Z zcus05 Jul 2010 10 0 Jul 24, 2010 #1 dx/dt = 1- sin[ ln(1+x )] Replace the right hand side by the linear Taylor's approximation about x0=0 and solve the resulting equation. Determine the particular solution for the initial condition x(0)=0.
dx/dt = 1- sin[ ln(1+x )] Replace the right hand side by the linear Taylor's approximation about x0=0 and solve the resulting equation. Determine the particular solution for the initial condition x(0)=0.
Jester MHF Helper Dec 2008 2,470 1,255 Conway AR Jul 24, 2010 #2 Taylor series for \(\displaystyle \ln(1+x)\) and \(\displaystyle \sin x\) \(\displaystyle \ln (1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} \cdots\) \(\displaystyle \sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} \cdots\) so to first order the taylor series would be \(\displaystyle \sin (\ln(1+x)) = x\)
Taylor series for \(\displaystyle \ln(1+x)\) and \(\displaystyle \sin x\) \(\displaystyle \ln (1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} \cdots\) \(\displaystyle \sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} \cdots\) so to first order the taylor series would be \(\displaystyle \sin (\ln(1+x)) = x\)
Z zcus05 Jul 2010 10 0 Jul 25, 2010 #3 Hadn't learnt the Taylor series so had to do a quick crash course, this helped a lot, thanks heaps.