How do I separate N DOUBLE numbers from a string without using split?
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How to prove the Fibonacci sum $\sum \limits_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}$ How can I find the number of partitions of $n$ into exactly $k$ distinct parts, where each part is at most $M$?  The number of partitions $p_k(\leq M,n)$ of $n$ into at most $k$ parts, each of size at most $M$, is given by the generating function: $$  \binom{M+k}{k}_{x} = \prod_{j=1}^{k}\frac{1-x^{M+k-j+1}}{1-x^j}= \sum_{n=0}^{kM} p_{k}(\leq M,n) x^n $$  For the number of the partitions $p_k(\mathcal{D},n)$ of $n$ into at most $k$ parts there is the recurrence relationship: $$   p_{k}(\mathcal{D},n) = p_{k}(\mathcal{D},n-k) + p_{k-1}(\mathcal{D},n) $$  But what, if I want to count only the partitions with distinct parts and restricted number of parts and restricted part size?  Update: Now I know the generating function for the number of distinct restricted partitions $p_k(\leq M, \mathcal{D},n)$ of $n$ into exactly $k$ distinct parts, all at most $M$ is $$    \prod_{j=1}^{M} (1+xq^{j}) = \sum_{k,n=0}^{\infty}p_k(\leq M, \mathcal{D},n)x^{k}q^{n} $$ and there is also a recurrence relation $$    p_k(\leq M, \mathcal{D},n) =         p_{k-1}(\leq M-1, \mathcal{D},n-k) +         p_k(\leq M-1, \mathcal{D},n-k) $$ How can I prove this? Could you recommend a book, where I could read about this? How is the authenticity of Debian packages guaranteed? I know that one can use "they" / "them" / "their" in place of "he" / "him" / "his" or "she" / "her" / "hers" when the subject's / direct object's gender is not known; for instance, just looking at the default user's about me section on Stack Exchange sites:     Apparently, this user prefers to keep an air of mystery about them   My question is what happens when someone wants to reference the subject using a pronoun instead.  Should the verb be conjugated accordingly to the pronoun? E.g.:     Apparently, they prefer to keep an air of mystery about them   Or should the conjugation be retained? E.g.:     Apparently, they prefers to keep an air of mystery about them   The former looks unclear because it wouldn't allow the reader / listener to understand if the writer / speaker is talking about a single person or about a group of people, however the latter sounds very weird, at least just to the non-native speaker which I am. What does `exec {STDIN}>&0` do? What is the advantage of using wp_mail? What is the advantage of using wp_mail() over mail(). Codex says they're similar, but they seem to be very similar. We were taught that an independent clause must have a complete thought/idea.     This is the only mammal that can fly.     How can "This is the only mammal" be treated as an independent clause?            It was a grammar lesson and there was no context there. In this situation how can I decide that it is an independent clause? Though for sure, there should be an independent clause in a correct sentence, and the clause starting with 'that' is not independent. But by this way, I can't fit it with the rule that every independent clause must have a complete thought. A simple sentence must have an independent clause. If the sentence has no independent clause how we can call it a sentence? Can even two dependent clauses form a sentence? I'm trying to solve a question where I need to write $\frac {1}{1+z^2} $ as a power series centered at $z_0=1$  I'm not allowed to use taylor expansion. So my first thought was to rewrite the function in a form where I can apply the basic identity:  $$ \frac{1}{1-x} = \sum_{n=0}^\infty x^n $$  So let's rewrite the original function $f(z) = \frac {1}{1+z^2} $ and use above identity:  $$\frac {1}{1+z^2} = \frac {1}{1-(-z^2)} $$ But observe that I need to center it at $z_0=1$ so I rewrote it to:  $$\frac {1}{1-(-z^2)}=\frac {1}{2z-(-(z-1)^2)} $$  I can factor out $\frac {1}{2z}$ and use the basic identity to get : $$\frac {1}{2z}\frac {1}{1-\frac{(z-1)^2}{2z}} = \frac {1}{2z}\sum_{n=0}^\infty \left(\frac{-(z-1)^2}{2z}\right)^n =\frac {1}{2z}\sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^n} = \sum_{n=0}^\infty \frac{(-1)^n(z-1)^{2n}}{(2z)^{n+1}}$$  The problem is if I ask wolfram alpha: power series$ \frac{1}{1+z^2} $centered at z = 1 it will give me a different answer that doesn't match the sum I found.  