MaplePrimes Questions

Hello

It has been years since I last used maple so I apologize if my question makes no sense and my code is outdated (and wrong!).   

I need to write a function (proc) that uses a functional operator inside.  Something like

test:=proc()

local f,vars, func, aux, res;

f:=arg[1]; # a list

vars:=arg[2]: # a list

aux:=op(vars):

func:=aux -> f:  # a function from () to []

res:=func(op(func(aux))):

return(res):

end:

 

This is the basic idea of the proc.  If f:=[y, y*z-x, -15*x*y-x*z-x] and vars:=[x,y,z], the function of a function does not return as it supposes to.  Please tell me what I am doing wrong and how to improve and update the code.

 

many thanks

 

Ed

 

 

 

im just getting a generic recursive assignment plus first unused arg error

thats what the maple engine says im doing wrong ill get the number of recursives counted but i guess it just makes me the most angry because its like the same type of error for me every ******* time

hy 
need help 
i made this code but i can not get the answer ,help me to find out where i did wrong.

thanx in advance




restart;
f:=x->(x^3+3*x^2-1);
n:=30;
tol:=1e-9;
a[0]:=0;
b[0]:=10;
Digits :=15;

 

printf("No root F(x) abs(x[i+1]-x[i])\n");

for i from 1 to n do
t[i-1] :=evalf( (b[i-1]-a[i-1])/(f(b[i-1])-f(a[i-1])));
c[i-1] := evalf((a[i-1]*f(b[i-1])-b[i-1]*f(a[i-1]))/(f(b[i-1])-f(a[i-1])));
x[i] :=evalf( x[i-1]-t[i-1]*f(x[i-1])^2/(f(x[i-1])-f(c[i-1])));

printf("%d %10.15f %10.15f %10.15e \n",i,x[i],f(x[i]),abs(x[i]-x[i-1]));
if f(a[i-1])*f(c[i-1])<0 then
a[i]:=a[i-1];
b[i]:=c[i-1];
else
a[i]:=c[i-1];
b[i]:=b[i-1];
if abs(f(x[i]))<tol then
print("approximate solution"= x[i]);
print("No of iterations"= i);
break;
end if;
end if;
end do:

Hey 

 

I just went from maple 18 to maple 2016. I use maple as a writing program aswell for notes and school.

 

I realized that maple 2016 does output lines when writing math in a textbox. This is kind of annoying since notes in chemistry is not allways a true mathematical expression. Therefore maple complains. Is there anyway to change this? 

Andreas

Hi all,

I was wondering how to go about validating some airfoil designs for my Formula SAE team's CFD results.  I know this is more common with simplier calculations but I'm hoping using Maple and maybe the new algebraic manipulation of non-comunitive differential operators, I could achive what I am after.   The two calculations of interest are the drag force and downforce.  Can someone shed some light? Thanks

Lets say we have to vectors u := Vector(3,[0,a_2,a_3]) and v :=Vector(3,[a_1,a_2,a_3]), in which a_1, a_2 and a_3 are arbitrary constants. It is clear that if we set a_1=0 we could see that u is contained in the vector space of v. Is there a function in Maple like isSubspace(u,v) which returns a boolean true or false?

An alternative interpretation could be that the image of u is a subset of the image of v.

Thank you alot for reading my question. 

 

When I try to use CodeTools:-Profiling:-Profile() (with no arguments), I get "kernel connection lost" after about a minute. Has anyone used this sucessfully, with no arguments? If so, would you please post a worksheet? I'm using Windows 8, if that makes a difference.

I think that I'll need to revert to the older kernelopts(profile= true).

In Maple V, Release 4 (1996):

 

_EnvX:=0; b:=0; proc() global b; print(_EnvX, b); assign('_EnvX=1, b=1'); print(_EnvX, b); end();
_EnvX:=0: b:=0: proc() global b; print(_EnvX, b); _EnvX:=1; b:=1; print(_EnvX, b); end();;

gives:
                                 0, 0
                                 0, 1

In newer Maple:

 

_EnvX:=0; b:=0; proc() global b; print(_EnvX, b); assign('_EnvX=1, b=1'); print(_EnvX, b); end proc();
_EnvX:=0: b:=0: proc() global b; print(_EnvX, b); _EnvX:=1; b:=1; print(_EnvX, b); end proc();

?


Hello! 

I'm trying to plt something, I can't see why Maple isn't multiplying these two rows as it should. 

Can anyone see the problem?

