Taylor Series Approximation for Solving Initial Value Problems

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SUMMARY

The discussion focuses on using Taylor Series approximation to solve the initial value problem defined by the ordinary differential equation (ODE) \(\frac{ds}{dt} = 10 - 9.8t\) with the initial condition \(s(0) = 1\). Participants confirm that the first derivative \(s'(0)\) is calculated as \(10 - 9.8(0) = 10\), and subsequent derivatives are \(s'' = -9.8\) and \(s''' = 0\). The conclusion is that the solution \(s(t)\) is a quadratic function of \(t\), highlighting the importance of understanding higher-order derivatives in Taylor Series expansion.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with Taylor Series expansion
  • Basic calculus, including differentiation
  • Knowledge of initial value problems
NEXT STEPS
  • Study the application of Taylor Series in solving ODEs
  • Learn about higher-order derivatives and their significance in approximation
  • Explore numerical methods for solving initial value problems
  • Investigate the implications of non-linear terms in ODEs
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are working with differential equations and seeking to understand approximation techniques for initial value problems.

cscott
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With a simple ODE like [itex]\frac{ds}{dt} = 10 - 9.8t[/itex] and you're given an initial condition of s(0) = 1, when doing the approximation would s'(0) = 10 - 9.8(0), s'' = ... etc?
 
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Yes, you are given s(0) so just plug it into the given equation and solve for s'(0). Actually, the crucial point is under your "etc." Since
s'= 10- 9.8t, s"= -9.8, s"'= 0, etc. It should be no surprise that s is a quadratic function of t.

It would be more interesting if there were an "s" on the right side of the equation.
 
Thanks for the help.
 

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