Items tagged with sum

Feed

when i evaluate this summation if Q[h],S[h],R[h] are big floating numbers it take along time how i decrease this time 
P := simplify(sum(sum((t+1)*Q[h]^2*(1-Q[h])^(t+T)/(t+1+(R[h]/S[h])^sigma*(T+1)), t = 0 .. infinity), T = 0 .. infinity))

The link below has my code for generating 2 matrices.  The 1st one does not generate flfoating point numerical data; whereas, the 2nd one does.  What is wrong with the 1st case?  I am attempting to single out one harmonic which works in the 2nd case.  Also, is there a way I can generate a spectrum of S2(k= 1 to 100, t= 0 to 1)?

?untitled6.mw

 

Why do I get different answers for the same command?

 different_answers.mw


 

``

restart

x := proc (n) options operator, arrow; cos(n*Pi) end proc;

proc (n) options operator, arrow; cos(n*Pi) end proc

(1)

"f:=k->"sum(k*x(n), n = 2 .. 3);f(1);

0

 

f(1)

(2)

f := proc (k) options operator, arrow; sum(k*x(n), n = 2 .. 3) end proc;

proc (k) options operator, arrow; sum(k*x(n), n = 2 .. 3) end proc

 

0

(3)

``


 

Download different_answers.mw

unknown := sum(exp(product(sum(kk, kk=1..jj),jj=1..y))*1/y!*x^y, y=0..infinity):
evalf(subs(x=1, unknown));
plot(unknown, x=-3..3, numpoints = 5);
 

g3 := 2*(1+exp(4*x))/(exp(4*x)-1);
a:=eval(diff(g3,x$n)/n!, x=0) assuming n>=0:
hello := sum(a*x^n, n=0..infinity):

it run a very long time like endless
 

 =  =

In the above output I  am geting trig functions. This is taking the derivative of f(x).  I would like to keep the answer in rational output

ie.

updated

after refer from

https://en.wikipedia.org/wiki/List_of_representations_of_e

exponential1 := sum((1/n!), n=0..infinity);
exponential1 is not a decimal number, it is exp(1)
 
hoyeung1:= sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x)), x=0..infinity);
 
hoyeung2:= sum((Int(exp(LambertW(1/(-1+x!))*(-1+x!)), x)), x=0..infinity);
 
how to evalute hoyeung1 or hoyeung2 as a decimal number?
 
how to evalute hoyeung^x as a decimal number function is func1 := proc(x) return hoyeung^x end proc:
 
but i do not know whether sum((Int(exp(LambertW(1/(-1+x))*(-1+x)), x))*m^x, x=0..infinity) = hoyeung^x
 
can limit(1+(Int(exp(LambertW(1/(-1+x))*(-1+x)), x)))^x, x=infinity) = hoyeung^1 ?

I am trying to evaluate the following summation which gives the result 0. But I think answer is not correct.
 

restart; x := 0; evalf(Sum(Sum(x^(q-p), p = 0 .. q), q = 0 .. 10))

0.

(1)

``


 

Download zero.mw

Problem.mw

Hi,

I want to know the partial derivative of

where

in order to 

ρ

If I write 

I obtain this

 

which is incorrect  since I did not have any indexes in sigma. However if I do this next command the result seems correct although no substituion of rhocrf

What I would like to obtain was the solution in order to rho and not to rhocrf and additionally having the result as a summation over p and q and not the extended result. How should I write it in Maple then? 

 

Thank you very much!

Good day everyone,

I'm trying to convert a series solution back to its original function but its given me "Error, (in rsolve/linindex) invalid subscript selector".

Anyone with useful informations please.

