I want to draw this equation either in Matlab or R (Matlab is preferred) :
f = p+(1-p)*(T-S)
where 0 < S < 1, 0 < p < 1 and T is a constant. I want to draw the function and find the min, max based on S and p. My basic problem is defining the span of the graph as a symbol. Since S changes from 0 to T.
Use ezsurf to plot. For example:
f = 'p + (1 - p) * (5 - S)'
ezsurf(f, [0 5 0 1])
Then use regular calculus to find critical values, double differentiate to find their type, and so on...
This is all explained in the online documentation (diff, solve, etc.). Also, this external example covers all of the points you want very nicely: http://msemac.redwoods.edu/~darnold/math50c/matlab/maxmin/index.xhtml
Related
Hello
The graph for y=sin(x*pi/5)*2*sin(x*pi4) looks like this:
Some y=0 are 4,5,8,10,12,15,16
Is there a general technique (imagine the equation having more components) - that can give us a new equation where the integers that satisfy y(x)=0 will be an arbitrary nonzero number, and the integers with a nonzero numbers will be set to zero?
In python I can easily find the nonzero with this code:
#for 100 first integers
for i < 100
if y != 0:
print x
1 += 1
But I wonder if this can be done mathematically - preferably as a trigonometric function.
Background
I read here that newton method fails on function x^(1/3) when it's inital step is 1. I am tring to test it in julia jupyter notebook.
I want to print a plot of function x^(1/3)
then I want to run code
f = x->x^(1/3)
D(f) = x->ForwardDiff.derivative(f, float(x))
x = find_zero((f, D(f)),1, Roots.Newton(),verbose=true)
Problem:
How to print chart of function x^(1/3) in range eg.(-1,1)
I tried
f = x->x^(1/3)
plot(f,-1,1)
I got
I changed code to
f = x->(x+0im)^(1/3)
plot(f,-1,1)
I got
I want my plot to look like a plot of x^(1/3) in google
However I can not print more than a half of it
That's because x^(1/3) does not always return a real (as in numbers) result or the real cube root of x. For negative numbers, the exponentiation function with some powers like (1/3 or 1.254 and I suppose all non-integers) will return a Complex. For type-stability requirements in Julia, this operation applied to a negative Real gives a DomainError. This behavior is also noted in Frequently Asked Questions section of Julia manual.
julia> (-1)^(1/3)
ERROR: DomainError with -1.0:
Exponentiation yielding a complex result requires a complex argument.
Replace x^y with (x+0im)^y, Complex(x)^y, or similar.
julia> Complex(-1)^(1/3)
0.5 + 0.8660254037844386im
Note that The behavior of returning a complex number for exponentiation of negative values is not really different than, say, MATLAB's behavior
>>> (-1)^(1/3)
ans =
0.5000 + 0.8660i
What you want, however, is to plot the real cube root.
You can go with
plot(x -> x < 0 ? -(-x)^(1//3) : x^(1//3), -1, 1)
to enforce real cube root or use the built-in cbrt function for that instead.
plot(cbrt, -1, 1)
It also has an alias ∛.
plot(∛, -1, 1)
F(x) is an odd function, you just use [0 1] as input variable.
The plot on [-1 0] is deducted as follow
The code is below
import numpy as np
import matplotlib.pyplot as plt
# Function f
f = lambda x: x**(1/3)
fig, ax = plt.subplots()
x1 = np.linspace(0, 1, num = 100)
x2 = np.linspace(-1, 0, num = 100)
ax.plot(x1, f(x1))
ax.plot(x2, -f(x1[::-1]))
ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
plt.show()
Plot
That Google plot makes no sense to me. For x > 0 it's ok, but for negative values of x the correct result is complex, and the Google plot appears to be showing the negative of the absolute value, which is strange.
Below you can see the output from Matlab, which is less fussy about types than Julia. As you can see it does not agree with your plot.
From the plot you can see that positive x values give a real-valued answer, while negative x give a complex-valued answer. The reason Julia errors for negative inputs, is that they are very concerned with type stability. Having the output type of a function depend on the input value would cause a type instability, which harms performance. This is less of a concern for Matlab or Python, etc.
