mehdi jafari

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7 years, 83 days

MaplePrimes Activity


These are replies submitted by mehdi jafari

@Carl Love thank you for your answer and curiosity. I will share the correct conditions whenever possible. 

@Carl Love thank you carl for the time you spend on this. everytime i learn a lot form your knowlegde and personlaity.
Sincerely Yours.

@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 ?

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