when solving a system of equations, I want to get rid of all the absolute functions.

for example, |y-2|=x，I don't want maple solve this equation directly...because maple may has difficulties when dealing with absolute values. Instead, I want to transform this equation by squaring the both sides at the same time which end up this equation: (y-2)^2=x^2.

The example I provide is kind of simple...what if there are multiple absolute term in the equations? Is there a general way to get what I need? Or Is it practical to use Maple to achive such thing?

thanks in advance.

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To get this update, you can use Tools>Check for Updates from within Maple, or visit Maple 2015.1 Downloads.

If you are a MapleSim 2015 user, you already have this update, as it was part of the MapleSim 2015 installation.

Kim

Hello Everyone

I have an expression which I wish to integrate. I would be grateful if you could please help me with it. I have uploaded the maple file for your refrence.

Thanks a lot for your time.

IntegrationExample.mw

restart;
pp:=-55471918776960000*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)+5350094400*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)*cp^2-129*tanh((1/3220)*sqrt(10368400-cp^2)*Pi*x/cp)*cp^4+2670899840*tanh((1/6450)*sqrt(41602500-cp^2)*Pi*x/cp)*sqrt(41602500-cp^2)*sqrt(10368400-cp^2);
Student[Calculus1]:-Roots(subs(x=8000,pp),cp=1..3220,numeric);
p1:=proc(v)
option hfloat;
local a;
a:=Student[Calculus1]:-Roots(subs(x=v,pp),cp=1..3220,numeric);
if nops(a)>=1 then seq([v,a[i]],i=1..nops(a));
end if;
end proc:
SS1:=[seq(p1(i),i=3500..20000,200)]:
plot(SS1,style=point,gridlines);

The final figure is different between maple12 and maple17.

On 17, unwanted points apprear.

is it a bug?

Dear Colleges

I have a problem with the following code. As you can see, procedure Q1 converges but I couldn't get the resutls from Q2.

I would be most grateful if you could help me on this problem.

Sincerely yours

Amir

restart;

Eq1:=diff(f(x),x$3)+diff(f(x),x$2)*f(x)+b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)^2*x^2,x$1);
Eq2:=diff(g(x),x$3)+diff(g(x),x$2)*g(x)+c*a^2*sqrt(2*reynolds)*diff(diff(g(x),x$2)^2*x,x$1);
eq1:=isolate(Eq1,diff(f(x),x,x,x));
eq2:=subs(g=f,isolate(Eq2,diff(g(x),x,x,x)));
EQ:=diff(f(x),x,x,x)=piecewise(x<c*0.1,rhs(eq1),rhs(eq2));
Eq11:=diff(theta(x),x$2)+pr*diff(theta(x),x$1)*f(x)+pr/prt*b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(theta(x),x$1)*x^2,x$1);
Eq22:=diff(g(x),x$2)+pr*diff(g(x),x$1)*f(x)+pr/prt*a^2*c*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(g(x),x$1)*x^1,x$1);
eq11:=isolate(Eq11,diff(theta(x),x,x));
eq22:=subs(g=theta,isolate(Eq22,diff(g(x),x,x)));
EQT:=diff(theta(x),x,x)=piecewise(x<c*0.1,rhs(eq11),rhs(eq22));
EQT1a:=eval(EQT,EQ):
EQT2:=eval(EQT1a,{f(x)=G0(x),diff(f(x),x)=G1(x),diff(f(x),x,x)=G2(x)}):
bd:=c;
a:=0.13:
b:=0.41:
pr:=1;
prt:=0.86;
reynolds:=12734151.135786774055543653356602;     #10^6;   #1.125*10^8:

c:=88.419896050808975395120916434619:
;
Q:=proc(pp2) local res,F0,F1,F2;
print(pp2);
if not type(pp2,numeric) then return 'procname(_passed)' end if:
res:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=pp2},numeric,output=listprocedure);
F0,F1,F2:=op(subs(subs(res),[f(x),diff(f(x),x),diff(f(x),x,x)])):
F1(bd)-1;
end proc;
fsolve(Q(pp2)=0,pp2=(0..1002));
se:=%;
res2:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=se},numeric,output=listprocedure):
G0,G1,G2:=op(subs(subs(res2),[f(x),diff(f(x),x),diff(f(x),x,x)])):
plots:-odeplot(res2,[seq([x,diff(f(x),[x$i])],i=1..1)],0..c);



Q2:=proc(rr2) local solT,T0,T1;
print(rr2);
if not type(rr2,numeric) then return 'procname(_passed)' end if:
solT:=dsolve({EQT2,theta(0)=1,D(theta)(0)=-rr2},numeric,known=[G0,G1,G2],output=listprocedure):
T0,T1:=op(subs(subs(res),[theta(x),diff(theta(x),x)])):
T0(bd);
end proc;
fsolve(Q2(rr2)=0,rr2=(0..100));


shib:=%;
sol:=dsolve({EQT2,theta(0)=1,D(theta)(0)=-shib},numeric,known=[G0,G1,G2],output=listprocedure):
plots:-odeplot(sol,[x,theta(x)],0..c);
#fsolve(Q2(pp3)=0,pp3=-2..2):

Amir

could you help me about maple
i try to calculating using chevypade rational approximating and the answer for cos(x) xe is(-.221091073962959*T(1, x-1)+.7710737338*T(0, x-1)-0.4212446689e-1*T(2, x-1))/(0.836360586596837e-1*T(1, x-1)+T(0, x-1)+0.3360079945e-1*T(2, x-1)) i can not to convert to rational form as x^^n .maple is not very friendship
Thanks

