I need to find the transfer function of two RC circuits connected together.  This is the schematic:   Here is how I tried solving it, using maple, the two equations for nodal tensions are found at (1) as the solve({eq1, eq2}) parameters:   eq1 is the sum of currents at node A, eq2 is the sum of currents at node B.  Well my problem is that I get a result at (3) that is not the answer from the solutions booklet (which is at (4)).  I am quite rusty into solving circuits, could anyone take the time to explain me how to get to the correct answer? I must find the transfer function \$ G(s) = Y_{2}(s)/U_{1}(s) \$. \$ u_{1}(t) \$ is a voltage excitation, and \$ u_{2}(t) \$ is the same voltage as \$ V_{C_{1}}(t) \$.  Thanks!  P.S. If anyone could tell me what I did wrong in nodal analysis it would be great.
I'm working on a 2nd order passive low pass filter, consisting of two passive low pass filters chained together.         &ndash; Schematic created using   Let \$ H(s) = H_1(s)H_2(s) \$ where \$ H_1(s) \$ and \$ H_2(s) \$ are the transfer functions for each separate filter stage.  Then \$ |H(s)| = |H_1(s)||H_2(s)| \$  Knowing the magnitude of a passive low pass filter,   $$|H(s)| = \dfrac1{\sqrt{ (\omega R_1C_1)^2 + 1} } \times \dfrac1{\sqrt{ (\omega R_2C_2)^2 + 1} } = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$  Then trying to find the cutoff frequency:  $$ \left(\dfrac1{\sqrt{2}}\right)^2  = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$ $$ 2  = \sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)} $$ $$ 4  = ((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1) $$ $$ 4  = (\omega R_1C_1)^2(\omega R_2C_2)^2 + (\omega R_1C_1)^2 + (\omega R_2C_2)^2 + 1 $$  And I'm stuck. Research on the web tells me \$ \omega_c = \dfrac1{\sqrt{R_1C_1R_2C_2}} \$, but I can't find why? Can anybody show my the derivation to find this?
Linear Discriminant Analysis for $p=1$ Fantasy novel with amnesiac witch, impotent soldier and cursed animals The story is set in a vaguely medieval world in which a young woman has been cursed by the evil witch who has enslaved her and caused her to lose her memory. With the aid of her talking animal friends who have also been cursed (toad has a huge stone under his tongue meaning he can't feed and a crow with a gem embedded in its wing and there's a little grey kitten too who never grows up) she sets out on a journey to break the curse for herself and her friends. Along the way she meets an injured unicorn and a horny-but-moral soldier rendered impotent by a curse after refusing the disguised witch's sexual advances.  After adventures there's a 'happy ever after' for all of the heroes and the witch meets a sticky end.    In the UK the book had a deep red cover with an illustration. I read it in either the 1990's or the early 2000's. Cracking book, definitely for adults given some of the situations, language and humour. In C++, what is the purpose of the scope resolution operator when used without a scope? For instance:  ::foo(); On the uniqueness of the Riemann-Christoffel tensor According to Section 6.2, Gravitation and Cosmology by Weinberg, the Riemann-Christoffel tensor is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives. The reasoning behind the same goes like this:   In a class of frames where $\Gamma^{\lambda}_{\mu\nu} =0$, the transformation rule for $\dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\kappa}}$ involves an inhomogeneous term which is symmetric in $\mu$, $\nu$, and $\kappa$. Thus, if one is to construct a tensor which is a linear combination of the first order derivatives of the Christoffel symbol then the only way to do so is by eliminating the inhomogeneous part of the transformation and this could be done only by making the combination explicitly antisymmetric in $\mu$ and $\kappa$. Since in these frames, $R^{\lambda}_{\mu\nu\rho}$ $ = \dfrac{\partial \Gamma^{\lambda}_{\mu\nu}}{\partial x^{\rho}} -  \dfrac{\partial \Gamma^{\lambda}_{\mu\rho}}{\partial x^{\kappa}} $, $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated using the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.  