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UNDER CONSTRUCTION

We derive the natural response of the inductor-capacitor, $\\text{LC}$, circuit . \n\n[[ \f1 \uc0\u9731 \f0 image 4]]\n\nThis is where sine waves are born.\n\n\n\n”,\ “images”: {},\ “widgets”: {\ “image 4”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 185,\ 150\ ],\ “backgroundImage”: {\ “url”: “https://ka-perseus-images.s3.amazonaws.com/e11e853050cfcb3b21fb71159ee18f39f490f1f2.svg”,\ “width”: 185,\ “height”: 150\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “—\n\n### Background\n\nThis article walks you step-by-step through the solution to a $2$nd-order differential equation. I don’t assume you have previous experience with this type of equation. Sal has videos on 2nd-order equations, too. $1$st-order differential equations are solved step-by-step in the articles on $\\underline{\\text{RC}}$ and $\\underline{\\text{RL}}$ natural response. You can also check out Sal’s videos on 1st-order differential equations.\n\n—\n\n### What we’re building to\n\nThe natural response of an $\\text{LC}$ circuit is described by this homogeneous second-order differential equation: \n\n$\\text L \\,\\dfrac{d^2 \\,i}{d \\,t^2} + \\dfrac{1}{\\text C}i = 0$\n\nThe solution for the current is: \n\n$i(t)=\\sqrt{\\dfrac{\\text C}{\\text L}}\\,\\text V_0\\sin \\omega_\\circ t$\n\nWhere $\\omega_\\circ =\\sqrt{\\dfrac{1}{\\text{LC}}}$ is the natural frequency of the $\\text{LC}$ circuit and $\\text V_0$ is the starting voltage on the capacitor.\n\nIn electrical engineering, we use the letter $j$ as the $\\sqrt{-1}$. \n (The letter $i$ is already taken for current.)\n\n—\n”,\ “images”: {},\ “widgets”: {}\ },\ {\ “content”: “## Introduction\n\n### First-order systems\n\n[[ \f1 \uc0\u9731 \f0 image 6]]\nUp to now we’ve looked at first-order circuits, $\\underline{\\text{RC}}$ and $\\underline{\\text{RL}}$, that have one energy-storage element, $\\text{C}$ or $\\text{L}$. The natural response of first-order circuits has an exponential shape that \“slumps\” to its final value. The energy in its storage element is dissipated by the resistor.\n\n”,\ “images”: {},\ “widgets”: {\ “image 6”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 225,\ 225\ ],\ “backgroundImage”: {\ “url”: “web+graphie://ka-perseus-graphie.s3.amazonaws.com/1c981e9e08057bb40b2fb59e0092bccd046a16be”,\ “width”: 225,\ “height”: 225\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “Natural response of an $\\text{RC}$ circuit, described by a $1$st-order differential equation.”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “### Second-order systems\n\nNow we look at a circuit with two energy-storage elements and no resistor. Circuits with two storage elements are second-order systems, because they produce equations with second derivatives.\n\nThis article covers the $\\text{LC}$ circuit, one of the last two circuits we will solve with full differential equation treatment. The last circuit to get this treatment is the $\\text{RLC}$ (in the next article). The mathematics of differential equations keeps getting harder. Fortunately, after we are done with the $\\text{LC}$ and $\\text{RLC}$, we learn a really nice shortcut to make our lives simpler.\n\nWe stick with differential equations instead of going straight to the shortcut because I want to show you where sine waves come from in electronics. Sine waves emerge from the solution to second-order equations. Sine waves are important. They are the building block for all other types of signals. \n\nSecond-order systems are the first systems that rock back and forth in time, or oscillate. The classic example of a mechanical second-order system is a clock with a pendulum. In electronics, the classic second-order system is the $\\text{LC}$ circuit.\n”,\ “images”: {},\ “widgets”: {}\ },\ {\ “content”: “## Natural response\n\nWe want to find the natural response of the $\\text{LC}$ circuit. The natural response is what a circuit does when there is no external driving force. Natural response is always an important part of the total response of a circuit. \n”,\ “images”: {},\ “widgets”: {}\ },\ {\ “content”: “## Natural response of a $2$nd-order circuit\n\nTo get going on a precise answer for the natural response, let’s set up the circuit with some initial energy. The components are labeled with careful attention to the sign convention for passive components. The inductor has an initial current of $0\\,\\text A$ because the switch starts in the open position. We assume the capacitor has an initial voltage before the switch closes, $v_{\\text C} =-\\text V_0$. (\nNotice how $v_\\text C$ has its $+$ sign at the bottom.) We let the switch close at $t=0$.\n\n\n[[ \f1 \uc0\u9731 \f0 image 10]]\n\nAs with every circuit analysis, we begin by writing one of Kirchhoff’s Laws. In this case, we’ll go with Kirchhoff’s Voltage Law (KVL) around the loop, starting at the lower left and going around clockwise. \n\n$v_{\\text{L}}+v_{\\text{}C} = 0$\n\n$\\text L\\,\\dfrac{di}{dt} + \\dfrac{1}{\\text C}\\,\\displaystyle\\int i \\,dt = 0$\n\nThis KVL equation contains an integral, which is awkward to deal with. The way to get rid of an integral (also known as an anti-derivative) is to take its derivative. We take the derivative of every term in the equation. \n\n$\\dfrac{d}{dt} \\left ( \\,\\text L\\,\\dfrac{di}{dt} + \\dfrac{1}{\\text C}\\,\\displaystyle\\int i \\,dt\\, \\right ) = \\dfrac{d}{dt}\\,0$\n\nThis gives us the second derivative of the $\\text L$ term, gets rid of the integral in the $1/\\text C$ term, and still leaves us with $0$ on the right side.\n\n$\\text L \\,\\dfrac{d^2i}{dt^2} + \\dfrac{1}{\\text C}\\,i = 0$\n\nThe equation is tidier if the first term has no coefficient, so we divide through by $\\text L$. This second-order differential equation captures the essence of our circuit. \n\n$\\dfrac{d^2i}{dt^2} + \\dfrac{1}{\\text{LC}}\\,i = 0$\n\n”,\ “images”: {},\ “widgets”: {\ “image 10”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 300,\ 136\ ],\ “backgroundImage”: {\ “url”: “https://ka-perseus-images.s3.amazonaws.com/9e161675df34dae6db77f7357ca96a963e429de5.svg”,\ “width”: 300,\ “height”: 136\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “### Propose a solution\n\nWhen we solved the first-order $\\text{RC}$ and $\\text{RL}$ circuits, we guessed at an exponential solution for $i(t)$. Guessing works with second-order equations, too. Our second-order equation has similar requirements: we want the function and its derivatives to look like each other so they can all add up to $0$. The exponential function fits the description. We propose an exponential function with some adjustable parameters:\n\n$i(t) = Ke^{st}$\n\n$K$ is an amplitude factor that scales the current big or small. \n \n$s$ is up in the exponent sitting next to time $t$. Since exponents don’t have dimensions, $s$ has to have units of $1/t$, which is also known as frequency. Since we are solving a natural response, $s$ is called the natural frequency. \n\nNow we substitute our proposed function into the differential equation and check to see if it makes the equation true.\n\n$\\dfrac{d^2 \\,i}{d \\,t^2} + \\dfrac{1}{\\text{LC}}i = 0$\n\n$\\dfrac{d^2}{d \\,t^2}(Ke^{st})+ \\dfrac{1}{\\text{LC}}(Ke^{st}) = 0$\n\nLet’s work on the first term by taking two derivatives. The first derivative is:\n\n$\\dfrac{d}{dt}(Ke^{st}) = sKe^{st}$\n\nAnd now the second derivative:\n\n$\\dfrac{d^2}{d \\,t^2}(Ke^{st}) = \\dfrac{d}{dt}(sKe^{st}) = s^2Ke^{st}$\n\nWe plug our new second derivative back into the equation:\n\n$s^2Ke^{st} + \\dfrac{1}{\\text{LC}}Ke^{st} = 0$\n\nAnd do some factoring to pull $Ke^{st}$ to the side:\n\n$Ke^{st}(s^2 + \\dfrac{1}{\\text{LC}}) = 0$\n\nHow many ways can we make this equation true?\n\n$K=0$ is pretty boring. $0=0$, who cares?\n\n$e^{st}$ never becomes zero for a finite amount of time. “,\ “images”: {},\ “widgets”: {}\ },\ {\ “content”: “That leaves the interesting solution when the ($s+1/\\text{LC)}$ term equals $0$: \n\n$s^2+ \\dfrac{1}{\\text{LC}} = 0$\n\nThis equation is called the characteristic equation of our circuit. \nWe want to find the roots of the characteristic equation (the value(s) of $s$ that make left side equal zero).\n\n$s^2 = - \\dfrac{1}{\\text{LC}}$\n\nWhoa, look what’s about to happen. We’re about to take the square root of a negative number. We are about to generate an imaginary number. \n\n$s$ has two possible values:\n\n$s_1 = +j\\,\\sqrt{\\dfrac{1}{\\text{LC}}}$\n\n$s_2 = -j\\,\\sqrt{\\dfrac{1}{\\text{LC}}}$\n\n\nElectrical engineers use the letter $j$ to indicate the imaginary unit, $\\sqrt{-1}$, since we already use $i$ for current. \n\nAs a shorthand, we give a name to the square root term:\n\n$\\omega_\\circ = \\sqrt{\\dfrac{1}{\\text{LC}}}$\n\n\n\n[[ \f1 \uc0\u9731 \f0 explanation 5]]\n\nThe roots of the characteristic equation can be expressed in terms of $\\omega_\\text o$ as:\n\n$s_1 = +j\\,\\omega_\\circ$\n\n$s_2 = -j\\,\\omega_\\circ$\n\nHow about that! The $\\text{LC}$ circuit produces two complex natural frequencies, $s_1$ and $s_2$. And one of the natural frequencies is negative. So curious. This will turn out to be very interesting. \n\nEither $s_1$ or $s_2$ by itself is a root of the equation. For our proposed solution we allow the possibility of both natural frequencies, $s_1$ and $s_2$. So we write a general solution as a linear combination of two terms, with two adjustable $K$ constants. \n\n$i(t) = K_1e^{\\large{\\raise{0.2em}{ \\,+j\\,\\omega_\\circ t}}} + K_2e^{\\large{\\raise{0.2em}{ \\,-j\\,\\omega_\\circ t}}}$\n\nAt this point you might be thinking, \“Complex exponents? Negative frequency? Is this really happening?\” The answer is, yes. So please hang in there while we work with these expressions.\n\n”,\ “images”: {},\ “widgets”: {\ “explanation 5”: {\ “type”: “explanation”,\ “alignment”: “default”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “showPrompt”: “What is the curvy w symbol?”,\ “hidePrompt”: “hide”,\ “explanation”: “$\\omega$ is the lowercase Greek character omega. $\\omega$ is commonly used as the variable name for frequency measured in radians per second. The angle of a full circle is $360^{\\circ}$ or $2\\pi$ radians. An angle of $1$ radian is $360^{\\circ}/2\\pi \\approx 57.6^{\\circ}$. Whenever you see $\\omega$, think \“radians per second.\” \n\nWhen we talk about the natural frequency or resonant frequency, it is often called $\\omega_{\\circ}$, pronounced, \“omega naught.\””,\ “widgets”: {}\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “## Euler’s identities\n\nTo work with these complex exponents, we resort to an important identity. \n\nUsing Maclaurin series expansions for $e^{jx}$, $\\sin jx$, and $\\cos jx$, it is possible to derive these Euler identities:\n\n$e^{+jx} = \\cos x + j\\sin x$\n\n$e^{-jx} = \\cos x - j\\sin x$\n\nIn the linked video, anytime Sal says $i$, we will say $j$. \n\nThese identities let us turn the strange $e^{imaginary}$ thing to a normal complex number. The real and imaginary parts come from a cosine or sine function, so both the real and imaginary components are somewhere in the range $-1$ to $+1$.\n\n## Use Euler’s identities\n\nWe can use Euler’s identities in our proposed solution. \n\n$i(t) = K_1e^{\\large{\\raise{0.2em}{ \\,+j\\,\\omega_\\circ t}}} + K_2e^{\\large{\\raise{0.2em}{ \\,-j\\,\\omega_\\circ t}}}$\n\n$i(t) = K_1(\\cos \\omega_\\circ t + j\\sin \\omega_\\circ t) + K_2(\\cos \\omega_\\circ t - j\\sin \\omega_\\circ t)$\n\nMultiply through the constants:\n\n$i(t) = K_1\\cos \\omega_\\circ t + jK_1\\sin \\omega_\\circ t + K_2\\cos \\omega_\\circ t - jK_2\\sin \\omega_\\circ t\\,,$\n\nand gather the cosine terms and sine terms together:\n\n\n$i(t) = (K_1+K_2)\\cos \\omega_\\circ t + j(K_1-K_2)\\sin \\omega_\\circ t$\n\nWe don’t know $K_1$ or $K_2$, or their sum or difference. It seems perfectly ok to replace the unknown $K$’s with different unknown $A$’s, just to make things appear a bit simpler. \n\nIf we let $A_1 = (K_1 + K_2)$, and $A_2 = j(K_1-K_2$), then $i(t)$ is:\n\n$i(t) = A_1\\cos \\omega_\\circ t + A_2\\sin \\omega_\\circ t$\n\nWe used Euler’s identities to rearrange the complex exponentials into a sum of trig functions. This equation is the very first time in electronics we see a sine or cosine as a function of time (a sinusoidal waveform). \n\n(Notice how we defined $A_2$ to include $j(K_1-K_2)$, so $j$ no longer directly appears in the proposed solution.)