1
$\begingroup$
In[1]:= Sum[(-1)^k*Binomial[n, k]*Binomial[k, j], {k, 0, n}]

Out[1]= (j Binomial[0, j])/(j - n)

According to this output, for any j,n, such as j=n, the result of this sum should be indeterminate, thanks to the division by zero. And indeed:

In[2]:= Sum[(-1)^k*Binomial[n, k]*Binomial[k, j], {k, 0, n}] /. {j -> 2, n -> 2}

During evaluation of In[2]:= Power::infy: Infinite expression 1/0 encountered. >>

During evaluation of In[2]:= Infinity::indet: Indeterminate expression 0 ComplexInfinity encountered. >>

Out[2]= Indeterminate

This seems wrong. Because, if I manually substitute the j's and n's with 2's, Mathematica itself claims that:

In[3]:= Sum[(-1)^k*Binomial[2, k]*Binomial[k, 2], {k, 0, 2}]

Out[3]= 1

Well... 1 is far enough from Indeterminate, isn't it? Actually, if my calculations are right, the correct result is like this: $\left(-1\right)^{\left[2\nmid j\right]}\left[n=j\right]$ (the brackets here are the Iverson brackets)

I suspect a bug. However, I'd really like to ascertain I'm not wrong. Too many times Mathematica has been giving apparently wrong results to me because of my own lack of understanding its peculiarities.

So, could you kindly tell me, if the above results are because of my errors, or is it a bug that should be reported to WRI? Thanks!

$\endgroup$
1
  • 6
    $\begingroup$ Put a GenerateConditions -> True in your sum and it will tell you when the answer is valid. $\endgroup$
    – wxffles
    Mar 11 '15 at 23:50
7
$\begingroup$

Expanding on the comment by @wxffles

Sum[(-1)^k*Binomial[n, k]*Binomial[k, j], {k, 0, n}, 
 GenerateConditions -> True]

enter image description here

For the case when j == n,

Sum[(-1)^k*Binomial[n, k]*Binomial[k, n], {k, 0, n}, 
 GenerateConditions -> True]

enter image description here

Combining these results

sum[j_, n_] = 
 Piecewise[{{Sum[(-1)^k*Binomial[n, k]*Binomial[k, j], {k, 0, n}], 
    j != n && Element[n, Integers] && 
     n >= 0}, {Sum[(-1)^k*Binomial[n, k]*Binomial[k, n], {k, 0, n}], 
    j == n && Element[n, Integers] && n >= 0}}]

enter image description here

Table[sum[j, n], {j, 0, 5}, {n, 0, 5}] // TableForm

enter image description here

However, j need not necessarily be an integer

Plot[Evaluate[Table[sum[j, n],
   {n, 0, 5}]], {j, 0, 5},
 PlotLegends -> Range[0, 5]]

enter image description here

$\endgroup$
2
  • $\begingroup$ If MMA's output is correct only for some special cases... why doesn't MMA automatically test the separable cases and return a Piecewise by itself...? $\endgroup$
    – gaazkam
    Mar 12 '15 at 11:51
  • 1
    $\begingroup$ @gaazkam - presumably in the trade-off between overall execution efficiency versus explicitly handling every possible special case when providing results, WRI chose overall efficiency. $\endgroup$
    – Bob Hanlon
    Mar 12 '15 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.