# Markiyan Hirnyk

Markiyan Hirnyk
7 years, 186 days

These are answers submitted by Markiyan Hirnyk

### By numeric

9 hours ago by

0

0

Eq2 := diff(z(x), x, x) = piecewise(and(x >= 1, x < 10), 2, and(x >= 10, x < 11), 10, 2)*(1+(diff(z(x), x))^2)^.5;

sol := dsolve({Eq2, z(0) = 1, (D(z))(0) = 2}, numeric);

proc(x_rkf45)  ...  end;
sol(2);

### Long way

13 hours ago by

0

0

The idea is simply to express: to convert your g to list, to flatten this list, to convert the result to a piecewise function back.

Its realization is not short:

restart; f := x-> piecewise(x <= 1, (1/2)*x-1/2, x < 2, -3*x+6, 2*x-1);
g := unapply(piecewise(op(ListTools:-Flatten([op(solve(f(x) = y, x))]))), y);

PS. g(2);

...

### By cutting the tail

Yesterday at 12:15 PM by

1

0

Because the series in y diverges in the case x-C<0, without loss of generality we may assume x-C>0. The Fit command does not deal with your model function y, even in the form A*x +evalf( Sum ( B * exp(-n^2 * (x-C)) /n^2, n=1..infinity)) (see ?sum for info). But the series very rapidly converges if x-C>0. In view of it we replace the series by a finite sum,...

### By fsolve

Yesterday at 11:30 AM by

0

0

After assigning the parameters, the fsolve command works. For example,

p := (3*a^2+2*a*lambda+lambda^2)*(-sc*t^4*lambda^4+288*k1)-12*a^2*sc*t^4*lambda^4:
fsolve(eval(p, [a = 1, k1 = 2, sc = -1, t = 2]), complex);
-1.23367574991969 - 3.64700913994696 I,

-1.23367574991969 + 3.64700913994696 I,

...

### After replacing

May 22 2013 by

1

0

with*plots; by with(plots): your code works. See executed_code.mw . Could you clarify the problem you met? BTW, in the future present your code without the cursors. These hinder to execute a code.

Best regard, Markiyan Hirnyk

### By convert to rational

May 21 2013 by

1

0

The response gives a hint. This modification helps

### One way

May 21 2013 by

1

0

What you display is the set {A,B}={[0.5e-1, 2.7], [.1, 6.8], [.15, 12.7], [.2, 1100],[0.5e-1, 3.2], [.1, 7.4], [.15, 13.7], [.2, 1110.5]}. It becomes clear when we draw

pointplot({A, B}, axes = boxed, symbol = diamond, color = blue, symbolsize = 30, view = [0 .. .2, 0 .. 10]);

### By evalf

May 21 2013 by

0

0

The problem consists in the following: Maple finds the exact solution, using a lot of the resources of a comp.

This works.

### Another way

May 20 2013 by

1

0

In most cases it is useful to have a few solutions. Because of this reason

### Reducing to usual derivative

May 19 2013 by

0

0

diff(log(Nu(RE,M,TAU)),log(M))=diff(log(Nu(RE,M,TAU)),M)/diff(log(M),M)=

### One way

May 19 2013 by

0

0

For example, it can be done as follows.

restart; T := {{1}, {1, 5}, {4, 5}, {3, 4, 5}}:
with(combinat):
for j to nops(choose(T, 2)) do T := {choose(T, 2)[j, 1] union  choose(T, 2)[j, 2]} union T: end do:
T;
{{1}, {1, 5}, {4, 5}, {1, 4, 5}, {3, 4, 5}, {1, 3, 4, 5}}

### Workaround

May 15 2013 by

2

0

Your command is correct, but every soft has its limitations. As yet, Maple can't find it. See a similar example in ?limit/multi . There is a workaround:

limit(mtaylor(sin(x)+sin(y), [x, y], 4), {x = 0, y = 0});
0.

### Two ways and reference

May 12 2013 by

0

0

isolve(2275*s-2503 = 1563*t);

{s = 976 + 1563 _Z1, t = 1419 + 2275 _Z1}
chrem([0, 1563], [2275, 2503]);

4704700

See ?isolve and

### By Variational Calculus

May 11 2013 by

0

0

Up to Wiki (see http://en.wikipedia.org/wiki/Hamilton%27s_principle ), the Hamilton principle is equivalent to the Euler–Lagrange equations. These can be created by the ?VariationalCalculus/EulerLagrange command.

### This works

May 11 2013 by

1

0

 1 2 3 4 5 6 7 Last Page 1 of 93