\n”,\ “images”: {},\ “widgets”: {}\ },\ {\ “content”: “## Test the proposed solution\n\nNext, we check our proposed solution by plugging it into the second-order differential equation. If we can come up with values for the constants that make the differential equation true, the proposed solution is a winner. \n\n### Figure out the initial conditions\n\nThe initial conditions needed for a second-order circuit are a little more involved than for a first-order circuit.\nWhen we did this for first-order circuits, $\\text{RC}$ or $\\text{RL}$, we had to know a single value, a starting current or voltage. With a second-order $\\text{LC}$ circuit, we need to know two things: the current and the derivative of the current when the switch closes. \n\n[[ \f1 \uc0\u9731 \f0 image 13]]\n\nWe write down everything we know about $t=0^-$ (the moment just before the switch closes):\n \n* The switch is open, so $i(0^-) = 0$\n* The starting capacitor voltage is specified: $v_\\text C(0^-)=-\\text V_0$. \n\nIf $t=0^+$ is the moment just after the switch closes, our goal is to find $i(0^+)$ and $di/dt(0^+)$.\n\nWe know some properties of inductors and capacitors that will take us from $t=0^-$ to $t=0^+$:\n\n* Inductor current cannot change instantly, so \n$i(0^+)=i(0^-)=0$\n* Capacitor voltage cannot change instantly, so \n$v(0^+)=v(0^-)=\\text V_0$\n\n(After the switch closes there is only one $v$, so we’ll just call it $v$ from now on.)\n\n[[ \f1 \uc0\u9731 \f0 image 16]]\n\n\n\nNow we have $i(0^+)$, but not $di/dt(0^+)$, yet. Where can we get this derivative? How about from the inductor’s $i$-$v$ equation? \n\n$v = \\text L\\,\\dfrac{di}{dt}$\n\n$\\dfrac{di}{dt}(0^+) = \\dfrac{1}{\\text L}\\,v(0^+)$\n\n$\\dfrac{di}{dt}(0^+) = \\dfrac{1}{\\text L}\\,\\text V_0$\n\nNow we have our second initial condition. This says the moment just after the switch closes, the current in the inductor starts changing with a slope of $\\text V_0/\\text L$ amperes every second. \n\n### Summary of the initial conditions\n\n$i(0^+)=0$\n\n$\\dfrac{di}{dt}(0^+) = \\dfrac{1}{\\text L}\\,\\text V_0$\n\n”,\ “images”: {},\ “widgets”: {\ “image 13”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 300,\ 136\ ],\ “backgroundImage”: {\ “url”: “https://ka-perseus-images.s3.amazonaws.com/d4bcc8430e1306080aea87b05cf0f99578a5c287.svg”,\ “width”: 300,\ “height”: 136\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “Circuit conditions just before the switch closes, at $t=0^-$.”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ },\ “image 16”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 300,\ 137\ ],\ “backgroundImage”: {\ “url”: “https://ka-perseus-images.s3.amazonaws.com/07a1c12c2029434c9cd73c4cc8426f2e9f9cfb89.svg”,\ “width”: 300,\ “height”: 137\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “Circuit conditions just after the switch closes, at $t=0^+$. There is now just one voltage, $v = v_\\text L = -v_\\text C$. The initial voltage on the capacitor is $v=+\\text V_0$.”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “### Use the initial conditions to find $A_1$ and $A_2$\n\nWe use our initial conditions one at a time to solve for the constants. The first initial condition is $i=0$ at $t=0^+$. Let’s plug it into the proposed solution and see where it takes us:\n\n$i(t) = A_1\\cos \\omega_\\circ t + A_2\\sin \\omega_\\circ t$\n\n$0 = A_1\\cos (\\omega_\\circ \\cdot 0) + A_2\\sin (\\omega_\\circ \\cdot 0)$\n\n$0 = A_1\\cos 0 + A_2\\sin 0$ \n\n$\\begin{array}{r}\n1 \\\\\n0=A_1\\cancel{\\cos 0}\n\\end{array}\n\\begin{array}{r}\n0 \\\\\n+ \\,A_2 \\cancel{\\sin 0}\n\\end{array}$\n\n$0 = A_1$\n\n$A_1$ is $0$, so the proposed cosine term drops out of the solution. Our proposed solution now looks like:\n\n$i(t) = A_2\\sin \\omega_\\circ t$\n\n\nNow we go after $A_2$ using the second initial condition. The derivative of $i$ at $t=0^+$ is: \n\n$\\dfrac{di}{dt}(0^+) = \\dfrac{1}{\\text L}\\,\\text V_0$\n\nTake the derivative of the proposed $i(t)$:\n\n$\\dfrac{di}{dt} = \\dfrac{d}{dt}(A_2 \\sin \\omega_\\circ t)$\n\n$\\dfrac{di}{dt} = \\omega_\\circ A_2\\cos \\omega_\\circ t$\n\nEvaluating this expression at $t=0$:\n\n$\\dfrac{1}{\\text L}\\,\\text V_0 = \\omega_\\circ A_2\\cos(\\omega_\\circ \\cdot 0)$\n\n$\\dfrac{1}{\\text L}\\,\\text V_0 = \\omega_\\circ A_2\\cdot 1$\n\n\n$A_2 = \\dfrac{1}{\\omega_\\circ \\text L}\\,\\text V_0$\n\nWe can expand $\\omega_\\circ$ into $\\text L$ and $\\text C$ to get:\n[[ \f1 \uc0\u9731 \f0 explanation 1]]\n\n$A_2 = \\sqrt{\\dfrac{\\text C}{\\text{L}}}\\,\\text V_0$ \n\nAnd finally, after a lot of hard work, the solution for the current is:\n\n$\\large i(t) = \\sqrt{\\dfrac{\\text C}{\\text L}}\\,\\text V_0 \\sin \\omega_\\circ t$”,\ “images”: {},\ “widgets”: {\ “explanation 1”: {\ “type”: “explanation”,\ “alignment”: “default”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “showPrompt”: “show me the algebra”,\ “hidePrompt”: “hide algebra”,\ “explanation”: “$\\omega_\\circ = \\sqrt{\\dfrac{1}{\\text{LC}}}$\n\n$A_2 = \\dfrac{1}{\\sqrt{\\dfrac{1}{\\text{LC}}}\\,\\text L}\\,\\text V_0$ \n\n$A_2 = \\dfrac{1}{\\sqrt{\\dfrac{1}{\\text{LC}}\\text L^2}}\\,\\text V_0$ \n\n$A_2 = \\dfrac{1}{\\sqrt{\\dfrac{1}{\\cancel{\\text L}\\text{C}}\\text L^{\\cancel 2}}}\\,\\text V_0$ \n\n$A_2 = \\dfrac{1}{\\sqrt{\\dfrac{\\text L}{\\text{C}}}}\\,\\text V_0$ \n”,\ “widgets”: {}\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “## Real component values\n\nTo demonstrate what the solution looks like, we assign component values $\\text L=1$ henry and $\\text C = 1/4$ farad, and a starting voltage on the capacitor of $10\\,\\text V$.\n\n\n[[ \f1 \uc0\u9731 \f0 image 8]]\n\nThe natural frequency, $\\omega_\\circ$ is:\n\n$\\omega_\\circ = \\sqrt{\\dfrac{1}{\\text{LC}}} = \\sqrt{\\dfrac{1}{1\\cdot 1/4}} = 2\\,\\text{radians/second}$\n\nThe current as a function of time is:\n\n$i(t) = \\sqrt{\\dfrac{\\text C}{\\text L}}\\,\\text V_0 \\sin \\omega_\\circ t = \\sqrt{\\dfrac{\\text 1/4}{1}}\\,10 \\sin \\omega_\\circ t$\n\n$i(t) = 5 \\sin 2 t$\n\nThe current starts up the moment the switch closes:\n\n[[ \f1 \uc0\u9731 \f0 image 3]]\n\nThe current takes off in a sine wave pattern that continues forever. (There is no resistor in this ideal circuit, so the energy never dissipates. In a real-world circuit there would be small resistances that eventually dissipate the energy.)\n\nThe natural frequency of the sine wave is $\\omega_{\\circ} = 2 \\,\\text{radians}/\\text{sec}$. We can convert from radians per second to cycles per second, (also known as hertz, or $\\text{Hz}$) knowing that $1$ full cycle of a sine function corresponds to $2\\pi$ radians. We usually use the symbol $f$ for cycles per second The conversion is:\n\n$\\omega = 2\\pi f$\n\nThe natural frequency of the circuit in cycles per second, hertz, $\\text{Hz}$, is:\n\n$f = \\dfrac{2\\,\\text{radians/sec}}{2\\pi} = \\dfrac{1}{\\pi}\\,\\text{Hz,}$ \n\nor equivalently, the current completes a full cycle every $\\pi$ seconds.\n\n### A quick look back at the initial conditions\n\nWe can look in close to the origin to see how the solution accounted for the initial conditions. The sine wave starts at the origin, $i=0$. And notice how the slope of the blue sine wave near the origin matches the slope of the straight black line, $i=10\\,\\text A/\\text{sec}$. \n\n[[ \f1 \uc0\u9731 \f0 image 4]]\n”,\ “images”: {},\ “widgets”: {\ “image 8”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 300,\ 136\ ],\ “backgroundImage”: {\ “url”: “https://ka-perseus-images.s3.amazonaws.com/0e3e866cab333778923b3bb8e5f4f9e99e66b7da.svg”,\ “width”: 300,\ “height”: 136\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ },\ “image 3”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 425,\ 425\ ],\ “backgroundImage”: {\ “url”: “web+graphie://ka-perseus-graphie.s3.amazonaws.com/4fc5770243e2cc9c16a48b64adafd615f463c73f”,\ “width”: 425,\ “height”: 425\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ },\ “image 4”: {\ “type”: “image”,\ “alignment”: “block”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 425,\ 425\ ],\ “backgroundImage”: {\ “url”: “web+graphie://ka-perseus-graphie.s3.amazonaws.com/e7957ce9f2960f6311571e38b351794abf3ea67f”,\ “width”: 425,\ “height”: 425\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “”\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “## Voltage, $v(t)$\n\nAt this point we have solved the current. If you want to press on a little further, have a go at solving for the voltage, $v(t)$. \n\nFind an expression for $v(t)$ after the switch closes.\n\nProbably the quickest route is to use the inductor $i$-$v$ equation to solve for $v$ in terms of $di/dt$. \n\n[[ \f1 \uc0\u9731 \f0 graded-group 1]]\n\n[[ \f1 \uc0\u9731 \f0 explanation 1]]”,\ “images”: {},\ “widgets”: {\ “graded-group 1”: {\ “type”: “graded-group”,\ “alignment”: “default”,\ “static”: false,\ “graded”: true,\ “options”: {\ “title”: “”,\ “hasHint”: false,\ “content”: “$v(t) = \\quad$[[ \f1 \uc0\u9731 \f0 expression 1]]”,\ “images”: {},\ “widgets”: {\ “expression 1”: {\ “type”: “expression”,\ “alignment”: “default”,\ “static”: false,\ “graded”: true,\ “options”: {\ “answerForms”: [\ {\ “value”: “10\\cos2t”,\ “form”: false,\ “simplify”: false,\ “considered”: “correct”,\ “key”: 0\ }\ ],\ “buttonSets”: [\ “basic”\ ],\ “functions”: [\ “f”,\ “g”,\ “h”\ ],\ “times”: false\ },\ “version”: {\ “major”: 1,\ “minor”: 0\ }\ }\ },\ “hint”: false\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ },\ “explanation 1”: {\ “type”: “explanation”,\ “alignment”: “default”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “showPrompt”: “Show Answer”,\ “hidePrompt”: “Hide Answer”,\ “explanation”: “The current is: $\\quad i(t) = 5\\sin 2t$\n\nThe inductor $i$-$v$ equation gives $v$ in terms of $i$.\n\n$v = \\text L \\,\\dfrac{di}{dt}$\n\n$v = 1\\,\\text H \\cdot\\dfrac{d}{dt} (5\\sin 2t)$\n\n$v = 5 \\cdot 2\\,\\cos 2t$\n\n$v(t) = 10\\,\\cos 2t$\n\n\n[[ \f1 \uc0\u9731 \f0 image 1]]”,\ “widgets”: {\ “image 1”: {\ “options”: {\ “static”: false,\ “title”: “”,\ “range”: [\ [\ 0,\ 10\ ],\ [\ 0,\ 10\ ]\ ],\ “box”: [\ 440,\ 440\ ],\ “backgroundImage”: {\ “url”: “web+graphie://ka-perseus-graphie.s3.amazonaws.com/a04a055105b8b557745affdfa346bc247ed70c46”,\ “width”: 440,\ “height”: 440\ },\ “labels”: [],\ “alt”: “”,\ “caption”: “”\ },\ “type”: “image”,\ “version”: {\ “major”: 0,\ “minor”: 0\ },\ “graded”: true,\ “alignment”: “block”,\ “static”: false\ }\ }\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ },\ {\ “content”: “## Summary\n\nWe derived the natural response of an $\\text{LC}$ circuit by first creating this homogeneous second-order differential equation: \n\n$\\dfrac{d^2 \\,i}{d \\,t^2} + \\dfrac{1}{\\text{LC}}i = 0$\n\nThen we assumed a solution of the form $Ke^{st}$, which gave us the characteristic equation for the circuit:\n\n$s^2 + \\dfrac{1}{\\text{LC}}=0$\n\nWhen computing the roots of the characteristic equation we ran head on into a very strange expression: $e^{j\\omega_\\circ t}$, an exponential with complex exponent. We reached deep into our bag of tricks and pulled out:\n\nEuler’s identities\n\n$e^{+jx} = \\cos x + j\\sin x$ \n$e^{-jx} = \\cos x - j\\sin x$\n\nThese identities let us express the complex exponential as a combination of sine and cosine functions. (In electrical engineering, we use the letter $j$ as the name for $\\sqrt{-1}.$)\n\nThen we looked carefully at the circuit to find the initial conditions. For a second-order system, we found an initial $i$ and an initial $di/dt$. \n\nWe found a function for $i(t)$ that satisfied the differential equation:\n\n$i(t) = \\sqrt{\\dfrac{\\text C}{\\text L}}\\,\\text V_0 \\sin \\omega_\\circ t$\n\n$\\omega_\\circ \\equiv \\sqrt{\\dfrac{1}{\\text{LC}}}$ is the natural frequency of the $\\text{LC}$ circuit.\n\n $\\text V_0$ is the starting voltage on the capacitor. \n\n(This solution is applies when the starting current in the inductor is $0$.)\n\n\n”,\ “images”: {},\ “widgets”: {}\ },\ {\ “content”: “[[ \f1 \uc0\u9731 \f0 explanation 1]]”,\ “images”: {},\ “widgets”: {\ “explanation 1”: {\ “type”: “explanation”,\ “alignment”: “default”,\ “static”: false,\ “graded”: true,\ “options”: {\ “static”: false,\ “showPrompt”: “Copyright notice”,\ “hidePrompt”: “hide”,\ “explanation”: “This article is licensed under CC BY-NC-SA 4.0.”,\ “widgets”: {}\ },\ “version”: {\ “major”: 0,\ “minor”: 0\ }\ }\ }\ }\ ]}


