I have missed terms 10*x-20*

restart: p:=x->10*x-20*piecewise(x<2,0,x>=2,x-2);

I have missed terms 10*x-20*

restart: p:=x->10*x-20*piecewise(x<2,0,x>=2,x-2);

Thank you very much for your help

Thank you very much for your help

@hirnyk 

Dear hirnyk 783,

I have not done this calculation (for any value of b and h) and I'm searching for the better way to get a simple solution to my equation with no assignement for b and h

I asked a question in primes to get help

Thank you

@hirnyk 

Dear hirnyk 783,

I have not done this calculation (for any value of b and h) and I'm searching for the better way to get a simple solution to my equation with no assignement for b and h

I asked a question in primes to get help

Thank you

@hirnyk 

With this method, we must define an order to minimize this term and get a better approximation of the solution

@hirnyk 

With this method, we must define an order to minimize this term and get a better approximation of the solution

@hirnyk 

In help

?series

You can find:

If the series is not exact then an "order term" (for example,
                              / 6\
                             O\x /
 ) is the last term in the series.

@hirnyk 

In help

?series

You can find:

If the series is not exact then an "order term" (for example,
                              / 6\
                             O\x /
 ) is the last term in the series.

Yes, when fixing parameters h and b is more easy :

h:=10: b:=5:allvalues(1/b*RootOf(_Z+h*tan(_Z)*b)):

evalf(%);

-.6160138034, -2.464938906, 1.232118272, -1.232118272, 0.

But we can use solve with series as:

Order:=6:mu:=solve(series(h*sin(mu*b)+mu*cos(mu*b),mu)=0,mu);

to get

mu := 0, -1/60*(-60*h^2*((h*b+1)/h)^(1/2)-54*h^3*((h*b+1)/h)^(3/2)-20*O((h*b+1)^2/h^2)*3^(1/2)*h^(1/2))*3^(1/2)/h^(1/2), 1/60*(-60*h^2*((h*b+1)/h)^(1/2)-54*h^3*((h*b+1)/h)^(3/2)+20*O((h*b+1)^2/h^2)*3^(1/2)*h^(1/2))*3^(1/2)/h^(1/2)

Thanks

Yes, when fixing parameters h and b is more easy :

h:=10: b:=5:allvalues(1/b*RootOf(_Z+h*tan(_Z)*b)):

evalf(%);

-.6160138034, -2.464938906, 1.232118272, -1.232118272, 0.

But we can use solve with series as:

Order:=6:mu:=solve(series(h*sin(mu*b)+mu*cos(mu*b),mu)=0,mu);

to get

mu := 0, -1/60*(-60*h^2*((h*b+1)/h)^(1/2)-54*h^3*((h*b+1)/h)^(3/2)-20*O((h*b+1)^2/h^2)*3^(1/2)*h^(1/2))*3^(1/2)/h^(1/2), 1/60*(-60*h^2*((h*b+1)/h)^(1/2)-54*h^3*((h*b+1)/h)^(3/2)+20*O((h*b+1)^2/h^2)*3^(1/2)*h^(1/2))*3^(1/2)/h^(1/2)

Thanks

When a is a real, we can do:

a:=2.1; and get

crlb_2.1;

0

Why assuming a real ?

When a is a real, we can do:

a:=2.1; and get

crlb_2.1;

0

Why assuming a real ?

@hermitian 

You can also use %^%H assuming x1::real, x2::real, y1::real, y2::real;