# Tag Info

0

This is too long for a comment but I suppose it could be of some interest. You could have obtained $y'$ much faster considering the function $$F=x \cos(x y)+y-4=0$$ and used the implicit function theorem $$F'_x=\cos (x y)-x y \sin (x y)$$ $$F'_y=1-x^2 \sin (x y)$$ $$y'=\frac{dy}{dx}=-\frac{F'_x}{F'_y}=\frac{x y \sin (x y)-\cos (x y)}{1-x^2 \sin (x y)}$$ ...

1

They contribute by their pointwise product, though it is not easy to guess, as their graphs are cut above $0.4$. First look at the dot at $(0.8,0.4)$. Take a vertical dive (with a ruler, a sheet of paper), you will see that the corresponding point on the second dashed curve ($u^3/6$) has coordinates $(0.8,0.1)$. On the black curve, $u=0.8$ corresponds ...

2

For example, look at where the dotted lines intersect, the height appears to be .3 for both. Thus, $$.3\cdot .3 = .09$$ which is about the height of the solid line. Similarly as you examine farther out.

1

I think it's possible to get a pretty good idea of what this looks like intuitively. First off we know that $x^2e^x\ge 0$ and that it hits the point $(0,0)$. We also know that $e^x$ dominates $x^2$, so as $x$ becomes large, the shape of the curve will pretty much look like $e^x$, but for smaller values of $x$, the function will grow a bit faster. The ...

1

I think there's good reason to believe the answer is "no." The product rule itself is fairly complicated; you can barely tell the derivative "at a glance" by looking at the algebra and I'm not aware of any more compact geometric interpretation. You'd be able to infer a parabola $x^2$ from a line $x$, which sounds absurd. Not sure how someone might trump ...

3

The √ symbol is used to denote the principal square root of a number, i.e. the positive one.

5

Actually, the function $f(x)=\sqrt{x}$ usually refers to the principal square root which is defined as the nonnegative solution to $r^2=x$ for $x\ge0$. While it's true that in general there are two solutions to the equation $r^2=x$, we really like single valued functions.

8

By definition $\sqrt{x}$ indicates the positive (or principal) square root of $x$. If we want indicate all the two root we have to write explicitly $\pm\sqrt{x}$.

1

To stick with your notation: $(a,d)\ \{a,d\}$ $(d,f)\ \{a,d,f\}$ $(a,b)\ \{a,b,d,f\}$ $(b,e)\ \{a,b,d,e,f\}$ $(c,e)\ \{a,b,c,d,e,f\}$ $(e,g)\ \{a,b,c,d,e,f,g\}$ Making the total cost $5+6+7+7+5+9=39$. You went wrong in the third step, where you added $(f,e)$ instead of $(a,b)$. NB: Draw it yourself, check that you understand each step and verify that ...

1

If you first draw the complete tree from the matrix then using Prim's algorithm you just add the egde with the lowest value to the minimum spanning tree and continue doing so until all vertices are connected to the minimum spanning tree (of course you should'nt add any edge if it doesn't add another vertice to the tree). I personally think that it is a lot ...

2

Explaining David G. Stork's result, \begin{aligned} I &=\int_{\sqrt a}^\infty e^{-t^2+\beta t} \sin(\beta t) \, dt \\ &= \Im \, \int_{\sqrt a}^\infty e^{-t^2+\beta t + i\beta t} \, dt \\ &= \Im \left[ e^{(\beta + i\beta)^2/4} \int_{\sqrt a - \beta(1+i)/2}^\infty e^{-u^2} \, du \right] \\ &= \Im \left[ \frac{\sqrt\pi}{2} \, e^{i\beta^2/2} ... 2 Possible solution: Start by completing the square in the exponential part:-t^2 + \beta t = -\left(t - \frac{\beta}{2}\right)^2 + \frac{\beta^2}{4}$$So you have for the moment:$$I(t) = \int_{\sqrt{\alpha}}^{+\infty} e^{-\left(t - \frac{\beta}{2}\right)^2 + \frac{\beta}{4}} \sin(\beta t)\ \text{d}t$$Now operate the shift$$t - \frac{\beta}{2} = ...

