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Lee := (-1+Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x))/(Int(exp(LambertW(1/(-1+t))*(-1+t)), t=1..x));
sum(unknown, n=1..infinity) = Lee
 
how to find unknown?

The image is an extract from the help page on the function "series" my question is in regard to how a method is selected, ie either it be a taylor or laurent series expansion, or as it defines in the passage attached "a more generalised series".

im just curious to know what procedure maple uses to make this choice when the series function is called, and im also finding it hard to replicate and understand the procedure of computing coeffiecents as described in the extract for a generalised series.

 

Thanks.
 

Good morning sirs,

Anyone with the idea(s) on how to convert series back to its original form should please share with me. 

Take for example

a-(1/2)*beta*a^2*y^2+(1/24)*beta^2*a^3*y^4-(1/720)*beta^3*a^4*y^6+(1/40320)*beta^4*a^5*y^8-(1/3628800)*beta^5*a^6*y^10+(1/479001600)*beta^6*a^7*y^12-(1/87178291200)*beta^7*a^8*y^14+O(y^16)

is a series for a*cos(sqr(a*beta)*y)

Thanking you in anticipation for your answer.

Hey guys,

I have the following occurence:

ii_inf:=x^(2-s)*(x^(-s)*GAMMA(3-s)*GAMMA(2-2*s)/(GAMMA(2-s)*GAMMA(3-2*s))+x^(s-2)*GAMMA(3-s)*GAMMA(2*s-2)/GAMMA(s))/(2-s)+(1/2)*(2*s*x-x+1)*(x+1)/(((x+1)^s)^2*(2*s^2-3*s+1))+x^(1-s)*(x^(-s)*GAMMA(2-s)*GAMMA(-2*s+1)/(GAMMA(1-s)*GAMMA(2-2*s))+x^(-1+s)*GAMMA(2-s)*GAMMA(2*s-1)/GAMMA(s))/(1-s)+(x+1)/((2*s-1)*((x+1)^s)^2)+x/((-1+s)*x^s)-(x+1)/((-1+s)*(x+1)^s);

ii_inf=simplify(ii_inf);

asympt(ii_inf,x,3);

Multiseries:-asympt(ii_inf,x,1);
gives different results...the last one however seems to be the correct one...

What is happening here?

 

theta := a-(1/2)*beta*a*y^2+(1/24)*beta^2*a*y^4-(1/720)*beta^3*a*y^6+(1/40320)*beta^4*a*y^8-(1/3628800)*beta^5*a*y^10+(1/479001600)*beta^6*a*y^12-(1/87178291200)*beta^7*a*y^14+(1/20922789888000)*beta^8*a*y^16-(1/6402373705728000)*beta^9*a*y^18+(1/2432902008176640000)*beta^10*a*y^20-(1/1124000727777607680000)*beta^11*a*y^22+(1/620448401733239439360000)*beta^12*a*y^24

Pls, anyone with useful informations on how to convert a series just like the one above to trigonometry or hyperbolic form. Need response as soon as possible. Thankin you in anticipation for your favorable response. 

I am trying to calculate the integral

where

Maple cannot calculate the integral. I tried to expand theta in the series form and substitute in the integral, still cannot calculate it.

any suggestion to tackle this problem whould be helpful.

Thank you

 

so yep pretty self explainatory, i was just wondering why it works when i use the asymptomatic expansion but not the iterated reciprocal substitution code shown in the that help interface for asympt.

``

``

Zeta(1/2+I*y)

Zeta(1/2+I*y)

(1)

subs(y = 1/y, series(subs(y = 1/y, Zeta(1/2+I*y)), y = 0, 6))

Error, (in series/Zeta) unable to compute series

 

asympt(Zeta(1/2+I*y), y, 6)

1+exp(-I*ln(2)*y)+exp(-I*ln(3)*y)+exp(-(2*I)*ln(2)*y)

(2)

``


 

Download MAPLE_IS_FUNNY.mw

Download Sum_Sum.mw
 

Restart:

Digits := 10:

Ha := 2:

R := 2:

a := 0.1e-2:

Rt := 1:

Br := 1.5:

Xi := 0:

U[0] := 0:

U[1] := alpha:

U[2] := -6+(1/16)*Rt*Xi:

U[3] := (1/6)*beta:

T[0] := -(1/2)*Rt:

T[1] := phi:

