Predefined variables k, z? Ans Help
显示 更早的评论
Below is my Matlab code that I entered, and the response I got.
>> clear
>> syms x a
>> Func=(.45*cos(x/a)/((sin(x/a))^6))*(5/8*(1-cos(x/a))-(5/48*(1-cos(3*x/a)))+(1/80*(1-cos(5*x/a))))+.09
>>solve(Func,x)
ans =
-a*(2*pi*k + i*log(z))
What in the world are k and z? When I input k I receive an error saying Undefined, z... undefined.
Please help.
Thank you,
Jennifer
回答（3 个）
Roger Stafford
2012-12-8
The expression in your 'Func' is identically equal to
3/100*(6*cos(x/a)^3+18*cos(x/a)^2+17*cos(x/a)+3)/(1+cos(x/a))^3
and therefore c = cos(x/a) is one of the three roots of the cubic equation
6*c^3+18*c^2+17*c+3 = 0.
This gives the six roots of a*acos(c) and -a*acos(c):
a*( 2.6773079709553582 - 1.0064927801893431*i)
a*( 2.6773079709553582 + 1.0064927801893431*i)
a*( 1.7996177555153143)
a*(-2.6773079709553582 + 1.0064927801893431*i)
a*(-2.6773079709553582 - 1.0064927801893431*i)
a*(-1.7996177555153143)
plus any integral multiple of 2*pi*a.
I have no idea what the z represents in the solution you describe.
Roger Stafford
k is a integer constant: Any integer k gives another solution. This is common with periodic functions. z is probably any complex number, but I am guessing here as I do not get this same solution.
Something doesn't look right about that solution. What version of MATLAB are you using?
Walter Roberson
2012-12-7
The "k" probably stands for "an arbitrary integer", but the log(z) has no obvious purpose there, not unless a RootOf() showed up somewhere and you for some reason edited it out.
If we convert the 0.45 and 0.09 to rational values, one version of the analytic solutions are:
(arctan((1/2)*((36+2*322^(1/2))^(2/3)*(24*322^(1/2)-4*(36+2*322^(1/2))^(2/3)-(36+2*322^(1/2))^(4/3)+428+24*(36+2*322^(1/2))^(1/3)))^(1/2)*(36+2*322^(1/2))^(1/3)/(((36+2*322^(1/2))^(1/3)+2/(36+2*322^(1/2))^(1/3)-6)*(18+322^(1/2))))+2*pi*Z+pi)*a
-a*(arctan((1/2)*((36+2*322^(1/2))^(2/3)*(24*322^(1/2)-4*(36+2*322^(1/2))^(2/3)-(36+2*322^(1/2))^(4/3)+428+24*(36+2*322^(1/2))^(1/3)))^(1/2)*(36+2*322^(1/2))^(1/3)/(((36+2*322^(1/2))^(1/3)+2/(36+2*322^(1/2))^(1/3)-6)*(18+322^(1/2))))+pi*(-2*Z+1))
(arctan(2^(1/2)*((36+2*322^(1/2))^2*(1+i*3^(1/2))-(36+2*322^(1/2))^(2/3)*(24*322^(1/2)+8*(36+2*322^(1/2))^(2/3)+428-(24*i)*966^(1/2)+24*(36+2*322^(1/2))^(1/3)-(428*i)*3^(1/2)+(24*i)*(36+2*322^(1/2))^(1/3)*3^(1/2)))^(1/2)*(36+2*322^(1/2))^(1/3)/(432+24*322^(1/2)), (1/12)*(-1+i*3^(1/2))*(36+2*322^(1/2))^(1/3)-((1/6)*i)*3^(1/2)/(36+2*322^(1/2))^(1/3)-(1/6)/(36+2*322^(1/2))^(1/3)-1)+2*pi*Z)*a
(arctan(-2^(1/2)*((36+2*322^(1/2))^2*(1+i*3^(1/2))-(36+2*322^(1/2))^(2/3)*(24*322^(1/2)+8*(36+2*322^(1/2))^(2/3)+428-(24*i)*966^(1/2)+24*(36+2*322^(1/2))^(1/3)-(428*i)*3^(1/2)+(24*i)*(36+2*322^(1/2))^(1/3)*3^(1/2)))^(1/2)*(36+2*322^(1/2))^(1/3)/(432+24*322^(1/2)), (1/12)*(-1+i*3^(1/2))*(36+2*322^(1/2))^(1/3)-((1/6)*i)*3^(1/2)/(36+2*322^(1/2))^(1/3)-(1/6)/(36+2*322^(1/2))^(1/3)-1)+2*pi*Z)*a
