I still do not see what MAA should answer.

If I understand you correctly then you want to take 6 terms to get sin(70).

And it is not clear which one. Either the first and last 3 for the Taylor
series cut of at degree 201 and written down. Or the first and last 3 of
the coefficients of the Taylor series.

This is not the same, since odd degrees vanish.

  "poly = Normal[Series[Cos[x], {x, 0, 6}]]";
  convert(%, FromMma);
  %;
                            2         4          6
                   1 - 1/2 x  + 1/24 x  - 1/720 x

  # taking the coefficients 
  L:=[0,1,2, 198,199,200];
  [seq(coeftayl(cos(x),x=0, k)*x^k, k in L)]: evalf(%);
  eval(%, x=70.):
  convert(%, `+`); 
                               2                       -370  198
  [1., 0., -0.500000000000000 x , -0.504654827485423 10     x   , 0.,

                            -374  200
        0.126797695348096 10     x   ]

                          -2449.00000944819

  # or only those which actually occure, the even ones
  L := [0, 2, 4, 200-4, 200-2, 200];
  [seq(coeftayl(cos(x),x=0, k)*x^k, k in L)]: evalf(%);
  eval(%, x=70.); #sort(%);
  convert(%, `+`);
                            997967.666742993

  # likewise
  convert(series(cos(x),x = 0,201),polynom):
  sort(%, x, ascending): # do that !
  [op(%)]:
  [seq(%[i],i=[1,2,3,-1,-2,-3])]: evalf(%);
  
  eval(%, x=70.): 
  convert(%, `+`);

                           997967.666742993

May be the book (which one, page/chapter?)is false or means something else?

PS: the precision in MMA is not clear, it may increase internally (have no
reference at hand, but remember discussions at usenet in sci.math.symbolic).

Yes.

If I remeber correctly it is not that clear what MMA does w.r.t. "working precision" internally, so it is hard to compare to Maple.

And as Kitonum said: quite different magnitudes. From +1e30 down to 1e-26. To catch 6 leading decimal in Maple here one would need 30 + 6 Digits, already knowing the situation.

Here I would proceed by: x3 at last, knowing x1,x2. And for those I would feed equation 2 by x2 = (from equation 1). which gives a polynomial equation in x1 of degree = 4. Formally (in Maple's language one can use 'eliminate', I think)

To be a bit unfair (and do some bashing (?)): adding any piecewise function would do. And Maple actually does such in complicated situations (for example the antiderivative given by Maple may be not continous and has to be cleaned up for that)

(1/6930)*exp(-(1/7938)*(X[4]-933)^2)*exp(-(1/6050)*(X[2]-805)^2)/
  ((1+exp((1/50)*X[4]-(1/50)*X[2]))*Pi):
subs(X[2]=x, X[4]=k, %):
T:= unapply(%, x);

c := x -> 1/6930*1/Pi*exp(-1/6050*(x-805)^2);
f := x -> Sum(exp(-1/7938*(k-933)^2)/(1+exp(1/50*k-1/50*x)),k = 0 .. 3600);

'c(x)*f(x)':
'%=Sum(T(x), k = 0 .. 3600)';
combine(%): is(%);

                                    3600
                                    -----
                                     \
                        c(x) f(x) =   )   T(x)
                                     /
                                    -----
                                    k = 0

                                 true

If you feed the original by 844.090508283072 you should get 0.112909871519970e-2, Digits = 15

The range for DirectSearch is too small (I used -100 .. +1200 to plot)

I do not quite agree here: one solves the intersection of the 2 quadrics and from that x3 follows (or: 'eliminate'). The sketched way will miss the complex solutions, it is a bit too much simplified. I think.

I am aware that he is likely trying brute force (and 3600 seems to be 1 year, 10 times day?)

That f is convex, so its extrema are in the boundary, but his 'full' task is g(x) = c(x) * f(x), c := x -> 1/6930*1/Pi*exp(-1/6050*(x-805)^2) a Gaussian.

Now plotting ln + ln over -100 .. +1200 gives ~ parabola, maximum ~ 800 (through c), diff(%,x) = 0 and fsolve locates it in x = 844.090508283072, eval(ln(c(x))+ln(f(x)), x=%) and applying exp gives me .112909871519971e-2 as maximum

I think that is the only extremum ('exp in x dominates'), but have not looked deeper

After pulling out factors being constant w.r.t. summation that writes as Sum(a(k)/(1+b(k)*xi),k = 0 .. n), n=3600, xi=exp(-1/50*x), which is a rational function in xi of degree (n-1, n). And I also see no reason for a 'simple' form.

Sum(exp(-1/7938*(k-933)^2)/(1+exp(1/50*k-1/50*x)),k = 0 .. 3600);
f:=unapply(%, x);
plot(f, -100 ..100);

plots nicely, may be you try some approximation (in your domain of application)

Playing with it I see it as KummerM(1/2-1/4*x,1,x), scaling by exp(x/2)
plots in a very slow way ( use ln(abs(...))) ) but not that bad.

Ignoring "1/2" it is LaguerreL(1/4*x,x) (after convert/hypergeom) and
for that the FunctionAdvisor says LaguerreL(a,z) = sin(Pi*(1+a))/Pi*
Int(exp(z*_t1)/(_t1^(1+a))*(1-_t1)^a,_t1 = 0 .. 1), And(-1 < Re(a),Re(a) < 0)

So I would guess that _may_ give a guess for the periodic & solutions.

But that's just an idea (using M18), hope it helps

For me the definition in that english Wikipedia is wrong, a correct version can be found at http://fr.wikipedia.org/wiki/Int%C3%A9grale_curviligne

Actually that stems from looking at differential forms. And I guess one would get funny results in applying the false definition in complex analysis, say using Cauchy for residue of 1/z, and would say that this means to loose orientation and more

Whatever one prefers as 'namings' for that stuff or which one is used in Maple

Menu -> Tools/Options/Display, choose 'Maple Display', then it should work:

plot( KummerM(1/2-(1/4)*sqrt(2)*sqrt(N), 1, sqrt(2)*sqrt(N)), N = 0 .. 10 );

Your display (for me) indicates that you switched to text instead of input

PS: you may want to use different threads for asking different questions

Though Maple does not answer for numerical evaluation the above is (almost) a 'proof', the missing part reduces to show t^2 <= exp(t) for positive t (instead of the 3d-plot)

In Maple 18 I had to modify that a bit

plot3d(f-F^2,x=-0.9..0.9,y=-sqrt(1-x^2)..sqrt(1-x^2), view=-1..3, axes=boxed);

q:=int(F^2,y=-sqrt(1-x^2)..sqrt(1-x^2)) assuming -1<x,x<1;

For me it is unclear "how" your function F2 actually is defined. And I would 1) avoid to use r as integration variable (if you define F as a function of r) and 2) you should be aware that an anti-derivative is not unique (so I would not use it your way in defining a function).

Anyway I see no reason why Maple can find an anti-derivative here.