If you want a plot similar the above in Julia, you can define your function like this:
f = x -> sign(x) * abs(complex(x)^(1/3))
Edit: Actually, a better and faster version is
f = x -> sign(x) * abs(x)^(1/3)
Yeah, it looks awkward, but that's because you want a really strange plot, which imho makes no sense for the function x^(1/3).
I have two points which form one line: (1,4) and (3,6), and another two which form another line: (2,1) and (4,2). These lines are continuous and I can find their intersection points by finding the equation for each line, and then equating them to find the x value at the intersection point, and then the y value.
i.e. for the first line, the equation is y = x + 3, and the second is y = 0.5x. At the intersection the y values are the same so x + 3 = 0.5x. So x = -6. Subbing this back into either of the equations gives a y value of -3.
From those steps, I now know that the intersection point is (-6,-3). The problem is I need to do the same steps in Excel, preferably as one formula. Can anyone give me some advice on how I would start this?
Its long but here it is:
Define x1,y1 and x2,y2 for the 1st line and x3,y3 and x4,y4 for the second.
x = (x2y1-x1y2)(x4-x3)-(x4y3-x3y4)(x2-x1) / [ (x2-x1)(y4-y3) - (x4-x3)(y2-y1) ]
y = (x2y1-x1y2)(y4-y3)-(x4y3-x3y4)(y2-y1) / [ (x2-x1)(y4-y3) - (x4-x3)(y2-y1) ]
Note that the denominators are the same. They will be ZERO! when the system has no solution. So you may want to check that in another cell and conditionally compute the answer.
Essentially, this formula is derived by solving a system of equations for x and y by hand using generic points (x1,y1), (x2,y2), (x3,y3), and (x4,y4). Easier yet, is solving the system by hand using well developed linear algebra concepts.
Wikipedia outlines this procedure well: Line-line intersection.
Also, this website describes all the different formulas and lets you put in whatever data you have in any mixed format and provides many details of the solutions: Everything about 2 lines.
Here's a matrix based solution:
x - y = -3
0.5*x - y = 0
Written as a matrix equation (I apologize for the poor typesetting):
| 1.0 -1.0 |{ x } { -3 }
| 0.5 -1.0 |{ y } = { 0 }
You can invert this matrix or use LU decomposition to solve it to get the answer. That method will work for any number of cases where you have one equation for each unknown.
This is easy to do by hand:
Subtract the second equation from the first: 0.5*x = -3
Divide both sides by 0.5: x = -6
Substitute this result into the other equation: y = 0.5*x = -3
I have a scalar function which is obtained by iterative calculations. I wish to differentiate(find the directional derivative) of the values with respect to a matrix elementwise. How should I employ the finite difference approximation in this case. Does diff or gradient help in this case. Note that I only want numerical derivatives.
The typical code that I would work on is:
n=4;
for i=1:n
for x(i)=-2:0.04:4;
for y(i)=-2:0.04:4;
A(:,:,i)=[sin(x(i)), cos(y(i));2sin(x(i)),sin(x(i)+y(i)).^2];
B(:,:,i)=[sin(x(i)), cos(x(i));3sin(y(i)),cos(x(i))];
R(:,:,i)=horzcat(A(:,:,i),B(:,:,i));
L(i)=det(B(:,:,i)'*A(:,:,i)B)(:,:,i));
%how to find gradient of L with respect to x(i), y(i)
grad_L=tr((diff(L)/diff(R)')*(gradient(R))
endfor;
endfor;
endfor;
I know that the last part for grad_L would syntax error saying the dimensions don't match. How do I proceed to solve this. Note that gradient or directional derivative of a scalar functionf of a matrix variable X is given by nabla(f)=trace((partial f/patial(x_{ij})*X_dot where x_{ij} denotes elements of matrix and X_dot denotes gradient of the matrix X
Both your code and explanation are very confusing. You're using an iteration of n = 4, but you don't do anything with your inputs or outputs, and you overwrite everything. So I will ignore the n aspect for now since you don't seem to be making any use of it. Furthermore you have many syntactical mistakes which look more like maths or pseudocode, rather than any attempt to write valid Matlab / Octave.
