mehdi jafari

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11 years, 300 days

MaplePrimes Activity


These are replies submitted by mehdi jafari

@ecterrab can this type of assumption for functions that can test their arguments for assumptions be included for all type of functions in next versions ?! for example for trangular functions , exp , logarithm and ,,,, ?
i think it will be great for physics package !

@vv really tnx, it is working perfectly

@Preben Alsholm tnx for your help and answer .


actually Z and R are different directions and thus indendepnt parameters,

Xstar is a paramatere dependet to Boundary Y 
from the last BC we conclude :
Y(T,1)=(Xstar)^n -> Xstar=Y(T,1)^(1/n) .

can not solve this problem using maple ?! what is th idea for solving this problem with and witouth maple !?

@Carl Love for example ( not surely correct example ) i say if -0.12a<a<0.08  and -0.44<c<-0.02 and b in sth like them, how many inervals can be found for my three variables that my answers are real ? the varibale are allowed to vary between -1 and 1 , but what are the intervals for each varibale that in those intervals,the answers are real ?
how many intervals exist in which the result is real ? can we find group of intervals for each varibale that in those intervals the answers are real ? for example
group one ={ <a< , <b< , <c< } -> the answers are real 

group two = { .... } sth like this, -> the answers are real  

and ....
what are the groups of intervals in which the answers are real .

@acer @Carl Love can i find the exact intervals of each variable for which it has real answer ?
the varibale can change between -1 and 1 , but what are intervals for each varibale for which the answer is real ? how many intervals are there for the vrabiles for which the answer in them are surely real ?

please change e to exp() and specify the varibales, are your variables theta[1](t) or theta[1] ?
please make correction about the varibales . and then speciify the equation which are going to be solved simultaneously. after all . u can use the command like this : 
dsolve( {eq[1],eq[2],eq[3]});
or after correction , please upload your new sheet again.

@Kitonum it is exactly what i needed

what is your problem exactly ? post your mw code so that we can help

@Axel Vogt as u can see,the integration has not been solved , ! i want the integration with respect to x, to be solved. in your answers,there is no x, but in mine there are . whay i can not do that , what is your maple version ?

@Axel Vogt this does not answere in my question ! :( 

 

NULL

restart:

l := 16; m := .75; EIy := 0.2e5; EIz := 0.400e7; GJ := 0.1e5; mj := .1; Pz := 5000;

16

 

.75

 

0.2e5

 

0.400e7

 

0.1e5

 

.1

 

5000

(1)

eq1 := m*(diff(diff(w(x, t), t), t))+diff(diff(EIy*(diff(diff(w(x, t), x), x))+(EIz-EIy)*(diff(diff(w(x, t), x), x))*(theta(x)+phi(x, t))^2, x), x)-Pz = 0;

.75*(diff(diff(w(x, t), t), t))+0.2e5*(diff(diff(diff(diff(w(x, t), x), x), x), x))+0.398e7*(diff(diff(diff(diff(w(x, t), x), x), x), x))*(theta(x)+phi(x, t))^2+0.1592e8*(diff(diff(diff(w(x, t), x), x), x))*(theta(x)+phi(x, t))*(diff(theta(x), x)+diff(phi(x, t), x))+0.796e7*(diff(diff(w(x, t), x), x))*(diff(theta(x), x)+diff(phi(x, t), x))^2+0.796e7*(diff(diff(w(x, t), x), x))*(theta(x)+phi(x, t))*(diff(diff(theta(x), x), x)+diff(diff(phi(x, t), x), x))-5000 = 0

(2)

w := unapply(q[1](t)*cosh(1.8751*x/l),x, t);

proc (x, t) options operator, arrow; q[1](t)*cosh(.1171937500*x) end proc

(3)

theta := proc (x) options operator, arrow; 0 end proc;

proc (x) options operator, arrow; 0 end proc

(4)

