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Length of output exceed limit 1000000...

Asked by: Sofey 10
May 03 2015
0  3

can anybody help me? i want to check the consistency of my scheme. My equation is too long if i check manually, so i used maple 13 to simplify my equation. But it cannot simplify it because of length of output exceed limit 1000000

restart

eqn1 := u+(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2)+(u-(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2))-u-(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)-u+(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3))*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3));

(1-exp(-m))^2*u[tt]-(1-exp(-m))^2*u[xx]+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-16*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+4*(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3))

 (1)

a := simplify(eqn1);

`[Length of output exceeds limit of 1000000]`

 (2)

``


Download consistency_expmle_4.mw

.

how to prove with maple...

Asked by: laporte bernard 45
student
May 03 2015
0  1


> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

> 

"the set Q of rational numbers does not have the least-upper-bound property under the usual order. "
> 

Proof:
> 

"Consider the part A = {x in `&Qopf;`; 1<x^(2)<2<}, this part is not empty as [4/(3)] in A; It is bounded by 2 as if x^(2)<4 then x<2. THe set of greatest elements of A, belonging to `&Qopf;`, is not empty."
> 

Lemma:
> 

"If [p/(q)] in A with q in `&Nopf;`^(*), then p>q for p>0 and p^(2)-2*q^(2)<0; or p^(2)-2*q^(2) in `&Zopf;`, thus"
> 

p^2-2*q^2 <= -1

p^2-2*q^2 <= -1

 (1)
> 

"for all r in `&Nopf;`^(*), put y=(r*p+1)/(r.p), we have :"
> 

y > p/q
> 

Now:
> 

"y^(2)-2=supA/(r^(2)*q^(2)), with supA= r^(2)*p^(2)+2 r*p+1-2 r^(2)q^(2),"
> 

otherwise; -1; s = r(p^2-2*q^2)+2*p+1

s = r(p^2-2*q^2)+2*p+1

 (2)
> 

"a good choice for r, for instance r=2*p+1, we get from (1)"
> 

(2*p+1)(p^2-2*q^2) <= (2*p+1)(-1)

2*p(p^2-2*q^2) <= 2*p(-1)

 (3)
> 

thus:
> 

(2*p+1)(p^2-2*q^2)+2*p <= -1

2*p(p^2-2*q^2)+2*p <= -2

 (4)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p] <= -2*p-1

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p] <= -2*p-1

 (5)
> 

(2*p+1)*[(2*p+1)(p^2-2*q^2)+2*p]+1 <= -2*p

(2*p+1)*[2*p(p^2-2*q^2)+1+2*p]+1 <= -2*p

 (6)
> 

"finally, supA<0"
> 

`and`(thus*y^2 < 2, `in`(y, A*with*y^2) and A*with*y^2 > 1)
> 

"so, for any x in p/(q) in A, there exists y in A such that y>x: in conclusion A does'nt admit a greatest element."
> 

Now, given*m = p/q, `and`(not `in`(Typesetting:-delayDotProduct(a*greatest*element*of*A, Then)*m, A), we*get*thus):
> 

`and`(p > q, p^2-2*q^2 >= 0)
> 

as*the*equation*p^2-2*q^2 = (0*has)*no*solutions and `in`((0*has)*no*solutions, nonnegint*nonnegint), we*get:
> 

2*p^2-2*q^2 >= 1:
> 

"for any r in `&Nopf;`^(*)let's put m'=(r*p)/((r*q+1)); we have:"
> 

(diff(m(x), x))^2-2 = supA/(r*q+1)^2:
> 

With*supA = r^2*p^2-2*(r*q+1)^2 and r^2*p^2-2*(r*q+1)^2 = r[r[p^2-2*q^2]-4*q]-2:
> 

"for a good choice of r, for example r=4*q+1, we get (back to (2)):"
> 

supA > (0*thus)*(diff(m(x), x))^2 and (0*thus)*(diff(m(x), x))^2 > 2:
> 

"So, for any greatest element m of A, it exists m', greatest element of A such as:"
> 

diff(m(x), x) < m(x):
> 

"A does not admit a least upper bound."
> 

``


Download rational_numbers.mw

dynamic programming?...