I think the presented argument can only suffice to prove that $R^{\lambda}_{\mu\nu\rho}$ is the only tensor that can be formulated from the first derivatives of the Christoffel symbol and is linear in them. I can't figure out why this suffices to assert that $R^{\lambda}_{\mu\nu\rho}$  is the only tensor that can be constructed out of the second (or lower) order derivatives of the metric tensor and is linear in the second order derivatives.  Edit: As the metric is covariantly constant, $\dfrac{\partial g_{{\mu}{\nu}}}{\partial x^{\rho}} = \Gamma^{\kappa}_{{\mu}{\rho}}g_{{\kappa}{\nu}}+\Gamma^{\kappa}_{{\nu}{\rho}}g_{{\kappa}{\mu}}$.  Therefore,   $\dfrac{\partial^2 g_{{\mu}{\nu}}}{\partial x^{\xi}\partial x^{\rho}} = \bigg(\dfrac{\partial \Gamma^{\sigma}_{{\mu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\mu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\nu}}+\bigg(\dfrac{\partial \Gamma^{\sigma}_{{\nu}{\rho}}}{\partial x^{\xi}} + \Gamma^{\sigma_1}_{{\nu}{\rho}}\Gamma^{\sigma}_{{\sigma_1}{\xi}}\bigg)g_{{\sigma}{\mu}} + \bigg(\Gamma^{\sigma}_{{\mu}{\rho}}\Gamma^{\sigma_1}_{{\nu}{\xi}}+\Gamma^{\sigma}_{{\nu}{\rho}}\Gamma^{\sigma_1}_{{\mu}{\xi}}\bigg)g_{{\sigma}{\sigma_1}} $  Now, if I want to show that the Riemann-Christoffel tensor is the only (non-trivial and independent) tensor that can be formulated out of the linear combinations of the second order derivatives of the metric tensor then I can equivalently show that the above expansion can be expressed linearly in terms of the Riemann-Christoffel tensor. But I am stuck over how to do it. Also, I am a little bit unclear as to what a linear combination means in this context. The coefficient of the second order term can be only a scalar constant or can it be the metric or can it be the metric that is getting summed over any of the indices as well? Does kinetic friction increase as speed increases? Possible Duplicate:          Layman alert...I last did physics in high school, and am just trying to understand something my son is working on.   To keep a box moving across the floor at a constant velocity I need to apply a force equal and opposite to that of the  force working on it.  If I start applying a force greater than the kinetic friction force, the box will speed up. Will it keep speeding up indefinitely, or will the kinetic friction force I need to overcome increase as the box gets faster? Martingale that is not a Markov process Using $g = \frac{Gm}{r^2}$, the force on a point mass located at 1 AU from the Sun ($m = 2 \cdot 10^{30} \text{ kg}$) is about ~0.006 N/kg.  Does that mean that, e.g., a 70 kg person is ~42g lighter during the day, and ~42g heavier at night? That seems like it could make a big difference for things like measuring gold bars or other weight-sensitive items. (Gold arbitrage: buy your gold during the day and sell it during the night! Risk-free profit!)  This makes me suspect that I'm overlooking something obvious, because a ~0.05% weight difference seems like something everyone would have noticed long ago. So what am I missing?    Edit: A few answers below indicate that there shouldn't be any weight difference, because the Earth orbits the Sun in free-fall. But if that's the reason, does that mean that a 1:1 tidally-locked Earth-Sun system wouldn't experience differential gravity from the Sun on opposite sides? That doesn't seem right. Given $|z|&lt;1$ with $z \in \mathbb{C}$ prove $\lim_{n \to \infty} z^n=0$.  My attempt given $\epsilon &gt;0$ we wish to show that there exists $\delta &gt;0$ such that $n&gt; \delta$ means $|z^n-0| &lt; \epsilon$.  For $n \in \mathbb{N}$. We have $|z^n|=|z|^n$. We wish $|z|^n &lt; \epsilon$. This means,  $$n \ln |z| &lt; \ln \epsilon$$  But because $\ln |z|&lt;\ln 1=0$ dividing both sides by $\ln |z|$ gives:  $n&gt; \frac{\ln \epsilon}{\ln |z|}$  So that's what we choose for $\delta$.   My question is is how can we get away from the assumption that $n \in \mathbb{N}$. Or is this correct even with the assumption? Opening Raster stored in Esri Geodatabase using QGIS Is there a way to open a raster that is stored in ESRI V10.0 Geodatabase in QGIS V2.4?    