text:
# Uncountable chains

$P(\mathbb N)$ = power set of $\mathbb N$.

$A \subset P(\mathbb N)$ is a chain if $a,b \in A \implies$ either $a \subseteq b$ or $b \subseteq a$

That is we have something like this:

$$\ldots a \subseteq b \subseteq c \subseteq\ldots$$ where $a,b,c \in A$ are distinct.

We can show easy enough that there is an uncountable chain - this is done by noting $\mathbb N\sim\mathbb Q$ then using Dedekind cuts in $\mathbb Q$ to define $\mathbb R$ we see that a family of (nearly arbitrary) cuts satisfy the condition.

For instance the family $L_r=\{q \in \mathbb Q : q < r \}$ for $r > 0$ gives us the sets we need and obviously we can pick others.

I tried doing this for $\mathbb R$ and don't seem to be getting anywhere. To be more specific, does there exist a chain in $P(\mathbb R)$ with cardinality $2^ \mathbb R$ ? Given a set of cardinality $X$ (necessarily non-finite), is there a chain in $P(X)$ of the same cardinalty of $P(X)$?

-
Adam, I see that you used $\LaTeX$ in parts of your question. It is possible, and even better, to fully $\TeX$-ify your post as it makes it easier (much much easier) to read. –  Asaf Karagila Dec 12 '11 at 22:41
Thanks Brian M. Scott for point out the error. –  William Dec 12 '11 at 22:47
In the meantime, this MO question might be useful. –  Asaf Karagila Dec 12 '11 at 23:01
I deleted my comments on formatting - just to clear this section up + that link probably has some mileage great find. –  Adam Dec 12 '11 at 23:05
My hunch is that this depends greatly on the behavior of the continuum function (the cardinality of power sets). –  Asaf Karagila Dec 12 '11 at 23:05

Consider the tree $T=2^{\lt\omega_1}$, consisting of all countable ordinal length binary sequences. The number of such sequences is precisely the continuum, since there are $\omega_1$ many levels, each of size continuumn. The lexical order on this tree (or on any two ordinal-length binary sequences), compares them by the order of their least differing bit, placing the shorter one below when one is an initial segment of the other.

Notice that any path $p\in 2^{\omega_1}$ through the tree determines a cut in the lexical order, namely, the collection $X_p$ of $s\in T$ which preceed $p$ in the lexical order. Furthermore, if $p$ preceeds $q$ lexically, then $X_p\subset X_q$, and so we have a chain in $P(T)$ of size $2^{\omega_1}$.

Thus, without any additional assumption, this establishes chains in $P(\mathbb{R})$ of size at least $2^{\omega_1}$.

In particular, if the continuum hypothesis holds, then $|\mathbb{R}|=\omega_1$, and so this shows that there is a chain in $P(\mathbb{R})$ of size $2^{\mathbb{R}}$. So an affirmative answer to your question is a consequence of CH and thus consistent with the axioms of ZFC.