Asked by: brian bovril 914
optimization proc
May 03 2015
0  2

This problem has real world applicability: Three vampires and three maidens are at the foot of a tall building and wish to get to the bar on the top floor.  The lift only holds two people (for convenience I am classing vampires as people), and needs one person to operate it.  If ever the vampires outnumber the maidens at any place, they will do something unspeakable.  How can the vampires and maidens all safety get to the top in the minimum of moves?  http://tonysmaths.blogspot.com.au/2012/05/vampires-and-maidens-problem.html

can this be solved procedurally or using optimization package?

possible manual soln:

No. m>= No. v

3m+3v, 0   [0]

2m+2v,m+v [1]

3m+2v,v  [2]

     3m, 3v [3]

3m+v , 2v  [4]

m+v  , 2m+2v [5]

#then reverse the steps

2m+2v, m+v [6]

       2v, 3m+v [7]

       3v, 3m  [8]

        v , 3m+2v [9]

    m+v, 2m+2v [10]

        0, 3m+3v [11]

 

Using angular units with trig functions...

Asked by: Dr. Ian Williams 20
convert value
May 03 2015
1  5

I'm looking at Maple as a possible alternative to Mathcad (which I've been using for years, but is now very jaded compared to other options like Maple and Mathematica).  I'm a civil engineer and for what I do, one of the better features of Mathcad is the way it handles units.  For example, if I specify an angle in degrees (say phi=30 degrees) and then ask for sin(phi), I get 0.5.  At face value, I though Maple would do the same kind of thing.  However, this doesn't appear to be the case (see attached worksheet).  The only workaround that I can see is to specify the angle in degrees (but without assigning ) and then multiply the specified value by pi/180 (to convert to radians) before passing it to the sin function.  Which is all a bit messy and not at all an attractive solution.

Am I misunderstanding the way units work in Maple and is there a clean way of specifying angles in degrees (which is what engineers work with) and using these values directy in trig functions?

Thanks in anticipation,

Ian

How to solve infinite series...

Asked by: laporte bernard 45
May 03 2015
0  2

If a dosage Q units of a certain drug is administrated to an individual, then the amount remaining in the bloodstream at the end of t minutes is given by Q*exp^-ct, where c>0. Suppose this same dosage is given at successive T-minute intervals.

 

a) Show that the amount A(k) of the drug is given by A(k) = ∑n=0k-1 Q*exp(^-ncT).

b) Find an upper bound for the amount of the drug in the bloodsteam after any number of doses.

c) Find the smallest time between doses that will ensure that A(k) does not exceed a certain level M for M>Q.

solving the following optimization problem...

Asked by: GPY 80
optimization markowitz quadratic
May 03 2015
0  8

What is the right syntax to solve :

min {sum(i=1 to 10) sum(j=1..10) (a_i_j)*(x_i)*(x_j) 

s.t sum(i=1..n) (b_i)*x_i=p and sum(i=1..n)x_i=1 

if a_i_j is a constant, b_i is a constant, p is a constant and x_i, x+j are the decision variables?

I understand that this is a quadratic programming problem and an application of Markowitz optimization. I've tried to use the in-built minimize function but haven't got the right output.

Installing A User Package...

Asked by: sjh5403 5
May 02 2015
1  1

I am very new to Maple and am wondering how i would install a package. The package I am trying to install is located at

http://math.rutgers.edu/~russell2/papers/IdentityFinder 

I am using Maple 2015 on windows 8. Any advice would be useful

How can PDE be solved in terms of Bessel function?...

Asked by: emmantop 40
plot
May 02 2015
0  4

Good day everyone,

please how can one solve this pde in terms of Bessel function or any other analytic solution with the plot.

See the file ID.mw

Thanks.

Graph for stability analysis...

