Solving non-linear equations - not enough input arguments

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LuC
LuC 2016 年 5 月 24 日
コメント済み: John D'Errico 2016 年 5 月 24 日
Hello,
I have some data that I approximated with a polynomial of degree 9, p(x), and I have a non-linear function (with unknown coefficients that is supposed to fit the data). I found a Taylor series of that function, ts(x). In order to find the coefficients, I would like to solve the system p(x) = ts(x).
function f = ts1( x )
% Taylor series expansion
A = 70.94;
B = x(1);
C = x(2);
D = x(3);
E = x(4);
%system of equations
f(1) = B*C*D - 1161; % x^1
f(2) = -D*(C*((B^3*E)/3 + B^3/3) + (B^3*C^3)/6) - 22.21; % x^3
f(3) = D*(C*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5)) + (B^2*C^3*((B^3*E)/3 + B^3/3))/6 + B*C*((B^4*C^4)/120 + (B*C^2*((B^3*E)/3 + B^3/3))/3)) + 0.2839; % x^5
f(4) = -D*(C*(B*((2*B^6*E)/15 + ((4*B^7*E^2)/9 + (8*B^7*E)/5)/(12*B) + B^2*((2*B^4*E)/15 + B^4/7)) + (22*B^7*E)/105 + (B^3*E*((2*B^4*E)/9 + B^4/5))/3) +...
C*((B^3*E)/3 + B^3/3)*((B^4*C^4)/120 + (B*C^2*((B^3*E)/3 + B^3/3))/3) + (B^2*C^3*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5)))/6 + B*C*((B^3*C^4*((B^3*E)/3 + ...
B^3/3))/60 + B^2*C^2*((B^4*C^4)/5040 + (B*C^2*((B^3*E)/3 + B^3/3))/60) + (4*B*C^3*((B^3*E)/3 + B^3/3)^2 + 8*B^2*C^3*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5)))/(24*B*C))) - 0.001837; % x^7
f(5) = D*(C*((10*B^9*E)/63 + B*(((4*B^9*E^2)/5 + (12*B^9*E)/7)/(18*B) + B^2*((2*B^6*E)/21 + ((4*B^7*E^2)/9 + (8*B^7*E)/5)/(20*B) + B^2*((2*B^4*E)/21 + B^4/9)) + (B*((4*B^7*E^2)/9 + (8*B^7*E)/5))/20 +...
(2*B^4*E*((2*B^4*E)/15 + B^4/7))/3) + (B^3*E*((2*B^6*E)/15 + ((4*B^7*E^2)/9 + (8*B^7*E)/5)/(12*B) + B^2*((2*B^4*E)/15 + B^4/7)))/3 + (B^5*E*((2*B^4*E)/9 + B^4/5))/5) + B*C*((B*C*(4*B*C^3*((B^3*E)/3 + ...
B^3/3)^2 + 8*B^2*C^3*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5))))/480 + (12*B^2*C^3*(B*((2*B^6*E)/15 + ((4*B^7*E^2)/9 + (8*B^7*E)/5)/(12*B) + B^2*((2*B^4*E)/15 + B^4/7)) + (22*B^7*E)/105 + (B^3*E*((2*B^4*E)/9 + B^4/5))/3) + ...
12*B*C^3*((B^3*E)/3 + B^3/3)*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5)))/(36*B*C) + B^2*C^2*((B^3*C^4*((B^3*E)/3 + B^3/3))/2520 + B^2*C^2*((B^4*C^4)/362880 + (B*C^2*((B^3*E)/3 + B^3/3))/2520) + (4*B*C^3*((B^3*E)/3 + B^3/3)^2 + ...
8*B^2*C^3*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5)))/(480*B*C)) + 2*B*C^2*((B^3*E)/3 + B^3/3)*((B^4*C^4)/5040 + (B*C^2*((B^3*E)/3 + B^3/3))/60)) + C*((B^3*E)/3 + B^3/3)*((B^3*C^4*((B^3*E)/3 + B^3/3))/60 + ...
B^2*C^2*((B^4*C^4)/5040 + (B*C^2*((B^3*E)/3 + B^3/3))/60) + (4*B*C^3*((B^3*E)/3 + B^3/3)^2 + 8*B^2*C^3*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5)))/(24*B*C)) + (B^2*C^3*(B*((2*B^6*E)/15 + ((4*B^7*E^2)/9 + (8*B^7*E)/5)/(12*B) + ...
B^2*((2*B^4*E)/15 + B^4/7)) + (22*B^7*E)/105 + (B^3*E*((2*B^4*E)/9 + B^4/5))/3))/6 + C*((B^4*C^4)/120 + (B*C^2*((B^3*E)/3 + B^3/3))/3)*((14*B^5*E)/45 + B*((2*B^4*E)/9 + B^4/5))) + 4.552e-06; % x^9
end
My main looks like this:
x0 = [-0.0095, 0.0014, 4.0000, 0.0165];
f = solve(@ts1, x0, options)
and the error is:
Error using ts1 (line 5)
Not enough input arguments.
Error in sym>funchandle2ref (line 1211)
S = sym(x());
Error in sym>tomupad (line 1114)
x = funchandle2ref(x);
Error in sym (line 151)
S.s = tomupad(x);
Error in solve>getEqns (line 410)
a = formula(sym(a));
Error in solve (line 227)
[eqns,vars,options] = getEqns(varargin{:});
  1 件のコメント
ashish ahir
ashish ahir 2016 年 5 月 24 日
Please simplify your question, Make it more readable and understandable,

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採用された回答

Star Strider
Star Strider 2016 年 5 月 24 日
The way you’ve written your code, I believe you want the fsolve function instead.
Try this:
f = fsolve(@ts1, x0, options)
  1 件のコメント
John D'Errico
John D'Errico 2016 年 5 月 24 日
Definitely the case. Anyway, solve would surely fail to find a solution, though vpasolve might succeed.

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