If $f(z)$ is analytic function and satisfies $ \vert f(z)\vert &lt;1$ for $\vert z\vert&lt;1$. show if $f(z)$ has a zero of order m at $z_o$ , then $\vert z_o \vert^ m &gt; \vert f(0)\vert$.   here is what I know: if we let $\psi = (z-z_o)/ ( 1-\bar z_o z )$ then $\vert f(z) \vert &lt; \vert \psi(z)\vert^m$ but how can I prove the last inequality? i.e. how to prove that $\vert f(z) \vert &lt; \vert \psi (z) \vert^m$?   I just think that Im missing something?   Im interested specifically in solving it by using this way $\psi$  thanks
There's this exercise: let $\,f\,$ be analytic on $$D:=\{z\;\;;\;\;|z|&lt;1\}\,\,,\,|f(z)|\leq 1\,\,,\,\,\forall\,z\in D$$  and $\,z=0\,$  a zero of order $\,m\,$ of $\,f\,$.  Prove that $$\forall z\in D\,\,,\,\,|f(z)|\leq |z|^m$$  My solution: Induction on $\,m\,$: for $\,m=1\,$ this is exactly the lemma of Schwarz, thus we can assume truth for $\,k&lt;m\,$ and prove for $\,k=m&gt;1\,$ . Since $\,f(z)=z^mh(z)\,\,,\,h(0)\neq 0\,$ analytic in $\,D\,$ ,  put  $$g(z):=\frac{f(z)}{z}=z^{m-1}h(z)$$   Applying the inductive hypothesis and using Schwarz lemma $\,\,(***)\,\,$ we get that  $$|g(z)|=\left|\frac{f(z)}{z}\right|=|z|^{m-1}|h(z)|\stackrel{ind. hyp.}\leq |z|^{m-1}\Longrightarrow |f(z)|\leq |z^m|$$ and we're done...almost: we still have to prove $\,|g(z)|\leq 1\,$ for all $\,z\in D$ in order to be able to use the inductive hypothesis and this is precisely the part where I have some doubts: this can be proved as follows (all the time we work with $\,z\in D\,$):  $(1)\,\,$ For $\,f(z)=z^mh(z)\,$ we apply directly Schwarz lemma and get $$|f(z)|=|z|^m|h(z)|\leq |z|\Longrightarrow |z|^{m-1}h(z)|\leq 1$$ And since now the function $\,f_1(z)=z^{m-1}h(z)\,$ fulfills the conditions of S.L. we get  $(2)\,\,$ Applying again the lemma,  $$|f_1(z)|=|z|^{m-1}|h(z)|\leq |z|\Longrightarrow |z^{m-2}h(z)|\leq 1$$and now the function $\,f_2(z):=z^{m-2}h(z)\,$ fulfills the conditions of them  lemma so...etc.  In the step$\,m-1\,$ we get  $$|z||h(z)|\leq |z|\Longrightarrow {\color{red}{\mathbf{|h(z)|\leq 1}}}\,$$  and this is what allows us to use the inductive hypothesis in $\,\,(***)\,\,$ above.  My question: Is there any way I can't see right now to deduce directly, or in a shorter way, that $\,|h(z)\leq 1\,$ ?
Can't download any apps on iPad 1 I have an iPad 1 with iOS 5.1.1 that won't download any apps or update its OS. It was donated to my school so it was totally wiped before we got it.  I've tried the work around previously mentioned of downloading the apps in iTunes then dragging them over to my device that is also connected to iTunes. When I do this, a dialogue box comes up saying it was unable to install the app because the app requires a newer version of iOS. I'm really hoping Apple didn't make it so a perfectly good piece of tech is obsolete by not making iOS upgrades or apps compatible with their older technology. Citing two authors gets displayed incorrectly Layers with same coordinates not lining up Example of quasi-compact, non-quasi seperated scheme where qcqs fails? Moving from SQL 2005 [SQL_Latin1_General_CP1_CI_AS] to 2008 - will I lose any features by using 'backwards compatibility' We're moving from SQL 2005 [Instance and DB have collation of SQL_Latin1_General_CP1_CI_AS]  to SQL 2008 [which defaults to Latin1_General_CI_AS].  I completed a SQL 2008 R2 installation, and used default Latin1_General_CI_AS collation, with the restoration of the database still on SQL_Latin1_General_CP1_CI_AS. The excepted problems occurred - the #temp tables where in  Latin1_General_CI_AS whilst the db was in  SQL_Latin1_General_CP1_CI_AS and this is where I am now - I need advice on the pitfalls now please.  On installation of SQL 2008 R2, I have the option on installation to use 'SQL Collation, used for backwards compatibility' where I have the option to select the same collation as the 2005 database : SQL_Latin1_General_CP1_CI_AS.   This will allow me to not have problems with #temp tables, but are there pitfalls? Would I lose any functionality or features of any kind by not using a "current" collation of SQL 2008? What about when we move (e.g. in 2 years ) from 2008 to SQL 2012? Will I have problems then? Would I at some point be forced to go to Latin1_General_CI_AS? I read that some DBA's script complete the rows of complete databases, and then run the insert script into the database with the new collation - I'm very scared and wary of this - would you recommend doing this? I was having a discussion with a friend of mine about some normal group properties and then came up with the question "if G is not commutative, then is there always a subgroup that is not a normal subgroup?" It's probably more easy to solve this in the following form:$$\forall H \leq G : H  \lhd G \Rightarrow \forall a,b \in G : ab=ba$$ My question is, can anybody give a proof, or a counter-example (because I don't think it holds) of this theorem? Thanks! Proving that expression is equivalent to the definition of derivative Let $f$ be differentiable at $x=a$.   Prove that if $x_n \to a^+$ and $y_n \to a^-$ then:  $$\lim_{n\to \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(a).$$  Every option that I think about seems to my very trivial, so I believe that I am doing something wrong.  