diff eq (honors) - lab 5
series solutions



demonstration (follow along)

consider the IVP y'' + y = 0, y(0) = 1, y'(0) = 0. here the coefficients for the series satisfy (n+2)(n+1)a_{n+2} + a_n = 0, i.e., a_n = -a_{n-2}/(n*(n-1)).

look at approximation up to t^10 term:

For[a = List[1, 0]; f = 1;
  i = 2, i <= 10, i++, a = Append[a, -a[[i - 1]]/(i*(i - 1))];
  f = f + a[[i+1]]*t^i];
p1 = Plot[f, {t, 0, 10}, PlotRange -> {{0, 10}, {-2, 2}}]
now compare with approximations using up to t^20, t^50.

do on your own

  1. use series to approximate solutions to the IVP y'' - y = 0, y(0) = 1, y'(0) = 0. here the coefficients for the series satisfy (n+2)(n+1)a_{n+2} - a_n = 0. how many terms do you need to take to determine the shape of the solution?
  2. use series to approximate solutions to the IVP y'' - y = 0, y(0) = 0, y'(0) = 1. here the coefficients for the series satisfy (n+2)(n+1)a_{n+2} - a_n = 0. how many terms do you need to take to determine the shape of the solution?
  3. use series to approximate solutions to the IVP y'' + y' + t y = 0, y(0) = 1, y'(0) = 0. how many terms do you need to take to determine the shape of the solution?
  4. use series to approximate solutions to the IVP y'' + y' + t y = 0, y(0) = 0, y'(0) = 1. how many terms do you need to take to determine the shape of the solution?

extra: convergence issues

Remember that series approximations are only good in the region of convergence. Suppose you want to make a Taylor series for f(t) = log t. First, the function is not defined at t=0, so you cannot expand around t=0. If you expand around t=x for some positive x, then the radius of convergence can only be x. Here is code to generate Taylor approximations to log t about x. Try varying the values of x and see what happens.
For[a = List[0]; x = 1; f = Log[x];
  i = 1, i <= 10, i++, a = Append[a, (-1)^(i + 1)/(i*x^i)];
  f = f + a[[i + 1]]*(t - x)^i];
Plot[f, {t, 0, 10}, PlotRange -> {{0, 10}, {-5, 5}}]

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