Can anyone help me figure this one out? What am I doing wrong? I am a new user of Ubuntu 12.04.   I installed Ubuntu and Windows XP, and I want to use both operating systems. When I restart or boot my PC the GRUB menu does not appear.  How can I solve this problem? Google map only loading half of itself Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that:     Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$, and $x$ is an element of minimal degree among the elements that are not invertible, then $D/(x)$ is isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.   Now the exercise asks:     Prove that $A$ is not a Euclidean Domain.   Everything hints to an argument by contradiction: let $(A, d)$ be a ED and $x$ an element of minimal degree among the non invertibles we'd like to show that $A/(x)$ is not isomorphic to $\mathbb Z/2\mathbb Z$ or $\mathbb Z/3\mathbb Z$.  How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this $x$!  I know that the elements of $A/(x)$ are of the form $a+(x)$, with $a$ of degree less than $x$ or zero. By minimality of $x$ this means that $a\in \{0, 1, -1\}$. Now I'm lost: how do we derive a contradiction from this? I'm suppose to prove the following...  Let $A$ be a nonempty subset of $\mathbb{R}$. If $\alpha = \sup A $ is finite show that for each $ \epsilon &gt; 0 $ there is an $a \in A $ such that $\alpha - \epsilon &lt; a\leq \alpha $.  This is what I have so far...  Proof: We claim that $\alpha = \sup A $   Then by definition of a supremum $\alpha $ is the least upper bound of set A  Thus $\forall a \in A$,  $a \leq \alpha$   Suppose $\epsilon &gt; 0$   Then $\alpha - \epsilon &lt; \alpha $.   Thus $\alpha - \epsilon &lt; a\leq \alpha$.   I feel like I'm missing some step between my last and second to last step so what am I forgetting? Also is the rest of my proof right? How to show the solution of $\dot{x} = Ax $ is an invariant subspace? Consider the linear dynamical system $\dot{x} = Ax $ in $V$ a finite dimensional vector space.  The definition of an invariant subspace $U$ is as follows:  For all $x_0 \in U$, (initial condition), the solution $x(t)$ with initial condition $x(0)=x_0$ lies in $U$, i.e. $x(t) \in U$ for all $t \in \mathbb{R}$. Intuitively, it means if we start in a subspace we never leave that subspace.  Prove that a subspace $U \subset V$ is an invariant subspace if and only if $AU \subset U$. I've recently run into many questions where to author asks several related questions in one post, many of them asked before.   I'm wondering what out policy is on such questions:   Should I vote-close them as being not constructive? Should I vote-close them as duplicates? Should I leave them be? Or even suggest an answer?    - Each of the questions: Waiting for an AJAX request with selenium, and headless browsing in selenium have been asked here before, but he asked both in one.  I've done my best to search for similar questions here in meta but I was unable to find any, reading appreciated. No Dropbox top panel icon since update to 18.04 Evaluating the indefinite integral $ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $ I have been having extreme difficulties with this integral. I would appreciate any and all help. $$ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $$ Which packages should be loaded after hyperref instead of before? The USB has various purposes for data transfer, movement of input and output signals for hardware like connecting an Android device to another Display screen (USB on the go), charging the device etc.(during a device charge the voltage drawn from the USB is same irrespective of the device or OEM then Why are two USB Cables different ?  A USB of one manufacturer may not be supporting another manufacturer. We get problems if we do connect them like      or Value of $\frac{x^n -a^n}{ x-a}$? While playing Phase Ten, is it legal to pick up the top card of the discard pile and then immediately discard that exact card?