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Checking your answers: A) Correct, it looks true. B) Correct, it is definitely false. C) Correct, it looks true. D) Correct, it looks true. E) Correct, it is definitely false. Looks good.

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If we have have $y = f(x)$ and $y=g(x)$ [1], a solution will satisfy $f(x)-g(x)=0$, hence we need $f(x)-g(x)$ to have exactly one solution. Quadratic equations always have exactly two solutions, so either $f(x)-g(x)$ is linear, or it has a multiple root. If it's linear, it has the form $y = mx+B$ for some $m,B$ (I'm using a capital $B$ to distinguish it from ...

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I've looked at the picture and made a crude approximation, but this should allow anyone who knows how to derive the real values used here to have somewhere to start https://www.desmos.com/calculator/jbfrwrju0e

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I noticed that your graph looks like a sine curve that's been wrapped around a cylinder viewed from above. From that I was able to work out the equation $y = -\sin(e \sin(x))$. Feel free to play around in desmos to modify it to your liking: https://www.desmos.com/calculator/jfwbvrmj7d Is there any particular application you had in mind for this graph?

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If $z=x+\frac\pi4i$, then$$\exp(z)=e^x\left(\frac{1+i}{\sqrt2}\right).$$Since $\{e^x\mid x\in\Bbb R\}=(0,\infty)$, $\left\{\exp\left(x+\frac\pi 4i\right)\,\middle|\,x\in\Bbb R\right\}$ is the open ray with origin in $0$ and passing through $\frac{1+i}{\sqrt2}$.

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By OP's request, a short explanation of the matrix formula: The matrix $$\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$ is the $2 \times 2$ rotation matrix by $\theta$ radians, as can be observed by noting that the first column is the anticlockwise rotation of the unit vector $(1, 0)$ by $\theta$ radians,...

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The set $\{z \in \mathbb{C}: |z+1| <2\}$ is the set of complex number whose distance to -1 is less than 2, hence an open circle centered at $z_0=-1$ and radius 2.

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You calculated the magnitude incorrectly. Note that $1$ is real, so part of the real part of $z + 1$. Therefore, you get with $z = x+ iy$ $$| z + 1| = \sqrt{(x+1)^2 + y^2}.$$ Hence, all the points satisfying the inequality equivalently satisfy $$(x+1)^2 + y^2 < 4.$$ This gives you the interior of a circle with midpoint $-1$ and radius $2$ on the complex ...

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The function $y=a-(2+\sin x)be^{-cx}$ will work. The $2$ can be changed to any number $\ge1$ and the limiting value will still be an upper bound for the function; if it's changed to exactly $1$, the upper bound will be attained over and over again.

