In many problem or theorem in which I have to show that given function is polynomial, there is the argument that for some k show that $f^{(n)}(z)=0, \forall n&gt;k $ . I know that by definition of polynomial above characteristics is ok. But I am not convinced with fact that converse part. that there is no function other than polynomial which satisfies this property. Where am I  missing? Any help will be appreciated.
Let $f$ be a complex function with the property that $f^{(n)} = 0$.  Does this imply that $f$ is a polynomial?  If so, why?  Upon thinking about this problem myself, I can easily observe that every $n$ degree polynomial satisfies $f^{(n+1)} = 0$.  Still, this doesn't show that some non-polynomial function couldn't also have this property.
Noncoalescing and bouncing droplets if you set a dish filled with water onto a loudspeaker or other vibrator and make vibration, When it oscillates, it is possible to hold small droplets on its surface for a long time. I mean, when you put a droplet of water on vibrating water surface, it doesn't coalesce with the water and bounce on the surface of water.( see this video  )  As we know, drop reaches its minimal potential when coalescing with water ( that's why drops coalesce)  what exactly cause this? why doesn't the drop coalesce with the water? Why can't you use the keyword 'this' in a static method in .Net? Does a monk's Unarmored Movement speed increase apply to the fly speed gained from a Fly spell? Unix separation of specific numbers from Column How can I separate specific numbers from a column and add it to a new file? For example, take all numbers that are "134" from column 3, I tried using   awk '$3 == 134" file.txt   but it didn't work, also tried using   awk -F, '{print $3 == 134 }' file.txt    didn't work.  Pink          132           134 Blue          189           167 Yellow        2000          134 Green         678           145 Black         789           145 White         780           135   So the new file will contain just the following   Pink          132           134 Yellow        2000          134 List sorting by custom field I dont know how to sort a list of SObject Case.  I have a List&lt;Case&gt; newCases which i iterate over in for cycle, where i save Cases to my new List&lt;Case&gt; casesList.  for (Case newCase : newCases) {             if(newCase.Case_Content__c == '111' &amp;&amp; newCase.Status == '24'){                 String cuid = newCase.CUID__c;                 cuidSet.add(cuid);                 casesList.add(newCase);                 casesList.sort(); //TODO sorting by CUID__c             }         }   What i need to do is, to sort this new list by field CUID__c. So the Cases with same CUID are together.  Thank you for any help :) Can somebody give me an example of a graph on eight vertices that has a pair of pseudosimilar vertices? Or, is it not possible? If it is not possible, please do not give me a proof, as I would like to figure out how to prove it myself.     Definition: vertices $u$ and $v$ in a graph $X$ are similar if there is an automorphism of $X$ that maps $u$ to $v$. If $u$ and $v$ are similar, then the vertex-deleted subgraphs $X-u$ and $X-v$ are isomorphic. If $X-u$ and $X-v$ are isomorphic but $u$ and $v$ are not similar, we say that they are pseudosimilar. Is there a way to "auto login" in PuTTY with a password? Is there a way to configure a password for a stored session in ?  I know there is the capability to specify an "auto-login username" (under Connection/Data), but is there a way to do the same with the password? Command Shift Period (show hidden files) stop working On my MacBook with OS Mojave, Command Shift Period (show hidden files) stop working. How to restore it?? Thanks in advance! using nginx with SNI By now I've not used SNI with nginx yet. But as IP address pools are quite filled and commercial XP support is about to cease (finally) I'm thinking about converting a few sites to SNI.  I'm aware of the general limitations and pitfalls that might come along with SNI (XP issue, very old browsers). But beyond that is there anything I should be aware of?   Like - nginx related pitfalls when using SNI - issues/bugs with recent (notable!) browsers Identifying which part of a large document is being compiled Extract single file from tar to other folder I've searched for multiple forums for this questions and have seen different answers but so far, nothing have worked yet even from the ones I've seen here.  I want to extract a single file from a tar to a different directory.  I've tried this:  tar xvf file.tar -C /home/dir/ filename   or this:  tar -x filename -f file.tar -C /home/dir   But got these errors respectively:  x filename, 14826 bytes, 29 media blocks. File -C not present in the archive. File /home/dir not present in the archive.   and:  tar: /dev/rmt0: A file or directory in the path name does not exist.   The first extracted the file but in the same directory, not to the folder I wanted. How to prove that homomorphism from field to ring is injective or zero? Possible Duplicate:          $f:K\rightarrow R$ is a homomorphism from field to ring. To prove is that it is injective or zero. This is just a curiosity. I have come across multiple proofs of the fact that there are infinitely many primes, some of them were quite trivial, but some others were really, really fancy. I'll show you what proofs I have and I'd like to know more because I think it's cool to see that something can be proved in so many different ways.  