help me to solve for θ?

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Goitom Mekonnen
Goitom Mekonnen 2020 年 1 月 1 日
編集済み: John D'Errico 2020 年 1 月 1 日
(jθ)^3+( jθ)^2*(0.002+e^(j0.05* θ))+ jθ*(e^(j0.05* θ)+0.002*e^(j0.05* θ))+0.002*e^(j0.05* θ)=0

回答 (2 件)

Star Strider
Star Strider 2020 年 1 月 1 日
編集済み: Star Strider 2020 年 1 月 1 日
Use the contour function to find the required coordinates, then extract them and plot the extracted values in a second plot:
fcn = @(th) (1j*th).^3+(1j*th).^2.*(0.002+exp(1j*0.05* th))+ 1j*th.*(exp(1j*0.05* th)+0.002*exp(1j*0.05* th))+0.002*exp(1j*0.05* th);
theta = linspace(-2*pi, 2*pi); % Theta Range
[R,I] = ndgrid(theta); % Real & Imaginary Matrices For Contour (R + j*I)
figure
hr = contour(R, I, real(fcn(R+1j*I)), [0 0]); % Plot Contour Plot At Contour = [0 0]
xlabel('Re(\theta)')
ylabel('Im(\theta)')
axis equal
grid
title('Contour Plot')
idx = find(hr(2,:) > 2*pi); % Find Line Segment Indices
idx = [idx numel(hr(2,:))];
for k = 1:numel(idx)-1
xv{k,:} = hr(1,(idx(k)+1 : idx(k+1)-1)); % Extract X-Coordinates From Contour Pliot
yv{k,:} = hr(2,(idx(k)+1 : idx(k+1)-1)); % Extract Y-Coordinates From Contour Pliot
end
figure
hold on
for k = 1:size(xv,1)
plot(xv{k}, yv{k}) % Plot Extracted Vectors
end
xlabel('Re(\theta)')
ylabel('Im(\theta)')
axis equal
grid
title('Extracted (X,Y) Data From Contour Plot')
xlim([-1 1]*2*pi)
producing the contour plot and:
that plots all the roots between and .
EDIT — (1 Jan 2020 at 20:45)
This slightly changed version may be closer to being correct —
fcn = @(th) (1j*th).^3+(1j*th).^2.*(0.002+exp(1j*0.05* th))+ 1j*th.*(exp(1j*0.05* th)+0.002*exp(1j*0.05* th))+0.002*exp(1j*0.05* th);
theta = linspace(-2*pi, 2*pi); % Theta Range
[R,I] = ndgrid(theta); % Real & Imaginary Matrices For Contour (R + j*I)
figure
hrr = contour(R, I, real(fcn(R+1j*I))+imag(fcn(R+1j*I)), [0 0]); % Plot Contour Plot At Contour = [0 0]
xlabel('Re(\theta)')
ylabel('Im(\theta)')
axis equal
grid
title('Contour Plot')
idx = find(hrr(2,:) > 2*pi); % Find Line Segment Indices
idx = [idx numel(hrr(2,:))];
for k = 1:numel(idx)-1
xvr{k,:} = hrr(1,(idx(k)+1 : idx(k+1)-1)); % Real Part: Extract X-Coordinates From Contour Pliot
yvr{k,:} = hrr(2,(idx(k)+1 : idx(k+1)-1)); % Real Part: Extract Y-Coordinates From Contour Pliot
end
figure
hold on
for k = 1:size(xvr,1)
plot(xvr{k}, yvr{k}) % Plot Extracted Vectors
end
xlabel('Re(\theta)')
ylabel('Im(\theta)')
axis equal
grid
title('Extracted (X,Y) Data From Contour Plot: Real Part')
xlim([-1 1]*2*pi)
It produces a slightly different plot.

John D'Errico
John D'Errico 2020 年 1 月 1 日
編集済み: John D'Errico 2020 年 1 月 1 日
The trick is to use contour plots properly. I'll use the function as written by Star to save me some editting time.
fcn = @(th) (1j*th).^3+(1j*th).^2.*(0.002+exp(1j*0.05* th))+ 1j*th.*(exp(1j*0.05* th)+0.002*exp(1j*0.05* th))+0.002*exp(1j*0.05* th);
fcmp = @(realpart,imagpart) fcn(realpart + 1j*imagpart);
Hr = fcontour(@(realpart,imagpart) real(fcmp(realpart,imagpart)));
Hr.LevelList = 0;
Hr.LineColor = 'r';
hold on
Hi = fcontour(@(realpart,imagpart) imag(fcmp(realpart,imagpart)));
Hi.LevelList = 0;
Hi.LineColor = 'b';
grid
The intersections of the red and blue lines indicate the solutions.
It looks like exactly three solutions in the complex plane, thus one roughly at 0, as well as a pair that is symmetric across the complex axis. Since the contours seem to be hyperbolic in shape, I might conjecture there will be no others.
syms ts
funsym = (1j*ts).^3+(1j*ts).^2.*(0.002+exp(1j*0.05* ts))+ 1j*ts.*(exp(1j*0.05* ts)+0.002*exp(1j*0.05* ts))+0.002*exp(1j*0.05* ts);
vpasolve(funsym,ts,0)
ans =
0.002i
vpasolve(funsym,ts,2)
ans =
0.8384242982243315324357037709473233131717 + 0.4994418650812189178378966738320816417544i
vpasolve(funsym,ts,-2)
ans =
- 0.8384242982243315324357037709473233131717 + 0.4994418650812189178378966738320816417544i
So not exactly zero, but close.
Do analytical solutions exist? It appears not easily obtained, t least solve gives up and just resorts to vpasolve, where it finds the easy solution.
solve(funsym)
Warning: Unable to solve symbolically. Returning a numeric solution using vpasolve.
> In solve (line 304)
ans =
0.002i

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