I just found something weird and I have no idea where it is wrong. I used two different approaches.  $\int _0^x(t^3-2t+3)\ dt=4\implies \frac{1}{4}x^4-x^2+3x=4\implies$ approximately $x=1.63$ or $x=-3.09$.  By the fundamental theorem of calculus, $\frac{d}{dx}\int _0^x(t^3-2t+3)\ dt=\frac{d}{dx}4\implies x^3-2x+3=0\implies$ approximately $x=-1.89$.   Could anyone tell me where it went wrong?
Suppose we have $x^2+4x+3=0$  We differentiate both sides to get $2x+4=0$.  So $x=-2$  But $(-2)^2+4(-2)+3=0$   $4-8+3=-1$ is not equal to $0$.  Which tells that we cannot differentiate equalities. But in many proofs, like in integral calculus, differentiating both sides occurs. So, where is the fallacy?
n tasks assigned to n computers, what is the EX value of a computer getting 5 or more tasks? Orthogonal basis for infinite-dimensional vector spaces How to run 32-bit app in Ubuntu 64-bit? For $f :\mathbb{R} \rightarrow \mathbb{R}\displaystyle$ we have ${\lim_{n \to +\infty} f(x+n) =0}$ almost everywhere I want to show that for $f: \mathbb{R} \rightarrow\mathbb{R}$ Lebesgue integrable  $\displaystyle{\lim_{n \to +\infty} f(x+n) =0}$ almost everywhere.  It is clear to me that this statement is true and I have tried to prove it by contradiction, by assuming that either $\displaystyle{\lim_{n \to +\infty} f(x+n) =\infty}$ or$\displaystyle{\lim_{n \to +\infty} f(x+n) =K}$ with $K\in \mathbb{R}$ .  In the first case  we can follow that $|f|=\infty$ so that $\int_\mathbb{R}f=\infty$ and we have a contradiction.  In the other case we know that there must be an $a\in \mathbb{R}$ so that $\forall x&gt;a$ we have $f_+(x)&gt;K/2$ and by that:  $\int_{{\mathbb{R}}+}|f|&gt;\int_{\mathbb{R}_+}f_+=\int_{[0,a]}f_++ \int_{(a,\infty)}f_+&gt;\int_{[0,a]}f_++\int_{(a,\infty)}K/2=\infty$  Would this be enough to show the statement? Can I even make the assumption that the limit of $f(x+n)$ "exists"? Any hints would be great. Limit $\lim\limits_{x\to0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$ Without using L'Hopital's rule find: $$\lim_{x\rightarrow 0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$$ I found the first derivative because I planned to use Taylor series:  \begin{align} \frac{d}{dx}{\left(\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}\right)} &amp;=\frac{\left(\frac{1}{x+e}-e^x\right)\left(\cos^2{x} -e^x\right)-\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)} {(\cos^2{x} -e^x)^2}\\ &amp;=\frac{\frac{1}{x+e}-e^x}{\cos^2{x}-e^x}-\frac{\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}{(\cos^2{x} -e^x)^2} \end{align}  However, it seems I haven't gone so far. Should I start from the beginning and try a different method?  Source in Croatian: 2.kolokvij, matematička analiza Syntax error due to using a reserved word as a table or column name in MySQL A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.  And my question is: What is the probability that all three tosses result in the same outcome?   So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?  If so, we could use the Addition Rule which is $P(A \cup B) = P(A) + P(B)$  where A and B are disjoint events, i.e. $A \cap B = \emptyset$, A is the event of tossing all heads and B is the event of tossing all tails.  I'm not sure how to continue after that... would the complement be used? Greiner in his book "Field Quantization" page 173, eq.(7.11) did this calculation:     ${\mathcal L}^\prime=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu A_\nu\partial^\nu A^\mu-\frac{1}{2}\partial_\mu A^\mu\partial_\nu A^\nu $   $\space\space\space\space=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu+\frac{1}{2}\partial_\mu[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$        The last term is a four-divergence which has no influence on the field equations. Thus the dynamics of the electromagnetic field (in the Lorentz gauge) can be described by the simple Lagrangian        ${\mathcal L}^{\prime\prime}=-\frac{1}{2}\partial_\mu A_\nu\partial^\mu A^\nu$     Yes, if it is a four-divergence of a vector whose 0-component doesn't contain time derivatives of the field, indeed according to the variational principle this four-divergence will not influence the field equation.  