Bug in splitting sum

Question

Bug introduced in 9.0, persisting through 12.3.1

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I was trying to evaluate the following sum.

$$ \frac{2}{m}\sum_{\substack{\text{odd }k\\1\leq k\leq m-1}} f\left(\frac{m+2+\sqrt{m^2-4k+4}}{2}\right)+f\left(\frac{m+2-\sqrt{m^2-4k+4}}{2}\right). $$

And I wrote the following Mathematica code.

2/m Sum[f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])] +
        f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}]

Then I tested it with $f(x)=x^2$ and $m=60$. I defined

f[x_] = x^2

and evaluated

2/m Sum[f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])] +
     f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}] /.
  m -> 60 // N

which gave

(* 3664. *)

However, if I split the sum into two parts, it ends with a very different value:

 2/m Sum[f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}] +
 2/m Sum[f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}] /. m->60 //N

gives

(* 4.6048 + 1.12029*10^-13 I *)

What is the problem here? There seems to be no problem if $f(x)=x^p$ and $p$ is fractional. I tried $p=1.1$, $1.5$, $2.1$, etc., and both answers agree. Looks like the problem occurs when $p$ is an integer...

Answer 1

There is a symbolic calculation bug in there:

Let's define:

plus[k_, m_]  := f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])];
minus[k_, m_] := f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])];

While

(With[{m = 60}, Sum[minus[k, m], {k, 1, -1 + m, 2}]] // N) ==
               (Sum[minus[k, m], {k, 1, -1 + m, 2}] /. m -> 60 // N)

(* True *)

on the other hand:

(With[{m = 60}, Sum[plus[k, m], {k, 1, -1 + m, 2}]] // N) ==
               (Sum[plus[k, m], {k, 1, -1 + m, 2}] /. m -> 60 // N)

(* False *)

So the term with the HurwitzZeta[ ..] functions is in error.

Answer 2

This is interesting! Here's a partial answer (so more of a long comment):

Clear[f]
f[x_] = x^2;
Sum[f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}]

and the output is:

whereas

Sum[f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}]

doesn't evaluate (i.e. it returns itself), and

Sum[f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])] + f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}]
(* -(1 + Floor[1/2 (-2 + m)]) (-2 - 2 m - m^2 + 2 Floor[1/2 (-2 + m)]) *)

You will note that if you set the terms with the HurwitzZeta function to zero, then these two expressions are identical up to a factor of 1/2. The HurwitzZeta functions would cancel each other out if the arguments for two of those functions didn't have the Floor[m/2] pieces in them.

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Now, if we evaluate the sums symbolically first (the 2/m doesn't make a difference), we get

Clear[f, f1]
f1 = Sum[f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])] + 
         f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}];
f1 = f1 /. m -> 60 // Expand;
f[x_] = x^2;
f1 // N
(* 109920. *)

and

Clear[f]
f1 = Sum[f[1/2 (2 + m + Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}] + 
  Sum[f[1/2 (2 + m - Sqrt[4 - 4 k + m^2])], {k, 1, -1 + m, 2}];
f1 = f1 /. m -> 60 // Expand;
f[x_] = x^2;
f1 // N
(* 109920. *)

which indicates to me that they should be equal.

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On the other hand,

Sum[(1/2 (2 + 60 + Sqrt[4 - 4 k + 60^2]))^2 + (1/2 (2 + 60 - Sqrt[4 - 4 k + 60^2]))^2, {k, 1, -1 + 60, 2}] // N
Sum[(1/2 (2 + 60 - Sqrt[4 - 4 k + 60^2]))^2, {k, 1, -1 + 60, 2}] +
   Sum[(1/2 (2 + 60 - Sqrt[4 - 4 k + 60^2]))^2, {k, 1, -1 + 60, 2}] // N
(* 109920 *)
(* 71.3744 *)

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I perhaps suspect a bug. However, I also suspect that if the second sum evaluated, it would have terms that would cancel the HurwitzZeta functions in the first term, so perhaps it's instead some sort of problem there.