Thomas Calculus 11.2.19: how do i get Maple to show 2^1/2/3 ?

Posted on 2008-10-09 23:11 By KewlEugene (

20)

Categories:

How do I ...? with Maple (Student)

for Thomas Calculus 11.2.19, the sequence below Maple repeats back the sequence again. how do i get it to show an EXACT ANSWER like the 3rd statement below ?

thanks.

sum

Comment on 2008-10-09 23:47 from Robert Israel ( Maple Rating 4

1460)

Certainly not with evalf: that is for floating-point evaluation. I would have thought that sum would do it, or else something from the SumTools package. But none of them seems to work in this case.

evalf and identify

Comment on 2008-10-10 00:29 from Doug Meade ( Maple Rating 4

721)

This is not quite an accepted proof, but it does fulfill your request "how do I get Maple to show 2^(1/2)/3?". Except, I think the correct value is 2/3^(1/2). My approach is to let Maple make a numerical approximation and then use identify to find the corresponding closed-form expression that has this approximation.

a := 2/sqrt(n+2)-2/sqrt(n+3);
                              2              2      
                         ------------ - ------------
                                (1/2)          (1/2)
                         (n + 2)        (n + 3)     
S := sum( a, n=1..infinity );
                    infinity                             
                     -----                               
                      \                                  
                       )    /     2              2      \
                      /     |------------ - ------------|
                     -----  |       (1/2)          (1/2)|
                     n = 1  \(n + 2)        (n + 3)     /

evalf( S );
                                 1.154700538
identify( % );
                                  2  (1/2)
                                  - 3     
                                  3

I think identify is a pretty neat command. I've used it several times in other situations like this. I've not shown it to my colleagues, but students think it's pretty handy.

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.ed

SumTools and _EnvFormal := true

Comment on 2008-10-10 00:50 from djc (

64)

# in Maple 12:

restart;

with(SumTools):

_EnvFormal := true;

a := 2/sqrt(n+2)-2/sqrt(n+3);

Summation(a, n=1..infinity );

2/3*3^(1/2)

_EnvFormal

Comment on 2008-10-10 03:35 from Robert Israel ( Maple Rating 4

1460)

An interesting approach. I guess what it's doing is formally summing 2/sqrt(n+2) and -2/sqrt(n+3) separately, obtaining 2*Zeta(1/2)-2-2^(1/2) and -2*Zeta(1/2)+2+2^(1/2)+2/3*3^(1/2) respectively, then adding. I'd still like to see an indefinite summation here: sum(a, n) should produce -2/sqrt(n+2).

Telescoping

Comment on 2008-10-10 21:38 from jakubi ( Maple Rating 4

923)

does not seem to work for non integer exponents, but can be done with a "generic" function call:

sum(f(k+2)-f(k+3),k=1..n);
eval(%,f=(x->2/sqrt(x)));

                           -f(n + 3) + f(3)

                                          1/2
                              2        2 3
                        - ---------- + ------
                                 1/2     3
                          (n + 3)

clever avoidance of automatic simplification

Comment on 2008-10-10 21:50 from Alex Smith ( Maple Rating 3

294)

This is a clever approach. Nice!

I keep thinking that it would be so nice to be able to turn off automatic simplification. With the original expression 2/sqrt(n+2)-2/sqrt(n+3), I suspect that automatic simplification converts the difference into a quotient, thereby blinding Maple to the telescoping quality of the series.

But your approach cleverly works around automatic simplification.

nice

Comment on 2008-10-10 22:46 from Joe Riel ( Maple Rating 4

1258)

Automatic simplification doesn't convert the different to a quotient. I suspect sum is recognizing the sqrt function and handling it differently. You could hide it with a pair of forward quotes (but I like the generic approach of using an inert function):

sum(2/''sqrt''(k+2)-2/''sqrt''(k+3),k=1..n); 
                                                  2           2
                                           - ----------- + -------
                                             sqrt(n + 3)   sqrt(3)
limit(%, n=infinity);
                                                     1/2
                                                  2 3
                                                ------
                                                  3

with %sqrt

Comment on 2008-10-11 01:11 from jakubi ( Maple Rating 4

923)

Perhaps it is better to take profit of the new inert form for the commands:

value(sum(2/%sqrt(k+2)-2/%sqrt(k+3),k=1..n));

                                          1/2
                              2        2 3
                        - ---------- + ------
                                 1/2     3
                          (n + 3)

solve by hand

Comment on 2008-10-10 01:07 from Joe Riel ( Maple Rating 4

1258)

Solving this by hand is trivial. Just split the sum into two sums and adjust the second summand so that it cancels all all but the first term of the first. Of course, you need to demonstrate that the sum actually converges, but you should know how to do that (Maple can be helpful here).

SumationTools...

Comment on 2008-10-10 01:19 from jakubi ( Maple Rating 4

923)

thought this example is trivial, and you do not really need Maple to split, change index, etc., a package providing these tools for sums, as IntegrationTools does for integrals (ie 'Split', 'Change', 'Combine', etc), is still missing.

checking convergence with csum

Comment on 2008-10-10 01:25 from djc (

64)

Using the command csum (from the Maple Advisor Database by Robert Israel) the convergence of the summation can be checked:

restart;

a := 2/sqrt(n+2)-2/sqrt(n+3);

csum(a,n);

true

Thomas Calculus 11.2.19: how do i get Maple to show 2^1/2/3 ?

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