Hello

I have such problem:

I calculated rather complicated function, and i want to obtain the array or list of data with definite step.

For this purpose I used next expression:

Out:=[seq(t,F(t),t=0..tmax,step]

I need minimum 1000 points, but calculation for the 100 points requires about 300 second, for 1000 - it will be about 3000 seconds. It's too long for me.

How could I accelerate this process?

P.S.

Plotting of my function requires about 90 seconds for the entire range. So, this is very strange, thad converting to data list requres so much time.

Is there any webpage like Wolfram Library Archive but with Maple related resources? I've missed something or everything goes to mapleprimes? 

I need to complete the definition of bcount so that bcount(n) returns the total number
of odd coefficients 
n
k , 0 ≤ k ≤ n. For instance, the values of 
n
k for n = 6, with odd values highlighted, are:
1, 6, 15, 20, 15, 6, 1,

bcount:=
proc(n::TYPE)
description "Count odd binomial coefficients.";
---MORE STUFF HERE---
end proc; # bcount

Any help appreciated

Hi everybody i've to minimize a function of about 5-6 variables so i tried some of maple functions like: 'Minimize', 'minimize', 'NLPSolve' but with 'minimize' my pc goes in loop and it eats a lot of ram without giving some results, while with the Optimization package i got the error :'

Error, (in Optimization:-NLPSolve) abs is not differentiable at non-real arguments

'

code :

restart;
L := 1:
b := 1.1:
p := 1.8:
xa := 0:
xd := -b/2:
ya := -p+z:
yd := 0:

quadrilatero := proc(q,xa,ya,xd,yd,l1,l2,l3,modo,alpha0,alpha1)
local xb,yb,l5,c,alpha,theta5,theta2,xc,yc,theta3,xm,ym:

xb := xa+l1*cos(q + alpha0 + pi/2):
yb := ya+l1*sin(q + alpha0 + pi/2):
l5 := sqrt((xd-xb)^2+(yd-yb)^2):
c := (l2^2+l5^2-l3^2)/(2*l2*l5):
if modo = 1 then alpha := arccos(c) else alpha := -arccos(c) end if:
theta5 := arctan(yd-yb,xd-xb):
theta2 := theta5 + alpha:
xc := xb + l2*cos(theta2 + pi):
yc := yb + l2*sin(theta2 + pi):
theta3 := arctan(yd-yc,xd-xc):
return (theta3 - pi - alpha1):
end proc:
#funione sterzo
f := proc(q,ya,l1,l2,l3,yd,alpha0,alpha1)
local theta;
theta := quadrilatero(q, 0, ya,-b/2,0, l1, l2, l3, -1,alpha0,alpha1) - pi/2;
if theta < Pi and Im(theta) = 0 then theta := theta + 2*pi end if;
return theta;
end proc:
#funzione desiderata
Yd(q) := arctan(tan(q),(1 - b*tan(q)/2/p));
arctan(tan(q), 1 - 0.3055555556 tan(q))

#Definisco procedura dei minimi quadrati:
n := 1: i := 0:
q := Vector[n]: rms := 0:
for i from 0 by 1 to (n-1) do
q[i] := Pi*(i/(4*n)):
rms := rms + abs(f(q[i],ya,l1,l2,l3,yd,alpha0,alpha1) - yd(q[i])^2):
end do:
rms := sqrt(rms / n):
#MIN := (minimize(rms, z = -0.4..0,alpha0 = 0..Pi/3,alpha1 = Pi/2..5*Pi/4,l1 = 0.3..1,l2 = 0.001..0.3,l3 = 0.01..0.3,iterationlimit = 100));
with(Optimization):
Optimization[NLPSolve](((rms)));
Error, (in Optimization:-NLPSolve) abs is not differentiable at non-real arguments

thanks in advance for your help !

How to realize the Cantor staircase function in Maple? Mathematica has it.

I'd like to stress the three desired properties:

(i) The virtual Cantor function can be evaluated to arbitrary numerical precision.

(ii) For certain arguments, this function automatically evaluates to exact values.

(iii) Mathematical function, suitable for both symbolic and numeric manipulation.

It is given 

   xⁿ (n x - n - x) / (x-1)² + x / (x-1)²  , n is a (symbolic) positive integer.

I want to transform it into

n xⁿ⁺¹ / (x-1) - x (xⁿ-1) / (x-1)². 

How is it possible?

I tried the  simplify, convert (parfrac), collect, combine, expand,  with/without assuming.

Thanks in advance.

Hi everybody,

Ever since i updated to maple 2015, the output of calculations display a disproportionate amount of zeros. For instance:

If i write:

2.3*10^(-4)

i get:

0.000230000000000

Is there anyway to make maple NOT display the last 10 zeros? I've tried Tools ->Options ->Precision -> Round screen display to X. But then it rounds all results to that X amount of digits, wich is also anoying. Like, if i set X to be 10, it'll write 3.000000000 instead of just 3.

Hope someone can help me :-)


Anton

The CIELAB perceptual model of human vision can be used to predict the color of a wavelength in sRGB. I found the CIEDE2000 corrections to hue, chroma, and lighness were required to get the right values in the blue to violet region. The CIEDE2000 results seem pretty close to what I see looking through a diffraction grating considering I have no way of knowing how my eyes are adapted.