 

This preamble is loaded:

restart;

with(Units[Standard]);
with(ArrayTools);

with(LinearAlgebra);

with(Statistics);
with(plots);
with(CurveFitting);

 

 

 

 

 

with(plots):

with(LinearAlgebra):

 

asd := [3.5400000*10^5, 3.4700000*10^5, 3.3700000*10^5, 3.2700000*10^5, 3.1700000*10^5, 3.0900000*10^5, 3.0300000*10^5, 2.9600000*10^5, 2.9200000*10^5, 2.8900000*10^5, 2.8600000*10^5, 2.8500000*10^5, 2.8200000*10^5, 2.8100000*10^5, 2.7900000*10^5, 2.7800000*10^5, 2.7800000*10^5, 2.7700000*10^5, 2.7600000*10^5]:
 

NULL

asf := [0.2866400798e-1, 0.6112772793e-1, 0.9946549241e-1, .1349950150, .1645923341, .1877395591, .2054364684, .2189514789, .2293646837, .2374847679, .2439099369, .2490583805, .2532402792, .2566744211, .2595269747, .2619197962, .2639470478, .2656751966, .2671596322]:

 

asd*asf;

[354000.0000, 347000.0000, 337000.0000, 327000.0000, 317000.0000, 309000.0000, 303000.0000, 296000.0000, 292000.0000, 289000.0000, 286000.0000, 285000.0000, 282000.0000, 281000.0000, 279000.0000, 278000.0000, 278000.0000, 277000.0000, 276000.0000]*[0.2866400798e-1, 0.6112772793e-1, 0.9946549241e-1, .1349950150, .1645923341, .1877395591, .2054364684, .2189514789, .2293646837, .2374847679, .2439099369, .2490583805, .2532402792, .2566744211, .2595269747, .2619197962, .2639470478, .2656751966, .2671596322]

(1)

 

NULL

 

Thank you!

Download maple_what.mw

I did not understand fully some of the notation used in 2D when I had the tools->options->Display->Output display->2D. So I thought if I change it to Maple notation. I might see what the symbol actually mean.  But when I did so, the result was even more confusing. Full of typesetting:-mrow commands and hard to read.

Here is the output in 2D

restart;
int(1/( (x-a)*(x-b)),x=-infinity..infinity  );

And here is the output when I switched to Maple output:

I was expecting to see "normal" looking Maple commands, which I can understand. Even the Latex is easier to read than the above mumple jumple code:

 

Does this mean one should forget about using Maple notation for output from now on? Why is it the output so complicated?

I have a dataset:

NMP:=<0.530,0.555,0.572,0.592>:
ETOH:=<0.136,0.153,0.163,0.170>:

For these four data points [NMP,ETOH] I want to find the least square function in form of:

ln(ETOH/(1-ETOH-NMP))=a*ln(NMP/(1-ETOH-NMP))+b

also I need to find appropriate a,b constant values.

This function is implicite so I cannot use with(Statistics):NonlinearFit.

Can you help me how to determine a,b constants?

Hi 

friend i want fit a curve regarding some data and fnction and how we can find the values of a,b,c and d for the following 

f=1-(8*a*b+6*c*d/(b*k*x))/(2*a*b+c*b*(1/(b*k*x)))

X := Vector([200, 210, 220, 230, 240, 250, 260, 270, 280, 290])

Y := Vector([.4172, .3030, .4668, .3317, .1276, .1303, .1733, .1451, .3466, .4125])

Code & result:

> map((a::uneval,b)->'args',[a,b,c,d],1..4,x);       
['a, 1 .. 4, x', 'b, 1 .. 4, x', 'c, 1 .. 4, x', 'd, 1 .. 4, x']

> map((a,b::uneval)->'args',[a,b,c,d],1..4,x);           
[a, 1 .. 4, x, b, 1 .. 4, x, c, 1 .. 4, x, d, 1 .. 4, x]

Why is the output not:
['a', 1 .. 4, x, 'b', 1 .. 4, x, 'c', 1 .. 4, x, 'd', 1 .. 4, x]
and
[a, '1 .. 4', x, b, '1 .. 4', x, c, '1 .. 4', x, d, '1 .. 4', x]
?

In the process of simplification I have the following multi-variable polynomial:

y:=-8*C*d1^2*(-2+d1)*(-1+d1)^3*r*L*R^3+(d1^4*(-2+d1)^2*L^2-4*C*(-2+d1)*(4*d1^3-13*d1^2+16*d1-8)*(-1+d1)^2*r^2*L+4*C^2*(-2+d1)^2*(-1+d1)^4*r^4)*R^2+(2*d1^4*(-2+d1)^2*r*L^2-2*C*(-2+d1)*(5*d1^3-24*d1^2+32*d1-16)*(-1+d1)^2*r^3*L+4*C^2*(-2+d1)^2*(-1+d1)^4*r^5)*R+d1^4*(-2+d1)^2*r^2*L^2-2*C*(-2+d1)*(d1^3-6*d1^2+8*d1-4)*(-1+d1)^2*r^4*L+C^2*(-2+d1)^2*(-1+d1)^4*r^6

This polynomial contains several (-2+d1), (-1+d1) terms with varying powers in each term. My question here is how to take out common terms and then form compact multi-variable polynomial (without having physical inspection).

 

Thank you for your help.

 

MVC

 

 

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