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Solve for unknowns 'x' and 'y' from Equations (1) and (2) using MATLAB. (However, if MATLAB is not the appropriate software to solve these two simultaneous equations, please suggest other methods, thanks!)
Equation (1)
0.5*exp(y*-29)+2.175/x*exp(y*-27)+4.47/x*exp(y*-26)+1.659/x*exp(y*-24)+1.76/x*exp(y*-21)+4.26/x*exp(y*-20)+11/x*exp(y*-19)+4.02/x*exp(y*-18)+4.44/x*exp(y*-17)+4.44/x*exp(y*-14)+10.36/x*exp(y*-13)+3.713/x*exp(y*-12)+1.33/x*exp(y*-11)+2.175/x*exp(y*-9)+2.556/x*exp(y*-6)+10.36/x*exp(y*-5)+3.375/x*exp(y*-4) = 1.3671
Equation (2)
0.5*exp(y*-58)+2.175/x*exp(y*-56)+4.47/x*exp(y*-55)+1.659/x*exp(y*-53)+1.76/x*exp(y*-50)+4.26/x*exp(y*-49)+11/x*exp(y*-48)+4.02/x*exp(y*-47)+4.44/x*exp(y*-46)+4.44/x*exp(y*-43)+10.36/x*exp(y*-42)+3.713/x*exp(y*-41)+1.33/x*exp(y*-40)+2.175/x*exp(y*-38)+2.556/x*exp(y*-35)+10.36/x*exp(y*-34)+3.375/x*exp(y*-33)+1.659/x*exp(y*-29)+4.44/x*exp(y*-28)+4.56/x*exp(y*-27)+2.175/x*exp(y*-26)+4.26/x*exp(y*-23)+1.76/x*exp(y*-20)+4.56/x*exp(y*-18)+11/x*exp(y*-16)+2.175/x*exp(y*-15)+3.2/x*exp(y*-14)+2.556/x*exp(y*-12)+3.713/x*exp(y*-9)+10.36/x*exp(y*-8)+10.36/x*exp(y*-6)+5.77/x*exp(y*-4) = 1.8642
Use Taylor series to derive the form of the truncation error of the approximation Suppose f € C3. Given that the roundo error has the similar form as in the Forward D
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