(arctan((-2*(36+2*322^(1/2))^2*(-1+i*3^(1/2))+2*(36+2*322^(1/2))^(2/3)*(-24*322^(1/2)-8*(36+2*322^(1/2))^(2/3)-428-(24*i)*966^(1/2)-24*(36+2*322^(1/2))^(1/3)-(428*i)*3^(1/2)+(24*i)*(36+2*322^(1/2))^(1/3)*3^(1/2)))^(1/2)*(36+2*322^(1/2))^(1/3)/(432+24*322^(1/2)), (1/12)*(-1-i*3^(1/2))*(36+2*322^(1/2))^(1/3)+((1/6)*i)*3^(1/2)/(36+2*322^(1/2))^(1/3)-(1/6)/(36+2*322^(1/2))^(1/3)-1)+2*pi*Z)*a
(arctan(-(-2*(36+2*322^(1/2))^2*(-1+i*3^(1/2))+2*(36+2*322^(1/2))^(2/3)*(-24*322^(1/2)-8*(36+2*322^(1/2))^(2/3)-428-(24*i)*966^(1/2)-24*(36+2*322^(1/2))^(1/3)-(428*i)*3^(1/2)+(24*i)*(36+2*322^(1/2))^(1/3)*3^(1/2)))^(1/2)*(36+2*322^(1/2))^(1/3)/(432+24*322^(1/2)), (1/12)*(-1-i*3^(1/2))*(36+2*322^(1/2))^(1/3)+((1/6)*i)*3^(1/2)/(36+2*322^(1/2))^(1/3)-(1/6)/(36+2*322^(1/2))^(1/3)-1)+2*pi*Z)*a
In the above, Z stands for any arbitrary integer (including 0 and negative values)
Each of these analytic solutions can be converted from arctan() form to ln() form because arctan(a,b) = -i*ln((b+i*a)/(b^2+a^2)^(1/2))
So possibly what you are seeing is the Symbolic Toolbox answering in ln form, and trying to present the information more compactly... but forgetting to print out the full information. That sometimes happens because of the communications between the MuPAD engine and MATLAB.
3 个评论
Matt Fig
2012-12-7
I get the more succinct:
-1.7996177555153143754981447772652*a
In both MATLAB r2011b and also in Maple 16.
bym
2012-12-8
vpa(ans)
ans =
1.7996177555153143754981447772652*a
(-1.7996177555153143754981447772652)*a
a*(1.0064927801893429956269684627402*i + 2.6773079709553580625862294378728)
-a*(1.0064927801893429956269684627402*i - 2.6773079709553580625862294378728)
-a*(1.0064927801893429956269684627402*i + 2.6773079709553580625862294378728)
a*(1.0064927801893429956269684627402*i - 2.6773079709553580625862294378728)
Walter Roberson
2012-12-8
Matt, when you solve() a expression that has floating point values, using Symbolic Toolbox, then the first thing it does is convert the floating point values to rationals. (This does not apply to sym() constructed with the decimal flag.) The Maple equivalent is to convert(expression,rational) . If you do not do that in Maple, you get "floating point pollution" (that is, at each stage, rationals and floating point are automatically combined to give floating point values with only Digits working precision.)
The "k" in the MuPAD output suggested that it was trying to give the generalized output. The Maple form of that is to solve() with AllSolutions as an option.
The large output I gave was Maple, converting each of the two floating point values to rational, and solving with AllSolutions; but then afterwards I edited to MuPAD output form (e.g., Pi -> pi, I -> i, _Z1 -> Z)
类别
在 帮助中心 和 File Exchange 中查找有关 Common Operations 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
