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0

You could achieve simultaneous circles in two of the three planes but not in all three. E.g. $$ x = \cos t \\ y = \sin t \\ z = \cos t $$ which is a circle in the $x$,$y$-plane and the $y$, $z$-plane, but not in the $x$, $z$ plane.


0

Consider $y=a(x+4)^2-2$. Let $Y=y+2$ and $X=x+4$. Then $Y=aX^2$. This means that the curve $y=a(x+4)^2-2$ is a translation of the curve $Y=aX^2$: the origin $(0,0)$ in $XY$-space goes to the point $(-4,-2)$ in $xy$-space. Thus, the vertex of $y=a(x+4)^2-2$ is at $(-4,-2)$, the symmetry axis is $x=-4$, etc.


0

For k=2 squaring, $$ r^2 = x^2 + y^2 = \sin^2 (2 \theta ) = 4 \,\sin^2 \theta \cos^2 \theta = 4 \,x^2 \, y^2 / (x^2 + y^2) $$. For integral k such conversion using De Moivre expansion is possible . For non-integral k an infinite series results. The domain is restricted due to circular trig functions.


0

Calculate x and y in a table separately for $$ x = (1/2 + \sin \theta) \cos \theta, $$ $$ y = (1/2 + \sin \theta)\sin \theta. $$ You notice that not always we have a single $r$ for a given $ \theta$ for the entire domain. In this case double angle $\theta $ gives rise to another $r$ after further rotation. Likewise $ r = \sin 2\theta $ gives rise to four ...


1

Here's a few sufficient conditions for uniform continuity that you should be able to recognize just by looking at a graph: The function continuous and defined on a compact set. The function is everywhere differentiable, and the derivative is bounded. The function is periodic. You can draw a two-sided cone (like an hourglass) of some fixed opening angle. If ...


0

I don't know about necessary and sufficient conditions, but if a function is differentiable with bounded derivative, or even just Lipschitz continuous, then it will be uniformly continuous. You can often see this type of thing by looking at the graph.


1

If $f: [0,+\infty) \to \Bbb R$ is uniformly continuous, then there exists $A,B > 0 $ such that $|f(x)| \leq Ax + B $. In other words, the graph of an uniformly continuous function defined on the positive real numbers always stays under some line. For example, $x^2$ defined in $[0,+\infty)$ is not uniformly continuous, just by seeing it. Be aware of the ...


2

There are two things that could go wrong with your reasoning: If $f$ is defined for $x < -10$ or for $x > 13$, you have no idea how it behaves. You can't be sure that there is exactly one value of $x$ near $4$ so that $f(x)=15$. It could be the case that there is none, or one, or two. This depends on whether the value of the local maximum of $f(x)$ ...


1

I don't think that it is obvious that there is a real number $\alpha$ such that $3\lt \alpha\lt 5$ and $f(\alpha)=15$.


0

This type of map is called a Mobius map, and it is helpful with these to find out where the unit disk goes. Plug in $z=e^{i\theta}$ and see what it simplifies to. The image of the unit disk should split the complex plane into two regions. Then see where the origin goes, and you will know which region $|z|<1$ maps to. The other region is where $|z|>1$ ...


0

It's not clear why you want to do this without a computer. But the traditional route would involve finding certain key points, trends, gradients and such like. For example, what happens as z tends towards 1? If it approaches from above it tends to negative infinity. If it approaches from below it tends to positive infinity. What happens as z tends ...


4

Of course the equation $$\max\bigl\{|x|, \>|y|\bigr\}=a$$ describes the circumference $\partial Q$ of a square simply and accurately. But you have the feeling that $\partial Q$ is also the "limit" of the sets $$S_p:=\bigl\{(x,y)\>\bigm|\>|x|^p+|y|^p=a^p\bigr\}$$ when $p\to\infty$. There are several notions for limits of sequences of sets. For the ...


5

Probably the best way to think about what is going on is to think about the $p$ norms $\| x \|_p = (|x_1|^p+|x_2|^p)^{1/p}$ for $p \in [1,\infty)$, where we define $\| x \|_\infty = \max \{ |x_1|,|x_2| \}$. The motivation for the latter notation is that $\lim_{p \to \infty} \| x \|_p = \| x \|_\infty$. Here's how to see it: First suppose $|x_1|=|x_2|=x$. ...


