Items tagged with expression expression Tagged Items Feed

Dear all, 

I'm trying to extract the coefficients from the equation below, the fat expressions in the equation. I don't have any trouble seperating the sine or cosine functions. But the constants are a problem. Since t is the only variable in the function i tried, coeff(R, t, 0). This does not work apparantly. Any suggestions? 

 

R:=(1/12)*C2^3*cos(2*t) - (1/48)*C2^3*cos(4*t) - C2^3*sin(t) + C2^3*sin(3*t)-C2^2*C4*cos(t) + C2^2*C4*cos(3*t) - (1/8)*C2^2*C4*sin(2*t) + (1/16)*C2^2*C4*sin(4*t) + (1/16)*C2*C4^2*cos(4*t) - C2*C4^2*sin(t) - C2*C4^2*sin(3*t) - C4^3*cos(t) - C4^3*cos(3*t) - (1/24)*C4^3*sin(2*t) - (1/48)*C4^3*sin(4*t) - (1/16)*C2^3 - (1/16)*C2*C4^2 - (1/2)*C2*cos(2*t) + C2*sin(t) + C4*cos(t) + (1/2)*C4*sin(2*t) + (1/2)*C2

I wish to substitute Re=W[0]/V[0] in the expression V[0]/W[0] * diff(u(t),t). I tried using the subs command but it didn't work. Any help would be appreciated !

Hello those in Mapleprimes,

 

I want to know whether there is a good way to modify the first expression to the second one below.

first expression: 

> p+p^(-1/(theta-1))*sum(q[i]^(theta/(theta-1)), i = (1 .. n));

second expression:

> p^(-1/(theta-1))*(p^(theta/(theta-1))+sum(q[i]^(theta/(theta-1)),i=1..n));

First and Second are the same. But, I want to know how I can modify from the former to the latter.

 

Thank you in advance.

 

taro

 

For context, I'm designing a work sheet based around quantum tunneling. Currently I'm looking at the boundary conditions.

What I want to be able to do is to set the expression psi[1] equal to psi[2], but only for the value x = 0. Is this possible? I've tried using if statements, and I've considered converting these expressions into functions for this purpose, but I'm not having much luck. 

Thanks

Blanky

I have system of equation contain of two equations

eq1:=a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1:
eq2:=a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2: 

I want to find the number k so that the equation

a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1 + k*(a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2) = 0

can be factor, where k satisfy

a:=a1+k*a2:
b:=b1+k*b2:
c:=c1+k*c2:
d:=d1+k*d2:
e:=e1+k*e2:
f:=f1+k*f2:

I tried

A:=a*x^2+b*y^2+c*x*y+d*x+e*y+f:
collect(A,x);
B:=collect(discrim(A, x), y);
C:= discrim(B,y);

and got the expression

-16*(4*a*b*f-a*e^2-b*d^2-c^2*f+c*d*e)*a=0.

For example 

> restart:

a1:=14:

b1:=-21:

c1:=0:

d1:=-6:

e1:=45:

f1:=-14:

a2:=35:

b2:=28:

c2:=0:

d2:=41:

e2:=-122:

f2:=56:

a:=a1+k*a2:

b:=b1+k*b2:

c:=c1+k*c2:

d:=d1+k*d2:

e:=e1+k*e2:

f:=f1+k*f2:

P:=c*d*e+4*a*b*f-a*e^2-b*d^2

-f*c^2:

eq1:=a1*x^2+b1*y^2+c1*x*y+d1*x+e1*y+f1:

eq2:=a2*x^2+b2*y^2+c2*x*y+d2*x+e2*y+f2:

with(RealDomain):

Q:=solve(P=0,k);

factor(eq1+Q*(eq2));

solve([

eq1=0,eq2],[x,y]);

Where, Q is k which I want to find. 

My question is, if I have the system of equations

eq1:=a1*x^3+b1*y^3+c1*x^2*y+ d1*x*y^2 + e1*x+f1*y+g1:
eq2:=a2*x^3+b2*y^3+c2*x^2*y+ d2*x*y^2 + e2*x+f2*y+g2

How can I get a similar to the 

-16*(4*a*b*f-a*e^2-b*d^2-c^2*f+c*d*e)*a=0?

 

 

 

I am new user of Maple. I have an expression like f(x,y)/g(x,y) and I want to have Maple name f(x,y) as something simple like "h" so that later equations write h/g(x,y) instead of the complete f(x,y)/g(x,y), and also to be able to simplify other equations in terms of h.

For example, a=(x+1)(y-5)^2/(x+y-3). Can I get Maple to let h=(x+1)(y-5)^2, so that I get h/(x+y-3) when I ask for a?

Sorry if this question is basic (and my math terminology incorrect), but I have been trying for a while.

I would like to tell Maple to evaluate all numbers in an expression using floating point numbers (of a desired precision). Short example:

evalf[30](x^(1/12));

prints

x^(1/12)

while I want it to print 

x^(0.0833333333333333333333333333333)

Note that the real expression may be arbitrarily complicated. I know that it is possible easily because I have already seen this somewhere, but I don't remember it. 

I'd like to bind a symbol to some expression, but display the expression analytically before evaluating, and show the value on the same line (not new line).

Something like:

restart;
a:=10: b:=3:
f:='a+b^2';`=`,f;

But this display the expression and its numerical value on 2 lines. I'd like to get this output:

      f:=a+b^2 ,`=`,19  

This is just so I can see the expression before it is evaluated, and its numerical value on same line. Or if there is a better way do this. I tried using print, but could not get the assignment to work then.

