I want to know an idiom for something like if the thing is good it will have some negative also. OR something like with good comes bad.    Let's say for a sentence "As they say ...................., nuclear energy has some negative aspects as well."    I know there are a few expressions like "curator's egg" and "double edged sword" but they don't fit in perfectly.
I'm trying to transition from a list of pros to a list of cons, and would like to have some sort of transition quote. Something in the form of,  "But, alas, (((something something quote quote)))"  Everything that has a beginning has an end, something along the lines of that.
How can I unlock my phone? I locked it using Android Device Manager, but it claims I entered a wrong password I have an Samsung Galaxy S3 Neo.  Bus 002 Device 042: ID 04e8:6860 Samsung Electronics Co., Ltd GT-I9100 Phone [Galaxy S II], GT-I9300 Phone [Galaxy S III], GT-P7500 [Galaxy Tab 10.1]   The keyboard &amp; language setting is português.  I today installed Android Device Manager and to test it's functionality I locked the phone with the password: "abcd" (4 letters were required).  After entering the password and confirming with the "OK" button of my keyboard, the phone showed me that the password is wrong. "Senha incorreta".  I reset the password after some tries to "123456" and the phone reloaded the lock screen. Again the password showed up to be wrong. I also tried the alphabetic password: "abcdefg" - same error.  I changed the language of my Google account from German to Portuguese and reset the password to "123456" again and I tried to really reboot my phone and unlock the SIM card again. But anyways I can not get through my lock of the Android Device Manager.  I tried to enter the password 35 times wrong with hope to get a chance to authenticate with my Google Account login.  I also tried to connect the phone with adb but it does not show up in this state with:  adb devices   So what can I do to get access to my phone again? On finding polynomials that approximate a function and its derivative (extensions of Stone-Weierstrass?) The Stone-Weierstrass Theorem tells us that we can approximate any continuous $f:\mathbb{R}^n\to\mathbb{R}$ arbitrary well on a compact subset of $\mathbb{R}^n$ by some polynomial. Suppose that $f$ is continuously differentiable, under what conditions can we guarantee we can find a polynomial $p$ such that $p$ approximates arbitrarily well $f$ and the partial derivatives approximate of $p$ arbitrarily well those of $f$ (on a compact subset of $\mathbb{R}^n$)? References are welcome. See below for my (probably naive) approach to the question.    It is straightforward to find conditions if $n=1$: Let $K:=[a,b]\subseteq\mathbb{R}$ and let $K_+$ denote some open cover of $K$. Suppose that $f:K_+\to\mathbb{R}$ is continuously differentiable and let $f'$ denote its derivative. Then, by Weierstrass's approximation Theorem we have that there exists a polynomial $q$ such that   $$||f'(x)-q(x)||\leq \varepsilon$$  for all $x\in K$. Choose some $y\in K$. For any $x\in K$ define  $$p(x):=f(y)+\int_y^xq(t)dt.\quad\quad(*)$$  By the Fundamental Theorem of calculus we have that $p'=q$. So for any $x\in K$  $$||f(x)-p(x)||=\left|\left|\int_y^x (f'(t)-q(t))dt\right|\right|\leq \int_y^x ||f'(t)-q(t)||dt\leq \varepsilon||x-y||\leq \varepsilon (a-b).$$  However, the above argument doesn't work in higher dimensions because the analogue of $(*)$ does not imply that $\nabla p = q$ (since not every vector of polynomials spans a gradient field). Can anyone suggest a way around this? That is, a way to to show that for continuously differentiable $f:\mathbb{R}^n\to\mathbb{R}$ and for any $\varepsilon&gt;0$, we can always find a polynomial $p$ such that   $$||f(x)-p(x)||+||\nabla f(x)-\nabla p(x)||\leq \varepsilon$$  for all $x$ in some compact set $K$? Feel free to add extra conditions on $K$, if it makes life easier. Draw power from a portable mobile charger to power circuits I usually power my beginner prototype circuits using mb102 power supply board, but it needs an external elimeter to draw power and then regulate it.  I was looking for a portable solution for my 5 volt and 3.3 volt circuit needs and came up with using a portable battery charger for mobile phones.  I tried powering up a simple transistor and LED circuit with it but it wouldn't power-up.   I feel that there is some load sensing circuit in the portable charger which doesn't detect these low power circuits.  I also tried connecting different resistors to the charger, the charger outputs 5 Volts for a few seconds and then again shuts down and doesn't stay ON continuously.  It would be really comfortable to use a portable charger since it could be charged using a phone charger and would be very handy.  What can I do to make the charger output power on low loads? I'm colorizing the header of a table formatted with column -ts $'\t'  Works well without color codes, but when I add color codes to the first line column doesn't properly align the output.  