15

Fix $a > 0$. In each diagram the exponent is even; if you plot $$ x^{2k+1} + y^{2k+1} = a^{2k+1} $$ for some non-negative integer $k$, you'll get a substantially different picture (except in the first quadrant). So, let's interpret the diagrams as special cases of $$ |x|^{p} + |y|^{p} = a^{p},\quad \text{$p > 0$ real,} $$ i.e. $$ \left(|x|^{p} + ...


0

Suppose $0 \le x <a$ you have $$y(x)=\pm a \sqrt[p]{1-\left(\frac{x}{a}\right)^p}$$ And $$\lim\limits_{p \to \infty} y(x) = \pm a$$ This prove that for $x \in [0, a)$ the graph of the curve converges (pointwise) to two lines $y = \pm a$.


1

You want the Gudermannian function: it is defined by $$ \operatorname{gd}{x} = \int_0^x \operatorname{sech}{y} \, dy $$ In particular, this can be expressed, by changing variables, as $$ \operatorname{gd}{x} = (\operatorname{sgn}{x}) \arccos{(\operatorname{sech}{x})} = \arcsin{(\tanh{x})}, $$ so $$ \cos{(\operatorname{gd}{x})} = \operatorname{sech}{x},\\ ...


0

In order the derivative $$f'(x) = \dfrac{(a-1)x^4 + (2a - 3)x^2 + a}{(1 + x^2)^2}$$ be always positive, the numerator must not cancel for any value $x$ and moreover this would imply $a \gt 1$. So, as you did, setting $u=x^2$, you look for the roots of $$(a-1)u^2+(2a - 3)u + a=0$$ The discriminant is then $$\Delta=(2a-3)^3-4a(a-1)=9-8a$$ and it must be ...


0

Why is the first derivative of an extremum zero? The reason is same for the second derivative at inflection point as well... because ... slope is extremized at PI. The first derivative at an inflection point locally is either a maximum or a minimum. The second derivative at an inflection point vanishes. An inflection point is associated with a complex ...


2

Since you tagged "Linear Algebra" in your question, let's generate a coefficient matrix from a system of equations. For example if you have the 4 lines $$ y = 3x + 2 \\ y = 4x - 1 \\ y = 6 \\ y = 2x, $$ first fill in the "missing" parts of your linear form so you explicitly have $y= mx +b$. That is, we have $$ y = 3x + 2 \\ y = 4x - 1 \\ y = 0x + 6 \\ y = ...


1

The first think you need to decide on is what is a spike and what is not. Consider the following graphs, surely there are spikes on left-most one, but what about the right one? If these are spikes, then how would you smooth that graph without increasing the sampling frequency? If these are not spikes, then what about the graphs in the middle, where do you ...


-1

We require $ f \in C^2 $ By taylors theorem: $f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2 +R$ Let $ f'(a)=0$ Now if $f''(a)>0$ then by continuity there is a small ball around $a$ such that $f''(a)$ is positive so $f(x+\epsilon) > f(x) $ and $f(x-\epsilon) > f(x) $.We do the same for $f''(a)<0$. So now $f''(a)=0$ and inflection points ...


1

An inflection point is the point where the concavity changes. At the inflection point, the curve might go from concave down (with a decreasing slope) to concave up(with an increasing slope). A decreasing slope will give you a decreasing derivative and a negative second derivative. An increasing slope will give you an increasing derivative and a ...


9

Short answer: The second derivative at an inflection point may be zero but it also may be undefined. Longer answer: One definition of an inflection point is where the second derivative changes sign (from positive to negative or the reverse). The Intermediate Value Theorem for derivatives says that if a derivative is defined on a closed interval then it ...


4

By definition, inflection points are where a function changes concavity, or in other words where a function is neither concave up nor down but is (often) moving from one to the other. A positive second derivative corresponds to a function being concave up, and a negative corresponds to concave down, so it makes sense that it is when the second derivative is ...


3

Think about what the second derivative means. A positive second derivative means concave up, negative means concave down. Well, an inflection point is when the concavity switches. So naturally the second derivative has to equal zero at some point if our second derivative is going to switch signs. It's very analogous to a critical value.