Use the new, faster  version. It can run in under 1 minute;

nprocWavelength_Color_CIEDE2000.mw   and  WtoC_data.xlsx

You will need the data for the color matching functions in the WtoC Excel File. The data directory can now be controlled in the mw file.

Cn ={ 1, n = 0 ,                                                  }

      {Xn−1[sum of] k=0   C(k)C(n−1−k) , otherwise.  }

looking to complete the definition of catalan so that catalan(n) returns Cn whenever n is a non-negative integer. usin g the definition above...any help appreciated

catalan:=
proc(n::TYPE)
description "Print the n'th Catalan number.";
option remember;
---MORE STUFF HERE---
end proc; # catalan

I am working on something where I need to use a stack. In the Maple documentation I noticed there are two different stack implementations. There is "stack" and "SimpleStack". I am curious if anyone has any experience using either of these and if there is any advantage to one over the other. Any suggestions or tips are appreciated.

Thanks,

Steven

This is a post that I wrote for the Altair Innovation Intelligence blog.

I have a grudging respect for Victorian engineers. Isambard Kingdom Brunel, for example, designed bridges, steam ships and railway stations with nothing but intellectual flair, hand-calculations and painstakingly crafted schematics. His notebooks are digitally preserved, and make for fascinating reading for anyone with an interest in the history of engineering.

His notebooks have several characteristics.

• Equations are written in natural math notation
• Text and diagrams are freely mixed with calculations
• Calculation flow is clear and well-structured

Hand calculations mix equations, text and diagrams.

Engineers still use paper for quick calculations and analyses, but how would Brunel have calculated the shape of the Clifton Suspension Bridge or the dimensions of its chain links if he worked today?

If computational support is needed, engineers often choose spreadsheets. They’re ubiquitous, and the barrier to entry is low. It’s just too easy to fire-up a spreadsheet and do a few simple design calculations.

 Spreadsheets are difficult to debug, validate and extend.

Spreadsheets are great at manipulating tabular data. I use them for tracking expenses and budgeting.

However, the very design of spreadsheets encourages the propagation of errors in equation-oriented engineering calculations

• Results are difficult to validate because equations are hidden and written in programming notation
• You’re often jumping about from one cell to another in a different part of the worksheet, with no clear visual roadmap to signpost the flow of a calculation

For these limitations alone, I doubt if Brunel would have used a spreadsheet.

Technology has now evolved to the point where an engineer can reproduce the design metaphor of Brunel’s paper notebooks in software – a freeform mix of calculations, text, drawings and equations in an electronic notebook. A number of these tools are available (including Maple, available via the APA website).

 Modern calculation tools reproduce the design metaphor of hand calculations.

Additionally, these modern software tools can do math that is improbably difficult to do by hand (for example, FFTs, matrix computation and optimization) and connect to CAD packages.

For example, Brunel could have designed the chain links on the Clifton Suspension Bridge, and updated the dimensions of a CAD diagram, while still maintaining the readability of hand calculations, all from the same electronic notebook.

That seems like a smarter choice.

Would I go back to the physical notebooks that Brunel diligently filled with hand calculations? Given the scrawl that I call my handwriting, probably not.

Hello everybody

In the attached file, as you can see, 'u' is function of 'thet1(t)' explicitly, But when I want to differentiate from 'u' w.r.t 'thet1(t)', maple returns zero as the answer. How is it possible?

Thanks in advance

d.mw

I have an ODE with L(x,y), which includes partial derivatives of L(x,y) and coefficients y,_y1 and _y1^2. I want to split this equation into seperate equations of these coefficients. So I can solve for the unknown variable L(x,y).

ODE:=diff(L(x,y),x,x)+2*_y1(x)*diff(L(x,y),x,y)+_y1(x)^2*diff(L(x,y),y,y)+y(x)/x*diff(L(x,y),x)+(y(x)*_y1(x)/x+_y1(x))*diff(L(x,y),y)-y(x)/x^2*L(x,y)=0;

Could someone help me? 

I am currently trying to use the coeffs command but not really getting anywhere.

Thank you.

I a working with circuits, and was wondering whether or not it might be possible to shorten the proces of calculating parallel resistors. My idea is that using some symbol, such as || would find the equivalent resistor.

My idea is based on the fact that CrossProduct can also be done using &x command. Like so.

CrossProduct(a,b) = a &x b

So for resistors it would look something like:

Parallel(a,b) = a || b

For those interested the function for parallel resistors would be:

Parallel := (a,b) -> 1/(1/a+1/b);

the binomial coefficient  n k  can be defined recursively as follows for all nonnegative integers n, k:

(n)  = {0,      k>0

(k)  = {1       k=0, k=n

         {(n-1)+(n-1), otherwise.

          (k-1)   (k)

I need to complete a deinition of binom so that m so that binom(n,k) returns  n k  for all n greater than 0, and k greater than or equal to 0 using the definition of the binomial above..Any help appreciated..

binom:=
proc(n::TYPE1,k::TYPE2)
description "Compute a binomial coefficient";
option remember;
---MORE STUFF HERE---
end proc; # binom