Too many input arguments Fsolve

Question

Ana Royce 2016-8-18

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/299955-too-many-input-arguments-fsolve

评论： Ana Royce 2016-8-25

Error using ==> derofexpexp2
function[derofexpexp1b,derofexp1a,derofexpexp1a]=derofexpexp2(rvar,eqdif,eqpardif,indexp1,indexpexp1)
Too many input arguments.
Error in ==> fsolve at 254
            fuser = feval(funfcn{3},x,varargin{:});
Error in ==> paragexpexp at 56
zeroofder2p=fsolve('derofexpexp2',finnoearth(finpeakh,1),options2a,[],eqdif,eqdf3,indexpl1,indexexp1);

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Answer 1

Walter Roberson 2016-8-18

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https://ww2.mathworks.cn/matlabcentral/answers/299955-too-many-input-arguments-fsolve#answer_232105

在 MATLAB Online 中打开

The objective function for fsolve is currently not defined to take more than one input.

Historically, extra parameters in the fsolve call were added as arguments to the objective function. That syntax has been obsolete and undocumented since MATLAB 5.1 and might or might not work.

If it does work, then in your call

fsolve('derofexpexp2',finnoearth(finpeakh,1),options2a,[],eqdif,eqdf3,indexpl1,indexexp1);

the argument finnoearth(finpeakh,1) would correspond to x0, and the argument options2a would correspond to options, leaving [], eqdif, eqdf3, indexp11, and indexexp1 as extra arguments. That would be 5 extra arguments in addition to the x argument, which would lead to 6 arguments passed to the function. But your function is only defined to take 5 arguments.

Please read http://www.mathworks.com/help/matlab/math/parameterizing-functions.html . (Unless, of course, you are still using MATLAB 4.x, which would be pretty major information that you would have been expected to mention already.)

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Ana Royce 2016-8-25

编辑：Walter Roberson 2016-8-25

在 MATLAB Online 中打开

i have defined value for options2a as : and attached code of two function file ...please review and assist me . i need to submit my work by tomorrow. Advance thanking

my paragexpexp.m (file that im calling in main)

%This will calculate the derivatives of the exp functions in expexp.
format long e
eq4=0;
eq4a=0;
%load fileofsechpar
%Not needed since it is executed directly after the profile model.
syms Hmaxexpa Semexpa Nemaxexpa Hmpar Sempar Nempar 
rvar='r';
options2a=[0 0 0 0 0 0 0 0 0 0 0 0 0 30000 0 0 0 0];
     for ki3=1:1:indexpl1;
        eq1='[(rvar-Hmaxexpa)/(Semexpa)]';
        eq1=sym(eq1);
        eq2=subs(eq1,{Hmaxexpa,Semexpa},{Hmaxexpl(1,ki3),Semexpl(1,ki3)});
        eq2=vpa(eq2,32);
        %eq2=eval(eq1);
        eq2a='Nemaxexpa*(exp(-eq2^2))';
        eq2a=subs(eq2a,Nemaxexpa,Nemaxexpl(1,ki3));
       eq2a1=vpa(eq2a,32);
        eq2b=subs(eq2a1,eq2,'eq2');
        eqdf2=diff(eq2b);
        eq4=eq4+eq2b;
        eq4a=eq4a+eqdf2;
        sizeofeq(ki3,1)=length(eqdf2(1,:));
        eqdif(ki3,1)=eqdf2;
        eqmod(ki3,1)=eq2b;
    end
eq5=diff(eq4);
eq5=vpa(eq5,32);
save eq5s eq5
eq5a=eq4a;
eqp2bb=0;
eqp2aa=0;
for ki7=1:1:indexexp1;
   term1='[(rvar-Hmpar)/(Sempar)]';
   trmev=subs(term1,{Hmpar,Sempar},{Hmaxexpexp2(1,ki7),Semexpexp2(1,ki7)});
   eqp1='Nempar*exp(-trmev^200)';
   eqp2=subs(eqp1,'trmev',trmev);
   eqp22=subs(eqp2,'Nempar',Nemaxexpexp2(1,ki7));
   eqp2mod=subs(eqp2,{Hmpar,Sempar,Nempar},{Hmaxexpexp2(1,ki7),Semexpexp2(1,ki7),Nemaxexpexp2(1,ki7)});
   eqp2a=vpa(eqp22,32);
   eqp2b=diff(eqp2a);
   eqp2aa=eqp2aa+eqp2a;
   eqp2bb=eqp2bb+eqp2b;
   %eq1p='-50*Nempar/(Sempar)*term1^49*exp(-term1^50)';
  eq1p='-50*Nempar/(Sempar)*trmev^49';
   eq2p=subs(eq1p,'trmev',trmev);
   eq3p=subs(eq2p,{Hmpar,Sempar,Nempar},{Hmaxexpexp2(1,ki7),Semexpexp2(1,ki7),Nemaxexpexp2(1,ki7)});   
   eq4p=vpa(eq3p,32);
   eqdf3=diff(eqp2mod);
   eqpardif(ki7,1)=eqp2b;
   eqpardif1(ki7,1)=eq4p;
end
zeroofder2p = fsolve(@(x) derofexpexp2(x,eqdif, eqdf3, indexpl1, indexexp1), finnoearth(finpeakh,1), options2a); 
rfound2p=zeroofder2p;
totmaxval=0;
   for ki3=1:1:indexpl1;
      rv=rfound2p;
      valoffun(ki3,1)=subs(eqmod(ki3,1),'rvar',rv);
      totmaxval=totmaxval+(valoffun(ki3,1));
   end
   totmaxval=subs(totmaxval)
   %Subs also used to transform symbolic to numeric.
   totmaxvalp=0;
   for ku3=1:1:indexexp1;
      rvp=Hmaxexpexp2(1,indexexp1);
      valoffunp(ku3,1)=subs(eqp2mod(ku3,1),'rvar',rvp);
      totmaxvalp=totmaxvalp+(valoffunp(ku3,1));
   end
   totmaxvalp=subs(totmaxvalp)
   %Subs also used to transform symbolic to numeric.
   totmaxvalt=totmaxval+totmaxvalp
return
my derofexpexp2.m
function[derofexpexp1b,derofexp1a,derofexpexp1a]=derofexpexp2(rvar,eqdif,eqpardif,indexp1,indexpexp1)
derofexp1a=0;
for ki3=1:1:indexp1;
   derofexp1aa(ki3,1)=eval(eqdif(ki3,1));
   derofexp1a=derofexp1aa(ki3,1)+derofexp1a;
end
derofexpexp1a=0;
for ku3=1:1:indexpexp1;
   derofexpexp1aa(ku3,1)=eval(eqpardif(ku3,1));
   derofexpexp1a=derofexpexp1aa(ku3,1)+derofexpexp1a;
end
rvar;
derofexpexp1b=derofexpexp1a+derofexp1a;
return

Walter Roberson 2016-8-25

Your options2a is a numeric vector, not an options structure. Is it intended to act as the initial values of the vector being solved on? If it is, then what is the role of finnoearth(finpeakh,1) ?

Your function uses parameters named rvar, eqdif, eqpardif, indexp1, and indexpexp1. How do your calling parameters finnoearth(finpeakh,1), options2a, eqdif, eqdf3, indexpl1, and indexexp1 correspond to each of those? Some of the matches look pretty obvious, but you have 6 inputs and only 5 parameters.

Ana Royce 2016-8-25