A bug with the WolframAlpha computational search engine?

Question

I think I may have discovered a bug with WolframAlpha.

So I was trying to determine all $x$ such that

$$\sum_{i=0}^{5}{x^{-i}} < \frac{13}{12}.$$

WolframAlpha spit out $x > 13$ (see this link).

However, when I double-checked,

$$\sum_{i=0}^{5}{{13}^{-i}} = \frac{402234}{371293} \approx 1.0833331\ldots.$$

Of course, we have

$$\frac{13}{12} = 1.08\bar{3}.$$

This means that $x = 13$ is a solution to the inequality

$$\sum_{i=0}^{5}{x^{-i}} < \frac{13}{12}.$$

Here are some more examples:

WolframAlpha link 1

WolframAlpha link 2

Does this mean that we can't trust WolframAlpha?

Answer 1

Yes. A simple equivalent is trying to solve:

$$13x^5<x^6+12$$

So you are seeking solution to $f(x)=x^6-13x^5+12=0$. $x=13$ is a gross estimate, but it is not far off. If we take $x=13+\epsilon$ then $(13+\epsilon)^6-13(13+\epsilon)^5 + 12 = 0$ which, if we ignore exponents for $\epsilon$ greater than $1$ yields the equation:

$$13^6 + 6\epsilon \cdot 13^5 - 13(13^5+5\epsilon\cdot 13^4) + 12 =12 + \epsilon \cdot 13^5$$

So $$\epsilon\approx \frac{-12}{13^5}\approx -0.00003231948892$$

And even that is not perfect: $f(12.99996171721108)\approx -2.21$.

A more accurate answer is: $$x> 12.999967680123.$$

Then $$f(12.999967680123)\approx 0.0000050783$$

I don't know what WA used for rounding. It only appears to use 6 digits of accuracy elsewhere, which would result in $x>13$.

If I ask WA to:

solve x^6-13x^5+12

It returns $13.0000$.

When I ask WA to:

solve (13+x)^6-13(13+x)^5+12

it returns $-0.0000323199$.

Answer 2

13 is just a rounded output because you can find zeros of polynomial of grade grater then 4 just numericaly hence you can't get exact form. In fact, the more precise output is $x>12.9999677$ so 13 is still a good value. You can get more precise output just by clicking "more digits"

Answer 3

In Mathematica, it suffices to enter the command

Reduce[Sum[x^-i, {i, 0, 5}] < 13/12, x, Integers]

Which gives the output (which I have interpreted) $$(x \in \mathbb Z) \cap (( x \le -1 ) \cup (x \ge 13)).$$

Unfortunately, Wolfram|Alpha does not support Reduce[] (for obvious reasons--if it did, why buy Mathematica?). But if you enter this:

Case x = 13

You will get True.