Wikipedia:Reference desk/Archives/Mathematics/2010 June 13

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June 13

Meromorphic invertible functions

The Möbius transformations are the only invertible meromorphic functions on the complex plane. In the more general case with n complex functions of n complex variables, are there any invertible functions with only n-1 complex dimensional singularities where $\lim _{w,z,\cdots \to singularity}{\frac {1}{f(w,z,\cdots )}}$ exists other than those of the form $f(w,z,\cdots )={\frac {aw+bz+\cdots +c}{jw+kz+\cdots +l}}$? 74.14.111.223 (talk) 06:06, 13 June 2010 (UTC)

Limits / infinite series

To abridge the question somewhat:

${\text{Show that }}{\frac {\theta }{e^{\theta }-1}}<1{\text{for all }}\theta {\text{. Show also that it is near }}1{\text{ if }}\theta {\text{ is small and near }}0{\text{ if }}\theta {\text{ is large.}}$

Now, I'm afraid I could not really get anywhere with this, trying an approach which began $e^{\theta }>0\implies e^{\theta }-1>-1$, but then did not know where to go from there. The answer says, verbatim:

$e{\text{.}}g{\text{. Expression is }}{\frac {\theta }{\theta +{\frac {\theta ^{2}}{2!}}+...}}$ (1)

${\text{always }}<1$ (2)

${\text{and this is }}\approx 1{\text{ if }}\theta {\text{ is small}}$ (3)

$\approx 0{\text{ if }}\theta {\text{ is large}}$ (4)

Now, I understand how they got to (1), but (2), (3) and (4) are something of a mystery to me at present. Could anyone enlighten me, please? It Is Me Here ^{t / c} 13:54, 13 June 2010 (UTC)

Surely theta > 0 is implied? If you understand how they got to (1), then (2) is obvious: the denominator is numerically larger than the numerator. If theta is small, then the higher power terms in the denominator are dominated by the linear term so your expression is approx theta/theta, this gets you (3). For (4), divide numerator and denominator through by theta and consider whats happens as theta tends to infinity. Zunaid 14:34, 13 June 2010 (UTC)

I think that you want θ ≥ 0, because as θ grows negatively then e^θ becomes very small, and your function behaves like y = –x for very large, negative θ. In fact, the line y = –x is an asymptote as θ tends to negative infinity. When θ = 0 your function is indeterminate (we get zero divided by zero). We can use L'Hôpital's rule to evaluate the limit of your function as θ tends towards 0, and we can show that this limit is 1. Also, it's clear that your function tends to 0 as θ tends towards positive infinity. This half-answers your question. You just need to show that your function doesn't misbehave away from those values. Personally, I would work out the derivative, and try to show that it is always negative. If you draw a graph of your function, you see that it decrease constantly for all θ. So let's calculate the derivative of your function:

$f'(\theta )={\frac {(1-\theta )e^{\theta }-1}{(e^{\theta }-1)^{2}}}.$

Clearly (e^θ – 1)² ≥ 0 for all θ and it follows that the sign of the derivative is given by the sign of g(θ) = (1 – θ)e^θ – 1. Well, that's not strictly true. When θ = 0 then the derivative is indeterminate (we get zero divided by zero). We can use L'Hôpital's rule to evaluate the limit of g(θ) as θ tends towards 0, and we see that this limit is ½(1 – e)^–1 (which is negative!). Now that we know what happens at θ = 0; to show that the derivative of your function is negative for all θ, we simply need to show that g(θ) < 0 for all θ ≠ 0. To do this, we can look for the stationary points of g(θ). We see that $g'(\theta )=-\theta e^{\theta },$ and so $g'(\theta )=0$ if and only if θ = 0. Is this a maximum or a minimum? Well, $g''(\theta )=-(1+\theta )e^{\theta }$ and so $g''(0)=-e<0$ meaning that θ = 0 is a maximum of g(θ). In fact, it is a global maximum since it is the only stationary point. Notice that g(0) = 0 and so g(θ) < 0 for all θ ≠ 0. It follows that $f'(\theta )<0$ for all θ. So your function has value 1 when θ = 0 and constantly decreases towards 0 as θ increases towards positive infinity. •• Fly by Night (talk) 15:42, 13 June 2010 (UTC)

P.S. The sample solution is a bit hand-wavy. What it is saying is that close to zero, the lowest order terms in the Taylor Series dominate. So your function will behave like θ/θ = 1 for very small θ. For very large θ, the highest order terms will dominate. What this means though is unclear. What is the highest terms in an infinite Taylor Series? Your function behaves very differently as θ tends towards negative infinity than it does when θ tends towards positive infinity. Personally, the sample solution isn't at all rigorous; like I said: it's a bit hand-wavy. •• Fly by Night (talk) 15:59, 13 June 2010 (UTC)

For $\theta >0$, the denominator is $>\theta +\theta ^{2}/2$. This is enough to show that the limit is 0 - no need to muse over the meaning of "highest terms". Also, after you factor out the θ all terms are increasing, so the function is strictly decreasing. -- Meni Rosenfeld (talk) 17:27, 13 June 2010 (UTC)

Fly by Night, surely if $g\left(\theta \right)=\left(1-\theta \right)e^{\theta }-1=e^{\theta }-\theta e^{\theta }-1$ then $g'\left(\theta \right)=e^{\theta }-\theta e^{\theta }$? It Is Me Here ^{t / c} 19:16, 13 June 2010 (UTC)

