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03/27/97 09:08am
Reply to message #252 from Chuang Iyuan: > MathID:61345329722059 > License:L27648010 > Hi, > I tried to solve some questions. However > the student version gave me out of memory message after around one hour. > I did these questions on professional > version 2.2.3 on school computer lab, and got the results in just few minitues. Is this the limit of student version? > > The questions are as follow: > sigma[x_,T_]=(a+c(Tx))Exp[k(Tx)]+b > h1=Integrate[sigma[v,T]*Integrate[sigma[v,s],{s,t,T}],{v,0,t}] > H1[t_,T_,a_,b_,c_,k_]=Integrate[h1,{T,t,T}] > > Iyuan Chuang Without more information on the systems involved, I can not say much about the differences in the results that you report. It may well be that the speed and memory differences of the two machines would be sufficient to explain the difference in performance. Unless you are using a pre V2.2.4 student version, your student version is functionally identical to the professional version within releases. That having been said there is something that you can do to speed up your calculation. First if your functions are reasonably will behaved, e.g. no singularities, you can replace the definite integrals with indefinite integrals and evaluate the results at the limits of integration. Careful definite integration also requires checking for singularities in the range of integration, and taking limits at the endpoints, rather than just substituting the endpoints into the indefinite integral. This takes much more time. Here is your example with some rudimentary changes in that direction. First I evaluate the function definition for sigma. In[35]:= sigma[x_,T_]=(a+c(Tx))Exp[k(Tx)]+b Out[35]= \!\(b + E\^\(\(k\)\ \((T  x)\)\)\ \((a + c\ \((T  x)\))\)\) Now I look at your next step and break it up, evaluating the inner integal first as an indefinite integral. h1=Integrate[sigma[v,T]*Integrate[sigma[v,s],{s,t,T}],{v,0,t}] In[36]:= InnerInt = Integrate[sigma[v,s],s] Out[36]= \!\(b\ s + E\^\(\(k\)\ s\)\ \(( \(\(\(c\ E\^\(k\ v\)\ s\)\/k\)\) + \(E\^\(k\ v\)\ \((\(c\)  a\ k + c\ k\ v)\)\)\/k\^2)\)\) Now I evaluate this at the limits of the original definite integration. In[37]:= DefInnerInt= (InnerInt /. s > T)  (InnerInt /.s>t) Out[37]= \!\(\(b\)\ t + b\ T  E\^\(\(k\)\ t\)\ \(( \(\(\(c\ E\^\(k\ v\)\ t\)\/k\)\) + \(E\^\(k\ v\)\ \((\(c\)  a\ k + c\ k\ v)\)\)\/k\^2)\) + E\^\(\(k\)\ T\)\ \(( \(\(\(c\ E\^\(k\ v\)\ T\)\/k\)\) + \(E\^\(k\ v\)\ \((\(c\)  a\ k + c\ k\ v)\)\)\/k\^2)\)\) Now I evaluate the outer integral as an indefinite integral. This time I suppress the output for space considerations. In[38]:= OuterInt = Integrate[sigma[v,T]*InnerInt,v]; Again the limits are put in. In[39]:= DefOuterInt =( OuterInt/. v>t)  (OuterInt /. v > 0); This is your h1. The process goes on with the replacement of your expression H1[t_,T_,a_,b_,c_,k_]=Integrate[h1,{T,t,T}] with the following In[40]:= LastInt = Integrate[DefOuterInt,T]; In[41]:= H1[t_,T_,a_,b_,c_,k_]=(LastInt/. T>T)  (LastInt /. T>t); In[42]:= H1[6,1,1,1,1,3]//N Out[42]= \!\(1.21486611562424196`*^12\) Forum Moderator
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