 

I can't simplify the expression of 

(1/a[x])^(-k-1+epsilon)/(1/a[I])^(-k-1+epsilon)

to 

(a[x]/a[I])^(k+1-epsilon);

 

Please tell me how to do this.

 

Thank you in advance.

 

Taro yamada

I have this expression; f(a,c,n)=64a²n-16a⁴n-256c²n+32c⁴n-96a²c²+ 8a²c⁴+24a⁴c²-64a²n²+256a²-124a⁴+ 15a⁶+256n²+64a²c²n

such as "a" is between 0 and 2, "c" between 0 and 2 and "n" between 0 and 4.

How do I plot f(a,c,n) so that I can study its sign?

 

I have defined a function by doing

 

f:= (x1,x2,x3) -> sum( [complicated expression], [summation ranges]),

 

which Maple converts to an explicit polynomial in x1,x2,x3.

 

However, it seems Maple does not quite think of f as this polynomial, but always remembers the original definition of f. How should I do in order for Maple to really define f to be this explicit polynomial?

 

 

Contrast the behavior of functions and expressions by performing the following commands.
a)Define a function f equal to x^3 and Define an expression g equal to x^3
b) Evaluate f and g at 2.
c) Evaluate f and g at y.
d )Assign the value 2 to x. Evaluate f and g.

Is it possible that this expression has an elementary one (specifically the dilog's):

Y0:=(1/16)*(s*t*(exp(2*t)*s+exp(4*t)+1)*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^16*(1+(-s^2+1)^(1/2))^16/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^16*(1-(-s^2+1)^(1/2))^16))+s^3*t*(exp(4*t)+1)*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^8*(1+(-s^2+1)^(1/2))^8/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^8*(1-(-s^2+1)^(1/2))^8))+exp(2*t)*t*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^32*(1+(-s^2+1)^(1/2))^32/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^32*(1-(-s^2+1)^(1/2))^32))+4*((exp(4*t)+1)*s+2*exp(2*t))*(s^2+2)*dilog((-exp(2*t)*s+(-s^2+1)^(1/2)-1)/(-1+(-s^2+1)^(1/2)))-4*((exp(4*t)+1)*s+2*exp(2*t))*(s^2+2)*dilog((exp(2*t)*s+(-s^2+1)^(1/2)+1)/(1+(-s^2+1)^(1/2)))+((32*s^2*t+64*t)*exp(2*t)+16*(((t+1/8)*s^2+2*t+2)*exp(4*t)-(5/4)*s*exp(-2*t)-(1/8)*exp(-4*t)*s^2+(5/4)*s*exp(6*t)+(1/8)*s^2*exp(8*t)+(t-1/8)*s^2-2+2*t)*s)*arctanh((exp(2*t)-1)*(-1+s)/((-s^2+1)^(1/2)*(exp(2*t)+1)))+8*(-s^2+1)^(1/2)*((1/8)*s*(exp(4*t)+1)*ln((exp(4*t)*s+2*exp(2*t)+s)^12/s^12)+(1/8)*exp(2*t)*ln((exp(4*t)*s+2*exp(2*t)+s)^24/s^24)+(s^2-6*t-3)*exp(2*t)+((-(1/8)*s^2-3*t)*exp(4*t)+s*exp(-2*t)+(1/8)*exp(-4*t)*s^2+s*exp(6*t)+(1/8)*s^2*exp(8*t)-(1/8)*s^2-3*t)*s))/((s*exp(-2*t)+exp(2*t)*s+2)*(exp(4*t)*s+2*exp(2*t)+s)*((-s^2+1)^(1/2)+2*arctanh((-1+s)/(-s^2+1)^(1/2))))

Also I'm wondering since Y0 should solve the ode

-(diff(diff(y(t), t), t))+(4-12/(1+s*cosh(2*t))+8*(-s^2+1)/(1+s*cosh(2*t))^2)*y(t) = C/(1+s*cosh(2*t))

with some constant C but I only get rubbish.

I ask this because I found that in another context this seems to be correct:

f1:=-(1/12)*Pi^2*((-s^2+1)^(1/2)-arccosh(1/s))/(-s^2+1)^(3/2)+(1/12)*arccosh(1/s)^3/(-s^2+1)^(3/2)-(1/4)*arccosh(1/s)^2/(-s^2+1)

f2:=(1/2)*((-s^2+1)^(1/2)*(polylog(2, s/(-1+(-s^2+1)^(1/2)))+polylog(2, -s/(1+(-s^2+1)^(1/2))))-polylog(3, s/(-1+(-s^2+1)^(1/2)))+polylog(3, -s/(1+(-s^2+1)^(1/2))))/(-s^2+1)^(3/2)

and f1=f2

but maple doesnt convert it.

Also maple has trouble to convert

2*arctanh(sqrt((1-s)/(1+s)))=arccosh(1/s)

everywhere: 0<s<1

Hello, 

I have a trigonometric equation.

I would like to isolate gamma[1](t) and to determine gamma[1](t) in fonction of alpha(t), beta(t) and z(t). The others variables in the equations are fixed parameters.

I have tried to use isolate function. But it doesn't work.

Of course, my expressions should be complex but that is not a problem if i manage to expresse gamma[1](t) in fonction of alpha(t), beta(t) and z(t).

Here my program

constraints_2.mw

Thank you for you help

Hi!

How do I get Maple to factor to following long expression:

where one factor is (pp1^2+pp2^2+pp3^2)?

 

Thanks,

Arka

 

1 2 3 4 5 6 7 Page 1 of 7