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The parabola (quadratic Bézier) solution might serve your needs, but it won’t give you very “sharp” curves that get close to the corners of your rectangle. To get a sharper curve, you need to use a rational quadratic Bézier curve. This is very similar to the regular quadratic curve (the parabola), but it has an added parameter $w$ called a “weight”. If $A=(0,... 0 There are quite some redundancy in the way your implicit function was formulated. Note that the rational function in$g$can be simplified. $$\frac{x^2 - 2x + 1}{x + 1 + 2\sqrt x}= \left(\frac{x-1}{\sqrt{x} + 1} \right)^2 = \left(\frac{(\sqrt{x} + 1)(\sqrt{x} - 1)}{\sqrt{x} + 1} \right)^2 = (\sqrt{x} - 1)^2$$ Put this back into$g$we have $$g=x+1-(x-2\... 1 One thing you can try is you can assume your curve is a parabola passing through the two points (0, 0) and (a, a) for some a \gt 0 . The next thing you need is to select a point of the form (b , a - b ) which lies on the diagonal line y = a - x This point will serve as the point where tangents to the parabola at (0,0) and (a, a) meet. That's ... 0 Very late to the party, but I found the following helpful. Abramowitz and Stegun give the analytic continuation for I and K in Eq. 9.6.30 and 9.6.31$$I_\nu(ze^{m\pi i}) = e^{m\nu\pi i}I_\nu(z), \qquad m\in\mathbb Z \\ K_\nu(ze^{m\pi i}) = e^{-m\nu\pi i}K_\nu(z) - \pi i \sin(m\nu\pi)\csc(\nu\pi)I_\nu(z), \qquad m\in\mathbb Z.$$These results are in the ... 2 There are two things necessary to increase the definition of the image: Compute the modular form to higher accuracy, and Plot the modular form in a finer mesh. In sage, the way to do the latter is with the plot_points parameter in complex_plot. By default, sage will evaluate the function on a 100 \times 100 grid and interpolate between the points. This ... 2 I would agree with @Yalikesifulei comment there. You have seen the graphs of the normal distribution and a rectangular hyperbola, but unless you do vivid mathematical calculations, you can't prove anything. Just saying they look similar doesn't prove any similarity. For any probability distribution function, if take its integral over its entire domain, the ... 2 I don't get quite well what the author tried to say about approximation, but a few things come to mind. Let f(x) be an arbitrary (smooth enough) function with local maximum at x = x_0. Then, using Taylor approximation near x_0,$$ f(x) \approx f(x_0) + f'(x_0)\cdot (x-x_0) + \frac{1}{2} f''(x_0) \cdot (x- x_0)^2 $$If x_0 is the point of local ... 0 The original parabola (y=.001x^2) touches the upper right and upper left corners of the given graphing area. For the second parabola, we need the same behavior; but the upper right corner should be (k,k). For this to occur, you need k^2=k. So k = 0 or 1; but k=0 doesn't work. So k=1. 0 Your two students' answers are how I myself solved the problem. A more intuitive explanation? How about the observation that with the appropriate x- and y- scale, every upward-facing parabola looks identical; likewise, every downward-facing parabola? In other words, to sketch a quadratic function, I might start by sketching a prototypical upward/downward-... 0 Start with the graph of$$y=f(x).$$Applying a horizontal scale factor of \frac12 gives$$y=f(2x):=g(x).$$Translating the graph by 1 unit to the left gives$$y=g(x+1)=f(2x+2).$$Your above procedure resulted in the graph of \arccos(2x+2). On the other hand, the procedure below gives the desired \arccos(2x+1). Start with the graph of$$y=f(x).$$... 2 Hint: if v is the direction vector of the line, then it is perpendicular to both n1 and n2. Do you know a way to compute a vector that’s perpendicular to two given vectors? 1 Great question OP! I just wanted to add my contribution here. It might be too late, however, I think it is worthwhile to post anyway. I created a program for this a while back for a multivariable calculus course I was teaching and figured I would include it here. My approach is to use a scatterplot, but I made some changes to really make the graphics pop. ... 2 Wow it's have so many solutions... \left\{{x}^{\mathrm{2}} \right\}={x}^{\mathrm{2}} −\left[{x}^{\mathrm{2}} \right]=\left\{{x}\right\}^{\mathrm{2}} \\ {x}^{\mathrm{2}} −\left\{{x}\right\}^{\mathrm{2}} =\left[\left(\left[{x}\right]+\left\{{x}\right\}\right)^{\mathrm{2}} \right] \\ \left[{x}\right]\left(\left[{x}\right]+\mathrm{2}\left\{{x}\right\}\... 1 It's a parabolla for any given value of z where z=x+y^2 0 First, let x=-10 \log(t) and consider the norm$$\Phi=\Bigg[\frac{y}{t^6+1}-\frac {69}{10}\Bigg]^2+\Bigg[\frac{y}{t+1}-4\Bigg]^2$$that you want to minimize to have the "most probable" values of (t,y). Compute the partial derivatives$$\frac{\partial \Phi}{\partial t}=-\frac{12 t^5 y \left(\frac{y}{t^6+1}-\frac{69}{10}\right)}{\left(t^6+1\right)... 1 You might be interested in David Lowry-Duda's work on that. David Lowry-Duda. Visualizing modular forms. arXiv:2002.05234 David Lowry-Duda. Notes behind a talk: visualizing modular forms. Blog post, 2019-11-22. https://davidlowryduda.com/notes-behind-a-talk-visualizing-modular-forms/ David Lowry-Duda. phase_mag_plot, a Sage package for plotting complex ... 3 This curve looks to (be at least very close to) have this equation: $$y=-\ln(x)-\sin(x)$$ adding to a logarithmic "carrier" (taken with a minus sign) a regular sinusoidal pulsation. In fact, the pulsation of the sine wave should be slightly different. Indeed, by taking $$y=-\ln(x)-\sin(0.9*x)$$ one gets the exact number of maxima in the [0,160] ... 1 I wouldn't count turns. Think of the graph of$y=x^8.$If a zero is simple, the graph crosses the$x$-axis like a line. In your graph, this happens at$x=9$. If a zero has even multiplicity, the graph won't cross the$x$-axis, but touch it and turn back. This happens at$x=4$. So there must be at least two more zeros. If a zero has odd multiplicity ... 1 Short way. Write$x=n+\frac{k}{n}$where$n$is integer,$0 \leq k < n$,$k$not necessarily an integer. Then: $$\{x^2 \}=\{ (n+\frac{k}{n})^2 \}=\{ n^2+2k+(\frac{k}{n})^2 \}=\{2k+(\frac{k}{n})^2 \}$$ аnd $$\{x \}^2=\{ n+\frac{k}{n} \}^2=\{ \frac{k}{n} \}^2$$ That means that we need to have $$\{2k+(\frac{k}{n})^2 \}=\{ \frac{k}{n} \}^2$$ and this is ... 1 Let$x=i+f, 0\le f<1$(integer plus fractional parts). The equation turns to $$\{(i+f)^2\}=f^2$$ which simplifies to $$2if=n$$ for some$n$. Hence the solutions come with all fractions $$f=\frac{n}{2i}$$ with$0\le n <2i$. Now count the possible values of$n$for$i\in[1,10]$. 2 I would not consider$f(x)=x^{2/3}$to be concave down on$(-\infty,\infty)$, but rather on$(-\infty, 0)\cup(0,\infty)$. If we use the definition of concave down to be "A function$f$is concave down when$f'$is decreasing", then we can see this is not the case with$f(x)=x^{2/3}$. Its derivative,$f'(x)=\frac{2}{3\sqrt[3]{x}}$, is decreasing on ... 3 Stirling's approximation is enough to generate solutions. $$2^x \approx y! \approx \sqrt{2\pi y}\left(\frac ye \right)^y \implies x \approx \frac12 \log_2 (2\pi y) + y \log_2 \left(\frac ye \right)$$ These are all pairs$(x,y)$which satisfy your inequality with$y \le 100$: {{7, 5}, {70, 22}, {98, 28}, {133, 35}, {170, 42}, {203, 48}, {220, 51}, {290, 63}, ... 2 It is trivial to solve by computer. Here are the first few results (where$r$is the ratio$2^p/q!$): p=1 q=2 r=1 p=7 q=5 r=1.06666666666667 p=70 q=22 r=1.05034773691979 p=98 q=28 r=1.03943839001342 p=133 q=35 r=1.05379655617941 p=170 q=42 r=1.06517520295244 p=203 q=48 r=1.03557207701665 p=220 q=51 r=1.08631055044999 The ... 0 Note in your example of$x + x\frac{10\%}{100\%}$that it can be simplified to$x(1 + 0.1) = (1.1)x$. That is, instead of "adding 10% to$x$", this can be thought of as "multiplying$x$by$1.1$". More generally, if you want to increase by$p\%$, then you calculate $$r = 1 + \frac{p\%}{100\%} = 1+(0.01)p$$ And the increase operation is$x ...