No, I'm afraid not. Try using the product rule. In particular:

${\frac {d}{d\theta }}(\theta e^{\theta })=\left({\frac {d}{d\theta }}\theta \right)e^{\theta }+\theta \left({\frac {d}{d\theta }}e^{\theta }\right)=e^{\theta }+\theta e^{\theta }$

So you will find that $g'(\theta )=-\theta e^{\theta }$ as stated. •• Fly by Night (talk) 19:23, 13 June 2010 (UTC)

Silly me, thanks! It Is Me Here ^{t / c} 19:42, 13 June 2010 (UTC)

Meni Rosenfeld, could you elaborate on the last sentence of your reply, please? It Is Me Here ^{t / c} 19:19, 13 June 2010 (UTC)

As Zunaid suggested, you can write ${\frac {\theta }{e^{\theta }-1}}={\frac {\theta }{\theta +\theta ^{2}/2!+\cdots }}={\frac {1}{1+\theta /2!+\cdots }}$. For $\theta >0$, Every term in the denominator is increasing (gets larger when θ gets larger), so the sum is also increasing. The function is therefore decreasing. This is not really necessary for the questions asked, it was more of a reply to Fly by Night. -- Meni Rosenfeld (talk) 19:31, 13 June 2010 (UTC)

Oh, I see now. Thanks, everyone. It Is Me Here ^{t / c} 19:42, 13 June 2010 (UTC)

(edit conflict) I like your solution Meni; it's a nice simple solution... I was using a sledgehammer to crack a walnut. Full marks Meni! •• Fly by Night (talk) 19:44, 13 June 2010 (UTC)

Er sorry, one more thing. Meni Rosenfeld, if the function is strictly decreasing, does that necessarily mean that it tends to y=0? Could one have a strictly decreasing function tending to y=-2 or y=3 or something? It Is Me Here ^{t / c} 19:47, 13 June 2010 (UTC)

Since the denominator grows without bound whilst the numerator is constant, the fraction as a whole tends to zero. (The proof of this fact is one of the elementary results demonstrated when limits are introduced in under-grad maths. The proof would typically use a squeeze theorem and/or a suitable limit comparison test to simplify the denominator, then prove that the simplified fraction converges to zero by the usual epsilon-delta definition. Once the course moves on to more difficult problems you can typically take those elementary results as a given without further proof.) Zunaid 21:47, 13 June 2010 (UTC)

No and yes, respectively. A function which is strictly decreasing can tend to 0, or to -2, or 3, or any other real number, or negative infinity. My proof that the function is decreasing was just to show that what Fly by Night did could have been done simpler (in particular, that the sample solution was rigorous, in the sense that once you write the Taylor series, everything else follows trivially). My proof for what you were actually asked - that the limit is 0 - was in the first sentence: After θ is factored out, the denominator is $1+\theta /2+\cdots >1+\theta /2\to +\infty$, so the function tends to 0. -- Meni Rosenfeld (talk) 08:15, 14 June 2010 (UTC)

Not all strictly decreasing functions have a limit of zero. The function ƒ(θ) = k + e^–θ is a strictly decreasing function; i.e. if θ > φ then ƒ(θ) < ƒ(φ), and its limit as θ tends to positive infinity is k. Since k can be chosen arbitrarily, we can give ƒ any limit we choose. •• Fly by Night (talk) 19:04, 14 June 2010 (UTC)

Taylor series

In one of your articles there is:${\begin{aligned}(x+y)^{n}&={n \choose 0}x^{n}y^{0}+{n \choose 1}x^{n-1}y^{1}+\cdots \end{aligned}}$ <extended content>

What is meant by ${n \choose 0}$? —Preceding unsigned comment added by 76.229.192.126 (talk) 21:05, 13 June 2010 (UTC)

${n \choose 0}={\frac {n!}{0!(n-0!)}}={\frac {n!}{n!}}=1,$ meaning that there is one way to choose no element out of n elements. See article N choose k. DVdm (talk) 21:14, 13 June 2010 (UTC)

It's rather ${n \choose 0}={\frac {n!}{0!(n-0)!}}={\frac {n!}{1\cdot n!}}={\frac {n!}{n!}}=1,$ but the conclusion is correct. --CiaPan (talk) 21:33, 13 June 2010 (UTC)

Of course, a typo induced by correcting a typo. Sorry. Silly DVdm (talk) 21:44, 13 June 2010 (UTC)

If you tell us which article you are talking about we can insert the appropriate link. Bo Jacoby (talk) 21:51, 13 June 2010 (UTC).

Naturally, one place where it appears is Binomial theorem, but there there is already a link to Binomial coefficient. -- Meni Rosenfeld (talk) 08:20, 14 June 2010 (UTC)

I had pointed to article N choose k, which redirects to Binomial coefficient. DVdm (talk) 09:57, 14 June 2010 (UTC)

What are you replying to? Bo asked where did the OP encounter the formula. -- Meni Rosenfeld (talk) 10:53, 14 June 2010 (UTC)

Bo's indentation suggests he was asking me. DVdm (talk) 13:34, 14 June 2010 (UTC)

I see. Since Bo was clearly replying to the OP, I've taken the liberty to correct his indentation. -- Meni Rosenfeld (talk) 14:49, 14 June 2010 (UTC)