# Tag Info

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. ...

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 ...

5

Hint: Each graph forms the boundary of a convex region. The line segment of minimal distance between the two curves must therefore be unique. The picture is symmetric about the line $y = x$, so reflecting the line segment through this line must yield another line segment of minimal length. We can deduce that the slope of the common normal must ...

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.

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 $$f(x) = \begin{cases} a_l x^2 + b_l x + c & \text{for } x < 0 \\ a_r x^2 + b_r x + c ... 3 We have$$f(x,y)=z=\ln(x-y)$$For some level curve where z=k, we then have$$k=\ln(x-y)$$Here, there are only two variables, x and y. It is now possible to write x as a function of y, and vice versa. You should have a set of two-dimensional functions. All you have to do is graph them. I won't give you the answer in that case, but I can give you a ... 3 f''(x)=0 is not a necessary condition. Following two conditions must be met for inflection point to exist: 1) f(x) must be continuous 2) Concavity must change, that means sign of f''(x) must change In this case f(x) is continuous at x = 0 f''(x) = -\dfrac{2}{9x^{5/3}}, which changes sign at x=0 So (0,0) is indeed an inflection point ... 2 Here is a plot using the (arbitrary precision) calculator Pari/GP. I use 200 dec digits precision as default in my computations and got this plot without oscillation up to x=20: I tried it so far up to x=256; no oscillation. See here the image up to x=128 (just to have the left increase visible) 2 Well, \sin has range [-1,1]. So you're applying \sin to something between -1 and 1, so you need to first know how \sin looks like on [-1,1]. It's increasing on this range. So from -\pi/2 to \pi/2, \sin(\sin(x)) is increasing from -\sin(1) to \sin(1). Note that 1 is slightly less than \pi/3, so \sin(1) is a bit less than ... 2 Choose a gridstep s between 0 and \min (|min|, max) (this will be the distance between any two consecutive vertical gridlines). Let m = \lfloor \dfrac {|min|} s \rfloor and M = \lfloor \dfrac {max} s \rfloor, where \lfloor \cdot \rfloor is the floor function (also known as the integer part). Then draw a vertical gridline at each point of the ... 2 Round the low value to the nearest lower multiple of 10 and the high value to the nearest higher multiple of 10. With your example,$$[-107,858]\to\left[10\lfloor\frac{-107}{10}\rfloor,10\lceil\frac{858}{10}\rceil\right]=[-110,860].$$In programming, when the bounds are integer, this can be achieved by means of the \% operator, with$$[(\min-9) \% ...

2

Starting with $$f(x) = (c-\frac{1}{c}-x)(4-3x^2)$$ to make computation a bit more manageable, set $\boxed{\gamma=c-\frac{1}{c}}$, so $$f(x)=4\gamma-4x-3\gamma x^2+3x^3$$ You had the correct idea to set $f'(x)=0$ at the turning points. So $$f'(x)=-4-6\gamma x+9x^2=0\quad(\text{at turning points})$$ By the quadratic formula ...

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 ...

2

Minimum Distance between Two curve is Distance between two parallel tangents drawn at point $P$ and $Q$ on the curves. and Here $f(x)=e^x$ and $f(x)=\ln(x)$ are Inverse of each other . So it is Symmetrical about $y=x$ Line. Let We take any point $P(x_{1},y_{1})$ on $f(x) = \ln(x)\;,$ Then Slope of tangent at $P(x_{1},y_{1})$ to the curve ...

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 $z$.

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 ...

1

I see a black line from $(0,0,0)$ to $(1,1,1)$ The reason it does not appear to be orthogonal to the brown plane is the scale of the vertical axis not matching the scales of the other two axes either in range or in size. Stretch it and it looks better

1

A graphing calculator brought up a pinched square shape, but I just can't understand the logical way to get to this shape. $\qquad\quad$ Geometric shapes described by algebraic equations of the form $|x|^n+|y|^n=r^n$ are called superellipses. For $n=1$, we have a diamond square, determined by four straight line segments. For $n>1$, these $4$ lines ...

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 ...

1

I would like to illustrate my comments in the following figure. I hope that this will explain everything. the dark blue line is the graph of $\color{blue}{\sqrt{(x)}}$ -- note that $\sqrt{(x)}$ is not defined on $(-\infty,0)$. the purple line is the graph of $\color{purple}{g(x)=\sqrt{(-x)}}$ -- note that this function is not defined on $(0,\infty)$. ...

1

Solving graphically would involve sketching the graph o fthe function. But sketches are no solution after all. As $[x]$ and $4$ are integers, we conclude that $2x$ is an integer. Also $x-1<[x]\le x$ make $-x-1<[x]-2x\le -x$, so that either $-4\le -x<-3$ or $4\le -x<5$. These conditions leave us with $x\in\{-4, -3\tfrac12,4,4\tfrac12\}$ to ...

1

In German maths teaching in school, around 10th or 11th year, there is the subject Kurvendiskussion, which should be translated as "Discussion of [the properties of] a Curve". It is a systematic poking of a given function for characteristic properties of its graph. Domain Intersections with $x$- and $y$-axis Symmetries Extrema Inflection points Poles Gaps ...

1

Well, when it comes to graphing any sort of function, a very extensive analysis would be that involving its first and second derivatives. Everything pointed out by mvw is great, though perhaps it could be a bit more explained. Anyways, if you really want to get a intuitive feel of how different graphs look like, I recommend you download a graphing ...

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

You could think of it in this way If the modification is in the form $y=f(x)+a$, then the graph shifts up/down by $a$. If the modification is in the form $y=f(x+a)$, then the graph shifts left/right by $-a$. (Take note of the negative sign here. So for example, if we have $y=f(x-2)$, then the graph shifts to the right by 2 units. If the modification is in ...

1

One way to imagine a map $f : D \rightarrow \mathbb{C}$ with $D \subseteq \mathbb{C}$ is to think of it as a 2-dimensional vector field. Remember that $\mathbb{C}$ is just the vector space $\mathbb{R}^{2}$ equipped with a special multiplication $$* : \ \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ $$(v_1,v_2)*(w_1,w_2) = ... 1 I like @Joker123's suggestion of visualizing complex maps as vector fields, but I want to point out that this is not what is done in the tool you linked at davidbau.com. The plot at davidbau.com is drawing the preimage of a flat grid and unit circle under the input map. If you try the identity map (z), you'll see that flat grid and unit circle. If you ... 1 If you have two points in the plane A(x_a, y_a) and B(x_b, y_b), the distance between them is given by:$$d(A, B)=\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}. Some more detailed explanation you can find for example here: https://www.mathsisfun.com/algebra/distance-2-points.html

1

Something you could try is graph several different examples to see how each one is different. For example, graph $\sin(x)$ then graph $\sin(\sin(x))$. You will see it is the same graph only $\sin(\sin(x))$ has a slightly smaller amplitude and can't reach $1$ or $-1$ in the y axis. If you graph $\cos(x)$ and $\sin(\cos(x))$, $\sin(\cos(x))$ will be the same ...

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