But, essentially, you seem to be asking, "I have a function which for each (x,y) coordinate on a 2D grid, it calculates a scalar output L(x,y)", where the calculation leading to L involves multiplying two matrices and then getting their determinant. Here's how to produce such an array L:
X = -2 : 0.04 : 4;
Y = -2 : 0.04 : 4;
X_indices = 1 : length(X);
Y_indices = 1 : length(Y);
for Ind_x = X_indices
for Ind_y = Y_indices
x = X(Ind_x); y = Y(Ind_y);
A = [sin(x), cos(y); 2 * sin(x), sin(x+y)^2];
B = [sin(x), cos(x); 3 * sin(y), cos(x) ];
L(Ind_x, Ind_y) = det (B.' * A * B);
end
end
You then want to obtain the gradient of L, which, of course, is a vector output. Now, to obtain this, ignoring the maths you mentioned for a second, if you're basically trying to use the gradient function correctly, then you just use it directly onto L, and specify the grid X Y used for it to specify the spacings between the different elements in L, and collect its output as a two-element array, so that you capture both the x and y vector-components of the gradient:
[gLx, gLy] = gradient(L, X, Y);
I have been given some work to do with the fractal visualisation of the Mandelbrot set.
I'm not looking for a complete solution (naturally), I'm asking for help with regard to the orbits of complex numbers.
Say I have a given Complex number derived from a point on the complex plane. I now need to iterate over its orbit sequence and plot points according to whether the orbits increase by orders of magnitude or not.
How do I gather the orbits of a complex number? Any guidance is much appreciated (links etc). Any pointers on Math functions needed to test the orbit sequence e.g. Math.pow()
I'm using Java but that's not particularly relevant here.
Thanks again,
Alex
When you display the Mandelbrot set, you simply translate the real and imaginaty planes into x and y coordinates, respectively.
So, for example the complex number 4.5 + 0.27i translates into x = 4.5, y = 0.27.
The Mandelbrot set is all points where the equation Z = Z² + C never reaches a value where |Z| >= 2, but in practice you include all points where the value doesn't exceed 2 within a specific number of iterations, for example 1000. To get the colorful renderings that you usually see of the set, you assign different colors to points outside the set depending on how fast they reach the limit.
As it's complex numbers, the equation is actually Zr + Zi = (Zr + Zi)² + Cr + Ci. You would divide that into two equations, one for the real plane and one for the imaginary plane, and then it's just plain algebra. C is the coordinate of the point that you want to test, and the initial value of Z is zero.
Here's an image from my multi-threaded Mandelbrot generator :)
Actually the Mandelbrot set is the set of complex numbers for which the iteration converges.
So the only points in the Mandelbrot set are that big boring colour in the middle. and all of the pretty colours you see are doing nothing more than representing the rate at which points near the boundary (but the wrong side) spin off to infinity.
In mathspeak,
M = {c in C : lim (k -> inf) z_k = 0 } where z_0 = c, z_(k+1) = z_k^2 + c
ie pick any complex number c. Now to determine whether it is in the set, repeatedly iterate it z_0 = c, z_(k+1) = z_k^2 + c, and z_k will approach either zero or infinity. If its limit (as k tends to infinity) is zero, then it is in. Otherwise not.
It is possible to prove that once |z_k| > 2, it is not going to converge. This is a good exercise in optimisation: IIRC |Z_k|^2 > 2 is sufficient... either way, squaring up will save you the expensive sqrt() function.
Wolfram Mathworld has a nice site talking about the Mandelbrot set.
A Complex class will be most helpful.
Maybe an example like this will stimulate some thought. I wouldn't recommend using an Applet.
You have to know how to do add, subtract, multiply, divide, and power operations with complex numbers, in addition to functions like sine, cosine, exponential, etc. If you don't know those, I'd start there.
The book that I was taught from was Ruel V. Churchill "Complex Variables".
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m}d/r{rlineto}d/X -2 q 1{d/Y -2 q 2{d/A 0 d/B 0 d 64 -1 1{/f exch d/B
A/A z sub X add d B 2 m m Y add d z add 4 gt{exit}if/f 64 d}for f 64 div
setgray X Y moveto 0 q neg u 0 0 q u 0 r r r r fill/Y}for/X}for showpage