mehdi jafari

559 Reputation

13 Badges

6 years, 309 days

MaplePrimes Activity


These are replies submitted by mehdi jafari

@vv thank you for the clarification. now another problem arises.
Error, (in fsolve) procedures don't evaluate to numeric types. could please help me to solve the issue? sorry for the inconvenience.
Inflation-Inverse-1.mw

@vv i couldn't solve the problem could please check this code? tnx in advance
Inflation-Inverse-2.mw

@pik1432 
 

restart:with(LinearAlgebra):

printlevel:=1

1

(1)

x:=<x1,x2>:
W:=<<w11, w12>|<w21, w22>>:
Cal:= (W.x)^%T.(W.x):

eq1:=Array([seq](diff(Cal,t),t=x))^+;  # 1

Vector(2, {(1) = (2*w11*x1+2*w21*x2)*w11+(2*w12*x1+2*w22*x2)*w12, (2) = (2*w11*x1+2*w21*x2)*w21+(2*w12*x1+2*w22*x2)*w22})

(2)

eq2:=(2*W^%T.W.x);#2

Vector(2, {(1) = (2*w11^2+2*w12^2)*x1+(2*w11*w21+2*w12*w22)*x2, (2) = (2*w11*w21+2*w12*w22)*x1+(2*w21^2+2*w22^2)*x2})

(3)

seq(is(eq2[i]=eq1[i]),i=1..numelems(eq1))

true, true

(4)

 

 


 

Download 2.mw

@vv thank you

@Carl Love thanks for the comment. So the answer of the integration is undefined?


 

restart

f := ((1 - a)^2 + a^2*((1 - exp(-y))*(1 - exp(-x)) - 2 + exp(-x) + exp(-y)) + a*(2 - exp(-x) - exp(-y) + (1 - exp(-y))*(1 - exp(-x))))/(1 - a*exp(-x)*exp(-y))^3;

((1-a)^2+a^2*((1-exp(-y))*(1-exp(-x))-2+exp(-x)+exp(-y))+a*(2-exp(-x)-exp(-y)+(1-exp(-y))*(1-exp(-x))))/(1-a*exp(-x)*exp(-y))^3

(1)

a := 3/10:f:

int(f*exp(-x), x= 0..y + t, AllSolutions):

s := 2*(int((int(f*exp(-x)*exp(-y), x = 0 .. y + t,AllSolutions)), y = 0 .. infinity,AllSolutions));

s := 2*piecewise(Re(t) < 0, undefined, undefined)

(2)

 

 


 

Download stat2.mw

@Kitonum 
 

``

restart

f:=[evalf(solve(x^6-3*x-5))]

[1.451571465, .5973639664+1.275126159*I, -.7760033304+.9926461157*I, -1.094292737, -.7760033304-.9926461157*I, .5973639664-1.275126159*I]

(1)

remove(has,f,I);

[1.451571465, -1.094292737]

(2)

 

 

``


 

Download remove.mw

@Carl Love @Kitonum Thank you for the answers. unanswered question gains 0 point.

@Carl Love actually i need opposite function of the Heaviside function, the case where if it is negetive it is 1 and when it is positive it is zero. Does maple have any function for this?( i've got my answer from above responses, only ask this for more knowledge).
 

@vv then what should i do for tau=0? how can simplify accounting for tau=0? is it possible?

@dharr thank you for your attention and your time.But if you look more carefully the denominator is not zero with sigma or tau equals zero.

@acer actually i have do this in lists. i asked if it is possible with sets or not?

 

restart

L:=[]:

for j in [3,5,6,1,1] do
birth:=j:
L:=[op(L) ,op(j)];
od;

3

 

[3]

 

5

 

[3, 5]

 

6

 

[3, 5, 6]

 

1

 

[3, 5, 6, 1]

 

1

 

[3, 5, 6, 1, 1]

(1)

 


 

Download setoder2.mw

@acer if i store SS1 and SS2 in matrix, using simplify assuming real does NOT operates. is it a bug ?

@acer yes. Thank you 

@Kitonum yes you are right, maybe it is a bug in maple. I have another question, can this integral be solved symbolically as a function of T ?

1 2 3 4 5 6 7 Last Page 1 of 21