``

delta := proc (k) options operator, arrow; `if`(k = 0, 1, 0) end proc;

proc (k) options operator, arrow; `if`(k = 0, 1, 0) end proc

(1)

for k from 0 to 10 do U[k+4] := ((Ha^2+R)*U[k+2]+Xi*T[k+2])/((4*(k+3))*(k+4)); sum((1+(4/3)*R+4*a*R*T[i])*(k-i+1)*(k-i+2)*T[k-i+2], i = 0 .. k) := -4*a*R*(sum((i+1)*T[i+1]*(k-i+1)*T[k-i+1], i = 0 .. k))-Br*(sum((i+1)*U[i+1]*(k-i+1)*U[k-i+1], i = 0 .. k))+(1/4)*Ha^2*Br*(sum(U[i]*U[k-i], i = 0 .. k)) end do:

u := sum(U[j]*y^j, j = 0 .. 9)

alpha*y-6*y^2+(1/6)*beta*y^3-(3/4)*y^4+(1/80)*beta*y^5-(3/80)*y^6+(1/2240)*beta*y^7-(9/8960)*y^8+(1/107520)*beta*y^9

(2)

``


 

Download Sum_Sum.mw

 

I am dealing relatively often with vector functions of a variable vector. Specifically these are (in general nonlinear) mappings from R^6 to R^6, and often I have one or more extra parameter that controls the map.

Often I'd like to get a first-order expansion of these. The general scheme I use to do that is to map mtaylor over the components of the function vector. Like so:

map(mtaylor,F(<x1...x6>),[x1...x6,p1...pn],order);# F(<...>)  is  <F1(<..>)..F6(<...>)>

This works fine until I hit a situation where the Taylor expansion for one of the parameters p does not exist. Then it bombs. Actually, I often can use assumptions on the parameters to prevent bombing; however, in that case I often get the unchanged function F returned. This is Bad as the whole thing happens in a loop with many concatenated functions F, and now expression swell makes Maple lock up.

In at least a significant subset of these functions, the series command works where taylor fails. "mseries", however, does not exist in Maple. So my question is whether someone has written an "mseries" command for Maple. I know how to do it in principle (but caution is needed to make it work as intended), however, I wonder whether either someone has done it before and might share his/her code, or whether there is a reason why "mseries" actually cannot work.

TIA,

Mac Dude

 

how to solve  Fourier Series on maple ??

how to find the first 6 non-zero terms of MacLaurin series of the function erf(x)

Hi all, 
I was only wondering if there is a way how to trunk a solution using a maple command.

I want to use it to give a truncation error which is equal to :(14/45)(D^5)(y)(0) h^5

The result I am getting is as followed: (it is correct, I just want to cut off the bit with the power of 6)
........
......
>error:=expand(Yx[i+1]-Yx[i-3]-(4*h/3)*(2*f[i-2]-f[i-1]+2*f[i]));
                          
          error :=  (14/45)(D^5)(y)(0) h^5  + 7/10 (D^6)(y)(0) h^6
                  
thank you in advance

Find the first 6 non-zero terms of MacLaurin series of the function erf(x)

I have computed the eigenfunction expansion for f(x)=x on 0<x<1 in terms of the eigenfunctions exp(-x/2)*sin(n*Pi*x).

I wish to calculate the weighted L2 error in this expansion (the weight function is w(x)=exp(x)).

Specifically, I want to determine how many terms in the eigenfunction expansion are necessary for the error to be less than say 0.3.

Here is the code:

f := x -> x
w := x -> exp(x)
assume('n', integer);
y :=  (n, x) -> exp(-x/2) sin(n Pi x)       
c := n-> (int(f(x)*y(n, x)*w(x), x = 0 .. 1))/(int(y(n, x)^2*w(x), x = 0 .. 1))
Fourierf := (n, x) -> sum(c(j)*y(j, x), j = 1 .. n)

fsolve(Lerror(n) = 0.3, n);

This seems to run forever without giving a value of n.  I know this is a large computation, but it seems that Maple should be able to handle it.  Does anyone have any suggestions?

Heather

 

f=sum((2*q*cos(2* i*x)*(-1)^(i)*(-1)^((2*i-1)))/(i*Pi),i=1.3.5...35)

I want to write this series but getting error

the result is

2*q*cos(2*x)/Pi-2*cos(6*x)*q/(3*Pi)+2*q*cos(10*x)/(5*Pi)-2*q*cos(14*x)/(7*Pi)+2*q*cos(18*x)/(9*Pi)-2*q*cos(22*x)/(11*Pi)+2*q*cos(26*x)/(13*Pi)-2*q*cos(30*x)/(15*Pi)+2*q*cos(34*x)/(17*Pi)-2*q*cos(38*x)/(19*Pi)+2*q*cos(42*x)/(21*Pi)-2*q*cos(46*x)/(23*Pi)+2*q*cos(50*x)/(25*Pi)-2*q*cos(54*x)/(27*Pi)+2*q*cos(58*x)/(29*Pi)-2*q*cos(62*x)/(31*Pi)+2*q*cos(66*x)/(33*Pi)-2*q*cos(70*x)/(35*Pi)

can anybody help 

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