More generally, we obtain a positive answer under the much weaker assumption merely that $2^{\lt\frak{c}}=\frak{c}$, an assumption that follows from but is much weaker than CH. For example, this assumption is compatible with $2^\omega=\aleph_{15}$, if also $2^{\aleph_{14}}=\aleph_{15}$. In this case, the tree $T=2^{\lt\frak{c}}$ has size $\frak{c}$ and $2^{\frak{c}}$ many branches. And so the cuts determined by the lexical predecessors of paths through this tree gives rise to a chain of size $2^{\frak{c}}$ in $P(\frak{c})$, as desired.

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Similarly, if $2^{\lt\kappa}=\kappa$, then there is a chain in $P(\kappa)$ of size $2^\kappa$. –  JDH Dec 13 '11 at 0:36

Under the Generalized Continuum Hypothesis, this true by a slight generalization of the Dedekind cut argument:

Assume GCH, and let $\alpha$ be an initial ordinal. (Since GCH implies AC, this is enough to establish the result for an arbitrary infinite set). I will prove that $\mathcal{P} (\alpha)$ contains a chain of cardinality $2^\alpha$.

Let $A=\{X\mid X\subseteq \alpha\}$ with the lexicographic ordering -- that is, $X < Y$ iff $\beta\in Y$ where $\beta$ is the least element on which the two sets differ. (Intuition for $\alpha=\omega$: A is the set of binary fractions between 0 and 1).

Let $B=\bigcup_{\beta<\alpha}\{X \mid X\subseteq \beta\}\subseteq A$. (Intuition: dyadic rationals are countable yet dense).

Because $\alpha \preccurlyeq B = \bigcup_{\beta<\alpha} 2^\beta \preccurlyeq \alpha\times\alpha$, we have $|B|=|\alpha|$. (GCH used for $2^\beta \preccurlyeq \alpha$).

Define an order-embedding $(A,\leq) \to (\mathcal{P}(B),\subseteq)$ by $X \mapsto \{Y\in B \mid Y<X\}$. (Intuition: Dedekind cuts). It is trivial that this mapping is monotonic, and fairly easy to see that it is injective. Thus, the image of $A$ is a chain in $\mathcal{P}(B)\cong\mathcal{P}(\alpha)$ of cardinality the same as $A$, namely $2^\alpha$.

(Adapted from an old Wikipedia refdesk answer of mine.)

-
But under GCH everything is so simple and boring... :( –  Asaf Karagila Dec 12 '11 at 23:34
Well, as JDH remarks, this at least shows that the existence of long chains cannot be disproved. –  Henning Makholm Dec 12 '11 at 23:52
Well... duh :-) vision from the future... $\small\text{But under AC everything is so simple and boring...}$ ;-) –  Asaf Karagila Dec 12 '11 at 23:55

format: plain


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# Partial Fractions Expansion of Cotangent

 It has been suggested that this page or section be merged into Mittag-Leffler Expansion for Cotangent Function. (Discuss)
 It has been suggested that this page or section be merged into Series Expansion for Pi Cotangent of Pi Lambda. (Discuss)

## Theorem

Let $x \in \R \setminus \Z$, that is such that $x$ is a real number that is not an integer.

Then:

$\ds \pi \cot \pi x = \dfrac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2}$

## Proof 1

We have that:

$\cot \pi x = \dfrac {\cos \pi x} {\sin \pi x}$

has a denominator which is $0$ at $x = 0, \pm 1, \pm 2, \ldots$.

Hence the limitation on the domain of $x \cot \pi x$ to exclude integer $x$.

Having established that, we should be able to express $\cot \pi x$ in the form:

$\cot \pi x = \dfrac a x + \displaystyle \sum_{n \mathop = 1}^\infty \left({\frac {b_n} {x - n} + \frac {c_n} {x + n} }\right)$

using a partial fractions expansion.

By evaluating the coefficients $b_n$ and $c_n$ in the usual manner, they are found to be:

$\forall n \in \N: b_n = c_n = \dfrac 1 \pi$

The result follows.

$\blacksquare$

## Proof 2

From the Euler Formula for Sine Function:

$\ds \sin x = x \prod_{n \mathop = 1}^\infty \paren {1 - \frac {x^2} {n^2 \pi^2} }$

Taking the logarithm of both sides:

 $\ds \map \ln {\sin x }$ $=$ $\ds \ln x + \sum_{n \mathop = 1}^\infty \map \ln {1 - \frac {x^2} {n^2 \pi^2} }$ $\ds$ $=$ $\ds \ln x + \sum_{n \mathop = 1}^\infty \map \ln {\frac {n^2 \pi^2 - x^2} {n^2 \pi^2} }$
 $\ds \cot x$ $=$ $\ds \dfrac 1 x + \sum_{n \mathop = 1}^\infty \dfrac {\dfrac {- 2 x} {n^2 \pi^2 } } {\paren {\dfrac {n^2 \pi^2 - x^2} {n^2 \pi^2} } }$ Derivative of Composite Function $\ds$ $=$ $\ds \dfrac 1 x + \sum_{n \mathop = 1}^\infty \frac {- 2 x} {n^2 \pi^2 - x^2}$ $\ds \leadsto \ \$ $\ds \cot x$ $=$ $\ds \frac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 \pi^2}$ Move the $-1$ to the denominator $\ds \leadsto \ \$ $\ds \pi \map \cot {\pi x }$ $=$ $\ds \pi \paren {\frac 1 {\pi x } + 2 \paren {\pi x } \sum_{n \mathop = 1}^\infty \frac 1 {\paren {\pi x }^2 - n^2 \pi^2 } }$ multiplying by $\pi$ and entering $\pi x$ $\ds \leadsto \ \$ $\ds \pi \map \cot {\pi x }$ $=$ $\ds \frac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 }$

Hence the result.

$\blacksquare$


text:
# Some Tralics obsolete features

## module (environment)

Before 2007, the Raweb preprocessor changed

\begin{module}[3]{}{appAI}{Static Analysis}
...
\end{module}

into

\begin{module}[3] {TEAM} {4} {appAI} {module7} {Static Analysis}
...
\end{module}

As you can see, there is one more argument TEAM, the symbolic section (in the example, the empty group {} was replaced by a default value namely `domaine') has been replaced by its numeric value (here 4), and a second additional argument has been added before the title.

These hacks were removed in 2007. For compatiblity reasons, the preprocessor, when used, generates a modulex environment, translated as dedcribed here.

The translator ignores the first two arguments (that were added by the preprocessor). It uses the next argument as a label name (the real name of the label is mod:appAI; but you should really use \moduleref {TEAM} {section} {appAI}, see \moduleref). The next argument is used for a html attribute.

back to home page © INRIA 2007 Last modified $Date: 2007/08/02 12:26:18$


text:
# Is there any easy way to solve this lock puzzle?

https://js13kgames.com/games/lock-puzzle/index.html

The link provides an online puzzle game of $$16$$ tiles in the form of a $$4 \times 4$$ matrix of which some entries are in the form of vertical tiles and other in the form of horizontal tiles arranged randomly by the system. Once you press a tile the tile along with all the tiles with same rows and columns get reversed i.e. vertical tiles becomes horizontal and vice versa. In order to solve it all we have to do is to make each of the $$16$$ tiles vertical.

I have solved it yesterday taking about half an hour. But I don't remember all the methods I used. My question is "Is there any algorithmic way of solving this puzzle which needs small amount of time"?

There is an easy way to solve it in the minimum number of moves.

For any square you can look at the seven tiles in the same row or same column, and count how many horizontal and how many vertical tiles there are. You want the number of vertical tiles to be odd, or equivalently the number of horizontal tiles to be even. If that is not the case, click the square to flip the seven tiles.

Apply the above rule for each of the 16 squares, and the puzzle will be solved.

The reason this works is:

Consider the 7 tiles that lie in the row or column of a particular square. Clicking the square would flip all 7 tiles, clicking a different square in the same row or column would flip 4 of those seven tiles, and clicking any other square would flip only 2 of those seven tiles. Flipping 2 or 4 tiles cannot change the parities - if the number of vertical tiles is odd it remains odd, if it's even it remains even. The only way to change the number of vertical tiles from even to odd is by pressing the chosen square.

• After applying this technique to the 16 squares, you need one final flip to make all of them vertical. But why does it work? – Shivansh Sharma Apr 1 '20 at 6:09
• It seems to me that you have your parity backwards in your answer. The goal is to get all the tiles vertical, not horizontal. – Ian MacDonald Apr 1 '20 at 21:30
• Ah, I read your statement wrong. I had interpreted it as "find one where it's odd, ...". Yes. – Ian MacDonald Apr 1 '20 at 22:40
• Is there an intuitive way to see that the only configuration where every square's parity is odd is when all bars on the board are vertical? – Jafe Apr 2 '20 at 12:38
• @jafe Good question. I don't know. The only way I know is to show that (1) a single tile can be flipped by clicking the 7 squares in the same row or column (2) Therefore all tile patterns are solvable (3) Hence one-to-one relationship between $2^{16}$ tile patterns and $2^{16}$ clicking patterns. The solution method in my answer can create $2^{16}$ solutions, since each button press is determined independently of whether other buttons have been pressed. So with this method no two tile patterns can give the same solution, and the all-vertical tile pattern is the only one with zero clicks. – Jaap Scherphuis Apr 2 '20 at 13:09

I don't know why this works, but I tried my old strategy from other similar puzzles and it has worked in several cases so far:

Make a note of all squares with horizontal lines initially. Click each of these squares once.

Then,

Do the same thing until it is solved: make a note of all squares with a horizontal line now. Click each of them once. Repeat and eventually you will solve it.

One final note - since I don't know why it works, I also can't prove that it always works, unfortunately. But in my many years of solving puzzles in this genre, that strategy always seems to eventually land on a solution.

Edit: See Jaap Scherphuis's comment and answer for an explanation of why this works.

• Oh! What a brilliant way. I have solved it using your ways in less than $5$ minutes. Nice observation +1. – math maniac. Apr 1 '20 at 5:47
• Now it's a matter of fact why should the process work? Actually the process you are trying to do is to make all the horizontal tiles vertical after each move and this process gradually converge. – math maniac. Apr 1 '20 at 5:48
• It is easy to see that after the first round, every row and every column will have an even number of horizontal tiles. From my answer you can then see that a second round will solve the puzzle. – Jaap Scherphuis Apr 1 '20 at 5:51
• Oh, I proved this in one of my high school research projects! To me it's quite an interesting result. – greenturtle3141 Apr 8 '20 at 7:28
• I’d argue this way is fastest – Cotton Headed Ninnymuggins Feb 4 at 6:34

Each cells operates one of sixteen NOT switches, each one of which changes seven cells. Cells that are vertical need to be changed an even number of times, cells that are horizontal need to be affected by an odd number of switches.