Both numerator and denominator approach zero as $n\to\infty$ as the case of the formal definition of derivative, but it isn't guaranteed that the limits are equal (“$\frac{0}{0}$”).   Any direction? Calculations in quotient rings Can we please hold the hats? I really like the hats game. But I would love to keep the hats longer then only one winter. I want to leave them in my wardrobe during summer. Last year, I collected some cool hats, and I apparently lost them during summer.  Maybe we could make a system that allows hats to last for about 2 or 3 years, so we don't have to start over each winter. Or even specify per type of hat a number of years they last? Very very rare hats could last for ever.  I asked this, simply because I love hats! :D QT MySql connectivity using Windows XP, Qt Creator 4.5.2(windows 32 bit) Exclusive lighting Is (or will be) there any sort of light linking in Cycles, or a method to get a light to affect only certain objects that doesn't involve compositing two or more different render layers? Sort of what 3Dstudio does with the Include/Exclude methods. Why was mixing declarations and code forbidden up until C99? I am learning P&amp;S's Introduction of quantum field theory. My teacher said that if the mass of neutrino is exactly 0, then we should not observe any right-handed neutrinos and left-handed anti-neutrinos according to Weyl's theory. It is because $$ \left( i\gamma^\mu \partial_\mu - m \right) \psi = \left( \begin{array}{cc} -m &amp;  i\left(\partial_0 + \vec\sigma \cdot \vec\nabla\right) \\ i\left(\partial_0 - \vec\sigma \cdot \vec\nabla\right) &amp; -m \end{array} \right) \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right) =0. $$  If we set m=0, then we will have:  \begin{align} i\left(\partial_0 - \vec\sigma \cdot \vec\nabla\right) \psi_L &amp;= 0; \\ i\left(\partial_0 + \vec\sigma \cdot \vec\nabla\right) \psi_R &amp;= 0. \end{align}  And if these formulas can describe neutrinos we must conclude that there are two types of neutrinos.  My confusions are:   It is known that the mass of neutrino is not 0, therefore there are chances (depending on the mass) to find right-handed neutrinos and left-handed anti-neutrinos. But we don't. If I accept that there are only these two types of neutrinos (right-handed anti-neutrinos and left-handed neutrinos), this property of neutrino causes the parity violation, so parity violation must relate to angular momentum and therefore spinor space. Am I write? Are there any reasons why neutrino is so weird that is has mass but still has helicity + Or - 1. This goes against with what I have learned that only when a particle  moves at the speed of light can its spin aligned with its momentum. What type of path in shebang is more preferable? How to limit execution time of a function call? Overleaf does not compile after using Biblatex How to disable pagebreak on \hline in longtable? How to disable pagebreak on \hline in longtable?  This is an example illustrating my problem:  \documentclass[12pt,a4paper,oneside]{letter}  \usepackage{longtable}  \begin{document}  \begin{longtable}{ | p{5cm} | } \hline Table header\\* \hline Lorem ipsum dolor sit amet, consectetur adipiscing elit. Mauris pharetra porttitor turpis, id egestas neque congue in. Nam sed lorem lorem, ac tristique tellus. Nunc ornare feugiat orci, a blandit turpis facilisis ut. Aenean pharetra ornare ante. Sed tempus leo in dolor eleifend placerat. Integer at enim augue. Ut nisl tortor, interdum ac porttitor ac, sagittis quis libero. Vivamus nibh metus, ornare vel pulvinar nec, fringilla nec orci. Maecenas mollis nulla a nibh fringilla nec convallis augue varius. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere   cubilia Curae. \\ Lorem ipsum dolor sit amet, consectetur adipiscing elit. Mauris pharetra porttitor turpis, id egestas neque congue in. Nam sed lorem lorem, ac tristique tellus. Nunc ornare feugiat orci, a blandit turpis facilisis ut. Aenean pharetra ornare ante. Sed tempus leo in dolor eleifend placerat.  Integer at enim augue. Ut nisl tortor, interdum ac porttitor ac, sagittis quis libero. Vivamus nibh metus, ornare vel pulvinar nec, fringilla nec    orci. Maecenas mollis nulla a nibh fringilla nec convallis augue varius.   Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere   cubilia Curae. \\ \hline Table header\\* \hline Lorem ipsum dolor sit amet, consectetur adipiscing elit. Mauris pharetra porttitor turpis, id egestas neque congue in. Nam sed lorem lorem, ac    tristique tellus. Nunc ornare feugiat orci, a blandit turpis facilisis ut. Aenean pharetra ornare ante. Sed tempus leo in dolor eleifend placerat.    Integer at enim augue. Ut nisl tortor, interdum ac porttitor ac, sagittis  quis libero. Vivamus nibh metus, ornare vel pulvinar nec, fringilla nec    orci. Maecenas mollis nulla a nibh fringilla nec convallis augue varius.   Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere  cubilia Curae. \\ \hline \end{longtable}  \end{document}   I'd like to have second header on page 3. I do not control table contents, so I can not do page breaks manually. 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Did it become popularized in anime/manga because of the popularity of the artist's work or because it is a common belief in japanese folklore?