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Warning: non-maths answer! One option is to do a monte carlo simulation to find an approximate solution, and then encode that approximation into your simulation. e.g. A=1, B=1, C=0.4 You simulate this ten times and generate three casualties and conclude that P(X=0)=0.7 and P(X=1)=0.3 This result is clearly incorrect but it is approximately correct. Think ...

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The description underlined in purple is actually the two transformations in the wrong order. Translation by $\binom{3}{0}$ results in $f(x-3)$. Now following this with a reflection in the $y$ axis results in $f(-x-3)$. This is because in order to perform the reflection, every $x$ in the previous expression must be replaced by $-x$. This is not the same as ...

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You have got the order of operations mixed up. Indeed, $f(3-x) = f(-(x-3))$. However, this implies that you replace $x$ by $-x$, or graphically: reflect around the $y$-axis. you replace $x$ by $(x-3)$, or graphically: shift $3$ units to the right. If we follow your order of operations, we get $f(x) \rightarrow f(x-3) \rightarrow f(-x-3)$. It is also ...

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What creates those circular arches in the Riemann zeta function graph is actually the behavior around the pole s=1. Check this video out: transformation of function. As you get closer to s=1, the Riemann zeta function behaves a lot like $$f(z)=\frac{1}{z-1} + \gamma$$ These are the leading terms in the Laurant series expansion for $\zeta(s)$. So those ...

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