Given a particular state $$S$$, we need to find a solution to: $$S=\sum_\limits{k=1}^{16} \delta_{k}S_i \pmod 2$$ because each switch $$S_i$$ has the property that $$S_i^2=I$$, therefore $$\delta_k\in\{0,1\}$$. So we get $$16$$ equations in $$16$$ unknowns, of the (example) form: $$S_2+S_5+S_6+S_7+S_8+S_{10}+S_{14}=x_6$$ and this can be solved using Gaussian Elimination over $$GF_2$$.

Or, create a new blank board, and click on it all the positions of the 'on' switches from the original board. Take this new pattern back to the original board, and click every 'off' position from the new board onto the original board. All lights are on!

• Oh, the good old days of linear algebra! +1 for the mathematics. – ThatOneNerdyBoy Apr 1 '20 at 13:32


text:
December 4, 2013

# Hoxton Ventures - New Early Stage VC fund Targeting $1b European Exits London has a new early-stage venture fund, called Hoxton Capital. This is great news for us in London, but also for the European startup scene in general. Read the TechCrunch article for more details. One thing that caught my attention is their laser sharp focus on only the big exits, ignoring anything smaller than$ 1b: “We're entirely outlier driven, meaning we want to find companies that will become  $1b+ companies” Given that$1b+ companies aren't created so often, this sounds to me like a very harsh utility function, and it got me thinking what the implications of this statement is on the investment strategy Hoxton Ventures should follow.

Assume for a second that Europe will produce only a single  $1b+ exit during the lifetime of Hoxton's first$ 40m fund, from the pool of companies they may invest in. (Hoxton Ventures partner Hussein Kanji estimates there's 3-5, see comments section)

If so, we can reformulate Hoxton's investment strategy as follows: 'Find out which company will become the billion dollar hit, and maximise our ownership in that company'. Let us denote the percentage of ownership Hoxton will have in company $i$ as $a_i$. After observing which company becomes a billion dollar hit (call this company $j$), Hoxton's perceived utility of its investment portfolio $U(a,j)$ is simply the percentage of this company owned, that is

$$U(a,j) = a_j.$$

Note that this utility function indeed ignores the ownership levels in all other portfolio companies, only cares about $j$, the big winner.

How should Hoxton set the target ownership values $a$ given this utility function? By maximising their expected utility, which in this case is:

$$\hat{U}(a) = \sum_i p_i a_i$$

This expected utility should be maximised subject to the constraint that Hoxton have exactly $40m to spend, and therefore: $$40m = \sum_i v_i a_i$$, where$v_i$denotes the valuation of each startup$i$at the time when they invest (corrected for expected dilution for simplicity) The solution to this very simple linearly constrained linear optimisation problem has a trivial solution: invest all your money into a single startup, one for which the ratio$\frac{p_i}{v_i}$is maximal. In other words: if you have a startup idea with nonzero subjective probability that turn out positive, simply start your own company :) It is easy to extend the analysis to accommodate multiple$1b+ exits. In this simple model, as long as your goal is to maximise your total combined share in all these companies, the trivial strategy is still optimal. I'm not suggesting of course that Hoxton Ventures should follow this naive strategy, but I think it's good exercise to evaluate what their words mean in terms of strategies, expected behaviour and risk.

Good luck Hoxton Ventures, it will be interesting to watch which companies you invest in (and how much).


text:
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JH

# Determine whether the series converges or diverges.$\displaystyle \sum_{n = 1}^{\infty} \frac {1}{n \sqrt {n^2 -1}}$

## Convergent.

Sequences

Series

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University of Michigan - Ann Arbor

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### Video Transcript

Let's determine whether the Siri's converges laboratories here. I'LL use the limit comparison test. So let's call this given term here, not the whole Siri's. Just the term. Just a fraction Let's call that a M, and then I'll call being one over and radical and squared, but that could just simplified to be won over and square. So the choice when you're using woman comparison, says the choice for Bien could often be determined just by looking at Anne here. I just see one up top, so I just leave it in there and be and then in the denominator is and gets really, really big. This minus one doesn't really play a significant role compared to end. So I just dropped the minus one, and that's how I have my formula for being. Also, we know that the sum of the B end this will converge here. You can use the pee test by the pee test with P equals two. That's the two right there. That's the value of P. So hopefully we can use this test. But before we use limit comparison tests, we need to find this limit and let's just enough this limit by the value, see? So we need to look at a N over bian in the limit. So lim and goes to infinity. So am I. And then dividing by being well, that's dividing by one over and square. That's the same thing, is multiplying by and squared over one. And then in the denominator work, let me just go ahead and pull out and end from the radical. So I'm looking inside the radical out, pull out and square, and then I have one minus one over and square, and then I could pull the end square outside of the square roots, and it just becomes an end after I evaluate the square root. So we have this and that's already here. But then I have this whole term appear after I rewrite the radical. So this term it's coming from the radical, and that's radical was the term that was originally in the denominator. So now the reason for doing this is that we could cancel those and squares rooms and then evaluate the limit. So there's and square and then another and square we have won over radical one, minus one over and square. Now take that limit you have won over the square root of one minute zero, which is just one. And this is good, because any time the value of C satisfies this inequality in our case, C equals one. So this is true. This means that we can actually use the Lim comparison test. And so since we know the B end, the sum of one over and square converges so sense Deon converges and the limit as n goes to infinity of a n over bien equals one by the limit comparison test our Siri's, You also must converge. Okay. Yeah, and that's our final answer.

JH

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# Tag Info

48

Ice cubes have three distinct cooling effects: The cube, initially at sub-zero temperature, absorbs some heat to reach fusion point (0⁰C). The cube absorbs more heat to switch phase: it takes some energy to turn 1 kg of ice at 0⁰C into 1 kg of liquid water at 0⁰C. The water absorbs some heat to become warmer than 0⁰C. The three effects occur more or less ...

29

This is a very interesting question with a very interesting answer. The key lies in the reason for the stretchiness of the rubber band. Rubber is made of polymers (long chain molecules). When the elastic band is not stretched, these molecules are all tangled up with each other and have no particular direction to them, but when you stretch the elastic they ...

22

Yes, it's 11% hotter today than yesterday. Of the three temperature scales you discussed, only the Kelvin scale allows meaningful ratios to be calculated. Dividing two temperatures expressed in Celsius or Fahrenheit is simply a mistake. There are numerous physical examples where it makes sense to multiply or divide by a Kelvin temp, e.g., the ideal gas law ...

21

What a great question! And because anything that involves food is close to my heart I can answer with authority having done the experiments :-) There's a simple answer, a more complex answer and even an unexpected answer! The simple answer is that if you just want to boil off water you should leave the lid off. If you try the experiment of putting a known ...

18

The reason being closer to a heat source makes you warmer is the inverse square law. Think of it this way: If you have a $1~\mathrm{m}^2$ piece of material facing the Sun and located at Mercury's orbit, it will be quite hot. What does the shadow of this square look like at Earth's orbit (about $2.5$ times further away than Mercury)? Well, it will be $2.5$ ...

11

Several points that need addressing: The seasons are due to the tilt of the Earth, but not because of the atmosphere. When the sunlight is grazing the ground at a low angle the same amount of heating is spread over a larger area than when the sun is directly overhead, so the temperature drops. The atmosphere has a negligible effect on absorbing radiation ...

11

A fundamental principle of thermodynamics is that heat flows from warm places to cold ones, through either convection, conduction or radiation, and it will continue to do so until the temperature equalizes across the system. The stones are colder than the whiskey when you put them in the glass, so as the system heads towards equilibrium, the whiskey gets ...

10

Two reasons: the cheese has a higher specific heat capacity than the crust; the cheese has a higher thermal conductivity than the crust. When you cool a given weight of cheese or crust from the oven temperature to your mouth temperature, the amount of heat it gives up depends on its specific heat. So the cheese, with its high specific heat, gives up more ...

10

In order to build any thermal engine as envisioned by you, you need both a cold and a hot reservoir, such that heat can flow from the hot part to the cold part and the entropy doesn’t decrease while you’re making energy. The efficiency of such a machine has an upper limit of $(T_{\textrm{hot}} - T_{\textrm{cold}})/T_{\textrm{cold}}$ (as given by the perfect ...

9

TL;DR: Whiskey stones work by absorbing heat from the whiskey in an attempt to reach thermal equilibrium1. As Thomas mentioned, ice has three cooling effects: Ice itself takes 2.11 kilojoules of heat per g to have its temperature increased by 1 degree (Celsius). This number is known as "specific heat capacity" Ice takes 334 kilojoules of heat per kg at 0 ...

7

John's answer is a good one, I just wanted to add some equations and addition thought. Let me start here: Heating is really only significant when you get a shock wave i.e. above the speed of sound. The question asks specifically about a $200^{\circ} C$ increase in temperature in the atmosphere. This qualifies as "significant" heating, and the ...

6

The metal needs to be specified as a heavy metal, for Lithium, the answer is the opposite. The heuristic is that every atom gives you 3R per mole specific heat (3k per atom). The reason is the equipartition theorem, which is reasonably accurate for solids at room temperature. The amount of energy at temperature T is $.5kT$ for each quadratic term in the ...

6

I think your question is perfectly fine, I don't think this forum is only for advanced research level questions. Assuming you are not actually receiving small droplets of water, the air around the droplets is cooled by the water, which will then cool your skin. This is strongly accentuated by the fact the water create air currents. While the effect you ...

6

You're assuming that the propellor speed is dominated by the bearing friction, while in practice it's dominated by the drag on the propellor as it moves through the air. Typically the propellor will rapidly accelerate to a steady speed at which the force from the rubber band matches the aerodynamic drag, then the propellor speed will slow as the band unwinds ...

6

You also could harm an iron pan, yes, but the chances to do so are much smaller. If you add cold water to a hot pan, the temperature in the metal drops locally (where the water touches the metal) and since metal extend with heat, cooling them down contracts them. So, the region that is cooled down by the cold water is contracted, which results in internal ...

6

The water vapor in the bathroom will condense preferentially on the coldest spots around the bathroom. When you wipe the mirror with a towel the water vapor readily condenses on it again because you have not changed the mirror temperature. When you use the hair dryer, the temperature of the mirror is now higher and the water vapor will prefer to condense ...

6

The identity $E=mc^2$ is a universal law of physics. It says that the mass – that can be interpreted as the conserved mass; inertial mass determining the resistance to acceleration; or gravitational mass determining the strength of the gravitational field – is equivalent to the energy, a conserved quantity that was originally defined as the sum of the ...

