New answers tagged

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How about plotting a histogram of the A,B,C,D values for all objects, then for each object show where it lies in the histogram, something like https://commons.wikimedia.org/wiki/File:Distribution_of_Annual_Household_Income_in_the_United_States_2010.png


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One way to use transformations to graph $y=\dfrac{1}{\vert x\vert +1}$ is, because of the absolute value, to break it into two parts. Case I: $x\ge0$ $$ y=\dfrac{1}{x+1} $$ which is the graph of $y=\dfrac{1}{x}$ shifted left by $1$ but we only want the portion of that shifted graph for which $x\ge0$. Case II: $x<0$ $$ y=\dfrac{1}{-x+1}=-\dfrac{1}{x-1}...


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Not terribly sophisticated, but something like the following may be a nice visual aid when comparing the change of any two candidates: https://public.tableau.com/s/sites/default/files/rank11.png


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The only piece of information that you need to distinguish between the points $(1,3)$ and $(3,1)$ is a component in the direction of $u_1$. In particular: the centroid of our data is $\mu = (2,2)$ (or right around there, anyway). We can deduce whether a signal $s = (s_1,s_2)$ should be $(1,3)$ or $(3,1)$ by stating that a signal is $(3,1)$ if $(s - \mu)^...


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My students used to have trouble with this idea too, and it came down to this: you don't add the x-values of the point, you just add the y-values. That's really all you need to know, but if you want to think more deeply, this goes back to how functions are designed. All that the x-value tells us is where we are looking on the graph. The x-axis itself ...


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This is how points on elliptic curves are added. You can google "addition on elliptic curves" to get many explanations. This may also help: http://www.math.vt.edu/people/brown/class_homepages/elliptic_curve_addition.pdf


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To find the equation of f(x,y)=0 after reflection about $x=y$ we just swap $x,y$ positions to get f(y,x) =0 in the same relative arrangement/setting of the equation variables.


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Let $c\in(0,2]$. The equation $f(x,y)=c$ is: $$\sqrt{4-x^2-y^2}=c.$$ Squaring both sides: $$4-x^2-y^2=c^2.$$ Rearranging $$x^2+y^2=4-c^2.$$ This is the circle of radius $\sqrt{4-c^2}$ centred at the origin.


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There is a pretty natural relationship with physics, specifically with the notion of changing reference frames. If you add a symmetry, say the length of the lines, then Einstein Relativity is all about this. Step into a valid reference frame and take the world-line of a rigid body, call this the receiving function. The world line of any other particle in ...


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Another option is to assume the change between points is less than $\pm 180^\circ$ and add or subtract multiples of $360^\circ$ to each point to make that true. It means your plotted values may go outside the range $0-360$, but the variation from point to point will be represented correctly. You could give that a try.


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We have this: $$y=0.04x^{1.7}$$ Exchanging $x$s and $y$s: $$x=0.04y^{1.7}$$ Multiplying each side by $25$ and using logarithms: $$\ln25x=1.7\ln y$$ Dividing by $1.7$ and simplifying: $$e^{\ln25x/1.7}=y$$ Using this logarithm property: $$(25x)^\frac{1}{1.7}=y$$ And we're finished


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Let's call $f(x)=0.04x^{1.7}$. The function is one-to-one and therefore has an inverse. To find the equation of the inverse, solve the equation $$x=0.04y^{1.7}$$ for $y$. I will leave that part up to you ☺


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It can be an onto and hence an inverse function but it would depend on the Domain and Range of the function. The given function will be invertible if the domain is $\mathbb{R}-\{1\}$ and the range is $\mathbb{R}-\{0\}$. This is just one possible choice, there can be several others.


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Let us go with $y = \alpha \log (x+\beta)$. Let us say that you want the axis intersections to be $(a, 0)$ and $(0, b)$. Then: $0 = \alpha \log(a + \beta) \Rightarrow a + \beta = 1 \Rightarrow \beta = 1-a $. Similarly: $b = \alpha \log \beta \Rightarrow \alpha = \frac{b}{\log (1-a)}$. So, for your logarithmic function to intersect the axes at $(a,0)$ ...


