Markiyan Hirnyk

Markiyan Hirnyk
9 years, 104 days


These are answers submitted by Markiyan Hirnyk

Here are the arguments. Let us switch to the polar coordinates.

restart; f := (1-cos(x*y^3))/(y^6+x^2)^(1+a);
A := eval(f, [x = r*cos(phi), y = r*sin(phi)]);

Now we expand A as a series in r:

B := MultiSeries:-series(A, r, 3) assuming a >= 0, a <= 1, phi >= 0, phi < 2*Pi;

If the double limit exists the one is equal to the both  iterated limits. But

limit(B, phi = (1/2)*Pi) assuminga >= 0, a <= 1

is infinite.

 

 

The LagrangeMultipliers command produces the solutions of the system, not outputing the system.

with(Student[MultivariateCalculus]):
LagrangeMultipliers(sqrt(x^2+y^2+z^2), [x^2+y^2+x+2*z, y^2+z^2-1], [x, y, z], output = detailed);

Numerically

14 hours ago Markiyan Hirnyk 6332

In my personal opinion, this can be done only numerically in such a way.

 

restart;
f(.1)

[HFloat(38.76590010557003), [x = HFloat(0.9922532017992352), y = HFloat(52.60229179442929), z = HFloat(0.3464256428660829)], 3]

(1)

``

NULL

NULL

The DirectSearch package should be downloaded from http://www.maplesoft.com/applications/view.aspx?SID=101333 and installed in your Maple >=12.

Download numerically.mw

PS. The execution takes a lot of time and the output is not stable.

 

No solution

Yesterday at 2:37 PM Markiyan Hirnyk 6332

Almost surely, there is no solution of the problem under consideration in view of

The DirectSearch package should be downloaded from http://www.maplesoft.com/applications/view.aspx?SID=101333 and installed in your Maple >=12.

By extrema

Yesterday at 2:02 PM Markiyan Hirnyk 6332

This can be done as follows.

extrema(x+2*y*sqrt(z^2+1), {(y*z+x)*y-c}, {x, y, z}, s);
       
           

s;

The first point is the maximum and the second point is the minimum of  your f.                                     



 

This works well

February 27 2015 Markiyan Hirnyk 6332

This works well in Maple 13.02.




 

Download MVT.mw

Reference

February 25 2015 Markiyan Hirnyk 6332

Yes, that acer's application does the job in Maple 13.

References

February 23 2015 Markiyan Hirnyk 6332

That was asked and answered. See the results of the "total derivative" search in MaplePrimes at the top of this page, especially this link.

Yes

February 22 2015 Markiyan Hirnyk 6332

Making use of the linestyle option, one obtains

and

By DirectSearch

February 22 2015 Markiyan Hirnyk 6332

The Search command of the DirectSearch package  does the job

and the Minimize command  fails

Error, (in Optimization:-NLPSolve) no improved point could be found

To be sure, you may try the GlobalOptima command instead of the Search command, but it takes a lot of time.

primset-2.mw

 

 

This can be done as follows.

M := Matrix(2, 2, proc (i, j) options operator, arrow; u(x) = x^(i+2j) end proc);



map(rhs, M);


First of all, the answer depends on the definitions. If you allow +infinity = infinity and -infinity for the value of a multidimensional limit, then the answer is infinity. Here are the arguments. The numerator approaches 1 and the denominator approaches 0 , being nonnegative, if x->1, y->0 as follows from the taylor expansions

mtaylor((1+y)^(x-1), [x = 1, y = 0], 5);


mtaylor(1-cos((x-1)^2+y^2)^(1/4), [x = 1, y = 0], 5);
.

Therefore, the limit

limit((1+y)^(x-1)/(1-cos((x-1)^2+y^2)^(1/4)),{x=1,y=0});

is +infinity.

The point (1,0) is a limit point of the domain of the function under consideration so the twodimensional limit is correctly defined.

Also if the limit (under consideration) exists, the one is equal to the both iterated limits

limit(limit((1+y)^(x-1)/(1-cos((x-1)^2+y^2)^(1/4)), x = 1), y = 0);

                            infinity

and

limit(limit((1+y)^(x-1)/(1-cos((x-1)^2+y^2)^(1/4)), y = 0), x = 1);

                            infinity

If the infinite limits are not allowed, then the limit  does not exist.

in the OrthogonalExpansions package of Maple:

Reference

February 12 2015 Markiyan Hirnyk 6332

Any loop is not necessary to this end. Look in ?max , especially the last example. Its slight modification is

max([-1, 5, 6]);

                               6

One more way

February 09 2015 Markiyan Hirnyk 6332
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