Hello People in mapleprimes,

I want to know how to factor the denominator of the first term in the right hand side, using commands not clickable math.

And, I know applyop can be used in this case, but I want to know is other ways than using applyop. 

-tau*alpha*sigma*omega*(tau-omega^(-sigma))/(s*L*(1-gamma-tau^2+tau^2*gamma))

-(sigma*alpha)/s*(1-tau*omega^sigma)/(omega^(sigma-1)*(1-gamma)*L*(1/tau-tau));

Thanks in advance.

taro

Dear people in Maple Primes,

I have a question about how to solve a system of equations.

In the following equation, I want to eliminate D(a).

x := D(a*b*c = 3*d); 
y := D(a^2*b^3*c = 3*a);

For this purpose,

a code of 

d_a := isolate(x, D(a));

eval(y, d_a);

works well. But, for me, this code is a little laborious.

Is there any better way than the above way?

Thanks in advance.

taro yamada

Hello those in Mapleprimes,

What I want to know is whether this is possible or not, and if possible, how should I write a code?

The following code works properly:

U:=(x,y)->(x^theta+y^theta+X)^(1/theta);

diff(U(x,y),x)/diff(U(x,y),y)=p/q;simplify(%);

But, what I want to ask is this. As for the part of simplify(%), I want to do it with a way which has me 

feel more being from the former to latter.

That is, if ,for example, "diff(U(x,y),x)/diff(U(x,y),y)=p/q;@simplify;" works, it is better to me, though this does not work.

As the second part, @simplify, receives the result of the first part"diff(U(x,y),x)/diff(U(x,y),y)", it seems more 

natural to me than to write simplify(%).

Can't I do this, in a meaning, reversal of operator to argument?

taro

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

Hello those in Mapleprimes,

I want to obtain the result of the following codes,

shift:=(f::procedure)->(x->(f(x+1))):

shift(sin);

as x->sin(x+1)..................(1)

But, the result I could get was only

x -> f(x + 1)......................(2)

Please teach me how I can obtain the result as (1), not (2)?

Thanks in advance.

taro