Example 12.1-2 Heating of a Finite Slab: Part b using eq. 12.1-31 > restart; > Thetan := (tau,eta,n)->exp(-(n+1/2)^2*Pi^2*tau)*cos((n+1/2)*Pi*eta); The same function found in our last session.
> D1[n]:=4*(-1)^n/(2*n+1)/Pi; > Theta:=(tau,eta,N)->sum(D1[n]*Thetan(tau,eta,n),n=0..N); calculate Theta as in eq. 11.1-31, but the sum only includes a finite number of terms. > 1-Theta(0.6,0.2,10); This is what is shown in Fig. 12.1-1. It agrees with the graph at this point.
> eq := (T1-T)/(T1-T0)=Theta(0.2,0.6,10); This is the way eq. 12.1-31 reads, and it looks like the figure gives a result greater than 0.5 so the agreement is OK here too. > T0:=100; T1:=250; > solve(eq,T); solve for T