Attached is the series below

 

theta := convert(a-(1/2)*beta*a^2*y^2+(1/12)*beta^2*a^3*y^4-(1/72)*beta^3*a^4*y^6+(1/504)*beta^4*a^5*y^8-(5/18144)*beta^5*a^6*y^10+(1/27216)*beta^6*a^7*y^12-(95/19813248)*beta^7*a^8*y^14+O(y^16), polynom);
     1       2  2   1      2  3  4   1      3  4  6
 a - - beta a  y  + -- beta  a  y  - -- beta  a  y 
     2              12               72            

       1      4  5  8     5       5  6  10     1       6  7  12
    + --- beta  a  y  - ----- beta  a  y   + ----- beta  a  y  
      504               18144                27216             

         95        7  8  14
    - -------- beta  a  y  
      19813248             
L := [seq(coeff(theta, y, n), n = 0 .. 14)];
  [        1       2     1      2  3       1      3  4     
  [a, 0, - - beta a , 0, -- beta  a , 0, - -- beta  a , 0, 
  [        2             12                72              

     1      4  5         5       5  6       1       6  7     
    --- beta  a , 0, - ----- beta  a , 0, ----- beta  a , 0, 
    504                18144              27216              

         95        7  8]
    - -------- beta  a ]
      19813248         ]
with(gfun);
rec := listtorec(L, u(n));
[ //      2     2  3         2     2  2          2     2  \        
[{ \2151 a  beta  n  - 1434 a  beta  n  - 30592 a  beta  n/ u(n) + 
[ \                                                                

  /             3                  2                  
  \2476 a beta n  - 137904 a beta n  + 207248 a beta n

                  \         
   + 424320 a beta/ u(n + 2)

     /         3           2                      \           
   + \-153660 n  - 322920 n  + 1803360 n + 2545920/ u(n + 4), 

                               1       2          \      ]
  u(0) = a, u(1) = 0, u(2) = - - beta a , u(3) = 0 }, ogf]
                               2                  /      ]

rsolve(rec[1], u);
      / //      2     2  3         2     2  2          2     2  \ 
rsolve|{ \2151 a  beta  n  - 1434 a  beta  n  - 30592 a  beta  n/ 
      \ \                                                         

         /             3                  2                  
  u(n) + \2476 a beta n  - 137904 a beta n  + 207248 a beta n

                  \         
   + 424320 a beta/ u(n + 2)

     /         3           2                      \           
   + \-153660 n  - 322920 n  + 1803360 n + 2545920/ u(n + 4), 

                               1       2          \    \
  u(0) = a, u(1) = 0, u(2) = - - beta a , u(3) = 0 }, u|
                               2                  /    /
sum(%*y^n, n = 0 .. infinity);
Error, (in rsolve/linindex) invalid subscript selector

thanx

 

I am trying to evaluate the following function J(n,phi) which can be used to find out a*b(-A*J(3,Pi/6)+B*J(6,Pi/6)) but it takes too much of time whereas mathematica takes much less time for the same. The maple file is attached. Hope my problem get solve. Thank you

restart;

r := 2.8749; a := 0.7747; b := 0.3812; A := 17.4; B := 29000; R := 5.4813; Z := 2;

J := proc (n, phi) options operator, arrow; 8*Pi^(3/2)*r*R*(sum((2*r*R)^(2*i)*pochhammer((1/2)*n, i)*pochhammer((1/2)*n+1/2, i)*(sum((-1)^j*cos(phi)^(2*j)*(sum((2*r*cos(phi))^(2.*l)*pochhammer(n+2*i, 2*l)*hypergeom([2*j+2*l+1, .5], [2*j+2*l+1.5], -1)*(.5*Beta(l+.5, n+2*i+l-.5)-sin(arctan(-Z/sqrt(R^2+r^2)))^(2*l+1)*hypergeom([-n-2*i-l+1.5, l+.5], [l+1.5], sin(arctan(-Z/sqrt(R^2+r^2)))^2)/(2*l+1))/(factorial(2*l)*pochhammer(2*j+2*l+1, .5)*(R^2+r^2)^(n+2*i+l-.5)), l = 0 .. 100))/(factorial(i-j)*factorial(j)), j = 0 .. i))/factorial(i), i = 0 .. 100)) end proc;


evalf(a*b*(-A*J(3, (1/6)*Pi)+B*J(6, (1/6)*Pi)));
JJ.mw

I really want to use if condition inside of eval and sum, example as below:

sum(eval(y=x^(j),(if j=1 then x=2 else x=3 fi)),j=1..2)

The reason is: the value of x to be evaluated depends on the value of j which differs inside of the sum (in the context of B-spline functions).

