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Inconsistent results for inequality query

Posted 3 years ago

Hi All,

I've been trying to have Wolfram|Alpha help me out with some inequality calculations, but I can't seem to get consistent results. Here is an example. I'm looking for values that will give certain errors in this series, and when I use the variable n, I do get a list of possible values that meet the inequality:

enter image description here

Wanting a precise value, and thinking that the choice of variable may have had something to do with it, I put in x, however the answer is then wildly inaccurate (real answer 4<x<5 somewhere):

enter image description here

What could be happening here?

POSTED BY: Shaun Olafson
3 Replies
Posted 3 years ago

I've seen examples where Wolfram|Alpha seems like it tends to treat names w,x,y,z as real variables and names m,n, maybe even i,j.k as integer variables. What you are seeing looks like it might be another example of this.

POSTED BY: Bill Nelson
Posted 3 years ago

The recognition of integer vs. real variables is really interesting! What I'm most curious about here though is how when using n it's able to recognize that n>4 satisfies the inequality, but when using x it believes we need to go up beyond 19 to satisfy it..

POSTED BY: Shaun Olafson
Posted 3 years ago

I am not trying to defend what it did or to be critical in any way of what you did..

Your query didn't specify whether the solution should be real or integer. It didn't specify whether it should give the smallest value which satisfied the problem and include that all larger values also satisfied the problem. It didn't specify that you wanted all possible solutions or some solutions or even just one solution.

For problems over the real numbers you might know a variety of different ways of finding solutions. For problems over the integers there are perhaps fewer methods that can be used. Perhaps that might explain the result that it gave, but there is likely no way way of knowing exactly why it did what it did.

If you modified your question slightly then

(.5)^(2(x-1))/(2(x-1))!<10^-7 and 3<x<7

gives a variety of different solutions.

POSTED BY: Bill Nelson
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