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Hi All,

 

I have o problem with simplify. A variable cp1r has been assumed to be positive. Why simplify still has csgn(cp1r) for it? Here is my code:

tmp := subs(cp1t(t)=cp1r, cp2t(t)=cp2r, Ca[2]);
1 / 2 2
----------- |-cp2r sin(x[1]) sin(x[7]) cp1r
2 2 |
cp1r cp2r |
\

2
+ 2 cp2r sin(x[1]) cos(x[1]) cos(x[7]) sin(x[7]) cp1r +

1 / 2 2 /
-------------- \cp2r cos(x[1]) cos(x[7]) sin(x[7]) \
(1/2)
/ 2\
2 \cp1r /
2 \\\
-2 cos(x[1]) cos(x[7]) sin(x[1]) + 2 sin(x[1]) cos(x[1])//|
|
|
/
assume(cp1r > 0, cp2r > 0);
simplify(tmp);
1 / / 3 3
---------- \sin(x[1]) sin(x[7]) \-cos(x[1]) cos(x[7])
2
cp1r cp1r

+ 2 cos(x[1]) cos(x[7]) cp1r csgn(cp1r) cp1r

2 3 \ \

- cp1r csgn(cp1r) cp1r + cos(x[1]) cos(x[7])/ csgn(cp1r)/

 

should csgn(cp1r) be simplified to 1 already? What is wrong with my script?

 

Thanks 

Everett

restart:

f:=exp(-I*Im(s)*(t)); # I is iota, s is a complex number and t is time.

simplify(%);

Thanks

I want to reduce all solution of the equation sin(x)^2=1/4

restart:
sol:=solve(sin(x)^2=1/4, x, AllSolutions);

and

restart:
k:=combine((sin(x))^2);
sol:=solve(k=1/4, x, AllSolutions = true, explicit);
simplify(sol);

How can I reduce solution sol := -1/3*Pi*_B3+1/6*Pi+Pi*_Z3 ?

How can I get x= pi/6+k*pi and x= -pi/6+k*pi?

 

Is there a way to make the first part of this look like the second picture depicted here? Also after intigration is there a way to make the highlighted posrtion not have an "ln(e)" parts and just have the exponetials and there constants?

Obviously I dont want to have to manually input this section everytime, is there some command I can use to achieve this goal?

hi.please help me .why rule ''''evalf'''' dose not work properly?and I should copy the former answer in a another 'evalf' rule a gain??can I use only one rule 'evalf' instad twice useage??

this program attached blow.thanks alot

evalf.mw

 

When I do simplify(LegendreP(n, 1, cos(t))), Maple gives me -sqrt(1-cos(t))*sqrt(cos(t)+1). Isn't it the same thing as -sin(t)? How can I have Maple further convert/simplify it to -sin(t)? (I tried simplify(%, trig) but it didn't work). I am new on Maple. Thanks in advance for anyone's help!

 

I want to simplify an expression via simplify();

However since the expression is very long (length(expression)=2713899) Maple is taking too long...So far one hour.

I have already simplified several expressions of a similar kind with lengths of up to 400.000 which worked fine within minutes.

Is there any way to speed up the process. Am I making a mistake?

Thanks in advance

Hi,

I m trying to simplify an expression involving square roots in Maple. But instead of giving the expected output it is not simplifying it. Please find the attached maple file for reference. I calculated manually and the result is 1. Please help me out for this. Thanx in advance.

Regards

Sunit

restart

temp := (1/2)*(a*r*t-b*p)*sqrt(p+v*sqrt(a*r))*sqrt(b)/(sqrt(a*b*r*t)*sqrt(a*r*t*(p+v*sqrt(a*r))))+(1/2)*(a*r*v*t-v*b*p+p*t*sqrt(a*r)-b*v^2*sqrt(a*r))*sqrt(b)/(sqrt(a*b*r*t*(p+v*sqrt(a*r)))*sqrt(t*(p+v*sqrt(a*r))))+(1/2)*b*(p+v*sqrt(a*r))/(a*r*t)

(1/2)*(a*r*t-b*p)*(p+v*(a*r)^(1/2))^(1/2)*b^(1/2)/((a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2))+(1/2)*(a*r*v*t-v*b*p+p*t*(a*r)^(1/2)-b*v^2*(a*r)^(1/2))*b^(1/2)/((a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2))+(1/2)*b*(p+v*(a*r)^(1/2))/(a*r*t)

(1)

simplify(temp)

