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Hi;
what's the difference between "assuming" and "assume" commands in maple? tell me this problem with example. Thanks.
with regards...

Hi all,

I'am new to Maple and have some very basic questions. Maple can figure out that the exponential is positive.

> is(exp(x)>0) assuming x::real;
                                     true

But it cannot figure out that a sum of exponentials is positive too:

> is(sum(exp(x*i),i=1..N)>0) assuming x::real,N::integer,N>1;
                                     FAIL

What am I doing wrong? Moreover, what's wrong with this statement:

> is(ln(a+b)-ln(a)=ln(1-b/a)) assuming a::real,b::real,a>0,b>0,b<a;
                                     FAIL

Finally, is there a way to declare a generic fuction f and assume that its image is always a positive real?

Thanks

Nico 

Hi everybody,

 

Here is a very simple question :
I have two strictly positive reals r and s and the expression
a := r^2 – r*s + s^2
Obviously a is strictly positive too

But I’m not capable to prove this with Maple (2016, Windows 7), even after having declared r and s both positive which, I thought, would have prevent Maple to consider them as potentially complex.
I failed in using "is", "coulditbe", "use RealDomain ..." and so on

Totally distressed I tried to do something simpler ... which ultimately appeared still more upsetting

restart:
assume(r, ‘positive’):
coulditbe(r, ‘complex’) ;    # the answer is (surprising ?) true

I'm probably doing some mistake bigger than I but I don't find it !

Could you be compassionate enough to enlight me !!!

Mammoth thanks in advance

Hello,

I want to check via Maple whether the term is always bigger than c/4 (it is), but using verify(TERM, c/4, greater_than) just returns FALSE. I did assume(a > 0, b > 0, -1 <= c and c <= 1); It would be nice if you could tell me what exactly is causing the problem and how to solve it.

Thanks in advance

(9*a^3*b^3*c^2+4*a^6*c+16*a^5*b*c+47*a^4*b^2*c+82*a^3*b^3*c+47*a^2*b^4*c+16*a*b^5*c+4*b^6*c+16*a^5*b+80*a^4*b^2+132*a^3*b^3+80*a^2*b^4+16*a*b^5)*(a+2*b)^2*(a+b)^2*(2*a+b)^2/((4*a^6+28*a^5*b+77*a^4*b^2+106*a^3*b^3+77*a^2*b^4+28*a*b^5+4*b^6)*sqrt(-9*a^2*b^2*c^2+32*a^3*b*c+62*a^2*b^2*c+32*a*b^3*c+16*a^4+80*a^3*b+132*a^2*b^2+80*a*b^3+16*b^4)*sqrt(-9*a^4*b^4*c^2+24*a^6*b^2*c+28*a^5*b^3*c-14*a^4*b^4*c+28*a^3*b^5*c+24*a^2*b^6*c+16*a^8+112*a^7*b+276*a^6*b^2+296*a^5*b^3+220*a^4*b^4+296*a^3*b^5+276*a^2*b^6+112*a*b^7+16*b^8))

 

I can't get the Real and Imaginary parts of matrix to seperate out as required. It is an SU(2) matrix. Want to convert is to a 4 Vector (quaternion). I think because Maple doesnt know what psi(t) is being conservative so to speak. Tried assume, assuming...
 

restart

assume*{psi(t), 'real'}

assume*{real, psi(t)}

(1)

