Explore BrainMass
Share

Explore BrainMass

    Taylor Series Calculus

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    The starting point for this example is the Taylor series for sine:
    sin (x) = x - (1/3!)x^3 + (1/5!)x^5 ?(1/7!)x^7 + ...... + (-1)^n * 1/(2n + 1)!*x^(2n+1) + .......
    a) Let f(x) = { sin(x) / x if x doesn't equal 0
    {1 if x doesn't equal 0

    Show that f is infinitely differentiable on R (including x = 0) and determine its Taylor series in powers of x.
    b) Define the function Si by the rule,
    Si (x) = the integral from 0 to x
    sin(t) / t dt , x E R
    Determine the Taylor series for Si in powers of x, and the open interval of convergence of the resulting series.

    © BrainMass Inc. brainmass.com October 10, 2019, 6:04 am ad1c9bdddf
    https://brainmass.com/math/real-analysis/taylor-series-calculus-527465

    Solution Preview

    Please find the attachment for the solutions.

    The starting point for this example is the Taylor series for sine:
    sin (x) = x - (1/3!)x^3 + (1/5!)x^5 -(1/7!)x^7 + ...... + (-1)^n * 1/(2n + 1)!*x^(2n+1) + .......
    a) Let f(x) = { sin(x) / x if x doesn't equal 0
    {1 if x doesn't equal 0

    Show that f is infinitely differentiable on R (including x = 0) and determine its Taylor series in powers of x.
    b) Define the function Si by the rule,
    Si (x) = the integral from 0 to x
    sin(t) / t ...

    Solution Summary

    The solution provides an example of the Taylor series caluclus.

    $2.19