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It's an "order of operations" issue, definitely, like Zachary Selk mentioned. To add some detail to it: in mathematics, $-1$ is a number to itself. That is, the - is just a way to remind us that this number is negative; however, computers (well, expression parsers) don't see it that way. It will treat exponentiation first, and then negate that result, ...


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The problem I encountered was twofold: 1) $f(x)=-1^x$, won't show oscillatory motion because $=-1^x=-(1^x)-(1)=-1$ 2) $f(x)=(-1)^x$, won't plot because if $\{x|x \in \mathbb R\} \land x=a/2b$ where $b \in \mathbb N$ "$x$ is in the domain of all real numbers and $x=a/$some even number" where $a \neq 2b \lor a \notin \mathbb Q \lor a=2n+1$ where $...


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What you wrote according to order of operations is $-1^x=-1$. You meant $(-1)^x$ probably. See: http://m.wolframalpha.com/input/?i=%28-1%29%5Ex&x=0&y=0


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You can use the value at t=8 to calculate the value of r. $7.58=48\cdot e^{-r\cdot 8}$ Soving for r. $\frac{7.58}{48}=e^{-r\cdot 8}$ $ln\left(\frac{7.58}{48} \right)=-r\cdot 8$ $\frac{ln\left(\frac{7.58}{48} \right)}{8}=-r$ $r=-\frac{ln\left(\frac{7.58}{48} \right)}{8}$ $r=\frac{ln\left(\frac{48}{7.58} \right)}{8}\approx 0.230711$ I checked it for ...


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OK, so I think the point of this problem is to get you to understand some different kinds of behaviour that drug concentrations might have in the body. In fact these are also important forms of behaviour for all sorts of situations involving variables changing in time, such as population growth, nuclear decay, disease spread, and so on. Let's look at the ...


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To visualize the pdf of $|X-Y|$ you can draw a picture. After reading off the pdf it is not difficult to calculate the expected value. First the equation $|X-Y|\leq z$ has to be solved. For this purpose two cases have to be regarded. $a) \ X-Y\geq 0$ The absolute value signs can be dropped-without any manipulation. The inequality becomes $X-Y\leq z \...


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If you mean a "straight line", then it's because a function with such a graph is always linear (so, not non-linear). Linear functions are precisely those whose graphs are (nonvertical) lines. I'm assuming we're talking about real-valued functions of a single real variable (functions $\mathbb R\to\mathbb R$).


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You want $\int_0^1\int_0^1 |x-y| d(x,y)=\int_0^1(\int_0^1(|x-y|dx)dy=\int_0^1(\frac{y^2+(1-y)^2}{2})dy=\int_0^1(y^2-y+\frac{1}{2})=\frac{1}{3}$


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Take a unit square in the $xy$-plane of 3-space. Draw the diagonal from $(0,0)$ to $(1,1)$. Now above the point $(1,0)$ draw a point at height $z =1$; above the point $(0, 1)$ draw another point at height one. Connect each of these two points to the diagonal line you drew in the plane, forming two triangles. Those triangles are the graph of your function. ...


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If you want to plot the data below in a linear fashion, you could write $$\frac{C_v} T=A + B\, T^2$$ So define $y=\frac{C_v} T$ and $x=T^2$ to have $y=A+Bx$. But you must take care that the results would not be the same fitting the first and second models (except in the very ideal case where the data would be perfectly exact). For the second model, this ...


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A simple way a computer can make a graph of a function $f$ on an interval $[a,b]$ is to divide $[a,b]$ up in $n$ ascending points $x_1,\dots x_n$ with $x_1=a$ and $x_n=b$ and then draw straight lines between $\left(x_i,f(x_i)\right)$ and $\left(x_{i+1},f(x_{i+1})\right)$. The number $n$ and the ways the partition $x_1,\dots x_n$ is chosen depends on the ...


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The question you ask is a standard linear regression problem. Your function $C_v=AT+BT^3$ is indeed not linear in $T$, but it is linear in $A$ and $B$, which is all that matters. A transformation that maps your data to a straight line exists, but you will not find it without knowing $A$ and $B$, and you do not need to find it to learn the values of $A$ and $...


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Unsophisticated graphing software evaluates the function at a number of points, and then draws straight lines between the points it obtained. There is typically not one function evaluation per pixel, and even if there is, the odds of hitting the exact point where your function is undefined are quite low.