> restart;
> with*plots;
> Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2 = 0; 

> N := 1;

> blt := 10;
> Eq2 := (diff(theta(eta), eta, eta))/Pr+f(eta)*(diff(theta(eta), eta)) = 0; 
> bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(blt) = 0;
> bcs2 := (D(theta))(0) = -N*(1+theta(0)), theta(blt) = 0;
> L := [2.5, 3, 5, 7, 10];
> for k to 5 do R := dsolve(eval({Eq1, Eq2, bcs1, bcs2}, Pr = L[k]), [f(eta), theta(eta)], numeric, output = listprocedure); X1 || k := rhs(R[3]); X2 || k := rhs(R[4]); Y1 || k := rhs(R[5]); Y2 || k := -rhs(R[6]) end do;

 how I will draw the graph for Pr against theta   for Pr=2.5 until 7  taking rest of the parameter fix

I wish to evaluate the expression

knowing that

where a is a constant.  It is not hard to see, assuming enough differentiability,  that the expression evaluates to

I know how to do this when all the derivatives are expressed in terms of the diff() operator.  Here it is:

eq := diff(u(x,t),t) = a^2*diff(u(x,t),x,x);
expr := diff(u(x,t),t,t);
eval['recurse'](expr,[eq]);

However, I would prefer to do the computations when all derivatives are expressed in terms of the D operator but cannot get that to work.  What is the trick?

Hello. I have a question. If you can help me, i am pleasure.

Have nice day. :)

restart;
soru := proc(n,x)
local top::0;
for x from 2 to n do
top =top+(((x^(2+i))*top)^(1/(n+2-i)));
print(top);
od;
end proc;

soru(5,1);

Error, (in soru) illegal use of a formal parameter

Hello. I have a question. If you can help me, i am pleasure.

Have nice day. :)

Question: Enter a number after maple find(understand) to prime number or normal number.

restart;
> asal := proc(n)
> local m;
> if n<2 then
> m := s=n/2+1;
> print(m);
> else
> for i from 2 to n do
> if irem(n,i,0) then
> print("asal değil");
> if i = m then
> print("asaldir");
> end do;
> end if;
> end if;
> end proc;

Consider the following code, which just generates two "identical" matrices, differing only in their requested storage type, and then does some simple manipulations.

restart;
#
# Define matrix using sparse storage
#
   testM:= Matrix( 40,40,
                           (i,j)->`if`(j>=i,1,0),fill=0,
                           storage=sparse
                        ):
#
# Define identical(?) matrix with rectangular storage
#
   nm:= Matrix( 40,40,
                        testM,
                        storage=rectangular
                     ):
#
# Define procedure to return some matrix properties
#
   matData:= proc( myMat::Matrix)
                            return op(3, myMat)[2], # check storage type
                                      myMat[5, 1..-1], # get 5-th row
                                      add(myMat[5, 1..-1]); # add elements in 5-th row
                    end proc:
#
# Get properies of the two matrices - should be identical
# but check result of adding elements in the 5-th row
#
    matData(testM);
    matData(nm);

The matData procedure ought to produce the same results for the two matrices, with the exception of the storrage type. But the 'add()' command does not. The 'myMat[5, 1..-1]' command produces the same vector, the 5-th row - but stick an add() wrapper around it and all hell breaks loose.

Is this a bug or am I missing something?

Suggestions such as avoiding sparse data storage are not really acceptable: the above is a much simplified version of my original problem where I was using graph theory to play with a "cost function" and (with G a graph) the command,

WeightMatrix(MinimalSpanningTree(G))

returned a sparse-storage matrix - and I didn't notice. There appears to be no option on the WeightMatrix() command to control the storage tyoe of the returned matrix. Result was that all subsequent code based on slicing/dicing/and particularly 'add()ing' sub-blocks of this weight matrix fell apart

Don't get me wrong: I can sort of accept that the weight matrix of minimal spanning tree would (hopefully) be mainly zeros so sparse-storage might be a good default option but I don't see why the results of a command such as

add(myMat[5, 1..-1])

should vary depending on the internal storage used for the matrix, particularly when I have no control over the storage type being adopted

How to find the maximum natural number n s. t. the sum of its cubed digits is greater than or equal to n? Of course, with Maple. The same question for the sum of the  digits to k-th power. Here are my unsuccessful attempts:
1.Optimization:-Maximize(n, {n <= convert(map(c ->c^3, convert(n, base, 10)), `+`)}, assume = integer);

Error, invalid input: `convert/base` expects its 1st argument, n, to be of type {integer, list(integer)}, but received n


2. for n while n <= convert(map(c ->c^3, convert(n, base, 10)), `+`) do print(n) end do;

                               1
                               2
                               3
                               4
                               5
                               6
                               7
                               8
                               9

I am facing a problem in maple 17. When I am launching maple 17, each time  the evaluating process is automatic started in server 1 without any input. I could not interrupt this evaluating process. For removing this problem I reinstall maple 17, but the problem is exist.  Please inform me soon how remove this problem ?

Maple responds with FALSE, but for which real number is the statement false?

restart; assume(x, 'real');

verify(abs(x), sqrt(x^2), {'equal'});

kind regards,

Harry Garst

Hello,

I'm very interested by this function so as to export my code Maple as MATHML thanks to the function ExportContent.

However, this function can be applied only to a expression.

I would like to apply to all the code of my worksheet.

Do you have some ides to apply ExportContent or equivalent to obtain the MATHML transformation of a whole worksheet ?

Thanks a lot for your help.

Best regards,