Is any way to get coordinates from Vector in maple? For example if I would like have function f(V) = sin(V[0]) + cos(V[1]) + V[2]
Where V = (x,y,z). Is it possible in maple?
In Maple a Vector starts its indexing from 1 (not from 0). So the first entry is V[1], rather than V[0].
Also, a Vector can be constructed with either the Vector command or its angle-bracket shortcut notation. Round brackets do not construct a Vector, as they are delimiters for either grouping or arguments of function application.
restart:
f := V -> sin(V[1]) + cos(V[2]) + V[3]:
W := Vector([a,b,c]):
f(W);
sin(a) + cos(b) + c
Y := <3,7,11>:
f(Y);
sin(3) + cos(7) + 11
An Array is more flexible and can be constructed so as to start its indexing from 0.
g := V -> sin(V[0]) + cos(V[1]) + V[2]:
W := Array(0..2,[q,r,s]):
g(W);
sin(q) + cos(r) + s
Note that the LinearAlgebra package deals with Matrix and Vector. Also, some arithmetic operations (such as .) act in an elementwise manner for Array and not a way you might expect for doing computational linear algebra.
restart:
F := Array(1..3,[q,r,s]):
F . F;
[ 2 2 2]
[q , r , s ]
U := Vector[row]([q,r,s]):
U . U;
_ _ _
q q + r r + s s
Note the complex conjugation occuring in the last example. And in contrast with Matrix structures, 2-dimensional Arrays are also multiplied elementwise under the . operator.
Related
I am trying to reduce number of iterations required to calculate multiplication using the CORDIC algorithm because I am using this algorithm in a continuous function to calculate square function. Here is the algorithm assuming -1<x<1'
function z=square(x)
y=x;
z=0;
for i=1:15
if (x > 0)
x = x - 2^(-i);
z = z + y*2^(-i);
else
x = x + 2^(-i);
z = z - y*2^(-i);
end
end
return
end
I already know the close value to multiplication result (from the previous result (call it pr)) and value of x (the value of x is continuous) . Does it help in anyway to decrease number of iterations?
If you are multiplying twice by the same constant, say a.x and a.x', then you can multiply and add with the delta a.(x' - x), which has less significant digits.
In case both factors vary, you can still use
x'.y' = (x'- x).(y' - y) + x.(y' - y) + (x' - x).y + x.y
where maybe the first term is neglectible.
For a square,
x'² = (x'- x)² + 2.x.(x' - x) + x²
I am working on a Matlab project and I want to make the gradient of the following function in Matlab:
f(x) = c^T * x - sum (log(bi - (ai ^ T) * x)).
Where ai^T are the rows of a random A matrix nxm , where n=2, and m=20
c is random matrix nx1, and x is also random nx1.
b is random matrix mx1.
I've done the following but the results i get don't seem to be right..
function gc0 = gc(x, c, b, A)
for k = 1 : length(A(:,1))
f1(k) = sum(log(b - A(k,:)'*x(k)));
end
gradient(-f1)
gc0 = c - gradient(f1)';
Any ideas? I'd appreciate your help, I'm newbie in Matlab..
It seems that your loop contains a mistake. Looking at the formula above,
I think that the function evaluation should be
f1 = c'*x;
for k = 1 : length(A(1,:))
f1 = f1 - log(b(k) - A(:,k)'*x)
end
A shorter and faster notation for this in Matlab is
f = c'*x - sum(log(b - A' * x)) ;
The function 'gradient' does not calculate the gradient that I think you
want: it returns the differences of matrix entries, and your function f
is a scalar.
Instead, I suggest calculating the derivatives symbolically:
Gradf = c' + sum( A'./(b - A' * x) );
This question is connected to this one. Suppose again the following code:
syms x
f = 1/(x^2+4*x+9)
Now taylor allows the function f to be expanded about infinity:
ts = taylor(f,x,inf,'Order',100)
But the following code
c = coeffs(ts)
produces errors, because the series does not contain positive powers of x (it contains negative powers of x).
In such a case, what code should be used?