Asked by: Sofey 10
complexplot
May 01 2015
1  4


can anybody help me..? why my graph not come out? Is that any mistake in my coding?

> 

restart
> 

y := x^2-x*(exp(I*k*`&Delta;x`)+exp(-I*k*`&Delta;x`)-m^2+m^2*((4*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3))*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3)*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3))*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3)/(epsilon*((((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3)*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3))*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3+((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3))+(((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3)*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3))*((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3+((1^2-2)*cosh(1+1)-4*sinh(1+1)+1^6*cosh(1+1)^3)))))+1;

x^2-x*(exp(I*k*`&Delta;x`)+exp(-I*k*`&Delta;x`)-m^2+2*m^2*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(epsilon*(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3)))+1

 (1)
> 

subs(m = 1-exp(-m), %);

x^2-x*(exp(I*k*`&Delta;x`)+exp(-I*k*`&Delta;x`)-(1-exp(-m))^2+2*(1-exp(-m))^2*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(epsilon*(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3)))+1

 (2)
> 

subs(epsilon = .17882484, %);

x^2-x*(exp(I*k*`&Delta;x`)+exp(-I*k*`&Delta;x`)-(1-exp(-m))^2+11.18412856*(1-exp(-m))^2*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3))+1

 (3)
> 

subs(k = n*Pi, %);

x^2-x*(exp(I*n*Pi*`&Delta;x`)+exp(-I*n*Pi*`&Delta;x`)-(1-exp(-m))^2+11.18412856*(1-exp(-m))^2*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3))+1

 (4)
> 

subs(`&Delta;x` = m, %);

x^2-x*(exp(I*n*Pi*m)+exp(-I*n*Pi*m)-(1-exp(-m))^2+11.18412856*(1-exp(-m))^2*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3))+1

 (5)
> 

subs(m = 0.1e-2, %);

x^2-x*(exp((0.1e-2*I)*n*Pi)+exp(-(0.1e-2*I)*n*Pi)-(1-exp(-0.1e-2))^2+11.18412856*(1-exp(-0.1e-2))^2*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3))+1

 (6)
> 

j := subs(n = 1, %);

x^2-x*(exp((0.1e-2*I)*Pi)+exp(-(0.1e-2*I)*Pi)-0.9990006498e-6+0.1117295170e-4*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3))+1

 (7)
> 

complexplot3d(x*j, x = -2-I .. 2+I);````

complexplot3d(x*(x^2-x*(exp((0.1e-2*I)*Pi)+exp(-(0.1e-2*I)*Pi)-0.9990006498e-6+0.1117295170e-4*(-cosh(2)-4*sinh(2)+cosh(2)^3)^2/(-2*cosh(2)-8*sinh(2)+2*cosh(2)^3))+1), x = -2-I .. 2+I)

 (8)
> 

a := fsolve(x*j, x);

0., .9865070072, 1.013677544

 (9)
> 

b := fsolve(x*j, x = 1);

0., .9865070072, 1.013677544

 (10)
> 

with(plots):
> 

complexplot({a, b}, numpoints = 100, color = green, filled = true, title = "Stability Region");

 
> 

``

``

``


Download Stability_1.mw

Copy several elements from Vector...

Asked by: Spinosaurus 120
May 01 2015
1  2

Hi everyone!

I use two "ArrayTools:-Copy" commands to copy elements from Vector A to Vector B.

restart;
A := LinearAlgebra:-RandomVector(10); A := convert(A, Vector[row]);
B := Vector[row](10); ArrayTools:-Copy(2, A, 1, B, 1); ArrayTools:-Copy(2, A, 7, B, 7); B;

My question is: Could I obtain the same result using only one command?

P.S. I tried to use "ArrayTools:-BlockCopy", but didn't get correct result.

 

Programmatic access to input commands...

Asked by: Athar Kharal 90
input document documenttools
May 01 2015
1  5

I want to retrieve all the commands input by user in a maple document saved on disk. How this can be done? DocumentTools:-Retrieve permits to receive only the labeled expressions but not the input commands.

looking for a maple V command that write the resul...

