Batman (triprc) wrote in algorithms,
I'm sorry allow me to rephrase this. If I'm graphing e^x is there a point when y will have a zero value?
  • Post a new comment

    Error

    Anonymous comments are disabled in this journal

    default userpic

    Your reply will be screened

    Your IP address will be recorded 

  • 15 comments

erad1cate

May 20 2009, 22:57:39 UTC 5 years ago

Only as you get closer to negative infinity will you be able to attain values that are close to 0. Concretely, there isn't a value for x in e^x that will equal to 0.

korean_guy_01

May 20 2009, 23:06:44 UTC 5 years ago

You should clarify what close to 0 means here.

erad1cate

May 20 2009, 23:17:52 UTC 5 years ago

I hate real analysis...

BUT:

Pick any positive value close to 0, you can find an x for e^x to give that value.

Example, for instance .001 = e^(-6.907755...)

If you're posting in algorithms, I'm assuming a value close to 0 at some point is 0 if you're looking for a numerical solution.

itman

May 21 2009, 00:21:33 UTC 5 years ago

log(eps), where epsilon is close to zero :-)
eps -> 0, log(eps) -> -infty

roadriverrail

May 20 2009, 22:59:48 UTC 5 years ago

You're right. You're rephrasing it. You're asking the same question. It has the same answer.

joee_girl

May 20 2009, 23:14:21 UTC 5 years ago

1. This isn't really an algorithms question.

2. e^k (for k>0) increases as k increases. This should be easy to see from the graph.
e^0 = 1
e^-k (k>0) = 1/(e^k)
so your question is, is there a number M such that 1/M = 0? And hopefully you can answer this for yourself.

triprc

May 21 2009, 05:45:05 UTC 5 years ago

Oh my mistake, I didn't mean to interrupt the fervent use of this community. I'm clearly wasting precious space and your very important time.

Knob job.

roadriverrail

May 21 2009, 17:55:10 UTC 5 years ago

There is no reason to cop an attitude like that.

triprc

May 21 2009, 20:06:47 UTC 5 years ago

That's weird, it seemed like you folks were the ones who initially copped a 'tude.

roadriverrail

May 21 2009, 20:13:20 UTC 5 years ago

We're not a homogeneous group (i.e. "you folks"), and it is worth noting that you essentially asked the same question twice, in two posts, that both posts were off-topic, and that both posts could have been answered using Wikipedia, a basic web search, or even an online graphing calculator.

I'll also note that nobody here called you anything pejorative, that several people answered your question in multiple ways, and that the worst you ever got was people telling you you were off-topic.

So, I fail to see why your histrionics and calling someone a "knob job" was appropriate or even necessary.

triprc

May 21 2009, 20:40:14 UTC 5 years ago

I'm making referring to the fact that his/her comment wasn't the first. Only at the end of the day after twenty or so comments of how foolishly off topic I was did I find myself annoyed. Either no one reads the comments to make sure his/her contribution isn't redundant or several members this community get their jollies off telling people that they are wrong.


roadriverrail

May 21 2009, 20:44:13 UTC 5 years ago

Either way, calling other people "knob jobs" isn't useful or necessary. I'd imagine you're capable of reasonable maturity even when you're not being shown it by others.

triprc

May 21 2009, 21:09:50 UTC 5 years ago

“When angry, count to four. When very angry, swear.”-Mark Twain

What can I say? I was raised in the city on the East Coast. Sticks and stones.

czarandy

May 21 2009, 02:23:06 UTC 5 years ago

Since this is algorithms and not mathematics, the answer is "yes", since any finite precision number representation will eventually round to 0.

For example: printf("%lf\n", exp(-1e20)) may print 0, depending on the compiler/platform.

roadriverrail

May 21 2009, 19:10:37 UTC 5 years ago

Well, algorithms do not imply their target architecture. A Turing Machine, for example, has the benefit of an infinite tape, and could give you infinite precisions.