assigning leads to empty value
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I construct some functions depending on w. But if I assign some value w (of type double) with nonnegativ y df_v returns just a empty value [ ].
func = parabolicCylinderD(x(w),y));
df = diff(func)
....
f_v = func(w);
df_v = df(w);
2 个评论
chicken vector
2023-5-15
You need to share the code of your function otherwise it's impossible for us to help you.
Sef
2023-5-15
回答（2 个）
The function fe containts a function called whittakerM which is not explicit and it can't be derived.
The problem originates from these two expression included in df:
diff(whittakerM(E.^2./2.0-M.^2./2.0+1.0./4.0,-1.0./4.0,k),E)
diff(whittakerM(E.^2./2.0-M.^2./2.0+1.0./4.0,1.0./4.0,k),E)
When you substitute values for E, M and k, the diff operation becomes numerical but is performed on only one double thus returning an empty value:
diff([1 2])
diff(1)
In general you would have two options: either you derive a close form of whittakerM so you can perform analytical derivation, or you opt for numerical methods.
In your case you can't do neither of these because, as you can see from the expresions above, your variable E is also the degree of the derivative that has to be computed [diff(f,n) computes the n-th derivative].
This means that your function is defined for non-negative integers only and it is not continuous, thus the derivative has no mathematical meaning.
7 个评论
Sef
2023-5-15
Now that you edited the question is even more clear that you can't use Newton-Raphson with analytical derivatives to solve this.
Your variable w is an index and is impossible to perform the derivative over an index.
If you riformulate your problem by eliminating w and using directly x as variable you can't still perform analytical derivatives for the same reason, but you can do it numerically.
In the following I used central derivatives, but you can use more accuate schemes or use matlab built-in method to solve your problem such as solve or fsolve.
f = @(x,y) parabolicCylinderD(x,y);
step = 5e-2;
X = 0:step:6;
dX = X(2:end) - step/2;
Y = 1:5;
M = length(Y);
N = length(X);
Z = zeros(M,N);
dZ = zeros(M,N-1);
for m = 1 : M
for n = 1 : N
Z(m,n) = f(X(n),Y(m));
end
dZ(m,:) = diff(Z(m,:)) / step;
end
figure;
tiledlayout(1,2);
nexttile;
plot(X,Z);
grid on;
title('Function');
legend(cellstr(compose('y = %d',1:M)),'Location','EastOutside');
nexttile;
plot(dX,dZ);
grid on;
title('Derivative');
function [ Y1 ] = ypsilon1(a,z)
Y1 = exp(-0.25*z*z)*hypergeom(0.5*a+0.25,0.5,0.5*z*z);
end
function [ Y2 ] = ypsilon2(a,z)
Y2 = z*exp(-0.25*z*z)*hypergeom(0.5*a+0.75,1.5,0.5*z*z);
end
function [ U ] = Ufunction(a,z)
zeta = 0.5*a+0.25;
U = 1/(sqrt(pi)*2^(zeta))*(cos(pi*zeta)*gamma(0.5-zeta)*ypsilon1(a,z)-sqrt(2)*sin(pi*zeta)*gamma(1-zeta)*ypsilon2(a,z));
end
function [ D ] = parabolicCylinderD(v,z)
b = -v-0.5;
D = Ufunction(b,z);
end
Sef
2023-5-24
Sef
2023-5-30
chicken vector
2023-5-30
Attack what x and y are for you, because if x is a vector, you can just loop over each element of x and check if the result is zero.
If you don't share any data I still have the feeling your problem should be formulated differently and I can't help you more than this.
Sef
2023-6-5
Sef
2023-7-6
Walter Roberson
2023-5-15
0 个投票
The basic problem is that there are two important functions named diff.
The primary diff() function calculates x(2:end) - x(1:end-1) which is numeric differences.
In the special case that the first parameter to diff() is symbolic or symfun then you instead get symbolic derivative (calculus).
Your code is working purely numerically, so diff() is numeric differences.
Note that numeric diff() does not do a numeric estimate of derivative: you need gradient() for that.
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