0

Assume $M=\overline{M}$ If $d(x,M)=0$ then $x\in \overline{M}$. So $x \in M$. Now if $x\in M$. $d(x,M)=inf\{ |x-w|; w \in M|\} \leq |x-x|=0$ so d(x,M)=0

2

If it's only real numbers you're working with, then no. If you're familiar with complex logarithms then $\log(z)=\log|z|+\mathrm{i}\theta$ might help, where $\theta$ is the argument of $z$.

0

Let's look at $r=1-\sin\theta$ the curve we're looking for is obtained by a dilation of a factor $3$. We first notice that $r(\pi-\theta)=r(\theta)$ and therefore the curve is invariant by a y-axis reflection. We can limit our study to the interval $[-\pi/2,\pi/2]$ Next we note that $r(-\pi/2)=2$, $r(0)=1$ and $r(\pi/2)=0$ and therefore the curve passes ...

1

The engineer's approach The simplest approach (for graphing) is the following $$x(\theta)=r(\theta)\cos(\theta)=3(1-\sin(\theta))\cos(\theta)$$ and $$y(\theta)=r(\theta)\sin(\theta)=3(1-\sin(\theta))\sin(\theta).$$ Now, list the values of $\theta$ from $0$ to $2\pi$, say, in steps of $0.05$ and put marks at the points determined by $(x(\theta),y(\theta)).$ ...

1

The cardioid cannot be represented as graph of an implicit function. However if we consider only the region in the first quadrant, we can represent the graph as the graph of an imlplicit function we will have $sin\theta = {y\over \sqrt{x^2+y^2}}$ and so you'll get $$\sqrt{x^2+y^2} = 3 - 3{y\over \sqrt{x^2+y^2}}$$ i.e. $${x^2 + y^2\over 3} + y = \sqrt {x^2 + ... 0 That's a good question, but the answer is not so satisfying. In the situation of an inverse function, y = f^{-1}(x) if and only if x = f(y). So, naturally, you're switching the x and y when passing from the graph of f to the graph of f^{-1}. As you point out, this is a reflection across the line y=x, which is a particularly nice type of ... 1 Yes you check the condition on x and use the formula that applies. By the way, your function is not well defined in general since it doesn't specify what happens if x=0, but maybe you want (-\infty,0)\cup (0,\infty) as your domain. Assuming you didn't want to include zero$$ f(x)=\left\{\begin{array}{lcl} 2x+3, &\mathrm{ if } & x <0 \\ & ...

1

The derivative has to be zero at 3 points, and at one of them, it has to have a double zero to get that saddle point. This is a total of 4 zeros for the derivative, meaning it has to be a fourth order polynomial where the zeros are at known locations, meaning I can write it as: $$f'(x) = (x-x_1)(x-x_2)(x-x_3)^2,$$ where $x_1$ is the max point, $x_2$ is ...

1

WLOG let $Y = [0,1]$. Let $X = [a, a+1]$. The over lap of $X$ and $Y$ is of length $1/2$, so $X\cap Y$ is some interval of length $1/2$. Now say $a > 0$. Then $a < 1$ other wise there would be no intersection. So $a+1 > 1$ and the overlap is $[a,1]$. The only way that can have length $1/2$ is if $a = 1/2$. Likewise you get only one possibility when ...

0

The function have not asymptote when $x \to +\infty$. The only asymptote is $y=0$ when $x \to -\infty$. $\displaystyle \lim_{x \to -\infty} x^{-3}e^{\frac{x^3}{3}} = \lim_{x \to -\infty} x^{-3} \times \lim_{x \to -\infty} e^{\frac{x^3}{3}}=0 \times 0 =0$. The formula $\lim_{x \to \infty}{\frac{f(x)}{x}} = k \in R$ is good also for $k=0$ but if directly ...

2

Here is a plot using the (arbitrary precision) calculator Pari/GP. I use 200 dec digits precision as default in my computations and got this plot without oscillation up to x=20: I tried it so far up to x=256; no oscillation. See here the image up to x=128 (just to have the left increase visible)

8

32bit vs. 64bit affects which integer types are used by default, which is of no interest here. Rather, the floting point computations are made (by default) with IEEE double type. With this double precision (53 bit mantissa), the relative error of $(1+\frac1{x^{16}})$ is approximately $2^{-53}$. Raising to the $x^{16}$th power roughly multiplies the relative ...