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The limit is identical from the left and from the right, so there is only an infinitely small point at $x=2$ where there is nothing to plot. But since you don't have infinite precision in your plot, it "looks" as though the plot just goes "right through" as you say.


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That graph is wrong. There should be a hole at $x=2$, as you correctly guessed. The correct plot are two lines with arrows at the ends where they touch the point $(2,5)$.


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The built in functions you refer to are using linear regression and least squares. This pdf gives a decent overview. For your case if you construct a matrix X = [1 9; 1 2;1 9; 1 5; 1 8; 1 7; 1 5; 1 10; 1 4; 1 4] and Y = [9;6;81;28;65;53;27;100;21;17] then b = (X'X)^-1*X'Y=[-15.9017; 8.9844] The first number is b, the second your m.


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There is a typo. $$b=\frac{\displaystyle\sum _{i=1} ^ny_i-m\sum_{i=1}^nx_i}{n}$$ Are you able to compute $b$ now?


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The result you're thinking of does not work for $\ln x$ since it takes nonreal values at $x<0$. However, note that $\mathfrak{R}(\ln x)$ is even on $\mathbb{R}\setminus\{0\}$ (where $\mathfrak{R}z$ is the real part of $z$)


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From geography, we know that Variation in daylight time: at equator, and latitudes near equator, is minimum. The daytime is approximately constant throughout the year. So most likely, none of the graphs correspond to this. is maximum near poles, so most likely, plot C corresponds to a place near arctic circle (not near antarctic circle, since graph C is ...


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Showing the limits at infinity is straightforward. Simply consider the related rational function $$g(x) = \frac{3-x}{1+x^2}$$ which, by definition of the floor function, satisfies $$f(x) \le g(x) < f(x) + 1$$ for all $x$, and for which $$\lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} \frac{\frac{3}{x^2} - \frac{1}{x}}{\frac{1}{x^2} + 1} = 0,$$ and ...


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OK, I think I got it... Prerequisites I will be calling the graphs by letters A-D. (i.e. A corresponds to Graph A). I will be calling the cardinal directions N, E, S, W. (i.e. N corresponds to north). Solution A is given to be N. D is A flipped horizontally. It is fairly obvious that D is S. The rest of the solution is where I was stumbling. ...


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Here's a general answer: rotation of the graph of a function $f$ around a point $(a,b)$ is but the symmetry of the graph w.r.t. this point, hence it is $$y=g(x)\stackrel{\text{def}}{=}2b-f(2a-x).$$ To see this, denote $x,y=f(x)$ a point on the original graph, and $(X,Y=g(X)$ athe symmetric point. By definition, we have $X=2a-x$, $Y=2b-y$ (writing the ...


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The graph of the function is shown below: It is clear that if we rotate the part $\{(x,f(x)):x \geq 1\}$ we get the portion of the graph $\{(x,f(x)): x \leq 1\}$. Hence $g = f$. This can also be seen as follows: If we rotate the point $P(x,f(x))$ on the curve by $180^\circ$ about $(1,-1)$, then we obtain the point $(2-x, -2-f(x))$. Note that $f(2-x)+f(x) = ...


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We can do a linear least square best fit if you know (or if we assume) the order of the polynomial. If the fit is not "good enough", we can increase the order. Here's how it works. Suppose we have the points $(x_1, y_1), ..., (x_n, y_n)$ and we want to interpolate it approximately with $y = a_m x^m + ... + a_0$. Then we want to solve: $$\mathbf y \approx \...


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Contour line is defined as $\{ (x,y) | f(x,y) = constant \}$. So for $e^{x+y}=C$, you get $x+y = \ln C$, which is a straight line. The answer is correct. Another strange example would be $(15x+6y)^{100}$.


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Here's a "modular" explanation, in the hopes that you can easily modify (if necessary) it to suit your needs. Let $S$ (for speed) be a function satisfying $0 \leq S(t) \leq 1$ for all real $t$; $S(t) = 0$ for $t \leq 0$; $S(t) = 1$ for $1 \leq t$. The "cubic interpolation" is probably smooth enough to suit your needs: $$ S(t) = 10t^{3} - 15t^{4} + 6t^{5} = ...


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Like continued fractions, if it's not periodic (or some sort of nice pattern), you have to do the innermost first. See my trial using Mathematica below:



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