Items tagged with assuming

Feed

Why does the following statement not evaluate, or better yet, how can I make it do so?

 

A:=value(floor(p)) assuming p>0,p<1,p::real;

or

A:=simplify(floor(p)) assuming p>0,p<1,p::real;

or any one of a lot of different attempts along the above lines, all of which seem (to me) that they should yield

A:=0

rather than

A:=floor(p)

which is what I get.

Thanks in advance

``

-(-2*N__1*`&omega;__2`*`&omega;__1`^2*lambda-8*N__2*lambda^3*`&omega;__2`-sqrt(4*N__1^2*lambda^2*`&omega;__1`^2*`&omega;__2`^2+16*N__1*N__2*lambda^4*`&omega;__2`^2+N__1*N__2*`&omega;__1`^2*`&omega;__2`^4+4*N__2^2*lambda^2*`&omega;__2`^4)*`&omega;__1`)/(4*N__1*lambda*`&omega;__1`^2*`&omega;__2`+16*N__2*lambda^3*`&omega;__2`)

-(-2*N__1*`&omega;__2`*`&omega;__1`^2*lambda-8*N__2*lambda^3*`&omega;__2`-(4*N__1^2*lambda^2*`&omega;__1`^2*`&omega;__2`^2+16*N__1*N__2*lambda^4*`&omega;__2`^2+N__1*N__2*`&omega;__1`^2*`&omega;__2`^4+4*N__2^2*lambda^2*`&omega;__2`^4)^(1/2)*`&omega;__1`)/(4*N__1*lambda*`&omega;__1`^2*`&omega;__2`+16*N__2*lambda^3*`&omega;__2`)

(1)

`assuming`([simplify(-(-2*N__1*`&omega;__2`*`&omega;__1`^2*lambda-8*N__2*lambda^3*`&omega;__2`-(4*N__1^2*lambda^2*`&omega;__1`^2*`&omega;__2`^2+16*N__1*N__2*lambda^4*`&omega;__2`^2+N__1*N__2*`&omega;__1`^2*`&omega;__2`^4+4*N__2^2*lambda^2*`&omega;__2`^4)^(1/2)*`&omega;__1`)/(4*N__1*lambda*`&omega;__1`^2*`&omega;__2`+16*N__2*lambda^3*`&omega;__2`), 'size')], [all, positive])

(1/4)*(4^(1/2)*((N__1*lambda^2+(1/4)*N__2*`&omega;__2`^2)*`&omega;__2`^2*(N__1*`&omega;__1`^2+4*N__2*lambda^2))^(1/2)*`&omega;__1`+2*lambda*`&omega;__2`*(N__1*`&omega;__1`^2+4*N__2*lambda^2))/(lambda*`&omega;__2`*(N__1*`&omega;__1`^2+4*N__2*lambda^2))

(2)

`assuming`([combine((1/4)*(4^(1/2)*((N__1*lambda^2+(1/4)*N__2*`&omega;__2`^2)*`&omega;__2`^2*(N__1*`&omega;__1`^2+4*N__2*lambda^2))^(1/2)*`&omega;__1`+2*lambda*`&omega;__2`*(N__1*`&omega;__1`^2+4*N__2*lambda^2))/(lambda*`&omega;__2`*(N__1*`&omega;__1`^2+4*N__2*lambda^2)), 'size')], [N__1 > 0, N__2 > 0, `&omega;__1` > 0, `&omega;__2` > 0, lambda > 0])

(1/4)*(`&omega;__1`*`&omega;__2`*((4*N__1*lambda^2+N__2*`&omega;__2`^2)*(N__1*`&omega;__1`^2+4*N__2*lambda^2))^(1/2)+2*lambda*`&omega;__2`*(N__1*`&omega;__1`^2+4*N__2*lambda^2))/(lambda*`&omega;__2`*(N__1*`&omega;__1`^2+4*N__2*lambda^2))

(3)

 

``

``


Download question_13.12.06.mw

I have a situation

restart:
with(IntegrationTools):
with(PDEtools):
declare(w(r));
assume(delta::constant, R::constant, K:: constant, U::constant):
ODE_1:=diff(w(r),r)+R*(diff(w(r), r))^3-K/r;
 
declare(u(r));
ODE_T_1:=collect(algsubs(w(r)=u(r)+(r-1)*(U-0)/(delta-1), ODE_1),diff(u(r), r)) ;
eq1:=int(phi[i](r)*ODE_T_1,r=1..delta) assuming delta > 1;
eq2:=Expand(eq1);
eq3 := applyop(u->Parts(u,phi[i...