7

Your problem is the finite precision of floating-point arithmetic. There are only so many numbers near $1$ that can be represented by the computer's floating-point format, and the larger your $x$ is, the more of the difference between $1$ and $1+\frac{1}{x^{16}}$ (which is what really matters when raising to a huge power) will be lost to rounding of the ...

0

Because your blue line is cos(x) and your green line is sin(x). This reminds me of someone who forgot once that geodesics are taken at unit speed and spent the next two months trying to find the error in the equations.

1

Because the height of these opposite sides equals the sine of the angles, OK, $\sin\alpha = y / 1 = y$ for one but $\cos\alpha = x / 1 = x$ for the other opposite site. these can be mapped onto a sine graph (x-axis is the angles in degrees, y-axis is opposite side height), OK. $F = (\alpha, y(\alpha)) = (\alpha, \sin(\alpha))$ and should ...

3

You are treating the height as a function of the $x$ position of the base of that vertical leg. But $\sin$ is a function of the angle. Or alternatively, a function of how much circumference has been traced out. It's not (directly) a function of the $x$ position of that vertical segment.

18

It's not standard to answer a question with an image, but I think the image says more than 1000 words in this case: The point is that what you are drawing on the x axis is the angle, not the length of one of the sides of the triangle. The angle is proportional to the length of the circle section. Image Source. Credit for the image goes to Lucas V. ...

1

It seems that to generate a continuous surface you will have to use either mesh or surf, but both of them only work for grid data points. So what you can do is to firstly interpolate and/or extrapolate your (x,y,v) data points such that they form a grid. griddata might be a good function to try first. Here is just an example I made up: % first, make up some ...

0

I don't really know the background on why looking for Fourier transform for a non-periodic function, but it seems the result you got is what you can really get. On a separate note, I always find that using anonymous functions in Matlab makes these things a lot easier. With your example: % original function f=@(t) exp(-t) % approximation function ...

0

Assuming you know the limit $$\lim_{t\to+\infty}\frac{\ln(t)}{t}=0$$ rerwrite $x(\ln(x))^2$ as $(\sqrt{x}\ln(x))^2$ and then $\left(-2\frac{\ln\left(\frac{1}{\sqrt{x}}\right)}{\frac{1}{\sqrt{x}}}\right)^2$. Now take $t=\frac{1}{\sqrt{x}}$. Note that the function $x\to x^2$ is continious on $\mathbb{R}$ and since $x\to 0^+$ so we have $t\to+\infty$.

0

Hints: Re-write $x$ln$(x)^2$ as $\frac{x}{\frac{1}{lnx^2}}$ and apply L'Hopital's Rule. Re-write $x^x$ as $e^{xln(x)}$ and solve. Alternatively, you could do as you said (graph y= the expressions with a graphing calculator), and find the y value as x approaches 0 for both problems.

3

$f''(x)=0$ is not a necessary condition. Following two conditions must be met for inflection point to exist: 1) $f(x)$ must be continuous 2) Concavity must change, that means sign of $f''(x)$ must change In this case $f(x)$ is continuous at $x = 0$ $f''(x) = -\dfrac{2}{9x^{5/3}}$, which changes sign at $x=0$ So $(0,0)$ is indeed an inflection point ...

0

I have found the error ;-) I misread a variable $X$ in [1, p. 46] as being the standardized binomial distribution, but it must be the standardized Bernoulli distribution. Thus we have $$P(\tilde B_n\le x) = \Phi(x) + \frac 1{6\sqrt n} k_3 (1-x^2) \phi(x) + \frac{R\left(np+x\sqrt{npq}\right)}{\sqrt{npq}}\phi(x) + O\left(\frac 1n\right)$$ Because ...

2

Start with $f(x) = ax^{3} + bx^{2} + cx + d$. Since you want this polynomial to have critical points at $x = \pm 1$, we require that $f'(\pm 1) = 0$. This yields the two equations \begin{align*} 3a + 2b + c & = 0\\ 3a - 2b + c & = 0. \end{align*} It is then obvious that $b = 0$ and $c = -3a$. From here, one can obtain another two equations from the ...

