# Tag Info

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After a very few attempts but very much guided by your underlying ideas, I effectively found that the reciprocal of the square root of the average light intensity varies linearly with $d$; its is almost perfect (I guess and hope) for your plot requirement. So, you keep $d$ for the $x$ axis and you change what was $y$ to $\frac{1}{\sqrt{y}}$. The least ...

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Try in http://people.hofstra.edu/Stefan_Waner/newgraph/regressionframes.html with the data: 0.04 766 0.01 217 0.0044 90 0.0025 50 0.0016 29 0.0011 21

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Equation of the two asymptotes: $$(2x+11-y)(-2x-1-y)=0$$ Equations of hyperbolas with these asymptotes: $$(2x+11-y)(-2x-1-y)=c$$ for nonzero constant $c$. For one sign of $c$ it "opens vertically" and for the other sign it doesn't.

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These are rounding errors. In those regions, the function is going from an extremely large positive number, to an extremely large negative number. The software is assuming that there must be a zero somewhere in between, and it just doesn't have the numerical resolution to find it. This does bring up an interesting point though, the parts where the fuzzy ...

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You will get a very different graph from the one you have drawn for intensity against distance if you plot intensity against $\cfrac 1{\text{distance}^2}$ What you seem to be trying to do is to use the numbers in your final column on the $x-$ axis, which will then run up to a maximum value of $\cfrac 1{25}=0.04$, so you might choose a scale which goes from ...

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I assume $K>1$. If you only want to know $a, b, c$, you can diagonalize the matrix $$A=\pmatrix{K & 1 & 1\\1 & K & 1\\ 1 & 1 & K}$$ the eigenvalues are $K-1, K-1$ and $K+2$. So $a, b, c$ would be $$\frac{1}{\sqrt{K-1}}, \frac{1}{\sqrt{K-1}}, \frac{1}{\sqrt{K+2}}$$

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As noted in the comments your ellipsoid is rotated in such a way that its coordinate axes are no longer given by the standard basis. So it can not be written in your form. But it is possible to describe exactly in which way the coordinate axes are tilted. I assume $K>2$. Then there exists $c>0$ such that $K=c^2+\frac{1}{c^2}$. Then ...

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You want to start off by initializing an array of all of your inputs n=3:2000 You also know that you will have an array of all your answers, so let's initialize an array for that x=zeros(2000) now let's go through every element of n and fill in the appropriate x value for i=n x(i)=.6530*x(i-1)-.7001*x(i-2)+v(i) end now we can plot everything ...

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Easily generalised: $f(x)=2\frac{(x-2)(x-3)(x-4)}{(1-2)(1-3)(1-4)}+4\frac{(x-1)(x-3)(x-4)}{(2-1)(2-3)(2-4)}+6\frac{(x-1)(x-2)(x-4)}{(3-1)(3-2)(3-4)}+\pi\frac{(x-1)(x-2)(x-3)}{(4-1)(4-2)(4-3)}$

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Hint. There are any number of answers to this, but perhaps you could try $$f(x)=2x+c(x-1)(x-2)(x-3)\ .$$ If you find the right value of $c$ it will work.

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$f(x)=2x$ if $x\neq4$ and $f(4)=\pi$

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The standard way to combine such functions is to use the R Formula ( http://www.oocities.org/maths9233/Trigonometry/RFormula.html) : Specifically, if we wanna combine your function into : $$a\sin{\theta} + b\cos{\theta} = R\sin{(\theta + \alpha)}$$ Then it is possible to compute $R,\alpha$ : $$R = \sqrt{a^2 + b^2}, \alpha=\tan^{-1}{\frac{b}{a}}$$ In ...

