,  mouse or chameleon! 

   I understand later versions of Maple are able to import graphic images.   I'd appreciate some feedback as to how easy this is, and the quality of the resulting images in Maple output.

Thanks,

    David  .  .    

 

restart:

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

# Horse tethered to barn - what area of grass?

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

with(plots):

with(plottools):

macro(palegreen=COLOR(RGB, .5607, .9372, .5607)):

col1:=`black`:

print(`A horse is tethered to the corner of a barn, 10 m wide and 20 m long.  Find the area`);

print(`the horse can graze, if the length of rope is:`);

print(` i.)   5 m   ii.)  25 m  amd iii.)  50 m`);

#Length of rope

L:=10:

#Dims of barn

len:=20:wid:=10:

#Position of bottom left corner of barn

x_barn:=50:y_barn:=50:

len:=20:wid:=10:

 

#dimensions of field

flen:=120:fwid:=110:

 

rect_barn:=rectangle([x_barn,wid+y_barn], [len+x_barn,y_barn], color=brown):

rect_field:=rectangle([0,fwid], [flen,0], color=palegreen):

 

#Position of horse

x0:=37:y0:=32:

a := 4: b := 2.5:

belli := ellipse([x0,y0], a, b, filled=true, color=col1):

legf:=line([x0-1,y0-2], [x0-2,y0-6], color=col1, linestyle=1, thickness=1):

legf2:=line([x0-2,y0-2], [x0-3,y0-6], color=col1, linestyle=1, thickness=2):

rleg:=line([x0+1,y0-2], [x0+2,y0-6], color=col1, linestyle=1, thickness=2):

 

rleg2:=line([x0+2,y0-2], [x0+3,y0-6], color=col1, linestyle=1, thickness=2):

head := polygon([[x0-6,y0+3],[x0-5,y0+4], [x0-2,y0+5], [x0-3,y0+2]], color=brown, linestyle=3, thickness=2):

tail:=line([x0+6,y0-4], [x0+4,y0+1], color=col1, linestyle=1,thickness=2):

 

a := arc([x0+13,y0+3], 15, Pi/2..Pi+.1, color=blue):

plots[display](a,tail,head,belli,legf,legf2,rleg,rleg2,rect_barn,rect_field, scaling=constrained, axes=none);

how i can simplify

(f(x[n])/Df(x[n]));
in code

restart;
taylor(f(x), x = gamma, 8);
f(x[n]) := subs([x-gamma = e[n], f(gamma) = 0, seq(((D@@k)(f))(gamma) = factorial(k)*c[k]*(D(f))(gamma), k = 1 .. 1000)], %);

1 2
f(gamma) + D(f)(gamma) (x - gamma) + - @@(D, 2)(f)(gamma) (x - gamma)
2

1 3 1 4
+ - @@(D, 3)(f)(gamma) (x - gamma) + -- @@(D, 4)(f)(gamma) (x - gamma)
6 24

1 5 1 6
+ --- @@(D, 5)(f)(gamma) (x - gamma) + --- @@(D, 6)(f)(gamma) (x - gamma)
120 720

1 7 / 8\
+ ---- @@(D, 7)(f)(gamma) (x - gamma) + O\(x - gamma) /
5040
2 3
c[1] D(f)(gamma) e[n] + c[2] D(f)(gamma) e[n] + c[3] D(f)(gamma) e[n]

4 5 6
+ c[4] D(f)(gamma) e[n] + c[5] D(f)(gamma) e[n] + c[6] D(f)(gamma) e[n]

7 / 8\
+ c[7] D(f)(gamma) e[n] + O\e[n] /

taylor(D(f)(x), x = gamma, 8);
Df(x[n]) := subs([x-gamma = e[n], f(gamma) = 0, seq(((D@@k)(f))(gamma) = factorial(k)*c[k]*(D(f))(gamma), k = 2 .. 1000)], %);

D(f)(gamma) + @@(D, 2)(f)(gamma) (x - gamma)

1 2 1 3
+ - @@(D, 3)(f)(gamma) (x - gamma) + - @@(D, 4)(f)(gamma) (x - gamma)
2 6

1 4 1 5
+ -- @@(D, 5)(f)(gamma) (x - gamma) + --- @@(D, 6)(f)(gamma) (x - gamma)
24 120

1 6
+ --- @@(D, 7)(f)(gamma) (x - gamma)
720

1 7 / 8\
+ ---- @@(D, 8)(f)(gamma) (x - gamma) + O\(x - gamma) /
5040
2
D(f)(gamma) + 2 c[2] D(f)(gamma) e[n] + 3 c[3] D(f)(gamma) e[n]

3 4
+ 4 c[4] D(f)(gamma) e[n] + 5 c[5] D(f)(gamma) e[n]

5 6
+ 6 c[6] D(f)(gamma) e[n] + 7 c[7] D(f)(gamma) e[n]

7 / 8\
+ 8 c[8] D(f)(gamma) e[n] + O\e[n] /

(f(x[n])/Df(x[n]));
this last term did not use f(x[n]) value from above to solve it. plxx help if any one can solve it...

how i can find order of convergence of newton method by expanding taylor series?? plz send me code???

 

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