Since the Taylor Expansion around infinity was likely performed with the substitution y = 1/x and expanded around 0, I would explicitly make that substitution to make the power positive for use on coeffs:
syms x y
f = 1/(x^2+4x+9);
ts = taylor(f,x,inf,'Order',100);
[c,ty] = coeffs(subs(ts,x,1/y),y);
tx = subs(ty,y,1/x);
The output from taylor is not a multivariate polynomial, so coeffs won't work in this case. One thing you can try is using collect (you may get the same or similar result from using simplify):
syms x
f = 1/(x^2 + 4*x + 9);
ts = series(f,x,Inf,'Order',5) % 4-th order Puiseux series of f about 0
c = collect(ts)
which returns
ts =
1/x^2 - 4/x^3 + 7/x^4 + 8/x^5 - 95/x^6
c =
(x^4 - 4*x^3 + 7*x^2 + 8*x - 95)/x^6
Then you can use numden to extract the numerator and denominator from either c or ts:
[n,d] = numden(ts)
which returns the following polynomials:
n =
x^4 - 4*x^3 + 7*x^2 + 8*x - 95
d =
x^6
coeffs can then be used on the numerator. You may find other functions listed here helpful as well.
In Maple I have a polar expression
and I need to convert it to a Cartesian expression. Though the convert function of Maple doesn't seem to have an option for this.
I currently have a by-hand conversion:
Though there has to be a proper automatic version, isn't there?
Thanks!
The general mechansisms are,
G:=a+b*I;
G := a + I b
H:=convert(G,polar);
H := polar(|a + I b|, argument(a + I b))
evalc(H);
a + I b
So, for your example with given operator specifying the modulus,
r := (theta,a,epsilon) -> a*(1-epsilon^2)/(1+epsilon*cos(theta)):
evalc( polar( r(theta,a,epsilon), theta ) );
/ 2 \ / 2 \
a \-epsilon + 1/ cos(theta) I a \-epsilon + 1/ sin(theta)
---------------------------- + ------------------------------
1 + epsilon cos(theta) 1 + epsilon cos(theta)
I would like to create a function that finds the parameters p and q of Bass diffusion model, given the data of two time periods.
The model (equation) is the following:
n(T) = p*m + (q-p)*n(T-1) + q/m*n(T-1)^2
where
n(T) = number of addoptions occuring in period T
n(T-1) = number of cumulative adoptions that occured before T
p = coefficient of innovation
q = coefficient of imitation
m = number of eventual adopters
for example if m = 3.000.000
and the data for the years below is the following:
2000: n(T) = 820, n(T-1) = 0
2005: n(T) = 25000, n(T-1) = 18000
then the following equation system has to be solved (in order to determine the values of p and q):
p*m + (q-p)*0 + q/3.000.000 * 0^2 == 820
p*m + (q-p)*18000 + q/3.000.000 * 18000^2 == 25000
By following Matlab documentation I tried to create a function Bass:
function F = Bass(m, p, q, cummulativeAdoptersBefore)
F = [p*m + (q-p)*cummulativeAdoptersBefore(1) + q/m*cummulativeAdoptersBefore(1).^2;
p*m + (q-p)*cummulativeAdoptersBefore(2) + q/m*cummulativeAdoptersBefore(2).^2];
end
Which should be used in fsolve(#Bass,x0,options) but in this case m, p, q, cummulativeAdoptersBefore(1), and cummulativeAdoptersBefore(2) should be given in x0 and all variables would be considered as unknown instead of just the latter two.
Does anyone know how to solve the system of equations such as above?
Thank you!
fsolve() seeks to minimize the function you supply as argument. Thus, you have to change your equations to
p*m + (q-p)*0 + q/3.000.000 * 0^2 - 820 == 0
p*m + (q-p)*18000 + q/3.000.000 * 18000^2 - 25000 == 0
and in Matlab syntax
function F = Bass(m, p, q, cumulativeAdoptersBefore, cumulativeAdoptersAfter)
F = [p*m + (q-p)*cumulativeAdoptersBefore(1) ...
+ q/m *cumulativeAdoptersBefore(1).^2
- cumulativeAdoptersAfter(1);
p*m + (q-p)*cumulativeAdoptersBefore(2) ...
+ q/m *cumulativeAdoptersBefore(2).^2
- cumulativeAdoptersAfter(2)];
end
Note: There is a typo in your Bass function (multiplication instead of sum).
Now you have a function, which takes more parameters than there are unkowns.
One option is to create an anonymous function, which only takes the unknowns as arguments and to fix the other parameters via a closure.
To fit the unkowns p and q, you could use something like
cumulativeAdoptersBefore = [0, 1800];
cumulativeAdoptersAfter = [820, 25000];
m = 3e6;
x = [0, 0]; %# Probably, this is no good starting guess.
xopt = fsolve(#(x) Bass(m, x(1), x(2), cumulativeAdoptersBefore, cumulativeAdoptersAfter), x0);
So fsolve() sees a function taking only a single argument (a vector with two elements) and it also returns a vector value.