-(1/2)*(-(p+v*(a*r)^(1/2))^(1/2)*b^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a^2*r^2*t^2+(p+v*(a*r)^(1/2))^(1/2)*b^(3/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*p-b^(1/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a^2*r^2*t^2*v+b^(3/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*v*p-b^(1/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t^2*p*(a*r)^(1/2)+b^(3/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*v^2*(a*r)^(1/2)-b*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*p-b*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*v*(a*r)^(1/2))/((a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t)

(2)

``


Download question.mw

can anybody help me? i want to check the consistency of my scheme. My equation is too long if i check manually, so i used maple 13 to simplify my equation. But it cannot simplify it because of length of output exceed limit 1000000

restart

eqn1 := u+(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2)+(u-(1-exp(-m))*u[t]+(1-exp(-m))^2*u[tt]/factorial(2))-u-(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)-u+(1-exp(-m))*u[x]-(1-exp(-m))^2*u[xx]/factorial(2)+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3))*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+(t+1-exp(-m)))-(4*(x+1-exp(-m)))*sinh(x+1-exp(-m)+(t+1-exp(-m)))+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+(t+1-exp(-m)))^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3));

(1-exp(-m))^2*u[tt]-(1-exp(-m))^2*u[xx]+(1-exp(-m))^2*u+(1-exp(-m))^2*u^3-(1-exp(-m))^2*(4*((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-16*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+4*(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)/((((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3)*((x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3+(x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)+(((x+1-exp(-m))^2-2)*cosh(x+1-exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+1-exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+1-exp(-m)+t)^3)*((x^2-2)*cosh(x+1-exp(-m)+t)-4*x*sinh(x+1-exp(-m)+t)+x^6*cosh(x+1-exp(-m)+t)^3)*(((x+1-exp(-m))^2-2)*cosh(x+2-2*exp(-m)+t)-4*(x+1-exp(-m))*sinh(x+2-2*exp(-m)+t)+(x+1-exp(-m))^6*cosh(x+2-2*exp(-m)+t)^3+(x^2-2)*cosh(x+t)-4*x*sinh(x+t)+x^6*cosh(x+t)^3))

(1)

a := simplify(eqn1);

`[Length of output exceeds limit of 1000000]`

(2)

``


Download consistency_expmle_4.mw

.

 

Its pretty obvious that i would like the to get the answer -exp(-n(cos(phi)+sin(phi))) from the 'simplify' command. What am i doing wrong? the problem

When solving a nonlinear differential equation on some variable x, but using some other parameter w, I am finding on Maple some complicated solution, which I would like to simplify by making evident what is the x dependence, and where I can compact complicated functions of the parameter w alone into new constants. How can I do that automatically?

 

For example, to have

 

(sinh(w) + ln(w))*x 

 

to be automatically called

 

c*x

 

Thank you in advance.

Hello.

given this expression

T:=unapply((1/6930)*exp(-(1/7938)*(X[4]-933)^2)*exp(-(1/6050)*(X[2]-805)^2)/((1+exp((1/50)*X[4]-(1/50)*X[2]))*Pi),X[2]);

U := unapply(sum(T(X[2]), X[4] = 0 .. 3600), X[2]):

I want to display U, but not all 3600 terms. is there anyway to simplify/reduce this sum?

kind of like geo series a+ar+ar^2+ar^3+...+ar^(n-1)=sum(ar^k,k=0..n-1) can be reduced to a*(1-r^n)/(1-r)

 

hello
i have a problem that you could help me
i have an expression that i want convert it to an expression according to the expression q[d] with maple
i have bellow expressions
B[d]:=(-d*w[1]+w[3])/(((-2*w[1]+w[3])^2)-((-d*w[1]+w[3])^2))
B[o]:=(-2*w[1]+w[3])/(((-2*w[1]+w[3])^2)-((-d*w[1]+w[3])^2))
A[o]:=w[1]*(alpha[o]-c[o]-t[o])+2*w[2]*e[o]
A[d]:=w[1]*(alpha[d]-c[d]-t[d])+2*w[2]*e[d]
q[o]:=B[o]*A[o]-B[d]*A[d]
q[d]:=B[o]*A[d]-B[d]*A[o]
i want simplify expression U[d] such as this one
U[d]:=w[1]*(q[d]*(alpha[d]-q[d]-d*q[o]-c[d]-t[d])-C)+w[2]*(e[d]*q[d]+e[o]*q[o])+w[3]*((1/2)*(q[d]+q[o])^2)
I'm looking to simplify U[d] according to the expression q[d]
please please help me

I am trying to simplify sums of a few LaguerreL polinomials of different n using the identities in the function advisor such as recurrsion relations. How does one go about in using the FunctionAdvisor identities when trying to simplify expressions containing orthogonal polynomials? 

 

 

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