There are many ways of coming up with the Taylor Series of a given function f(x). Here are some of the possible recipes:
From scratch: Given the formula for f(x), directly compute an = (f(n)(c))/(n!) to obtain the nth Taylor series coefficient an for f(x) at c. Then the Taylor series for f(x) is:
(ii) Substitution: Write your function f(x) as g(2x), or as g(x2), or as g(h(x)) for some other simple function h(x), where g(x) is a function whose Taylor series you already know, and get the Taylor series for f(x) by plugging in u=h(x) into the Taylor series for g(u).
(iii) Differentiation or Integration: If f(x) = d/dx(g(x)), that is, if the given function f(x) is the derivative of some function g(x) whose Taylor series you know, then the series for f(x) can be obtained from differentiating term-by-term the series for g(x). If f(x) is an antiderivative of a function h(x) whose series you know, you can integrate the series for h(x) term-by-term; however, remember to also solve for the constant of integration "C" in this case.
(iv) Creative Combos: If the Taylor series for f(x) and g(x) are known, with nth terms an and bn respectively, what is the Taylor series for h(x) if h(x)=f(x)+g(x)? If h(x) = x f(x)?
Note: in these recipes, all Taylor series are centered at the same point, c.
Cook up a storm by finding the Taylor series for the following cornucopia of functions, as well as determining the radii and intervals of convergence of each Taylor series you find.
|1/[ 1 + 4x2 ]||tan-1(2x)||1/(1 + x2)||1/(4 + x2)|
For an extra challenge, try f(x) = ln [(1+x)/(1-x)] -- try rewriting it first -- and f(x) = 2/(3x+4).
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