You can think of this as follows, the zero order Taylor approximation provides a “constant” function approximation. That is because the fourth order Taylor series approximation of a fourth order polynomial function is identical to the function itself. It is important to note that the error reduces to zero when using the fourth order Taylor series approximation. The fourth order Taylor series expansion around has the following form: The third order Taylor series expansion around has the following form: The second order Taylor series expansion around has the following form: The first order Taylor series expansion around has the following form: The zero order Taylor series expansion around has the following form: The true value of the function at is given by: Calculate the error associated with each expansion. Use zero through fourth order Taylor’s series expansion to approximate the value of the function defined as at. There are some analytical conditions that would indicate the radius of convergence of a Taylor series however, this is beyond the scope of this course! However, for functions with square roots, the Taylor series converges when is relatively close to. For some functions, like, , and, the Taylor series always converges. In general, the Taylor series works best if the distance between and is as small as possible. This is a large upper bound and indicates that using six terms is not giving a good approximation. The maximum value will be obtained when : The error term in the theorem gives an upper bound for when six terms are used as follows: It is clear that the Taylor series is diverging. In fact, the following table gives the values up to 21 terms. Indeed, the actual value of the error is less than the upper bound:įor, the Taylor series for this function around doesn’t give a very good approximation as will be shown here but rather keeps oscillating. The error term in the theorem gives an upper bound for as follows: The terms are getting closer to each other and four terms provide a good approximation. The derivatives of the function are given by: If, then, the upper bound of the error is: If terms are used (including ) and if, then the upper bound of the error is: The Taylor approximation around is given as: Solutionįirst, we will calculate the numerical solution for the and : View Mathematica CodeĪpply Taylor’s Theorem to the function defined as to estimate the value of and. The following code provides a user-defined function for the Taylor series having the following inputs: a function, the value of, the value of, and the number of terms including the constant term. The actual error in this case is indeed less than the upper bound: Which satisfies that the actual error is less the upper bound: Therefore, the error when two terms are used is: Using Mathematica, the square root of approximated to 4 decimal places is. In this case, the upper bound for the error is: The derivatives of the function evaluated at the point can be calculated as: The derivatives of have the following form: That’s because the absolute values of the derivatives of attain their maximum value in the interval at 4. With, an upper bound for the error can be calculated by assuming. If terms are used (including ), then the upper bound for the error is: The Taylor approximation of the function around the point is given as follows: Open Educational Resources Taylor Series:Īpply Taylor’s Theorem to the function defined as to estimate the value of.
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