6

Once the core has solidifed it can no longer generate a magnetic field. There may be some frozen in field, but I would guess that the strength of the magnetic field with decrease as more and more of the core freezes, so any residual field is likely to be small. The core of Mars is thought to be frozen and it's magnetic field is negligable. Anyhow, the main ...

6

We already store waste by combining it with molten glass. It doesn't take very much energy - obviously no more than making glass! The main reason for not doing it is political, it cost a lot to make that plutonium and it might come in handy to build bombs one day - so we don't want to waste it by reprocessing it. (We also have a small problem of persuading ...

5

Heat is not caused by thermal energy being radiated from particles due to their energy heat is the ramdomized (i.e. neglecting bulk flows) energy of motion in any material (including, for instance, photon gases). Any vacuum that we can make or have access too includes a small amount of matter, and the temperature of that stuff can be measured. Not ...

5

When first coming into contact with the water, it is conduction. The skin feels the water colder than air because water is a better conductor of heat than air. So the skin cools faster in water than in air. For longer intervals convection will enhance the effect bringing cooler water next to the skin and removing the water already heated by the skin. The ...

5

I suspect this is a formidably difficult question to answer from first principles. The cooling depends on the details of the air flow around your cylinder, and also on the temperature difference. An aeroplane flies at an altitude where the temperature is -50C, so the cooling is obviously going to be very different to say a baseball travelling at sea level. ...

5

Assuming 1, you live somewhere that is colder outside than in 2, the curtain has finite thermal resistance (ie some insulating value) 3, the curtain is close enough to the window to reduce convection Then yes. Try measuring the air temperature on the window side of the curtain, it should be lower than the room.

5

In this example both the systems are of the same type of particles (with two energy states) and same number of particles. Therefore thermal equilibrium is defined when energy is equally shared between the systems, but the particles are still not allowed to be exchanged. The particles, although of the same kind, are distinguished as being in system A or B. ...

5

Whiskey stones aren't necessarily designed to keep the drink cold, instead they are designed to allow flavor profiles to come out in the drink that might not be present at room temperature. Some whiskeys open up at a slightly cooler temperature and using stones allows you to experiment without diluting the flavor of the beverage. There are better math ...

5

Double the ammount of water does not need doulbe the ammount of time to heat, since while the energy needed is doubled indeed, losses due to vaporization and radiation from the kettle should be approximately constant. You can plot the time needed for a given ammount of water to boil and try to fit a function into that. With two data points you can manage to ...

4

Fire is a plasma. There are two kinds of plasmas: hot plasmas relevant to astrophysics or fusion are indeed a mixture of totally ionized gas. In cold plasmas ( northen lights, Neon tubes,flamme) the ionization degree is less than one but the mixture typically exhibit collective behaviour and a zoo of waves one do not encounter in gases. The most famous is ...

4

Heat is the thermal motion of particles. Hot object's atoms vibrate more than cold object's atoms. Heat is transfered by 3 main ways: Conduction: Heat flows from hot objects to cold objects. If you have an electric stove, the heat flows from the coils to the pan. Convection: Heat flows by bulk motion of a fluid. If you heard "hot air rises" this is the ...

4

Toilet paper is made from cellulose and cellulose molecules contain many hydroxyl (-OH) groups. These groups have a strong dipole and interact strongly with water molecules, so toilet paper is hydrophilic. When you breathe on toilet paper the water vapour in your breath condenses on the cooler toilet paper, and because toilet paper is so hydrophilic the ...

4

The motion of molecules that is responsible for heat content in water is random motion; that is there are molecules moving in all directions. The directed motion that you are considering ( all molecules moving in the same direction) from the flowing water does possess kinetic energy, but it is not heat energy. However, if the water flow encountered some ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# exponential functions

• July 29th 2010, 10:01 PM
nando29
exponential functions
I'm having trouble with the following:

Attachment 18382

i cannot find a problem similar in my book and my friend is stump too. there is an option for no solution. (Bow)
• July 29th 2010, 10:08 PM
pickslides
Can i suggest

$4^{1-7x}= 5^{x}$

$\ln 4^{1-7x}= \ln 5^{x}$

$(1-7x)\ln 4= x\ln 5$

$\ln 4-7x\ln 4= x\ln 5$

$\ln 4= x\ln 5+7x\ln 4$

$\ln 4= x(\ln 5+7\ln 4)$

can you finish it?
• July 30th 2010, 05:49 PM
nando29
ln4 / 7ln4 + ln5 could imply only $\frac{\ln 4 }{7\ln 4}+\ln5 = \frac{1 }{7}+\ln5$


text:
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Rejoice Obiekwe
The age of a young man and his wife sum to $52$ years. In seven years time, the man will be twice as old as his wife. Find the age of the man and his wife
0 0 comment
dev

## Solution 1

Let age of the man and his wife be $x$ and $(52-x)$ respectively.

$x+7= 2(52-x+7)\\\Rightarrow x+7=118-2x\\\Rightarrow 3x=111\\\Rightarrow x=37$

Present age of the man $=37$
Present age of his wife $=52-37=15$

## Solution 2

Sum of their ages after $7$ years $=52+7+7=66.$ At this time their ages are in the ratio $2:1$ and hence man's age is $66×\dfrac{2}{3}=44$ and wife's age is $66×\dfrac{1}{3}=22$

Present age of the man $=44-7=37$
Present age of his wife $=22-7=15$

0 0 comment


text:
# A parallelogram has sides A, B, C, and D. Sides A and B have a length of 6  and sides C and D have a length of  4 . If the angle between sides A and C is (5 pi)/12 , what is the area of the parallelogram?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
May 19, 2018

Area of the parallelogram color(purple)( = 23.18 sq units

#### Explanation:

Area of a parallelogram is given by the formula

$A r e a = a b \sin \theta$

Given : $a = 6 , b = 4 , \theta = \frac{5 \pi}{12}$

$\therefore A r e a = 6 \cdot 4 \cdot \sin \left(\frac{5 \pi}{12}\right)$

color(purple)(Area = 23.18 sq units

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# A cylindrical mould of height 2.5 meter with outer and inner radius 14 cm and 7 cm respectively. The approximate total curved surface area of cylinder will be:

1. 2.2 cm2
2. 3.3 m2
3. 5.2 cm2
4. 1.1 m2

Option 2 : 3.3 m2

## Detailed Solution

Formula used:

Curved surface area of cylinder = 2 × πRh

Where,

h = height of cylinder

Calculation:

Given that,

Inner radius r = 7 cm

Outer radius R = 14 cm

Height of cylinder h = 2.5 m

Hence, total curved surface area of the cylindrical mould

A = Area of inner curved surface + Area of an outer curved surface

⇒ A = 2π(14 + 7) × (2.5 × 100)

⇒ A =  32970 cm2

⇒ A ≈ 3.3 m2

Hence, the total surface area of cylinder 3.3 m2.


text:
## Bulletin of the Belgian Mathematical Society - Simon Stevin

### A new proof of extreme amenability of the unitary group of the hyperfinite II$_1$ factor

#### Abstract

We provide an alternative proof for the extreme amenability of the unitary group of the hyperfinite II${}_1$-factor von Neumann algebra, endowed with the strong operator topology.

#### Article information

Source
Bull. Belg. Math. Soc. Simon Stevin, Volume 22, Number 5 (2015), 837-841.

Dates
First available in Project Euclid: 17 December 2015

Permanent link to this document
https://projecteuclid.org/euclid.bbms/1450389251

Mathematical Reviews number (MathSciNet)
MR3435085

Zentralblatt MATH identifier
1347.46046

#### Citation

Dowerk, Philip A.; Thom, Andreas. A new proof of extreme amenability of the unitary group of the hyperfinite II$_1$ factor. Bull. Belg. Math. Soc. Simon Stevin 22 (2015), no. 5, 837--841. https://projecteuclid.org/euclid.bbms/1450389251


text:
# Least Angle Regression (LAR)¶

## A Basic Example Using the Boston House Prices Dataset¶

### Overview¶

Linear regression (i.e. ordinary least squares) is one of the most commonly used statistical modeling technique. In this example, we execute a linear regression model (lr) and then compare it to a Least Angle Regression (lar). We will not go into the mathematical details of either model. A few resources are listed below if you are interested in a deeper dive.

### Data Exploration¶

First we import a the boston house prices dataset, and print a description of it so we can examine what is in the data. Remember in order to execute a 'cell' like the one below, you can 1) click on it and run it using the run button above or 2) click in the cell and hit shift+enter.

We randomly select a third of our data to be the 'test' dataset. This way we can train our model on 2/3 of the data, and test it on the remainder. Once we are confident that our model is generalizing well (i.e. there is not a HUGE different in the training/testing performance, or in other words, not obviously overfitting), then we can use all of our data to train the model.

### Basic Linear Regression¶

In linear regression we assume that the relationship between the independent variables (X) and the dependent variable (Y) is linear and then go about finding one that minimizes the squared error between the predicted Y and the actual Y.

$${y}_i = \beta_0 + \beta_1 {x}_i + \epsilon_i$$

We now import the LinearRegression method from the sklearn library. Note that the process of creating the model involves the very simple command lm.fit(X,Y). This runs the model and we find the intercept-term, $\beta_0$, and the coefficient $\beta_1$ that minimizes the squared errors.

### Least Angle Regression¶

From SCIKIT:

Least-angle regression (LARS) is a regression algorithm for high-dimensional data, developed by Bradley Efron, Trevor Hastie, Iain Johnstone and Robert Tibshirani. LARS is similar to forward stepwise regression. At each step, it finds the feature most correlated with the target. When there are multiple features having equal correlation, instead of continuing along the same feature, it proceeds in a direction equiangular between the features.

1. It is numerically efficient in contexts where the number of features is significantly greater than the number of samples.
2. It is computationally just as fast as forward selection and has the same order of complexity as ordinary least squares.
3. It produces a full piecewise linear solution path, which is useful in cross-validation or similar attempts to tune the model.
4. If two features are almost equally correlated with the target, then their coefficients should increase at approximately the same rate. The algorithm thus behaves as intuition would expect, and also is more stable.
5. It is easily modified to produce solutions for other estimators, like the Lasso.

The disadvantages of the LARS method include:

1. Because LARS is based upon an iterative refitting of the residuals, it would appear to be especially sensitive to the effects of noise. This problem is discussed in detail by Weisberg in the discussion section of the Efron et al. (2004) Annals of Statistics article.

