Axel Vogt

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17 years, 218 days
Munich, Germany

MaplePrimes Activity

These are answers submitted by Axel Vogt

You have many terms, 196*109. Here is a suggestion: write the inner Sum as function, evaluate the outer additions and then feed the function with its numerical values.

I changed your variable l, because it is bad to read it and reduced addition to 20*10 (well, I lost patients). Also note that you may have to do some more work if you want to care for rounding errors of 20000 terms of quite different magnitude and signs.

Hope it helps.

Edit: I need roughly 200 sec to get 5.76900063193468*10^29 with an old Win PC, summing up over your original range.

Likewise you can look into the Digital Library of Mathematical Functions,

I would do it in a paper & pencil style (it is a mess how this board displays it)




f := 2*y^3*z - y^2 -2*m;
#F:=unapply(f, y);

f := 2*y^3*z-y^2-2*m


assume(m<0, 0<z);

discrim(f,y); signum(%); convert(%, piecewise, m);



piecewise(m < -1/(54*z^2), -1, m = -1/(54*z^2), 0, -1/(54*z^2) < m, 1)


# translate "0 < discrim(f,y)"
additionally(-1/(54*z^2) < m);


diff(f, y);
Q:=[solve(%)]; # local extrema


Warning, solve may be ignoring assumptions on the input variables.

Q := [{y = 0, z = z}, {y = 1/((3*z)), z = z}]


eval(f, Q[1]); signum(%);




eval(f, Q[2]); normal(%); # = discrim(f,y) * factors with known signs





limit(f, y=-infinity);
limit(f, y=+infinity);





Thus f comes from -oo, takes a positive value in y=0 and a negative value

in the positive y= 1/(3*z) and runs to +oo.

Hence it has 3 real zeros.



Download MP_234277_CubicSolve.mws


As far as I see you have a linear system 4 x 4. The according matrix has rank <= 3. Hence the solution is not a point

For your example the following would do:

sin(x)/x; convert(%, hypergeom, include=sin); convert(%, BesselJ);

or even shorter

sin(x)/x; convert(%, BesselJ, include=sin);


convert(%, MeijerG);
#convert(%, hypergeom); # needed if evalf is used
convert(%, Int): value(%);
                          \      Beta(1/2, k)
                           )     ------------
                          /                2
                         -----    (2 k + 1)
                         k = 1

    1/4 Pi    MeijerG([[-1/2, 0, 0], []], [[0], [-3/2, -3/2]], -1)

                            4 - 4 Catalan


You may try assuming, not assume. Or use "simplify(%) assuming ..." for the result.

BTW: Maple 2021 returns /(A^2-B^2)^(1/2)*EllipticK(B*2^(1/2)*(-1/(A^2-B^2))^(1/2))

I posted an answer at   but that thread disappeared (was it deleted?)

indets(%, symbol);
                                {E, T}

So your VT depends on E = income and T = time and the following shows the values:

plot3d(VT, E=10^5 ... 10^6, T = 1 ..5, axes=boxed);

plot3d(VT/E, E=10^5 ... 10^6, T = 1 ..5, axes=boxed); shows you the values relative to income

Remark: your r seems to stand for rates, 0.1 = 10% is somewhat high, perhaps you want to see it for 1% or 3%

You may consider to use Excel and - if really needed - connect Excel and Maple

well ... hm ...


convert(v, exp); simplify(%); abs(%);

gives 1/2*2^(4/5)

The easy way would have been to do it during installation of Maple, one of the final steps ask whether it should open those files as default application.

I guess you are on Windows (because you say "PC").

After selecting one *.mw file through the context menu (=right mouse key) you can say "always open with" and can select the application through a dialog (you should however know where your maple is installed, for example "C:\Program Files\Maple2021\bin.X86_64_WINDOWS\maplew.exe".

Edit: Likewise you can use Windows' System Control to define which (standard) programs handle specific file extensions. That will depend on your Win version - and since I have a non-concurrent German version you better google for that way yourself (works for any pairing program & file extension)

1/12 + arbitrary constant is still an arbitrary constant

Your command says "cut at +-5 for the y-axis"

Try: plot((x^3 - 4*x^2 - 9*x + 36)/(x^3 - 9*x),  x = -3 .. 3);


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