And actually I calculated the time-derivative dependence of 0-component of $[A_\nu(\partial^\nu A^\mu)-(\partial_\nu A^\nu) A^\mu]$, in which only $[A_0(\partial^0 A^0)-(\partial_0 A^0) A^0]$ could possibly contain time-derivative, which vanishes fortunately, so whatever the general case it doesn't matter in this present case.  But how can he seem to claim that it holds for a general four-divergence term,The last term is a four-divergence which has no influence on the field equations?  EDIT: I only assumed the boundary condition to be $A^\mu=0$ at spatial infinity, not at time infinity. And the variation of the action $S = \int_{t_1}^{t_2}L \, dt$ is due to the variation of fields which vanish at time, $\delta A^\mu(\mathbf x,t_1)=\delta A^\mu(\mathbf x,t_2)=0$, not having the knowledge of $\delta \dot A^\mu(\mathbf x,t_1)$ and $\delta \dot A^\mu(\mathbf x,t_2)$, which don't vanish generally, so the four-divergence term will in general contribute to the action, $$\delta S_j=\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \partial_\mu j(A(x),\nabla A(x),\dot A(x))^\mu =\delta \int_{t_1}^{t_2}dt\int d^3\mathbf x \dot j^0 =\int d^3\mathbf x [\delta j(\mathbf x, t_2)^0-\delta j(\mathbf x, t_1)^0]$$ which does not vanish in general! Going into a directory linked by a link How do I set the grub timeout and the grub default boot entry? In Ubuntu 12.04 (or above), how do I set the GRUB time and the default OS (that I see at boot time) as I'm dual-booting Windows (7/8) and Ubuntu (12.04 or above)? What exactly is a kilogram-meter? Can't append  element Any idea why the piece of code below does not add the script element to the DOM?  var code = "&lt;script&gt;&lt;/script&gt;"; $("#someElement").append(code); How do moving charges produce magnetic fields? Here is my question:  Is a homeomorphism, that is, a continuos function whose inverse function is also continuous, always bijective? I haven't logged onto my blog for years, and have, of course, forgotten the password. I click the 'Forgot' button and get a mail with a link, but that mail already says     Sorry, that key does not appear to be valid.   And sends me round the click to reset, receive a mail, click the link again etc. As if the timeout for password resets has been set to 0.5s or something.  This is a shared hosted blog, so I have full access to cPanel and all of the WP files via FTP, if that helps anything. Problem starting Gnome terminal An excited electron loses energy in the form of radiation. The radiation constitutes photons which move at a speed $c$. But is the process of conversion of the energy of the electron into the kinetic energy of the photon instantaneous? Is there a a simple way to visualize this process rather than math? How can I stop iPhone prompting me to setup Apple Pay in Settings? Who is this running kid wearing a shirt with Uchiha symbol? There's a kid wearing a shirt with Uchiha symbol, but I have no idea who s/he is.    Who is this kid? Meta Theory when studying Set Theory What exactly serves as the meta theory in the study of set theory (for example in Kunen's text). It seems to involve a certain amount of number theory and a certain concept of what it means to be finite.   Also: We usually use $\bigcup_{\alpha\in{ON}}R_{\alpha}$ as our set theortic universe $V$ (in the sense we believe that all sets we are interested in live there) where $R_{0}=\emptyset$, $R_{\alpha+1}= \mathcal{P}(R_{\alpha})$ and $R_{\alpha}=\bigcup_{\beta&lt;\alpha}R_{\beta}$ when $\alpha$ is a limit ordinal. The study of independence results then seem to be based on relatavizing notions with this $V$ as our set theoretic universe. It seems that it is implicitly assumed that the $\omega$ in $V$ has no non-standard elemements (in the sense that there is no $n&gt;a$ in $\omega$ where $a$ is obtained be repeatedly applying the successor operation to $0$.) Is this true and can it be formalized or is it something that is taken for granted?   e.g. If the above were true then $ZFC\vdash{\neg{Con(\ulcorner{ZFC}\urcorner)}}$ would imply that $ZFC$ is inconsistent since we may look at the $ZFC$ axioms $\phi_{1},...,\phi_{n}\vdash\neg{Con\ulcorner{ZFC}\urcorner}$, then look at a ctm these axioms (which exists by reflection and downward Lowenheim Skolem) and recover the proof of the contradiction.