We measure the accuracy of the new approach using R-squared metric, and compare it to linear regression. The performance of this algorithm on this dataset is inferior to that of linear regression (on both training and testing data). However the benefit of LAR is that instead of linear regression, it results in a simpler model (compare coefficients of linear regression to lar, and note the zeroed coefficients for variables that were not selected). LAR is similar to forward selection method.

You can explore the linear regression model further, here: https://predictivemodeler.com/2019/08/19/py-ols-boston-house-prices/

### Feedback¶

If you have ideas on how to improve this post, please let me know: https://predictivemodeler.com/feedback/


text:
Report Number: CS-TR-79-765
Institution: Stanford University, Department of Computer Science
Title: Relation between the complexity and the probability of large numbers
Author: Gacs, Peter
Date: September 1979
Abstract: H(x), the negative logarithm of the apriori probability M(x), is Levin's variant of Kolmogorov's complexity of a natural number x. Let $\alpha (n)$ be the minimum complexity of a number larger than n, s(n) the logarithm of the apriori probability of obtaining a number larger than n . It was known that $s(n) \leq\ \alpha (n) \leq\ s(n) + H(\lceil s(n) \rceil )$. We show that the second estimate is in some sense sharp.
http://i.stanford.edu/pub/cstr/reports/cs/tr/79/765/CS-TR-79-765.pdf


text:
# What is an "Effective interest" ?

By | 13/06/2011

Many people do not know what is an effective interest, and how to calculate it.

Actually, it is a very simple concept, and it is very easy to calculate.

First, you need to understand that interests are expressed in annual terms. When the bank says that it will loan you a sum of money at an interest rate of 4%, this is the annual interest rate.

Now, interest is a logarithmic function. Hey, no need to close the window!!! I did not say a dirty word!

In a laymen language – you cannot just add or subtract interest rates, because interest rates are expressions of a product, not a sum. For instance, if the bank lends you 5,000,000$for one year, at an interest rate of 4%, the interest will be: $5,000,000\\cdot4\%=5,000,000\cdot0.04=200,000\$ In order to calculate the interest we multiplied the principal by the factor 0.04 which represents $4\%=\frac{4}{100}$. In all the following examples I will use a loan for demonstrations, but you can do the same calculations on deposits as well, in this case, you earn the interest rather than paying it. Now the bank tells you that the interest rate is 4% per annum, same as before, but now it states that the interest is calculated on a quarterly basis. This means, that for every quarter you will pay 1%. Remember, you cannot add interest rates ! this means, that paying 1% per quarter is not the same is paying 4% per annum. The reason is that after the first quarter, you will owe the bank 5 million dollars plus interest of 1%, or 50 thousand dollars. For the next quarter, you will also pay 1%, but on an initial principal of 5 million dollars and 50 thousand dollars. So after the second quarter you will owe the bank: $5,050,000\\cdot101\%=5,050,000\cdot1.01=5,100,500\$ There's another 500$ there ! You'd expect that after half a year you will owe the bank half the interest, 2%, or 100,000$, over the 5 million principal, but because you cannot sum interest rates, you would owe 100,500$ and not 100,000$. This residual element, in our case 500$, is called the logarithmic error, it is the "error" added when adding interest rates using simple addition.

In the year-end, you will need to pay the bank:

$5,000,000\\cdot101\%\cdot101\%\cdot101\%\cdot101\%=5,000,000\cdot1.01^4=5,203,020.05\$

The additional 3020.05$we incurred over the basic calculation of 4% per annum on 5 million dollars are the logarithmic error. This logarithmic error will increase when: • Dividing the year to smaller periods of time. • Interest rate is higher Pay attention that the increase in the logarithmic error is also not linear. If the logarithmic error on a 5M$ loan at 4% was 3020.05\$, on 8% it will not be twice as much.

Now for the punch:

The effective interest is the interest, expressed in annual terms, that will take this logarithmic error into account.

To calculate this interest we divide the interest the bank say by the periods of time, and then we multiply the percentages.

When the bank tells you that you will pay 4% annually calculated on per quarter basis, it actually tells you that you will pay:

$I_{effective}+1=101\%\cdot101\%\cdot101\%\cdot101\%=1.01^4=1.04060401= 104.060401\%$

or 4.060401% per annum, effectively.

Now what will happen if the bank will calculate the interest on per month basis:

$I_{effective}+1=(100+\frac{4}{12})\%\cdot(100+\frac{4}{12})\%\cdot(100+\frac{4}{12})\%\cdot(100+\frac{4}{12})\%+...=(1+\frac{4}{12})^{12}=1.00333334^{12}=1.0407415429=104.07415429\%$

or 4.074% effective per annum, higher than on per quarterly basis.

Lets take annual interest of 12% calculated monthly. Try to guess how much will the effective interest be:

$I_{effective}+1=(100+\frac{12}{12})^{12}\%=1.01^{12}=1.1268250301=112.68250301\%$

12.68%! the bank "forgot" to mention 0.68% !

So the basic formula for calculating effective interest, or the real interest you will pay for a loan:

$I_{effective}=(1+\frac{I_{per-annum}}{n})^n$

where "n" is the number of periods in a year (for quarters, n=4, for months, n=12, for days, n=365).

That's it. You can stop here. You know all there is about effective interest. You got what you want… now before you leave:

The more the bank will divide the year, the higher the interest rate will be, but not to infinity.

For the math geeks (like me), lets see what happens if the bank decides to divide the year to seconds, or even milliseconds, or even nanoseconds, or…. ok you got the idea.

$\lim_{n\to\infty}I_{effective}=\lim_{n\to\infty}(1+\frac{I_{per-annum}}{n})^n=\lim_{n\to\infty}((1+\frac{I_{per-annum}}{n})^\frac{n}{I_{per-annum}})^{I_{per-annum}}=e^{I_{per-annum}}$

The limit of the effective interest is $e$, the natural logarithm (2.71828…), by the power of the annual interest stated by the bank.

See? Wasn't that hard!


text:
# Borsuk fixed-point theorem

Let $B$ be an open bounded symmetric subset of $\mathbf{R}^n$ containing the origin. Here, symmetric means that if $x \in B$, then $-x \in B$ also. Let $\phi : \partial B \rightarrow \mathbf{R}^m$ be a continuous mapping and let $m < n$. Then there is an $x \in \partial B$ such that $\phi(x) = \phi(-x)$.
The original version (K. Borsuk, 1933) was for $B = D^n$, the $n$-dimensional ball in $\mathbf{R}^n$, $n = m$. The result is also known as one of the Borsuk antipodal theorems (see Antipodes) or as the Borsuk–Ulam theorem.
The central lemma for the Borsuk–Ulam theorem is that an odd mapping $f : S^n \rightarrow S^n$ has odd degree (see Degree of a mapping). A mapping $f : S^n \rightarrow S^n$ is called odd if $f(-x) = - f(x)$. Many people call this odd-degree result itself the Borsuk–Ulam theorem. For a generalization, the so-called Borsuk odd mapping theorem, see [a1], p. 42.


text:
# tsql write multi line parameter to text file sql 2000

I am reading the body of an email into a parameter (multi lined) i want to write this out to a text file. I have the following which will only write one line but i need to keep the lines split as they appear in the email, can anyone help?

Declare @Script      varchar(8000)
Set @Script = 'Echo test > c:testfolder\testfile.txt'
exec master..xp_cmdshell @Script, no_output
END
###### Who is Participating?

Ummm....

If you are reading the body, then one would assume it is more like

"Hi 9772885,
How is it going ?
kind regards,
Mark"

and using "echo" will not handle that type of message.

What you might need to do is run a BCP type export into a flatfile... Also depends a bit on the datatype that the body is being stored in.

for example:

-- OK suppose we have a variable that represents the email body

DECLARE @EmailBody nvarchar(4000)
Set @emailbody = 'Hi 9772885,
How is it going ?
kind regards,
Mark'

-- now if ou temp table does not exist, then create it

if object_id('tmp_email_output','U') is null Create table tmp_email_output (id int identity primary key, body nvarchar(4000))

-- Now let's populate our table (and we probably need to have update flags etc depending on concurrency)

insert tmp_email_output (body)
select @emailbody

-- lets then export the contents

EXEC master..xp_cmdshell 'bcp "select body from mydb..tmp_email_output x order by x.id" queryout "c:\ee\my_email_flatfile.txt" -T -c -CACP'

GO
0

Billing EngineerCommented:
Declare @Script      varchar(8000)
Set @Script = 'Echo test > c:testfolder\testfile.txt'
exec master..xp_cmdshell @Script, no_output
Set @Script = 'Echo test2 >> c:testfolder\testfile.txt'
exec master..xp_cmdshell @Script, no_output
END

0

Author Commented:


text:
# How do you differentiate f(x)=ln2x * tan3x using the product rule?

Mar 25, 2016

I found: $f ' \left(x\right) = \tan \frac{3 x}{x} + \frac{3 \ln \left(2 x\right)}{\cos} ^ 2 \left(3 x\right)$

#### Explanation:

The Product Rule tells us that if we a function $f$ that is the product of 2 other functions:
$f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$
differentiating we get:
$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$
in our case:
$f ' \left(x\right) = \frac{2}{2 x} \tan \left(3 x\right) + \ln \left(2 x\right) \frac{1}{\cos} ^ 2 \left(3 x\right) \cdot 3$
Where I used the Chain Rule to differentiate each component function;
finally:
$f ' \left(x\right) = \tan \frac{3 x}{x} + \frac{3 \ln \left(2 x\right)}{\cos} ^ 2 \left(3 x\right)$


text:
# Game mechanic player following path looks very roughly

I want my player entity to follow a path, which works, but it looks very roughly. Like it would stuck everytime.

    bool arrivedX = false;
bool arrivedY = false;

float speed = 0.06f;

if(bestway.empty()) {
//arrived
} else {

if(entity->x < bestway.front()->x * gridSize) {
entity->x += speed * machine->getDeltaTime();
} else if(entity->x > bestway.front()->x * gridSize) {
entity->x -= speed * machine->getDeltaTime();
} else {
arrivedX = true;
}

if(entity->y < bestway.front()->y * gridSize) {
entity->y += speed * machine->getDeltaTime();
} else if(entity->y > bestway.front()->y * gridSize) {
entity->y -= speed * machine->getDeltaTime();
} else {
arrivedY = true;
}

if(arrivedX && arrivedY) {
bestway.pop_front();
}
}

the vector contains grid coords from 0 to WIDTH or HEIGHT / gridSize. And the entity has pixels coords. I have tried already:

entity->x / gridSize < bestway.front()->x

which runs smoothly, but when it goes up or right it will omit one grid tile. I think it is because of when the player is at pixel 19 and the gridSize is 20 it would calculate 0 . 19/20 = 0... I tried to ceil and floor it, but it didn't work out. Hope you guys can help me.

edit:

        int ix = int( x + 0.5f );
int iy = int( y + 0.5f );

if(ix < bestway.front()->x * gridSize) {
x += speed * machine->getDeltaTime();
} else if(ix > bestway.front()->x * gridSize) {
x -= speed * machine->getDeltaTime();
} else {
arrivedX = true;
}

Video

http://youtu.be/-DugHbMdODg

• What are the types of entity->x and gridSize and bestway.front()->y? – akaltar Mar 12 '15 at 14:52
• hi. they're all integers. – Kronos Mar 12 '15 at 15:00
• And what is deltatime? What unit and what type? Also, what is your update frequency? – akaltar Mar 12 '15 at 15:04
• Deltatime is float most time between 30 and 60. But now I saw that when it stucks deltatime ist above 100. I use the SDL_iPhoneSetAnimationCallback() method for updating. So I don't know how often it updates. I have to figure this out first. – Kronos Mar 12 '15 at 15:33

## 1 Answer

I believe your error arises from rounding errors.