Asked by: c4meleon 30
May 01 2015
0  9

Hi
how can we write a same result n-times near each other


x:=3:
for i from 1 to 4 do       # here the 4 can be a variable and can change...
    x;
    od;
                                  3
                                  3
                                  3
                                  3

I need the result to come in that form  :    3  ,  3  ,  3  ,  3
and because the number of times can change i can't just easily write x,x,x,x;

thanks in advance

Gerschorin's theorem Linear Algebra...

Asked by: laporte bernard 45
May 01 2015
0  6

worksheet/expressions/copypasteMaple

Gerschgorin := proc (A::Matrix) local Delta, m, n, AA, R, C, i, c, eig, P, Plt; Delta := proc (i, j) if i = j then 0 else 1 end if end proc; m, n := LinearAlgebra[Dimension](A); AA := Matrix(m, n, proc (i, j) options operator, arrow; Delta(i, j)*abs(A[i, j]) end proc); R := evalm(`&*`(AA, Vector(m, 1))); C := {seq(('plottools[circle]')([Re(A[i, i]), Im(A[i, i])], R[i], color = violet), i = 1 .. m)}; c := {seq(('plottools[point]')([Re(A[i, i]), Im(A[i, i])], color = blue, symbol = diamond), i = 1 .. m)}; eig := evalf(LinearAlgebra[Eigenvalues](A)); P := {seq(('plottools[point]')([Re(eig[i]), Im(eig[i])], color = red, symbol = box), i = 1 .. m)}; Plt := `union`(`union`(C, c), P); plots[display](eval(Plt), scaling = constrained) end proc

 

A := Matrix([[5, 8, 4, -3], [8, -9, 7, 5], [0, 4, 4, 2], [5, -5, 9, -9]]); evalf(LinearAlgebra[Eigenvalues](A), 3); Gerschgorin(A)

worksheet/expressions/pasteMathML

 

F := Matrix([[2, -1/2, -1/3, 0], [0, 6, 1, 0], [1/3, -1/3, 5, 1/3], [-1/2, 1/4, -1/4, 4]]); evalf(LinearAlgebra[Eigenvalues](F)); Gerschgorin(F)

Could you print A & F ?

 

regards

 

 

How to insert a plot in word with maple functions ...

Asked by: 85951971 30
msword plotting
April 30 2015
0  1

Hello i would love some help with this code i need Maple to make an exercise guide in word automaticly the thing is that i don't know how to automaticly insert the plot in the word, i think there is no fprintf format to add images or plots maybe with exportplot? i really don't know; here is the Code is Basic for a starter student i just need the comand to insert the plot just that.
-------------------------------------------------------------------
fd:fopen("Graphs.doc", WRITE);
 for i from 1 by 5 to 11do
  if i=1 then plotsetup(bmp,plotoutput=plot1);
 a:= -50;
 b:=30;
 elif i=6 then
 plotsetup(bmp,plotoutput=plot2);
 a:= 50;
 b:=-30;
  elif i=11 then  plotsetup(bmp,plotoutput=plot3);
  a:= 2;
 b:=-3;
 end if;
 h:=(i*x^(2)-((3*a)/(i-42))*x-(i*b)/(-i+42));
  fprintf(fd,"the quadratic equation is: ")
fprintf(fd,"%q \\n",h);
 #`HERE IS WHERE I NEED THE PLOT TO BE`
  plot(i*x^(2)-((3*a)/(i-42))*x-(i*b)/(-i+42),x)
   end do; 
 fclose(fd);
--------------------------------------------------------------------
pd. i can't do it manually i need it to be automatic later i will add some inputs to make it even faster.

how to copy and paste from maple 12 a quite big qu...

Asked by: laporte bernard 45
April 30 2015
0  4

Hi, as I can't manage to copy and paste on mapleprimes, I would be glad to get a hint ...

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