0

In your example $$y=4-\sqrt{16-\frac{x^2}{16}}$$ $y$ has both a ceiling and a floor. The ceiling is $y\le 4$ since it is found by subtracting a nonnegative number from $4$. Remember, the principal square root is never below zero, so subtracting it from something always keeps it the same or reduces it: it never increases. So $y$ is at most $4$. In your ...


0

Horizontal asymptotes for the graph of a function $f$ occur when $f(x)$ has a finite limit $\ell$ when $x$ tends to $+\infty$ or $-\infty$. Here $\,\lim\limits_{x\to\pm\infty}\dfrac x{x^2+1}=\lim\limits_{x\to\pm\infty}\dfrac x{x^2}=0^+$ or $0^-$. Hence the $x$-axis is a horizontal asymptote to the graph. We even can say the graph is above its asymptote for ...


0

If your function has limits $\lim \limits _{x \to -\infty} f(x) = l_1$ and/or $\lim \limits _{x \to \infty} f(x) = l_2$, then the lines $y=l_1$ and $y=l_2$ are horizontal aymptotes toward $-\infty$ and, $\infty$, respectively. In your case, both limits are $0$ so the $Ox$ axis is a horizontal asymptote towards both $- \infty$ and $\infty$.


0

The first equation is a cubic, $ -(x+1)(x-1) (x-3) = 0, $ with roots $ x = \pm 1, 3 $ at points of intersection.


2

$x(-x^{2}+3x+1)=3\quad$ doesn't imply $x=3$ or $-x^{2}+3x+1=3$. Instead of that we have $$x(-x^{2}+3x+1)=3\iff -x^3+3x^2+x-3=0\iff-(x+1)(x-1)(x-3)=0$$ Therefore the graphs of the functions mets at $x=-1$, $x=1$ and $x=3$.


2

Dividing the two cases: $$ x-2>0 \iff x>2 \Rightarrow |x-2|=x-2 $$ and $$ x-2<0 \iff x<2 \Rightarrow |x-2|=2-x $$ your function become: $$ f(x)= \begin {cases} y=-x-1 \quad , \quad x<2\\ y=0 \quad,\quad x=2\\ y=x+1\quad,\quad x>2 \end{cases} $$ the graph is done by two open half-lines ( decreasing for $x<2$ and crescent for $x>2$) ...


0

Note that $\frac{x}{|x|}=1$ for $x>$ and $\frac{x}{|x|}=-1$ for $x<0$. Thus, $$\bbox[5px,border:2px solid #C0A000]{\frac{(x-2)(x+1)}{|x-2|}=\text{sgn}(x-2)(x+1)}$$ where $\text{sgn}(x-2)=1$ for $x>2$ and $\text{sgn}(x-2)=-1$ for $x<2$. We can define the sign function as $0$ when its argument is $0$. And we're done!


6

Hint: recall (ignoring the $x = 2$ case): $$|x-2| = \begin{cases} x-2, & x-2 > 0 \\ -(x-2), & x-2 < 0 \end{cases} = \begin{cases} x-2, & x > 2 \\ -(x-2), & x < 2\text{.} \end{cases}$$ So $$\dfrac{x^2-x-2}{|x-2|} = \begin{cases} \dfrac{x^2-x-2}{x-2} = \dfrac{(x-2)(x+1)}{x-2} = x+1, & x > 2 \\ \dfrac{x^2 - x - 2}{-(x-2)} = ...


0

Let $f$ be a function and $[a,b]$ be the domain in which you want to plot the function. In most programming languages with type conversion between booleans and numbers you could define the desired function as f_restricted(x) = f(x)*(x >= a && x <= b),since (x >= a && x <= b) will be converted to 0 or 1 if it is evaluated to be ...


1

Rory's suggestion is quite good, and it should work for Desmos. But apparently what you want is to stealthily include domain information; that's what it will boil down to, as we'll see. To this end, you can cook up auxiliary "characteristic functions" that are $1$ at least on the domain you want, and $0$ certain places where you don't want to plot the ...


