Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hello guys,

I want to find exact form of a(t) in following differential equation:

-diff(a(t), t)^2*_C1*6^(-1/(-1 + 2*alpha))*((diff(a(t), t, t)*a(t) + diff(a(t), t)^2)/a(t)^2)^(-1/(-1 + 2*alpha))/(4*(diff(a(t), t, t)*a(t) + diff(a(t), t)^2)*(-1/2 + alpha)) = (6*diff(a(t), t, t)*a(t)^2*alpha + 6*a(t)*diff(a(t), t)^2*alpha + k^2)/a(t)^3

please guide me,

Has someone tried to connect Grasshopper with Maple yet?

Grasshopper is a visual programming environment on Rhino.

Dear esteem Colleagues,

Please how do I modify the following two files (though similar) to get consistent errors? I am not sure where I made the mistake.

Any modifications would be appreciated.

Thank you all for your time and mentorship. Best regard

Contour integration notation

" (∫)[+infinity]^(+infinity)((-x)^(z))/((e)^(x)-1). (ⅆx)/(x)"


The limits of integration are intented to indicate a path of integration which begins at + ∞, moves to th e left down the positive real axis, circles the orign once in positive ( counterclockwise) direction, and returns up to the positive real axis to  +∞

-How does this contour look like  in a  graph ?
- the "(ⅆx)/(x)" notation  ?
- calculating this complexe contour integral?

Seems that the concept of the contour integration is similar wit a line integral in real calculus ?

Some more information needed about singularities ( first en second order ..more?)

Nieuwe pagina 1 (



Can someone confirm, that autosave is not implemented for workbooks?

The only thing I can restore form a workbook session are specific worksheets of the workbook that I was working on.

None of the Maple codes are saved for example.

This is a major issue, folks. If you haven't implemented it, you need to tell us about it in the help document.


I following a example of products multiplication like this one


Calculating with  this with maple 1d input is correct, but when i convert a maple 1d input  to 2D input ( i did somewhere) and use this then there is difference with the maple 1d calculation

Seems to be not a advisable to use converted maple 1d to 2 D input for calculation : ( for a mixed calculation(maple input/2D input)  or solely 2d input) , but only for purpose of seeing what the expression in maple input is standing for.   

Note: i did the calculation again with mixed input and now the correct sequenze of answers shows up ?

Just wanted to post that I had some data loss because I opened two different workbooks with the same name from different locations.

This could lead to loss of data.

I had code attachments from the "old" workbook implemented in the "new" workbook, while the new code was gone.

Thought always that the round d is reserved for function of two variables x,y , but  that seems to be not the case here ?


Comparing Different Answers


Een antwoord ergens gegeven is

Int(sqrt(x^2+1), x) = (1/2)*x*sqrt(x^2+1)+(1/2)*ln(x+sqrt(x^2+1)) + C                                                             (vb)


Mple geeft



Int((x^2+1)^(1/2), x) = (1/2)*x*(x^2+1)^(1/2)+(1/2)*arcsinh(x)+C[1]



De twee antwoorden lijken nog niet opelkaar !
In het gegeven antwoord staat er een ln en in Maple kan een expressie omgezet worden in ln termen


Int((x^2+1)^(1/2), x) = (1/2)*x*(x^2+1)^(1/2)+(1/2)*ln(x+(x^2+1)^(1/2))+C[1]


(2)  is hetzelfde (vb)

Dezelfde integraal i sook gegeven als


Int((x^2+1)^(1/2), x) = (1/8)*(x+(x^2+1)^(1/2))^2+(1/2)*ln(x+(x^2+1)^(1/2))-(1/8)/(x+(x^2+1)^(1/2))^2+C[2]



een effectieve manier om twe antwoorden t evergelijken voor hetzelfde probleem is het verschil te berekenen van een vergelijking met de twee integralen








(x^2+1)^(1/2)-(1/4)*(x+(x^2+1)^(1/2))*(1+x/(x^2+1)^(1/2))-(1/2)*(1+x/(x^2+1)^(1/2))/(x+(x^2+1)^(1/2))-(1/4)*(1+x/(x^2+1)^(1/2))/(x+(x^2+1)^(1/2))^3 = 0



0 = 0


Strange that  diff(lhs(%)-rhs(%)=0,x);  is translated by 2 d input with round d notation for functions with two variables ?
The two integrals are functions of one variable
diff(f, x)


I have defined a function II1norm of one variable. The variable has units "microns". It plots perfecting using a range defined in microns, but gives an error when I try to find the root using NextZero. If I just leave off the "microns" in the second argument, NextRoot just reports "FAIL". If I rewite the worksheet without units, then the NextZero executes fine. Why? How to I use units when finding roots?

This is so useful to see geometrical mapping diagram to visualize Complex analysis

Something that also can be made for Maple 

Mapping Diagram for Cauchy Integral Formula – GeoGebra

Using GeoGebra for visualizing complex variable. (

I highly encourage everyone interested in complex variable to read Tristan Needham „Visual Complex Analysis” and try to solve problems with or without aid of GeoGebra. I hope that in this workshop we will manage to get a feeling of complex functions and as a final point understand how complex integration works. It is a common misconception that complex integration can't be visualized, and using Tristan Needham's ideas we will try to explore this idea. It's a pity that we don't have a lot of time, thus we will skip a lot of important information and construct only some graphs. 