1

In order for $f(x, y) = \exp(y \log x)$ to be well-defined you have to fix a branch of $\log$ on the real axis. A standard choice is $$\log x = \begin{cases} \ln x & x > 0, \\ \ln(-x) + i\pi & x < 0. \end{cases}$$ With this definition, $$f(x, y) = \exp(y \log x) = \begin{cases} x^{y} & x > 0, \\ (-x)^{y} \exp(i\pi y) & x ... 0 no. look at the graph of y = x^{2/3}. this has a cusp at (0,0) but concave down on (-\infty, \infty) and (0,0) is certainly not a point of inflection. 1 There is no quicker way, I think, than graphing the curves as accurately as possible, finding the points of intersection around the area enclosed...and see the limits /extreme point positions for x,y. 0 You can just reflect \sin(x) in the line y=x thus: web.geogebra.org/?command=Reflect[sin(x),y=x] 1 You can probably do it easily in Mathematica and turn it into an interactive demonstration that can be played with the free CDF player. For examples, see the Wolfram Demonstrations Project, which includes source code for many examples. The example below should get you started: Graph and Contour Plots of Functions of Two Variables These pages should help ... 3 We have$$f(x,y)=z=\ln(x-y)$$For some level curve where z=k, we then have$$k=\ln(x-y)$$Here, there are only two variables, x and y. It is now possible to write x as a function of y, and vice versa. You should have a set of two-dimensional functions. All you have to do is graph them. I won't give you the answer in that case, but I can give you a ... 0 A useful way to think about the multiplication of complex numbers is that if some complex number z is multiplied by a complex number w then we may consider their product zw to be a rotation and a dilation (stretching) of the original complex number z. In this case if we first have a complex number z = e^{i \theta} then this is simply the equation ... 0 Your curve lacks the symmetry of the sigmoidal curve you have referenced. You could try fitting y=Ae^{-k_1x}+B to the first three hours' worth of data and y=C-De^{k_2x} to the last 3.2 hours, then combining the two. This won't give a \sinh-like curve because k_1 \ne k_2. 1 Programmatically, you'd run through a nested loop for x = -10 to 10 do for y = -10 to 10 do z[x,y] = f(x,y) Then comes the harder part: Convert this z-array into a perspective display with shading and occlusion and all ... 0 I'm going assume that you mean symmetric along a fixed x value. What you can do to construct a parabola which does not have that is to start with one which has one and then apply a rotation of all points in the plane. Basically exchanging the old x and y coordinates with new ones according to the rotation:$$x_{old} = \cos(\phi)x_{new} + ...

3

You can stitch a Frankenbola together like this. $$f(x) = \begin{cases} a_l x^2 + b_l x + c_l & \text{for } x < 0 \\ a_r x^2 + b_r x + c_r & \text{for } x > 0 \\ c & \text{for } x = 0 \end{cases}$$ You can require continuity for $f$ then you get $$f(x) = \begin{cases} a_l x^2 + b_l x + c & \text{for } x < 0 \\ a_r x^2 + b_r x + c ... 0 You can construct a function whose parts are all parabolas but the resulting graph will not be a parabola. For example:$$f(x)= \begin{cases} x^{2} &\text{if}\,\,x\le 0\\ 2x^{2} &\text{if}\,\, x> 0 \end{cases}$$In a more general way, you can define$$p(x)=\sum_{i}\left(\mathbb{I}_{[a_{i},b_{i})}(x)\right)(\alpha_{i} x^{2}+\beta_{i}) where ...

1

If you need something skew, looking roughly like a parabola, you could use higher polynomials: y(x) = $x^4 + 2x^3 + 3x^2$, or with other coefficients.

1

I like @Joker123's suggestion of visualizing complex maps as vector fields, but I want to point out that this is not what is done in the tool you linked at davidbau.com. The plot at davidbau.com is drawing the preimage of a flat grid and unit circle under the input map. If you try the identity map ($z$), you'll see that flat grid and unit circle. If you ...

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