Proof 1 : Euclid's. If there are finitely many primes then $p_1 p_2 ... p_n + 1$ is coprime to all of these guys. This is the basic idea in most proofs : generate a number coprime to all previous primes.  Proof 2 : Consider the sequence $a_n = 2^{2^n} + 1$. We have that  $$ 2^{2^n}-1 = (2^{2^1} - 1) \prod_{m=1}^{n-1} (2^{2^m}+1), $$  so that for $m &lt; n$, $(2^{2^m} + 1, 2^{2^n} + 1) \, | \, (2^{2^n}-1, 2^{2^n} +1) = 1$. Since we have an infinite sequence of numbers coprime in pairs, at least one prime number must divide each one of them and they are all distinct primes, thus giving an infinity of them.  Proof 3 : (Note : I particularly like this one.) Define a topology on $\mathbb Z$ in the following way : a set $\mathscr N$ of integers is said to be open if for every $n \in \mathscr N$ there is an arithmetic progression $\mathscr A$ such that $n \in \mathscr A \subseteq \mathscr N$. This can easily be proven to define a topology on $\mathbb Z$. Note that under this topology arithmetic progressions are open and closed. Supposing there are finitely many primes, notice that this means that the set $$ \mathscr U \,\,\,\, \overset{def}{=} \,\,\, \bigcup_{p} \,\, p \mathbb Z $$ should be open and closed, but by the fundamental theorem of arithmetic, its complement in $\mathbb Z$ is the set $\{ -1, 1 \}$, which is not open, thus giving a contradiction.   Proof 4 : Let $a,b$ be coprime integers and $c &gt; 0$. There exists $x$ such that $(a+bx, c) = 1$. To see this, choose $x$ such that $a+bx \not\equiv 0 \, \mathrm{mod}$ $p_i$ for all primes $p_i$ dividing $c$. If $a \equiv 0 \, \mathrm{mod}$ $p_i$, since $a$ and $b$ are coprime, $b$ has an inverse mod $p_i$, call it $\overline{b}$. Choosing $x \equiv \overline{b} \, \mathrm{mod}$ $p_i$, you are done. If $a \not\equiv 0 \, \mathrm{mod}$ $p_i$, then choosing $x \equiv 0 \, \mathrm{mod}$ $p_i$ works fine. Find $x$ using the Chinese Remainder Theorem.   Now assuming there are finitely many primes, let $c$ be the product of all of them. Our construction generates an integer coprime to $c$, giving a contradiction to the fundamental theorem of arithmetic.  Proof 5 : Dirichlet's theorem on arithmetic progressions (just so that you not bring it up as an example...)  Do you have any other nice proofs? Prove that for every $a, b\in R\setminus\{0\}$ is correct this inequality:  $$\frac{|a+b|}{1+|a+b|}&lt;\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}.$$ Countable axiom of choice: why you can't prove it from just ZF This is a follow-up question to the discussion about the finite axiom of choice .  Suppose we have a countable collection of non-empty sets $\{A_1, A_2, A_3,\cdots\}$ Reasoning as indicated in that discussion, we can prove the existence of a choice function $c_n$ such that $c_n(i)$ belongs to $A_i$ for $i = 1,\cdots,n$  Again reasoning as suggested and using induction, we can prove the existence of a function $C$ defined over $\{1, 2, \cdots\}$ such that $C(n)$ is a choice function defined for $i=1,\cdots,n$ with $C(n)(i)$ an element of A_i. We can require in addition that $C(n)(i)$ and $C(n-1)(i)$ are equal for $i=1,\cdots,n-1$.  Now it would it seem that we can prove the countable axiom of choice by taking the union of $C(1)$, $C(2)$, $\cdots$  The question is as follows: why is that proof wrong? I suspect the answer may have something to do with the axiom of union. What does "causally connected" or "causes" really mean? In a different thread, a user stated the following about events preceding or following other events:     However, if the two events are causally connected ("event A causes event B"), the causal order is preserved (i.e. "event A precedes event B") in all frames of reference.   My question is what does "causally connected" really mean? What does "causes" mean? Further, given that we know that we can have instantaneous effects in typical quantum processes (e.g. flip the polarizer, effect on another a reading light years away, even though we cannot transmit useful information with this), does that not constitute "causing" for the purpose of this statement? It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0,\forall x$.  Hence, we cannot have the wavefunction and its derivative vanishing at the same point.  My question is, how to prove the above statement? What is "it" in the following sentence: It is clear that Bob likes doughnuts I am very confused. Unless I am mistaken, I know "it" has to be a noun of some sort, but I am unable to figure out what noun "it" is referring to.   What is "it" in the following sentence:      It is clear that Bob likes doughnuts.   Heres another couple of examples:     It is impossible to fly. Duplicate explanation should encourage improvement of original The new closed-as-duplicate banner says:     This question has been asked before and already has an answer. If those answers do not fully address your question, please .   I think it should instead say     This question has been asked before and already has an answer. If answers there do not fully address your question, please  or  to the pre-existing one.   &mdash; to discourage re-re-asking the same question. (I also changed "those answers", which has no plural , to "answers there", which doesn't need one.)    As of September 19, 2013, the original text quoted above has returned here. Find all natural numbers $x,y$ such that   $$3^x=2y^2+1$$  solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$.