Testwiki:Reference desk/Archives/Mathematics/2007 September 21

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September 21

Sign Diagrams

Hi, semi-continuation of the Binomial question: I was looking at that sheet, and I'm wondering, how do I solve an inequality through sign diagrams?? Typing "sign diagrams" in wikipedia turns up nothing. --24.76.248.193 04:52, 21 September 2007 (UTC)

If you have to solve, for example, the inequation f(x) < g(x) for x, you can equivalently solve g(x) – f(x) > 0, or, putting h(x) := g(x) – f(x), the inequation h(x) > 0. So if you can find an argument for which function h returns a positive value, you're done. A web site explaining about sign diagrams can be found here.  --Lambiam 08:08, 21 September 2007 (UTC)

Thanks Lambiam. Tell me, how did you find websites like that? --24.76.248.193 02:25, 22 September 2007 (UTC)

By entering "sign diagram" into the Google search box and hitting the button labelled "Google Search".  --Lambiam 11:33, 22 September 2007 (UTC)

Was it Google?? I went up to page 20 and I still couldn't find it! Major props for finding it! --24.76.248.193 05:16, 25 September 2007 (UTC)

For me it was entry #18 in the results of this search.  --Lambiam 21:32, 25 September 2007 (UTC)

mathematical fallacy

erm, really I've got this obvious non-truth here, and it appears to stem from taking logs of something to the power ni but I wasn't aware that was a bad mathematical step to take, any explanations?
${𝑒}^{\pi 𝑖}=-1$
${𝑒}^{2𝑛\pi 𝑖}=1,𝑛\in ℝ$
taking natural logs leaves
$2𝑛\pi 𝑖=0,𝑛\in ℝ$
this obviously proves n = m for any integers n and m in the real number set, so whats wrong? ΦΙΛ Κ 20:53, 21 September 2007 (UTC)

Short explanation is that the exponential function maps multiple values in the complex plane to the same image, and so its inverse, the complex logarithm, is a multivalued function. Gandalf61 21:05, 21 September 2007 (UTC)

The even shorter explanation: When dealing with complex numbers, taking logs is a bad step. -- Meni Rosenfeld (talk) 18:21, 22 September 2007 (UTC)

Amongst other things - wheres m ?87.102.17.252 15:16, 23 September 2007 (UTC)

The OP has spared us some mundane details. Since $e^{2n\pi i}=1$ for every $n\in \mathbb{Z}$, for $n,m\in \mathbb{Z}$ we have $e^{2n\pi i}=1=e^{2m\pi i}$, so "taking logs" gives $2n\pi i=2m\pi i$ and $n=m$.

He did, however, have a different mistake; it should be $n\in \mathbb{Z}$, not $n\in ℝ$. -- Meni Rosenfeld (talk) 15:31, 23 September 2007 (UTC)

Such as ${𝑒}^{2𝑛\pi 𝑖}={𝑒}^{2𝑚\pi 𝑖}$ when n=1,m=2 so therefor n=m X

the error in resoning could be explained in terms failure to notice the oscillating value of e^ix..87.102.17.252 16:02, 23 September 2007 (UTC)