#theta(x);eq1;

phi := unapply(q[4](t)*sin(1.5708*x/l),x, t);

proc (x, t) options operator, arrow; q[4](t)*sin(0.9817500000e-1*x) end proc

(5)

eq2:=unapply(lhs(eq1),x,t);

proc (x, t) options operator, arrow; .75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)+3.772661154*q[1](t)*cosh(.1171937500*x)-302.9569534*q[1](t)*cosh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000 end proc

(6)

f[1] := value(expand(int(expand(eq2(x,t)*cosh(1.8751*x/l)), x = 0 .. l)));

int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

(7)

expand(eq2(x,t)*cosh(1.8751*x/l)):

Int(%, x = 0 .. l);

expand(%);

value(%);

 

Int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

 

Int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

 

int(.75*(diff(diff(q[1](t), t), t))*cosh(.1171937500*x)^2+3.772661154*q[1](t)*cosh(.1171937500*x)^2-302.9569534*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*sin(0.9817500000e-1*x)^2+2515.691179*cosh(.1171937500*x)*q[1](t)*sinh(.1171937500*x)*q[4](t)^2*sin(0.9817500000e-1*x)*cos(0.9817500000e-1*x)+1053.716523*q[1](t)*cosh(.1171937500*x)^2*q[4](t)^2*cos(0.9817500000e-1*x)^2-5000*cosh(.1171937500*x), x = 0 .. 16)

(8)

 

 

NULL

 

Download 1.mw

f(x)=x^2= 1/(1/x)^2 , thus what does this plot will have that the other one won't ?

@Preben Alsholm but now it is optional, and u should subscribe for receiving email notifications.

i have tried sth like this, but it has some problems, for example for an unkown number of rows and columns,i should update Matrix dimensions ! but i do not know them. i want the code be general .


restart:

M:=Matrix(2,9):N:=Matrix(3,4):

file1:="ELIST.txt":fclose(file1);

file2:="NLIST.txt":fclose(file2);

for i from 1 to 2 do

M[i]:=(fscanf(file1,"%{9}fr")[1]);

od:

for i from 1 to 3 do

N[i]:=(fscanf(file2,"%{4}fr")[1]);

od:

N;M;

Matrix(3, 4, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (2, 1) = 2., (2, 2) = 50., (2, 3) = 0., (2, 4) = 0., (3, 1) = 3., (3, 2) = 25., (3, 3) = 0., (3, 4) = 0.})

 

Matrix(2, 9, {(1, 1) = 1., (1, 2) = 1., (1, 3) = 1., (1, 4) = 1., (1, 5) = 0., (1, 6) = 1., (1, 7) = 1., (1, 8) = 3., (1, 9) = 0., (2, 1) = 2., (2, 2) = 1., (2, 3) = 1., (2, 4) = 1., (2, 5) = 0., (2, 6) = 1., (2, 7) = 3., (2, 8) = 2., (2, 9) = 0.})

(1)

``


Download matrix.mw

@Mac Dude it only imports one row ! not more !

restart:

FileTools[Text][ReadFile]("NLIST.txt");

"       1     0.00000000000       0.00000000000       0.00000000000    
       2     50.0000000000       0.00000000000       0.00000000000    
       3     25.0000000000       0.00000000000       0.00000000000
        

"

(1)

op(FileTools[Text][ReadFile]("ELIST.txt"));

"       1   1   1   1   0   1      1     3     0
       2   1   1   1   0   1      3     2     0

      
      "

(2)

M:=ImportMatrix("NLIST.txt", source=delimited, datatype=float, delimiter=" ");

Error, (in ImportMatrix) end of input encountered while reading Matrix

 

M:=ImportMatrix("NLIST.txt", source=delimited, datatype=float);

M := Matrix(5, 1, {(1, 1) = 1., (2, 1) = 0., (3, 1) = 0., (4, 1) = 0., (5, 1) = 2.})

(3)

 

``


Download convert_to_array.mw

@schloemilch please insert your original question here, and maybe i can help u and upload a working version !
actually i do not know u can not use these. until u insert your worksheet here. tnx

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