MaplePrimes - answers and comments on Question, Identity with Maple from Rhysick

For example

alec — Thu, 29 Jan 2009 10:46:40 Z

For example,

simplify(arctan(n + 1) - arctan(n - 1)) assuming Re(n)>1;

                                     2
                             arctan(----)
                                      2
                                     n

simplify(arctan(n + 1) - arctan(n - 1)) assuming Re(n)<-1;

                                     2
                             arctan(----)
                                      2
                                     n

simplify(arctan(n + 1) - arctan(n - 1)) assuming Re(n)>-1, Re(n)<1;

                                  2
                          arctan(----) + Pi
                                   2
                                  n

The last one is not necessarily true though. For example,

eq:=n->arctan(2/n^2)=arctan(n+1)-arctan(n-1):
evalf(eq(0.5+0.5*I));

     1.570796327 - 0.2554128119 I = 1.570796327 - 0.2554128119 I

Maple is rather confused about that, however,

evalf(eq(1/2+I/2));

     -1.570796327 - 0.2554128119 I = 1.570796327 - 0.2554128119 I

But sometimes it is definitely true. For example,

evalf(eq(I));

                  -1.107148718 = 2.034443936 + 0. I

Finally, a plot,

n := x+I*y;
plots:-densityplot(Re(arctan(n+1)-arctan(n-1)-arctan(2/n^2)), 
    x = -3 .. 3, y = -3 .. 3, style = patchnogrid, 
    grid = [100, 100], color = red, axes = boxed);

Alec

Ok Alex! But, still not what I want

lemelinm — Thu, 29 Jan 2009 23:43:34 Z

Thanks for analysing the région where the function exist. But, I need a proof of the identity, not simple say it is true because whent Maple simplify, the two sides agree. Le me put it in the situation.

My exploration is one of Ramanujan solution of:

arctan(2/n^2)=arctan(n+1)-arctan(n-1);
                       /2 \                                
                 arctan|--| = arctan(n + 1) - arctan(n - 1)
                       | 2|                                
                       \n /                                
S:=Sum(arctan(2/n^2),n=1..infinity);
                             infinity          
                              -----            
                               \               
                                )          /2 \
                               /     arctan|--|
                              -----        | 2|
                              n = 1        \n /
SN:=Sum(arctan(2/n^2),n=1..N)=Sum(arctan(n+1)-arctan(n-1),n=1..N);
            N                  N                                  
          -----              -----                                
           \                  \                                   
            )         /2 \     )                                  
           /    arctan|--| =  /    (arctan(n + 1) - arctan(n - 1))
          -----       | 2|   -----                                
          n = 1       \n /   n = 1                                
SN := arctan(N+1)+arctan(N)-arctan(1)-arctan(0);
                                                  1   
                      arctan(N + 1) + arctan(N) - - Pi
                                                  4   
S:=limit(SN,N=infinity)assuming N>1;
                                    3   
                                    - Pi
                                    4

You can see the advantage of using this identity. So what I need is to prove:

arctan((x-y)/(1+x*y))=arctan(x)-arctan(y);
                         / x - y \                        
                   arctan|-------| = arctan(x) - arctan(y)
                         \1 + x y/

Ok! thanks Alec...

lemelinm — Fri, 30 Jan 2009 01:29:41 Z

You are welcome

alec — Fri, 30 Jan 2009 05:00:43 Z

Mario,

You are welcome. You have a gift for starting interesting threads.

Alec

Well Alec, I try my best :-)

lemelinm — Fri, 30 Jan 2009 19:28:08 Z

The proof

alec — Fri, 30 Jan 2009 00:01:04 Z

Mario,

You said that you needed a proof in Maple, and the first simplify does that - for those values of n, except n=1, in which case it can be simply checked.

The proof "on the paper" for those values of n is rather simple - if one takes tan of both sides, (s)he will get an obvious identity. Now, for n>=1, the lhs is in the interval between 0 and Pi/2, and the rhs is in the same interval. On this interval, tan is increasing, so it is a 1-1 function. Thus, since tan(lhs)=tan(rhs), we have lhs=rhs.

In my post above I didn't analyze where the function exists. I analyzed where the identity holds. It holds in the region painted black on the plot, and it doesn't hold in the region painted red.

The same is true for your last suggested identity. Sometimes it is true, and sometimes the lhs and the rhs of it differ by Pi.

Alec