Hi all,

I have the following functions.

restart:
with(IntegrationTools):
N:=4:

i:='i':
for i from 1 to N do
assume(x[i-1]::constant):
assume(x[i+1]::constant):
assume(x[i-1]::constant):
assume(h::constant):
 phi[i](t):=piecewise(t>=x[i-1] and x[i]>t, (t-x[i])/(h), x[i]
(x[i+1]-t)/(h), 0);
end do;
## my goal is solve the integrals involving phi[i]'s as integrand and x[i]'s as its limits.

This example was reported to me after a Calculus II student encountered this ridiculous result:

f := (k+5)/sqrt(k^7+k^2):
Int( f, k=1..infinity ):
% = value( % );
/infinity
| k + 5
| -------------- dk = -infinity
| (1/2)
/1 / 7 2\
\k + k /

Dear Maple Users,

I'm solving quite a complicated task, so I tried to simplified an example.

There is an equation:

SOL := fsolve(Nz+int(int(StrssCctXY(x, y), x = -(1/2)*b .. (1/2)*b), y = -(1/2)*h .. (1/2)*h) = 0, {C1 = -(1/2)*h .. (1/2)*h})

 StrssCctXY(x,y) is piecewise function containing C1 variable, to solve an equation I had to use assumptions on C1 via assume(C1<num1, C1>-num2) command, after that C1 becomes C1~;

 

can you explain me when i solve equation beta-t*beta=0 with condition beta<>0. I write command:
r :=solve(beta-t*beta, t, UseAssumptions), assuming beta <> 0
It works well. But it will be a problem when command is:
r :=solve(beta-t*beta, t, UseAssumptions), assuming beta <> 0, gamma<>0;
Error: Error, (in assuming) when calling 'assume'. Received: 'cannot assume on a constant object'.
thank you very much.

How can I most succinctly and straightforwardly get Maple to simplify f below to g below?

> f:=(6*x^2-6*x+6)^(1/2)*(2*x^2-2*x+2)^(1/2);

                          (1/2)                 (1/2)
          /   2          \      /   2          \     
          \6 x  - 6 x + 6/      \2 x  - 2 x + 2/     

> g:=simplify(f) assuming x::real;

                        (1/2) / 2        \
                     2 3      \x  - x + 1/

> simplify(g-f...

I need maple to perform the following:

"int((1+m^2*(alpha-theta)^2*sin(theta)^2/sin(alpha)^2/alpha^2)^(1/2),theta = 0 .. alpha)"

but maple does not integrate. I have tried assuming that the term inside the square root is positive, with no result. What else can I do?

I really need a result to the integration below but Maple 13 just won't return one. Could you please help me or advise me as to what might be wrong or what I might try ? I'm integrating on the real line in x but even when I alter the limits of integration maple just returns the integrand.

 

s1:= int((1/8)*sqrt(2)*exp((1/2)*k^2*cos(x)^2/sigma^2)*exp(-(1/2)*k^2/sigma^2...

Please help me to make sense of the ways to use the simplify function. In this particular case Maple does some computation and gives me some huge output which I paste below. When I try to simplify the huge output Maple just hangs. But if I use varied commands of simplify detailed below such as simplify(huge_output,symbolic)  or  simplify(huge_output,size)  Maple gives me an output but none of the output are equal to each other and I also noticed that in one instance...

Why does this happen to Maple 15?

    `assuming`([sum(k*p*Beta(k, p+1), k = 1 .. infinity)], [p > 1]); eval(%, p = 2)

Is it possible to show with Maple that for any real p>1 the series converges to p/(p-1), e.g.,

    `assuming`([sum(k*p*Beta(k, p+1), k = 1 .. 1000)], [p > 1]): subs(p = 2, %): evalf(%)

How do I show this symbolically? Thanks.

In the following examples I attempt to remove integer multiples of Pi from sin/cos. Nothing works.

Why? How do I make it work?

Thank you.

 

term1:=cos(Pi*(3*a-2*n));
> term2:=cos(3*Pi*a+2*Pi*n);
> term3:=cos(3*Pi*a-2*Pi*n);

                     term1 := cos(Pi (3 a - 2 n))


       ...

i need to calculate the radius of an arc. i have that the chord lenght is 96 inches and the distance between the chord and the arc is 6 inches perpendicular from the center of the chord. assuming that i am working with a circle.... how do i calculate the radius of the circle????

Thank you

Your answer makes sense if Maple evaluates from innermost to outermost, so the substitution in T2c takes place AFTER Test is evaluated with a symbolic value of "n".

 But it doesn't explain why the "assuming" examples work - or don't work. If "Test" is evaluated first, then none of the examples should work; if the assumption is passed through "simultaneously" (whatever that means), then the wording of the error message doesn't make sense...

1 2 3 Page 2 of 3