Without colored output it works as expected: printf "1\t2\t3\nasdasdasdasdasdasdasd\tqwe\tqweqwe\n" | column -ts $'\t'  But when adding color on the first line column doesn't align the text of the colored row: printf "\e[7m1\t2\t3\e[0m\nasdasdasdasdasdasdasd\tqwe\tqweqwe\n" | column -ts $'\t'  Observed this behaviour both on Ubuntu Linux and Mac OS X. I have Ubuntu on my laptop. Now I want install Windows 7 in a dual-boot. How can I do this? I can't lose my Ubuntu files, and I'm afraid that I might break . Go  for UEFI only! Why is the mol a fundamental physical quantity? I am starting to study physics in detail and as I read about physical quantities, I was puzzled why mol (amount of substance) is taken as a physical quantity.  A physical quantity is any quantity which we can measure and has a unit associated with it. But a mol represents the amount of substance by telling the number of particles (atoms, molecules, ions, etc.) present. So it is a pure number and numbers are dimensionless. So mol should not be considered a physical quantity.  Also, fundamental physical quantities should be independent of each other. I am wondering whether mass and mol are independent. This is so as they surely affect each other as we can evidently see while calculating the number of moles and using the mass of that sample for calculation.  So how is the mol a fundamental physical quantity and independent of mass? When and how did "momentarily" come to mean "in a moment", rather than "for a moment"? "Momentarily" used to mean "for a moment" only, and not "in a moment". Thus, newscasters could be divided into two clear groups: those who would say "we'll be back momentarily," and those who would not.  This restriction made sense to me, because having both definitions would promote ambiguity if a unique interpretation could not always be derived from the context. But in recent years it seems "momentarily" is regularly, maybe even more often, used to mean "in a moment" by newscasters of every caliber, and in fact this is even shown to be the definition when looked up in most dictionaries.  When did this word's meaning change? How did it come about? GOTO still considered harmful? Is Gauss's law wrong, or is it possible that $\int_s{\vec E} \cdot d\vec{s}=0$ does not imply $\vec E = 0$? This is a question from David J Griffith's Introduction to Electrodynamics.      A specified charge density $\sigma(\theta)=k\cos(\theta) $ is glued over the surface of a spherical shell of radius $R$. Find the resulting potential inside and outside the sphere.   The question was solved using legendre polynomials and the final answer for potential inside the sphere was : $V(r,\theta) = \frac{kr}{3\epsilon_0}cos\theta$  This final answer is confusing because The Electric field inside the sphere is coming out to be dependent on $r$ and $\theta$ whereas electric field inside a shell, no matter what the charge distribution is outside, is $zero$ from gauss's law.      my doubts:         why is the electric field inside non zero?    Can gauss's law explain this, or does it fail here?      Since solving using ordinary surface integral gave me the same result and since the. divergence inside the shell is $zero$, I concluded that legendre polynomials and gauss's law in differential form are correct. So the problem should be with the integral form of gauss's law : $\int_s{\vec E}.d\vec{s} = \frac{q}{\epsilon_0}$  The answer I got for this doubt is that  "since the charges are glued to the surface, and not evenly distributed, electric field inside need not be zero. "     This is not convincing because proof of gauss's law does not expect the charges to be free to move. presence of an exrernal force that would hold the charges in place does not change the theorem. That is    Say only a single charge $q_i$ is present outside      then $\int_s{\vec E_i}.d\vec{s} = \frac{q_{inside}}{\epsilon_0}=0$      Now if there are more charges, following any distribution, net electric field $\vec E = \vec E_1 + \vec E_2+\vec E_3+...$       So the net flux,       $\int_s{\vec E}.d\vec{s} = \int_s{\vec E_1}.d\vec{s}+\int_s{\vec E_2}.d\vec{s}+\int_s{\vec E_3}.d\vec{s}+. . .=0$   Or is it possible that $\int_s{\vec E}.d\vec{s}=0$ does not imply $\vec E = 0$? Do I lose anything converting to DNG? I'd like to convert all my ARW (Sony) files to DNG for several reasons, but the only thing keeping me back from doing it is the fear of losing useful metadata during the conversion. I know that the image it's self is completely safe during the conversion but what about proprietary maker noteS? Will those stay too? And even if they do, are they actually useful and can be actually used by something other then Sony's own RAW converter?  P.S. I'm using Ubuntu so I'd prefer to use KipiPlugins' DNG converter because from what I understand it does everything the Adobe converter does, but if I'm wrong I can use Adobe's instead.  