0
$\begingroup$

I submitted the following query to WolframAlpha:

Is sum Floor[n/k]Phi[k], k=1 to n equal to sum k, k to n

and the result (link) was:

$\sum^n_{k-1}\left\lfloor \frac{n}{k} \right\rfloor \phi(k)$ is not always equal to $\sum^n_{k=1}k$

The interesting thing is that it is not true according to the solutions presented in:

What happened? Is it a bug, or there are some edge cases that WolframAlpha is taking under account?

$\endgroup$
6
  • 2
    $\begingroup$ Mathematica (and therefore WolframAlpha I presume) is not able to simplify the sum down to $n(n+1)/2$ so thats probably the reason it returns false. Returning "I don't know" would probably have been a better answer. $\endgroup$
    – Winther
    Sep 2, 2016 at 15:44
  • $\begingroup$ @Winther This is why I am asking. WolframAlpha should not say that they are not always equal. $\endgroup$
    – Mateusz Piotrowski
    Sep 2, 2016 at 15:45
  • $\begingroup$ It is known that Worlfram alpha makes sometimes mistakes. $\endgroup$
    – N. S.
    Sep 2, 2016 at 15:46
  • 1
    $\begingroup$ I think that this mistake was pointed to them couple years back, still not fixed :wolframalpha.com/input/?i=derivative+floor+x $\endgroup$
    – N. S.
    Sep 2, 2016 at 15:48
  • 1
    $\begingroup$ @N.S. Interesting. Even the newest Mathematica (version 11) gives strange values for the derivative of the floor function at non-integer values. For example $\text{Floor}'(1.1) = -4.29$ apposed to $0$. $\endgroup$
    – Winther
    Sep 2, 2016 at 16:08

1 Answer 1

Reset to default
1
$\begingroup$

I've submitted this bug to the WolframAlpha team.

This is their response:

We appreciate your feedback regarding Wolfram|Alpha. The issue you reported has been passed along to our development team for review. Thank you for helping us improve Wolfram|Alpha.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .