# Tag Info

3

Probably the easiest way is to choose a third point $(a,b),\ \ b\in(0,100)$, what your curve also contains, and then construct a quadratic function to fit these $3$ points, using \begin{aligned} p_0(x) &:=\frac{(x-a)(x-100)}{(-a)(-100)} \\ p_a(x) &:=\frac{x(x-100)}{a(a-100)} \\ p_{100}(x) &:=\frac{x(x-a)}{100(100-a)} \end{aligned} These ...

2

The point at $0^0$ does not exist, however the points very near it and the limit $x^x$ as $x\to 0$ do exist. As for the behavior of the graph in the negative real numbers, $x^x=(-a)^{-a}$ for some positive $a$ in the real numbers. By Euler's identity, $e^{\pi i}+1=0$, so we can write $-a$ as $e^{\pi i}a$, and $(e^{\pi i}a)^{-a}=e^{-a\pi i}a^{-a}$. Since ...

2

To me, y = sqrt(x) * a + c seems to be very good in particular for large values of x (above x=25 the fit is excellent). To me, the real question is : what do you want to do ? If it is interpolation, I think that Lagrange polynomials (as suggested by Mark) would be the best idea since there is no much noise in your data. If it is extrapolation for values such ...

2

A simple and accurate solution could be : $$f(x):=\sqrt{432718+184284\;x}$$ \begin{array} {c|cc} x&y&f(x)\\ \hline\\ 1& 782&785\\ 2& 893&895\\ 3& 992&993\\ 4& 1081&1082\\ 5& 1164&1164\\ 6& 1241&1240\\ 7& 1313&1313\\ 8& 1382&1381\\ 9& 1447&1446\\ 10& 1510&1508\\ 20& ...

2

The $xy$-plane is also known as $\Bbb{R}^2$, and it forms a vector space over $\Bbb{R}$. So the operation $\lambda (a,b) = (\lambda a, \lambda b)$ is valid by definition of a vector space. Look up the definition of vector space to see. Now if $(a,b)$ is a coordinate of a graph of a function $y = f(x), \ f : \Bbb{R} \to \Bbb{R}$, where by graph you mean ...

2

I hope the following plot helps you. I made it by Maple environment using below codes: [> with(plots): a := ln(3); complexplot(2*exp(-a*t)*exp((1/2)*(2*I)*Pi*t+(1/2)*Pi), t = 0 .. 4);

1

I'd recommend Excel as a good way to experiment with curve fitting. Use your point data to make a chart. Then right-click on the graph, and choose "Add Trendline". A dialog appears that lets you fit several different types of curves to your data. By default, the displayed equation will show the coefficients accurate to 4 decimal places, which may not be ...

1

Plotting 3D things on 2D paper is problematic. Orbits are nice because they are 2D (in the right coordinate system, and over short periods-the orbit plane changes). It depends on what you want to show. You can: plot in the orbit plane. A nice representation of the orbit. Probably the most useful single plot. plot projected on the equatorial plane. ...

1

This is IMO 2006 Problem 5 A solution to it is this: Suppose $a$ is a solution to $Q(x) = x$. Let $a_0 = a$ and $P^m(a) = a_m$. Then we know $(a_z - a_{z-1})|(a_{z+1}-a_z)$ for all $z \ge 1$. In particular, $(a_k - a_{k-1})|(a_1 - a_0)$ so we quickly deduce that $P$ has order $2$ on $a$ or order $1$. This is because firstly we have $$\prod_{i=1}^k ... 1 You can see that the domain is \{x\in \mathbb{R} : x\geq 0 \}. To find the max/min (which eventually gets you the range), we find out the first derivative. f'(x)=\frac{4-x}{2\sqrt(x)(x+4)^2}. Here the critical point is x=4. Also f''(x)=-\frac{1}{4x^{3/2}}-\frac{4-x}{x^{1/2}(x+4)^2}, then f''(4)=-\frac{1}{32}. Thus the function has maximum at ... 1 Try not to think of a rational function ( f(x)/g(x) ) as being something completely different. The process is the same except that now the threat of a vertical asymptote is more imminent. Your process should be as follows: Find intercepts (where x=0 or y=0) Check for vertical asymptotes (where the denominator, or in your case g(x) is zero) Check for ... 1 Your first question is solved in a way similar to your previous question. We need to find all x such that h is defined. In other words, f(|x|) should be defined. So, we just need to look for values of x such that |x| evaluates to one of -3,-1,0,2. Can you do this? For an example to get you started, the solutions to |x|=7 is x=7 and x=-7. ... 1 You are right: this formula defines a triangular wave. It's certainly nothing new, though. An even shorter formula for the same function can be given using the concept of distance from a point to a set:$$f(x)=\operatorname{dist}(x,4\mathbb Z) But the above is not likely to be understood by a computer, if you want to use it for graphing. In that case, ...

1

You can think of a local minima as being a point on the graph where you can draw an arbitrarily small circle around the point and every point in the graph that is in the circle will be larger than the proposed minima. In the case where a minima is an end point, all that we examine is the side of the graph inside the interval we care about. The main idea here ...

1

Because you are asked to use a graph you should start by plotting $f(t)$ in mathematica, matlab, maple etc. If you don't have access to any of this software take a look at wolframalpha. Then, the time it is increasing most rapidly should be where the slope is greatest. You are right, you will need to find an inflection point which is the deravitave of the ...

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