Any lights? Thanks,

any idea for my problem?

 

> k1 := sum(X[h, t], t = 1 .. 23) >= 9;
9 <= X[h, 1] + X[h, 2] + X[h, 3] + X[h, 4] + X[h, 5] + X[h, 6]

   + X[h, 7] + X[h, 8] + X[h, 9] + X[h, 10] + X[h, 11] + X[h, 12]

   + X[h, 13] + X[h, 14] + X[h, 15] + X[h, 16] + X[h, 17]

   + X[h, 18] + X[h, 19] + X[h, 20] + X[h, 21] + X[h, 22]

   + X[h, 23], h = 1 .. 6

why 'h' still 'h'. from my textbooks the formula must be like this :
 

i want to solve the system of equation ( 1 )  , (2)  ,  (3)   under the assumation that x , y have the CDF in (4)  ,  (5)
 

diff(L(lambda[1], lambda[2], alpha), lambda[1]) = n/lambda[1]+sum(x[i], i = 1 .. n)-(sum(2*x[i]*exp(lambda[1])/(exp(x__i*`&lambda;__1`)-1+alpha), i = 1 .. n))

diff(L(lambda[1], lambda[2], alpha), lambda[1]) = n/lambda[1]+sum(x[i], i = 1 .. n)-(sum(2*x[i]*exp(lambda[1])/(exp(x__i*`&lambda;__1`)-1+alpha), i = 1 .. n))

(1)

diff(L(lambda[1], lambda[2], alpha), lambda[2]) = m/lambda[2]+sum(y[j], j = 1 .. m)-(sum(2*y[j]*exp(lambda[2])/(exp(y__j*`&lambda;__2`)-1+alpha), j = 1 .. m))

diff(L(lambda[1], lambda[2], alpha), lambda[2]) = m/lambda[2]+sum(y[j], j = 1 .. m)-(sum(2*y[j]*exp(lambda[2])/(exp(y__j*`&lambda;__2`)-1+alpha), j = 1 .. m))

(2)

diff(L(lambda[1], lambda[2], alpha), alpha) = (n+m)/alpha-(sum(2/(exp(x[i]*`&lambda;__1`)-1+alpha), i = 1 .. n))-(sum(2/(exp(y[j]*`&lambda;__2`)-1+alpha), j = 1 .. m))

diff(L(lambda[1], lambda[2], alpha), alpha) = (n+m)/alpha-(sum(2/(exp(x[i]*`&lambda;__1`)-1+alpha), i = 1 .. n))-(sum(2/(exp(y[j]*`&lambda;__2`)-1+alpha), j = 1 .. m))

(3)

G(x, lambda[1], alpha) = 1-alpha/(exp(lambda[1]*x)-1+alpha)

G(x, lambda[1], alpha) = 1-alpha/(exp(lambda[1]*x)-1+alpha)

(4)

G(y, lambda[2], alpha) = 1-alpha/(exp(lambda[2]*x)-1+alpha)

G(y, lambda[2], alpha) = 1-alpha/(exp(lambda[2]*x)-1+alpha)

(5)

``

``


 

Download internet.mw

Here is a problem from SEEMOUS 2017 (South Eastern European Mathematical Olympiad for University Students)
which Maple can solve (with a little help).

For k a fixed nonnegative integer, compute:

Sum( binomial(i,k) * ( exp(1) - Sum(1/j!, j=0..i) ), i=k..infinity );

(It is the last one, theoretically the most difficult.)

1 2 3 4 5 6 7 Last Page 1 of 12