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10

I am new to Complex Analysis but this is the impression I have so far. Functions of the form $ f:\mathbb{C} \to \mathbb{C} $ can be viewed as "warping" or "distorting" a Complex Plane. In fact the more "well behaved" Complex Functions are exactly the same as Conformal Mappings of a Plane. Given that, a Polynomial is just a specific way of distorting the ...


5

Well for $x<0$ you get $y=-x^2$ and for $x>0$ you get $y = x^2$. This is similar to the $y = x^3$ curve but not quite the same. You will notice when you plot all three simultaneously. See the plot here: green is $y=x^3$, blue is $y=x^2$ and red is $y=\frac{x^3}{|x|}$


3

If by "relative minimum", you mean "local minimum", then yes, you can have two minimums, since the derivative of the quartic polynomial is of order three and can have three roots. Your particular polynomial has two local minimums and one maximum, as seen on this graph


3

Here's another approach to visualizing the zeroes of a complex function $f(z)$. The idea is to plot contour diagrams of both the real and imaginary parts of $f(x+iy)$. The points of intersection of the zero contours are exactly the roots of $f$. Furthermore, the diagrams tend to look pretty cool as the real and imaginary contours meet at right angles. As ...


2

A period of a function $f$ is $b$ such that for all $x,f(x)=f(x+b)$. Geometrically you can just observe when the function repeats itself. For the second example you said that the period might be 15, which is not a bad guess, but at the point 0, your function proceeds to increase, and at the point 15, your function takes on a whole different value. Maybe you ...


2

Try using Aspect Ratio, Plot[{-2*Sqrt[1 - x^2], 2*Sqrt[1 - x^2], 2*x - 2*Sqrt[2]}, {x, -2, 2}, PlotStyle -> {Blue, Blue, Red}, AspectRatio -> Automatic]


2

I like the program Asymptote. See a tutorial and a gallery of examples. Also, you can see the drawing in Calculating line integrals via Stokes theorem.


2

In Matlab: >> x=linspace(-10,10,5000); >> y=linspace(-10,10,5000); >> t=y.^x-10; >> plot(x,real(t)) >> plot(x,imag(t)) Real: Imaginary: Mathematica 3D plots: Plot3D[Re[y^x - 10], {x, -10, 10}, {y, -10, 10}] Plot3D[Im[y^x - 10], {x, -10, 10}, {y, -10, 10}] Real: Imaginary: 2D Contour plots: ContourPlot[Re[y^x ...


2

$$q(t) = \cases{7.5t & \text{ , if } 0 \le t \le 2\\ 15 & \text{ , if } 2 < t < 4 \\ 15-5(t-4) &\text{ , if } 4 \le t \le 7 \\ 0 &\text{ , otherwise}}$$ You get this by considering the intervals $(-\infty, 0), [0,2], (2,4), [4,7]$ and $(7,\infty)$ individually.


2

A general method is to calculate the instantaneous phase for each of your discrete samples. The Hilbert transform, which can be represented in terms of the Fourier transform and it's inverse, can be used for this. In Matlab, using the hilbert, angle, and unwrap functions: dt = 1e-2; t = 0:dt:20; y1 = sin(t); h1 = hilbert(y1); y2 = sin(t+1); h2 = ...


2

This is done precisely so that we can make the statement that (For $a \neq 0$,) the vertex of the quadratic $$y = a(x - h)^2 + k$$ is simply $(h, k)$. If we instead use the form $$y = a (x + h')^2 + k,$$ the vertex would be at $(-h', k)$, which of contains a minus sign.


2

You don't choose any derivatives. The direction field consists of vectors $A\vec x$ where $\vec x$ ranges over the plane. For example, at $(2,4)$ you draw the vector $$ \begin{pmatrix} -1/2& 1 \\ -1 & -1/2 \end{pmatrix} \begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix} $$ (One usually scales down these vectors; ...


1

(a) To find the slope of the secant line, find the vertical distance between the two points by subtracting $y_2=1$ and $y_1=4$ to get -3. Divide that by the horizontal distance $x_2=2 - x_1=-1$ to get -3/3 or -1. To get the y-intercept, we can start at (-1, 4) and move forward along the x axis to (0, 4-1) = (0, 3). Thus, the equation of the secant line is ...


1

On the first two images, the boundary effects on the left look bad; they should not be there. The third image looks exactly right for the boundary value problem with $$\psi(x,0)=-\sin x,\quad \psi(x,1)=\sin x$$ $$\psi(0,y)=0=\psi(1,y)$$ Indeed, its graph shows anti-symmetry with respect to the horizontal line $y=1/2$: namely, $\psi(x,1-y)=-\psi(x,y)$. The ...


1

$y^x = 10$ is equivalent to $e^{x\ln y}=10$, i.e. $\ln y=\frac{\ln 10}{x}$, or (again) $y=e^{\frac{\ln 10}{x}}$ (in particular, from the very beginning observe that we must have $y > 0$ for the expression to be defined without ambiguity). The latter form can be easily plotted, e.g. via Mathematica.


1

Postal rates (as a function of weight, in ounces): Look at the price of a $1.9$ oz letter, then $1.99$ oz letter, then $2.01$ oz letter, etc.


1

"Relative minimum" means the same thing as "local minimum." A function could have infinitely many local minima. On the other hand, there can be at most one global minimum, though it could be attained at more than one point.


1

Without a turning ratio limitation you can't use the speed to determine the curved distance, so it would be a straight line calculation between points. The variation on the Pythagoras theorem which I believe you refer to is $$d((x_1, y_1),(x_2, y_2)) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 }$$ So to work out the distance for a series of points you just need to ...


1

Various thoughts that might help the sketch: There is a third possibility of $a=0$, though this leads to the simple $f(x)=x$ $f(x)-x \to 0$ as $x \to \infty$ You have $f(e^{-a})=0$ $f'(x)$ is close to $1$ for very small positive $x$ Looking at $f'(x)=0$: there are no real solutions for $\log_e(x)$ when $-4 \lt a \lt 0$, suggesting no local maxima or ...


1

See these: Visual Complex Functions by Wegert. Phase Plots of Complex Functions: A Journey in Illustration by Wegert and Semmler. The Fundamental Theorem of Algebra: A Visual Approach by Velleman. Try an interactive demo at http://www.math.osu.edu/~fowler.291/phase/.


1

I agree that this is, perhaps, directed to math.stackexchange.com. Just for fun and assuming an interpretation of the question: if the aim is to find the number of ways of selecting a triple (with replacement) from {0,0.1,0.2,...,1} that sums to 1. This is equivalent to integer partitioning 10 (just divide by 10). Some approaches to counting: tup = ...


1

It is from the definition. However you may be intereseting in that what it come from or why we called it by this name. In fact: Pick a point $(x,y)$ from the space $\Bbb R^2$, you will find that $x$ is just the projection of the point $(x,y)$, and $y$ is also!



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