3

The method of restricting the domain depends on your particular software. Here are some examples. Let's graph $y=x^2$ for $0\le x\le 1$. TI-84 Plus (or similar): $$y1(x)=x^2/(x\ge 0)/(x\le 1)$$ TI-Nspire: $$f1(x)=x^2\mid 0\le x\le 1$$ Geogebra: $$f(x)=\operatorname{If}[0<=x<=1,x^2]$$ Standard math notation: $$y=x^2\mid _{0\le x\le 1}$$ or ...


0

$$ (x-8)(x+5) = x^2-3x-40 $$ because $$ (x+a)(x+b) = x^2+(a+b)x + ab $$


0

$4f(x)=4x^2-12x-160=(2x-3)^2-\text{what?}$


1

There is a quite well known "quadratic formula", that given a polynomial $f(x)=ax^2+bx+c$, the roots of the polynomial are given by $$x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ The roots of the polynomial are the zeros of $f(x)$. Can you continue from here?


2

Only addressing Q#2, italicizing:           The result above was achieved by multiplying by the shear matrix below, with $\phi=30^\circ$: $$ M = \left( \begin{array}{ccc} 1 & \tan (\phi ) \\ 0 & 1 \\ \end{array} \right) $$ For $\phi=60^\circ$:           If you parametrize the sine-curve ...


0

The equations you have in your code assume constant acceleration. So, the plot of acceleration $a$ vs. time $t$ will be a horizontal line at $a$. The plot of velocity $v$ vs. $t$ will be a straight line with slope $a$ and y-intercept of the initial velocity $v_0.$ The plot of displacement $d$ vs. $t$ will be a parabola: $$d(t) = d_0 + v_0t + ...


1

The focal width of the parabola $(x-h)^2=4p(y-k)$ is $|4p|$. If you know the vertex, you must know how to transform it to this standard form $$(x-4)^2=y+34$$ So the focal width is $|4p|=1$.


2

We can represent 3D objects as a function of three variables. Here are some simple, finite geometric shapes, defined implicitly in Cartesian coordinates. The coefficients are used to scale the shape, and can be set to any value. You will end up with tall or flat prisms and pyramids by adjusting them away from their current values. Sphere : $x^2+y^2+z^2 = ...


2

Do you mean that no matter the number of zombies the zombies can convert one villager per second, or that each zombie can convert a villager per second? Let t be the number of seconds that have transpired, V(t) the number of villagers at t seconds, and Z(t) be the number of zombies at t seconds. Note that the number of overall people does not change so in ...


0

One possible curve: $$y=\frac1{a(x+\sqrt{\frac{50}{9a}+\frac{25}4}-\frac52)}-\frac1{a(\sqrt{\frac{50}{9a}+\frac{25}4}+\frac52)},a>0$$ Methodology: As equation $y=\frac1{ax},a>0$ has similar shape with the required curve, we may adopt it as an approximation measure. In the equation, larger the $a$, greater the "curvature". In order to make the ...


2

Since you said "mostly" fits the points, if you make this formal by assuming you have data that is function values with noise, for some underlying function, then you can do linear regression in any free tool like octave or R, where you estimate a polynomial fit of fairly low degree compared to your number of points, using linear regression to estimate the ...


0

A nearly-correct solution would be to use an exponential decay $N = N_{0}e^{-t/\tau}$. To get this to hit zero at $t=5$, you can apply a correction by subtracting the value at $t=5$, $N_{0}e^{-5/\tau}$. A further correction to $N_{0}$ is then required to bring the start point back towards the required point. These corrections are not quite perfect (the ...


0

Interesting question! This falls into the category of SIR(succeptible infected recovered) epidemic models, and is used by scientists to see how fast diseases can spread. For your problem, you want to only consider SI since youre saying you remain a zombie forever and you cant recover or die. This paper goes into some of the details and shows some plots ...


0

Try $y=90 \left(1-\sqrt[n]{1-\left(1-\frac{x}{5}\right)^n}\right)$ for varying $n$ (which does not need to be an integer). The curve is one quarter of a superellipse centered at $(5,90)$. It is tangent to the axes at the given points.


1

Assuming that you can approximate the curve with a polynomial and you have a sufficient amount of points, you can fit a curve into it using any software that solves you a system of equations: say you have points $P = {p_1,p_2, \dots, p_n}$. You construct a $n-1$ degree polynomial $P(x)$, and form $n$ equations in terms of the coefficients of the polynomial. ...



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