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This is a standard thing: \begin{align} & \phantom{{}={}}3\sin(\pi x) + 2\sqrt{3}\cos(\pi x) = \sqrt{\Big(3\Big)^2+\Big(3\sqrt{3}\Big)^2}\Big(\frac{3}{\bullet}\sin(\pi x)+\frac{3\sqrt{3}}{\bullet}\cos(\pi x)\Big) \\[10pt] & = 6\Big(\frac{3}{6}\sin(\pi x)+\frac{3\sqrt{3}}{6}\cos(\pi x)\Big) = 6\Big( \frac 1 2 \sin(\pi x)+\frac{\sqrt{3}}{2}\cos(\pi ... 0 Hint: f(x)=3\sin(πx)+3\sqrt{3}\cos(πx). What is A\sin(Kx+D)? It is A\sin(Kx+D)=A(\sin (Kx) \cos (D) + \cos (Kx) \sin (D)). Let A=1. Then, let K=\pi, then \sin (D)=3\sqrt{3}, \cos (D)=3, i.e., \tan(D)=\frac{1}{\sqrt{3}}. Can you find D now? 1 Hint: \quad\sin(a+b)=\sin a\cos b+\sin b\cos a,\qquad\dfrac12=\sin t,\qquad\dfrac{\sqrt3}2=\cos t,\qquad t=?\quad :-) 4 In this example, every two maxima have a unique distance, but the average distance stays constant:f(x)=\cos(g_\epsilon (x)), \qquad g_\epsilon (x)=x+\epsilon \sin(\sqrt{2}\cdot x), \qquad \epsilon=0.4$$Edit: To understand the idea, first draw the unperturbed inner function g_0(x)=x on a sheet of paper. Whenever this diagonal line crosses the ... 1 The instantaneous period (what you call the "trough spacing") at x is the reciprocal of the cyclic frequency at x. The frequency of y = \cos( f(x) ) is defined to be f'(x)/(2\pi), so all you need to do is choose f(x) such that f'(x) is not a constant. A common example would be$$ y = \cos( \pi \gamma x^2), $$such that f'(x) = 2\pi \gamma x. ... 1 For the function f:(0,\infty) \rightarrow \Bbb R ,f(x)=sin(1/x) the successive minimas and maximas keep on going far apart. 0 You can't really speak about rotating a function. What you're describing sounds like you're rotating the graph of a function. This will sometimes but not always result in a geometrical figure that is the graph of another function. (For example consider the function f(x)=0 whose graph is the x axis. If you rotate that by 90° around the origin you get the ... 4 First of all, you must parametrize your function. Let t the parameter. Then:$$ \displaystyle\left[\begin{array}{c}x \\ y\end{array} \right] = \left[\begin{array}{c}t \\ \cos(t)\end{array} \right]$$Now, consider to rotate the function of an angle \theta. Then you obtain the followings:$$ \displaystyle\left[\begin{array}{c}x' \\ y'\end{array} \right] ...

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Sorry but this is not the answer but too long for a comment: Probably the easiest verification is to type the equation on Google you'l be surprised : The easiest way is to Google :2 ...

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Hint Going, from definition, to natural logarithms, you have $$f(x)=\sqrt{\log_x2 - \log_2x}=\sqrt{\frac{\log (2)}{\log (x)}-\frac{\log (x)}{\log (2)}}$$ I am sure you can take from here.

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First $x$ should be positive and $x\ne1$. Moreoer $$\log_x2-\log_2x=\frac{\log2}{\log x}-\frac{\log x}{\log 2}=\frac{\log^22-\log^2 x}{\log 2\log x}=\frac{(\log2-\log x)(\log2+\log x)}{\log 2\log x}\ge0\\\iff x\in [0,\frac12)\cup(1,2]$$ so the domain of definition is $[0,\frac12)\cup(1,2]$.

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If you have an equation of the form $y = A \sin( \frac{2 \pi}{p}(x + b))$ then, it will have an amplitude of A, a period of length $p$, and be translated to the left by $b$. Therefore, for your wave, notice it has amplitude 1.5, and must be $B$ or $C$. Then, notice it has period 4, so, we can re-write $\frac{\pi x}{2} = \frac{2 \pi x}{4}$, and we see that ...

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You should look at it as f1=s f2 =1/(s+1)^2 f3=1/(0.1s+1) then tou graph each by itself and because it is linear you just attach the lines. You will be able to see where is the zero hope this helps. Nachum

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$x^2(x^2-1)=0\Rightarrow \text{either } x^2=0 \text{ or } x^2-1=0$

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I assume you know what the graph of $$y=e^x$$ looks like. The graph of $$y=b^x$$ is similar, only the slope of the tangent line at $x=0$ is $\log b$ instead of 1. The graph of $$y=b^{x-h}$$ is similar to that of $y=b^x$, only pushed $h$ units to the right. The graph of $$y=ab^{x-h}$$ is similar to that of $y=b^{x-h}$, just that the distances to the ...

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Just play a little with logarithms and you will quickly arrive to $$y=a e^{(x-h) \\log (b)}+k$$ which shows you a classical exponential function which will not make any problem to plot or evaluate (provided that $b$ be a real positive number).

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We have $f(4)=f(6)=1$, so it's not injective.

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