I can open the vector layer, but there does not seems to be a way to open a raster. Fly and Two Trains Riddle Show that a holomorphic function satisfying $|f(z)|\leq c|z|^n$ must be $f(z)=a\cdot z^n$ for some a. Let $f:\mathbb{C}\to\mathbb{C}$ be a holomorphic function such there exists a $c\in\mathbb C$ with $|f(z)|\leq c\cdot |z|^n$ for all $z\in\mathbb C$. Show that there exists an $a\in\mathbb C$ with $f(z)=a\cdot z^n$.   Proof.  My general idea is to show that the taylor series only has one non-zero coefficient:  $\displaystyle |a_k|\leq\frac{k!}{2\pi}\oint_{|w|=r}\frac{|f(w)|}{|w|^{k+1}}dw\leq \frac{k!}{2\pi}\oint_{|w|=r}\frac{c\cdot|w|^n}{|w|^{k+1}}dw\leq \\\displaystyle\leq \frac{c\cdot k!}{2\pi}\oint_{|w|=r}|w|^{n-k-1}dw$  Now using the standard estimate since $\oint=2\pi r$:  $\displaystyle \leq c\cdot k!\cdot r^{n-k}\longrightarrow\begin{cases}0,\ k&gt;n\\c\cdot n!,\ k=n\\ \infty,\ k&lt;n\end{cases}$ as $r\to\infty$  In conlusion the taylor series for now looks like $f(z)=\sum_{k=0}^{n-1}a_k z^k+ a_n z^n$ with $|a_n|\leq n!\cdot c$. What do I do with the remaining summands?  Edit: I would say that for $k&lt;n$ we have $|z|^k&gt;|z|^n$ for $z\in D_1(0)$. So since the estimate has to hold for all $z\in \mathbb C$ we can't have that either. Proving prime $p$ divides $\binom{p}{k}$ for $k\in\{1,\ldots,p-1\}$ Prove if $p$ is a prime then $p \mid \binom pk$ for $k\in\{1,\ldots,p-1\}$  I don't really know where to begin with this one.   I can see that I have to use the fact that $p$ is prime somewhere - the same is not true for composite numbers, for example $4\nmid 6=\binom42$.  I have checked that this is true for the first few primes:    $3$ divides $\binom 31=\binom32=3$ $5$ divides both $\binom 51=\binom 54=5$ and $\binom 52=\binom 53=10$ $7$ divides $\binom71=\binom76=7$, $\binom72=\binom75=21$ and $\binom73=\binom74=35$. If $a_0\in R$ is a unit, then  $\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$ How to use Parameterized MessageFormat with non-Value attributes of JSF components References for multivariable calculus Due to my ignorance, I find that most of the references for mathematical analysis (real analysis or advanced calculus) I have read do not talk much about the "multivariate calculus". After dealing with the single variable calculus theoretically, it usually directly goes to the topic of measure theory.    After reading the wiki article "", I'd like to find the rigorous proof for this test. The first book comes to my mind is Courant's Introduction to Calculus and Analysis which includes the multivariate case in the second volume.   Motivated by this, I'd like to put the question here:     What are the usual references for the   theoretical treatment for   ? Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$ Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$.    I'm not sure how to start this proof, if it should be done algebraically or if I should try to use a proof by induction. $A^TA+I$ is always invertible? I have an older Nvidia video card and my PC crashes when installing the driver for the video card in Ubuntu 16.04. When I get to the login screen but no further: after entering my password, the system went into a loop and returned always back to the login screen [with funny stripes present].   How do I fix this problem? This may be a simple question, but how do I use files starting with "-" with command line programs?   For example,  I am trying to use pdfgrep on a file named -2013-01-01.pdf but it complains that there are no options defined for 2, 0, 1, 3 and so on.  Changing the filename to something that does not start with "-" solves it, but this is not an option since the files aren't created by me. I simply want to check for a specific change I know is coming. So in the image below there are two circles, how can I remove the internal vertices between them without messing up the outside shape. I'm working on a gun and this is just a simple way to illustrate the issue I'm dealing with. If I where to go in and delete the internal faces it would leave holes on the outside of the object, is there a fast way to get around this?