V := Matrix(2, 2, {(1, 1) = (1/4)*cos((1/2)*psi(t))*sqrt(2)*sqrt(4-2*sech((1/6)*sqrt(3)*t))-I*cos((1/2)*psi(t))*sqrt(2)*tanh((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t))-((1/2)*I)*sin((1/2)*psi(t))*sqrt(2)*sqrt(3)*sech((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t)), (1, 2) = (1/2)*cos((1/2)*psi(t))*sqrt(2)*sqrt(3)*sech((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t))+((1/4)*I)*sin((1/2)*psi(t))*sqrt(2)*sqrt(4-2*sech((1/6)*sqrt(3)*t))-sin((1/2)*psi(t))*sqrt(2)*tanh((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t)), (2, 1) = ((1/4)*I)*sin((1/2)*psi(t))*sqrt(2)*sqrt(4-2*sech((1/6)*sqrt(3)*t))+sin((1/2)*psi(t))*sqrt(2)*tanh((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t))-(1/2)*cos((1/2)*psi(t))*sqrt(2)*sqrt(3)*sech((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t)), (2, 2) = ((1/2)*I)*sin((1/2)*psi(t))*sqrt(2)*sqrt(3)*sech((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t))+(1/4)*cos((1/2)*psi(t))*sqrt(2)*sqrt(4-2*sech((1/6)*sqrt(3)*t))+I*cos((1/2)*psi(t))*sqrt(2)*tanh((1/6)*sqrt(3)*t)/sqrt(4-2*sech((1/6)*sqrt(3)*t))})

Matrix(%id = 18446744074495800014)

(2)

``

``

v := Vector(4, [Re(V[1, 1]), Im(V[1, 2]), Re(V[2, 1]), Im(V[1, 1])])

Vector[column](%id = 18446744074531387982)

(3)

``


 

Download extract_R_and_I.mw

I'm trying to solve

deq := diff(theta(x), x) = -sin(theta(x));
                         
sol := dsolve({deq, theta(0) = -(1/2)*Pi});
 

The answer should be theta=-2*arctan(exp(-x))

but I get

theta(x) = arctan(2*exp(-x+I*Pi)/((exp(-x+I*Pi))^2+1), -((exp(-x+I*Pi))^2-1)/((exp(-x+I*Pi))^2+1))

Does it have to do with assuming?

Staffan Malmberg

 

#my testing for wildcard to one
#after testing, it loop a very long time and not stop
ppp := [[0,0,0,x],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
mmmeaght1 := [seq(0,ii=1..nops(ppp[1]))]:
bbb1 := [seq(0,ii=1..nops(ppp[1]))]:
emap := [(xx) -> if [xx < 0 assuming x > 0] then 0 else 1 end if, (xx) -> evalf(1/(1+exp(xx)))]:
#trace(perceptronrule1);
MM(ppp, ttt1, mmmeaght1, bbb1, check1, emap);
 

when test wildcard variable for input, would like to assume x > 0 then

i try assuming x > 0 , got error

 

Hi.

I have 2 expressions, the first expression cosist of x´s and y´s, the other expression consits only of x´s

I wanna test the relation between those 2 expression to check wether A>B giving the condition that 2y<x

I have tried this:

assume(2y<x);

is(A>B);

the problem is that maple returns FAIL, I could put in values to check and it works, but that is not really what im trying to acomplish.

 

Thanks

I am interested in the behaviour of a system of equations close to the origin- these equations are quite long, and there are a lot of them so i would like to have commands that i can use to assume products of variables are zero. 

here are the first two polynomials:


alpha*k[a1]*B[1]^2+(-alpha*k[a1]-alpha*k[a2])*B[2]*B[1]+2*alpha*k[a1]*B[1]*B[11]+alpha*k[a1]*B[12]*B[1]+2*alpha*k[a1]*B[1]*B[211]+alpha*k[a1]*B[221]*B[1]+2*alpha*k[a1]*B[1]*B[2211]+(-alpha*R[b]*k[a1]-k[d1])*B[1]+2*B[11]*k[d1]+B[12]*k[d2]+k[d1]*B[211]+k[d2]*B[221]

(-alpha*k[a1]-alpha*k[a2])*B[2]*B[1]+alpha*k[a2]*B[2]^2+2*alpha*k[a2]*B[2]*B[22]+alpha*B[2]*B[12]*k[a2]+alpha*k[a2]*B[2]*B[211]+2*alpha*k[a2]*B[2]*B[221]+2*alpha*k[a2]*B[2]*B[2211]+(-alpha*R[b]*k[a2]-k[d2])*B[2]+B[12]*k[d1]+2*B[22]*k[d2]+k[d1]*B[211]+k[d2]*B[221]

the varables are the terms with B and a subsript and everything else is a parameter.