EDIT:  It looks like the LensID does carry over! With the ARW and the DNG files converted by Kipiplugins' built in converter, info is stored as "Exif.Sony.0x___". With Adobe's official converter, it moves the info to more logically named fields in the XMP, such as "LensID" and "Lens".  Now that I've figured it out, I think I might use Adobe's Official converter, just because I prefer how it reorganizes the Exif Data.  I'm not fully convinced though and I'd like to hear others take on all this. Fancyhdr does not apply same header/footer on chapter and non-chapter pages How do I prove that $S^n$ is homeomorphic to $S^m \Rightarrow m=n$? This is what I have so far:  Assume $S^n$ is homeomorphic to $S^m$. Also, assume $m≠n$. So, let $m&gt;n$.  From here I am not sure what is implied. Of course in this problem $S^k$ is defined as:  $S^k=\lbrace (x_0,x_1,⋯,x_{k+1}):x_0^2+x_1^2+⋯+x_{k+1}^2=1 \rbrace$ with subspace topology. Allow the owner of a post to approve a suggested edit even if it was rejected by reviewers Why should I use the keyword "final" on a method parameter in Java? I am faced with the following problem: Servers get saturated since current load balancing strategy is based on client IP. Some corporate clients access our servers from behind large proxies so all clients appear with same IP to our load balancer. I think we are using some hardware load balancing device (can investigate further if necessary). We need to maintain session affinity (site is constructed in ASP), so all requests with same IP get routed to the same node.  Since all the communication goes over the HTTPS, no request data (like session Id) is available to balancer as a client discriminator. Is there a way to use some other data besides the IP to distinguish between clients and route the clients even when coming from same IP to different nodes?  Note: I need to maintain the traffic between the balancer and nodes safe (encrypted). Showing that the class of all sets of a particular cardinality is not a set. How to show that the class of all sets of a particular cardinality ,say $h$ is not a set.   My argument:  I assume that I've shown the following lemma.  Lemma: If $X$ is an infinite set of cardinality $h$ , $a\ne x$ for every $x\in X$ then $\{a\}\cup X$ is of cardinality $h$ too i.e. adding new element to an infinite set doesn't change cardinality.  Let $h$ be an infinite cardinal number. Let $T_h=\{x:\text{x is a set of cardinality h}\}$. Suppose $T_h$ is a set.   Let $a$ be any set and consider any $t\in T_h$. Let $R=t\cup \{a\}$. Using the above lemma, $R\in T_h$.   So, $a\in R\subset \bigcup T_h$. As $T_h$ is a set then $\bigcup T_h$ is also a set.  As $a$ was arbitrary set then we have, $a\in\bigcup T_h$ for every set $a$. Hence, the set $\bigcup T_h$ contains all sets which meas that the collection of all sets is a set, contradiction.    If $h$ is non-zero finite cardinal number. Define $T_h$ the same way as above. Let $a$ be any set, Let $A_1=a, A_{n+1}=A_n\times a$. Consider $A=\{A_1,A_2,...,A_h\}$. So, $A\in T_k$. Hence, $a=A_1\in \bigcup T_h=K$. So K is a set containing all sets $a$ since $a$ was arbitary,contradiction.  My main question is, Is there any flaws in my argument? If yes, point it out please.   If the argument is valid, Is there any standard proof to show the same fact? Is it possible that $a_n>0$ and $\sum a_n$ converges then $na_n \to 0$? (without assuming $a_n$ is decreasing) How is dark energy consistent with conservation of mass and energy? I initially thought that dark energy must in some way violate conservation of mass and energy since the component of the energy density of space that comes from dark energy is constant, and space is expanding. Therefore, as space expands the energy in the universe that comes from dark energy would increase. I presumed the source of this energy was not coming from the conversion of other types of energy to dark energy, so it must violate conservation.  I decided to Google this and came upon this article:    It says that dark energy does NOT violate conservation and quotes Carroll, Press, and Turner (1992):  "…the patch does negative work on its surroundings, because it has negative pressure. Assuming the patch expands adiabatically [i.e. without loss or gain of heat], one may equate this negative work to the increase of mass/energy of the patch. One thereby recovers the correct equation of state for dark energy: P = – ρ c2. So the mathematics is consistent."  Is there a way to explain this in layman's terms? (The blog attempted to do this, but it was very unclear to me.) More specifically, can you explain where my initial train of thought described above fails when I erroneously concluded that dark energy violates conservation?  Thank you. Finding an example of a discrete-time strict local martingale. How can I keep multiple live/bootable ISO images on a single USB drive?