I believe the problem could be solved by using float for pixel-coordinate storage. This way you will always get the exact increment needed for the given delta time, which is usually very small and thus the resulting per update increment is less than 1, and truncated to 0.

Don't forget to use proper rounding when converting back for rendering or comparison. For rendering I would use: int ix = (int)( entity->x + 0.5 ) Which will yield 'proper' rounding: up from 0.5 and down with anything less.

Edit:

From the video it seems you are overshooting the position you are aiming for. There are three ways to solve this:

• After moving, check if you overshot and correct for it:

entity->x += speed * dt;
if( entity->x > movelocation->x * gridSize )
entity->x = movelocation->x * gridSize;

• Lower the speed you are moving with.

• Use a higher update frequency.

The most solid option seems to be checking for overshoot and correcting. Note: The simple code I provided doesn't account for the movement speed lost when arriving at each tile.

• Thanks for the answer. I cannot easily use float since I'm using the SDL_Rect x, y coords which are integers. And I really will not use a Wrapper or tricky stuff. Any idea? – Kronos Mar 12 '15 at 16:21
• Still is shaky... that is the better description for it. looks like it would increment, then decrement and so on... edit: used floats for x,y coords and updating the SDL_Rect so I can use it as well – Kronos Mar 12 '15 at 16:43
• @Kronos could you post the new code too? I showed how to convert the float to an integer, it should be smooth. I see no real reason for the jumping forward and backward.. A gif or video of the issue would be helpful as well. Also: Are you warning-free? You should usually treat compiler warning as errors. – akaltar Mar 12 '15 at 17:56
• Hi. there are no Warnings. I edited the original post and added a video. Hope this will help. – Kronos Mar 12 '15 at 18:36
• @Kronos Edited, I guess this will solve your problem. – akaltar Mar 12 '15 at 19:45


text:
## CryptoDB

### Paper: On the Shortness of Vectors to Be Found by the Ideal-SVP Quantum Algorithm

Authors: Léo Ducas Maxime Plançon Benjamin Wesolowski DOI: 10.1007/978-3-030-26948-7_12 (login may be required) Search ePrint Search Google The hardness of finding short vectors in ideals of cyclotomic number fields (hereafter, Ideal-SVP) can serve as a worst-case assumption for numerous efficient cryptosystems, via the average-case problems Ring-SIS and Ring-LWE. For a while, it could be assumed the Ideal-SVP problem was as hard as the analog problem for general lattices (SVP), even when considering quantum algorithms.But in the last few years, a series of works has lead to a quantum algorithm for Ideal-SVP that outperforms what can be done for general SVP in certain regimes. More precisely, it was demonstrated (under certain hypotheses) that one can find in quantum polynomial time a vector longer by a factor at most $\alpha = \exp ({\widetilde{O}(n^{1/2})})$ than the shortest non-zero vector in a cyclotomic ideal lattice, where n is the dimension.In this work, we explore the constants hidden behind this asymptotic claim. While these algorithms have quantum steps, the steps that impact the approximation factor $\alpha$ are entirely classical, which allows us to estimate it experimentally using only classical computing. Moreover, we design heuristic improvements for those steps that significantly decrease the hidden factors in practice. Finally, we derive new provable effective lower bounds based on volumetric arguments.This study allows to predict the crossover point with classical lattice reduction algorithms, and thereby determine the relevance of this quantum algorithm in any cryptanalytic context. For example we predict that this quantum algorithm provides shorter vectors than BKZ-300 (roughly the weakest security level of NIST lattice-based candidates) for cyclotomic rings of rank larger than about 24000.
##### BibTeX
@article{crypto-2019-29865,
title={On the Shortness of Vectors to Be Found by the Ideal-SVP Quantum Algorithm},
booktitle={Advances in Cryptology – CRYPTO 2019},
series={Lecture Notes in Computer Science},
publisher={Springer},
volume={11692},
pages={322-351},
doi={10.1007/978-3-030-26948-7_12},
author={Léo Ducas and Maxime Plançon and Benjamin Wesolowski},
year=2019
}


text:
# Probability Proof

kuahji
We have three events A, B, and C such that
P(A|C)$\geq$P(B|C) and P(A|C')$\geq$P(B|C')

Prove that P(A)$\geq$P(B).

First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

From above,
P(AC)/P(C)$\geq$P(BC)/P(C) and since P(C)$\geq$0,
P(AC)$\geq$P(BC).

Similarly from above,
P(AC')$\geq$P(BC').

Here is where I am stuck. I tried to expand P(AC)=P(A)+P(C)-P(AUC) with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.

Last edited:

## Answers and Replies

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
How did you get P(A|C)=P(AB)/P(C)? Why is B there?

kuahji
How did you get P(A|C)=P(AB)/P(C)? Why is B there?

Typo, fixed it above.

drawar
If I'm not mistaken then P(A and C) + P(A and not C) = P(A).

Science Advisor
Homework Helper
We have three events A, B, and C such that
P(A|C)$\geq$P(B|C) and P(A|C')$\geq$P(B|C')

Prove that P(A)$\geq$P(B).

First I started with,
P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

From above,
P(AC)/P(C)$\geq$P(BC)/P(C) and since P(C)$\geq$0,
P(AC)$\geq$P(BC).

Similarly from above,
P(AC')$\geq$P(BC').

Here is where I am stuck. I tried to expand P(AC)=P(A)+P(C)-P(AUC)
You have this backwards. It should be $P(A\cup C)= P(A)+ P(C)- P(A\cap C)$

with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.

kuahji
Okay, just figured it out. Thanks everyone for the assistance.

Last edited:


text:
+0

# I thought I was smart till I saw this

0
58
3

Let x =                                                            What is $$(x+1)^{48}$$

4

--------------------------------------------------------
$$(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1) (\sqrt[16]{5}+1)$$

Dec 31, 2020

#1
+1

This question was posted before and the answer should have been 125

Dec 31, 2020
#2
+1

I just entered it all in a calculator....    answer is 125

Really no solving involved.....just enter it into a calculator.....to find 'x' then add 1   And 48th power it.

Dec 31, 2020
edited by Guest  Dec 31, 2020
#3
+112009
+1

I think the point of this question is that it is meant to be done without a calculator.

$$Let\;\;x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1) (\sqrt[16]{5}+1)}\qquad find\;\;(x+1)^{48}$$

Let    $$y=5^{1/16}$$

$$(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1) (\sqrt[16]{5}+1)\\ =(y^8+1)(y^4+1)(y^2+1)(y+1)\\ =y^{12}+y^{11}+y^{10}+y^9 ..... y^1+1\\ \text{This is the sum of a GP with a= 1, r=y, and n=16}\\ =\frac{a(r^n-1)}{r-1}=\frac{y^{16}-1}{y-1}=\frac{5-1}{y-1}=\frac{4}{y-1}$$

so

$$x=\frac{4}{\frac{4}{y-1}}=y-1\\~\\ (x+1)^{48}\\=(y-1+1)^{48}\\=y^{48}\\=(5^{1/16})^{48}\\=5^3\\=125$$

LaTex:

(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1) (\sqrt[16]{5}+1)\\
=(y^8+1)(y^4+1)(y^2+1)(y+1)\\
=y^{12}+y^{11}+y^{10}+y^9 ..... y^1+1\\
\text{This is the sum of a GP with a= 1,  r=y, and n=16}\\
=\frac{a(r^n-1)}{r-1}=\frac{y^{16}-1}{y-1}=\frac{5-1}{y-1}=\frac{4}{y-1}

x=\frac{4}{\frac{4}{y-1}}=y-1\\~\\
(x+1)^{48}\\=(y-1+1)^{48}\\=y^{48}\\=(5^{1/16})^{48}\\=5^3\\=125

Dec 31, 2020


text:
# Question 90a6b

Oct 22, 2016

$\frac{5}{8}$

#### Explanation:

If you have two variables, and changing one results in a the other changing as well, we say that those variables are proportional. If that change is just multiplying by a constant, then we call that constant the constant of proportionality.

If we look at the equation $y = \frac{5}{8} x$, then we know $x$ and $y$ are proportional, because changing one results in the other changing as well. Furthermore, we can see that if we change $x$, then we are changing $y$ by that amount multiplied by $\frac{5}{8}$. For example, it's easy to see that if $x = 0$, then $y = \frac{5}{8} \cdot 0 = 0$. Let's see what happens to $y$ when we add $n$ to $x$:

$y = \frac{5}{8} \cdot \left(0 + n\right) = \frac{5}{8} n = 0 + \frac{5}{8} n$

So, when we increased $x$ by $n$, we increased $y$ by $\frac{5}{8} n$. We can use a similar process to show that it doesn't matter where we start. A change of $n$ in $x$ will result in a change of $\frac{5}{8} n$ in $y$. This matches our description of a constant of proportionality.

You're also right about it being the slope. In a line, the slope is just another way of expressing the change in $y$ with relation to the change in $x$. That is often the first way students are introduced to it:

"slope" = "rise"/"run" = ("change in "y)/("change in "x)#

A very fast way to find the slope of a line given the equation is to put it into point-slope form.

If a line has the equation $y = m x + b$, where $m$ and $b$ are constants, then $m$ is the slope of the line and $b$ is the $y$-intercept (that is, the line passes through the point $\left(0 , b\right)$).

Note that the given equation is actually in that form already, with $m = \frac{5}{8}$ and $b = 0$.


text:
## Found 86 Documents (Results 1–86)

100
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Full Text:

Full Text:

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Numerical techniques for stochastic optimization, Springer Ser. Comput. Math. 10, 267-272 (1988).
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