Chapter 10
Polynomial and Series Representations of Functions





10.4 More Taylor Polynomials and Series

10.4.3 Taylor Polynomials and Series for `text[arctan(]xtext[)]`

We have seen that it is permissible to substitute for the independent variable in a function and in its approximating Taylor polynomials, and the resulting polynomials will be approximating Taylor polynomials for the resulting function — as long as the new independent variable has values in the right range. We also have seen that it is permissible to integrate (to a variable upper limit) a function and its approximating Taylor polynomials — and the resulting polynomials are approximating Taylor polynomials for the resulting antiderivative. We now use the same ideas to find approximating Taylor polynomials for the arctangent function — an antiderivative of `1text[/(]1+t^2text[)]`.

Activity 2

  1. Substitute `x = -t^2` in the geometric polynomials

    1 + x + x 2 + ⋅⋅⋅ + x n

    that approximate `1text[/(]1-xtext[)]` to find approximating polynomials for
    `1text[/(]1+t^2text[)]`.

  2. For what values of `t` would you expect these polynomials to converge to `1text[/(]1+t^2text[)]` as the degree `2n` becomes large?

  3. Integrate the typical polynomial of degree `2n` (from `0` to `x`) to find a polynomial of degree `2n + 1` that should approximate `text[arctan(]xtext[)]`.

Comment 2Comment on Activity 2

Example 1

Verify that the Taylor coefficients of degrees `0`, `1`, and `2` match the coefficients of the polynomials computed in Activity 2.

Solution   If `f text[(] x text[)] = text[arctan(] x text[)]`, then

f ′ ( x ) = 1 1 + x 2 and f ′ ′ ( x ) = - 2 x ( 1 + x 2 ) 2 ,

so `f text[(] 0 text[)] = 0, f^'text[(] 0 text[)] = 1,` and `f^('')text[(] 0 text[)] = 0`. Thus the coefficients of the first three terms in a Taylor polynomial for `text[arctan(] x text[)]` are `0, 1, 0`. That means the constant term and the quadratic term are both `0`, and the linear term is `x` — exactly what we saw in Activity 2.

Checkpoint 2Checkpoint 2

Checkpoint 2 suggests that it would be difficult to calculate more than the first few Taylor coefficients for `text[arctan(] x text[)]` by direct evaluation of `f^(k)text[(] 0 text[)]`, because the successive derivatives get more and more complicated — and only the odd-numbered ones contribute new information. However, Activity 2 shows that it is quite easy to find all of the Taylor polynomials for the arctangent function by substitution into and integration of a known set of Taylor polynomials. On the basis of this information, you should be prepared to believe that

arctan ( x ) = x - 1 3 x 3 + 1 5 x 5 - ⋅⋅⋅ = ∑ k = 0 ∞ ( - 1 ) k x 2 k + 1 2 k + 1 , if `text[|] x text[|] < 1.`
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