My intuition was to use coeffs but I couldn't get anything helpful

hi every one ! i want to use Assume option to simplify some expression ! but it is not working ! what should i do !?

i have assume that ( a+b+c=0) and i want maple returns me exp(a+b+c) =1 ! but it does not ! what should i do !?


restart:with(Physics):

Assume(a+b+c=0):

about(a+b+c)

a+b+c:

  is assumed to be: 0

 

simplify(exp(a)*exp(b)*exp(c))

exp(a+b+c)

(1)

simplify(exp(a+b+c))

exp(a+b+c)

(2)

 


Download assume.mw

Assume i got an equation like f(x):=2*x-tan(x); how to us solve or other function to give me that this equation got 2 root: x=1/2*(tan(x)) and x=arctan(2x).

^^ hope everyone can help.

Thank for spend time reading.

intersection in the geometry package does not seem to recognize assume.

restart: with(geometry):

assume(p[1]<>0, p[2]<>0, p[3]<>0);
assume(q[1]<>0, q[2]<>0, q[3]<>0);
point(T,[p[1],q[1]]);
point(U,[p[2],q[2]]);
point(V,[p[3],q[3]]);
point(Op,[0,0]);

line(OT,[Op,T]);
line(OU,[Op,U]);
line(OV,[Op,V]);

point(B,2*q[2],solve(subs(x=2*q[2],Equation(OU)),y));
coordinates(B);
IsOnLine(B,OU);

PerpendicularLine(AD,B,OT);
ArePerpendicular(AD,OT);
sol:=solve({Equation(AD),Equation(OT)},{x,y});
eval(x,sol);
point(A,eval(x,sol),eval(y,sol));  ## the intersection exists
intersection(xA,AD,OT); ## fails
about(p[1]),about(q[1]);

Hi guys,

I'm doing some heavy analytical calculations, as the calculations take hours to finish, I want to save the results, which is easily done, however when I load the saved results, it behaves differently, somehow I can no longer substitute the variables. I extracted the main problem into the following minimal working example.

Thanks for any useful insight,

Cheers, Sören

restart():

assume(`in`(n, real))

x := n+1

n+1

(1)

subs(n = 1, x)

2

(2)

save x, "./myfunction.m";

read "./myfunction.m";

x

n+1

(3)

Now we can't substitute n in x:

subs(n = 1, x)

n+1

(4)

subs(n = 1, n+1)

2

(5)

y := x:

Download assume_and_save.mw

Hey,

I think I found a bug concerning the useage of assume and alias:

restart:

alias(a=a(t),b=b(t));

a, b

(1)

assume(a(t),real);

getassumptions(a(t));

{((a(t))(t))::real}

(2)

assume(b(t),real);

getassumptions(a(t));

{((b(t))(t))::real}

(3)

getassumptions(b(t));

{((b(t))(t))::real}

(4)

 

 

 

Commenting out the alias command produces correct results. I am on linux with build 922027.

alias_bug.mw

I am having trouble removing assumptions that are stored within expresssions.

Example code:

assume(l1>0): # this assumptions later helps to find a solution for a geometric problem with two four-bar-linkages
a := sqrt(l1);
save a, "test.m";
restart;
read "test.m"
a; # the assumptions are stored within the saved data
l1:='l1'; # try to remove the assumption
a; # assumption in a still existing
subs({l1=2}, a); # nothing happens: I can not access l1 any more
subs({l1~=2}, a); # This does not work either, nothing changes in a

So my question is: How do I remove the assumption within a stored expression?

My main problem lies in the handling of the expression with assumptions. At some point, I want to generate Matlab code, and the codegen-command gives me:

Warning, the following variable name replacements were made: l1~ -> cg

 

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