There is so much experimenting with Geogebra software and doing too this in Maple ?

Hello there, 

Would you allow me to ask one question?

Is there any way to get a saturated water vapor pressure value with a temperature outside of the range?

The range here means [273.06 K, 647.096 K]. The pressure value certainly exists (T<273.06 K), but the API only comes up with an error. 

Here is the Maple worksheet where I got into this issue:






T2 := (-40.0 + 273.15) * Unit('K');



xbb := 1: # 100% steam, saturated.

Pg2 := PropsSI("P", "T", T2, "Q", xbb, "water");

Error, (in ThermophysicalData:-CoolProp:-PropsSI) Temperature to QT_flash [233.15 K] must be in range [273.06 K, 647.096 K] : PropsSI("P","T",233.15,"Q",1,"water")




I must feed the not trivial zeros numbers into this aproximation formula ?


Riemann hypothese and staircase of primes













numelems(select(isprime, [seq(1 .. 10000)]))



The prime counting function is approximated by Li(x) and x/ln(x).

plot([PrimeCounting(x), Li(x), x/ln(x)], x = 1 .. 500, legend = [pi(x), Li(x), x/ln(x)])

The staircase of primes approximated by two functions
Interesting is the video: How i learned to love and fear the Riemann Hypothesis


for n from 1 to 30 do
 if is(n,prime)
     then ps[n]:=plot([[n,y],[n,y+1],[n+1,y+1]]):
     else ps[n]:=plot([[n,y],[n+1,y]]):
 end if ;

plot([PrimeCounting(x)] ,x = 1 .. 35, legend = [pi(x)]):

plot([PrimeCounting(x), Li(x), x/ln(x)], x = 1 .. 35, legend = [pi(x), Li(x), x/ln(x)])





Prime counting function
What found RIEMANN for the prime counting function in relation to the zeta function after he defined the zeta function?


He found further a function what follows exactly the shape of the prime counting function

Final discovery v. Riemann.  

- step in the omhoog in de priemtelfunctie = log(p) (zie video)


Using the logarithmic primecount function( from Chebyshev) (approximation)
Further  analyse with this Chebyshev approximation formula in relation to the not trivial zero points from Riemann zeta function ( zeros) gives another real function for approximating the primecounting function what uses the non trivial zeros from Riemanns zeta function  in this function:


"(not trivial zeros ) u[k ] = "i*w[k]+v[k]   
Number now all nottrivial zeros in the upperhalfplane from down to bottom,  as u[1], u[2], u[3, () .. ()]

"`&varphi;`(x)  := x-ln(2Pi)-1/(2 )ln(1-1/(x^(2))) - (&sum;)2/(|u[k]|) x^(v[k]) cos(w[k] ln(x)-alpha[k])"


Its only alpha[k] that must be calculated out of the not trivial zeros and i must have a list of  serie of not trivial zeroes from the zeta function. => see Hardy's Z(t) ? from this ..... alpha[k]  can be calculated ?
All not trivial zeros are complex numbers laying on a line ,but  orginating from (0,0) in the complex plane as  vectors to the points    

varphi(x):= x - ln(2*Pi) - 1/2*ln(1 - 1/x^2) - sum(2*x^v[k]*cos(w[k]*ln(x) - alpha[k])/abs(u[k]), k = 1 .. infinity);

x-ln(2*Pi)-(1/2)*ln(1-1/x^2)-(sum(2*x^v[k]*cos(-w[k]*ln(x)+alpha[k])/abs(u[k]), k = 1 .. infinity))


This formula seems to be correct .
Now how to make a plot ?
Hardy's Z(t) function shows the not trivial zeros in the upperhalfplane of the critical strip of the Riemann zeta function  as zeros in this Z(t) real function : derived from a alternating serie ?


It's sad when a bug is found for such a simple problem!

sys := [x*y*z = 0, x*y*z + x*z + z = 0, x*y*z + x*y + x = 0]:
solve(sys, {x,y,z});    # OK
#         {x = 0, y = y, z = 0}, {x = x, y = -1, z = 0}

solve(sys, [x,y,z]);    # ???
#                               []



Happy New Years All

Can anyone explain two document formating features I am trying to implement?

1.)  Sections.  I select numbering with follow previous selected and I get three sections with a 1. in front of them vs 1, 2, 3.  


I would expect Technical to become 3.1 when properly following Section 3. header.   Also, I have seen how to manually type the section number, but that seems to defeat the document layout and formatting if it gets "jumbled up". 

2.)  Headers.   Can you suppress headers on pages 2 through X to have a unique cover page header?   This would be similar to word (headers and footers linked to previous section, or not(?) with section breaks as new pages)


i have a long proc i am defining in a loop inside the proc for some reason it calculates the value of a1[j+1] and a2[j+1] and assigns it(in the case of the picture j=1 and in j=2 it doesnt sub) but when it is called again in the next iteration  it doesnt sub the value it just calcluated ! 
i would be so thankful for any help ive been stuck on this for a while now 

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