Calculus II
Class Notes, 3/24/99
To expand f(x) = 1 / `sqrt(x) in a degree-3 Taylor polynomial about a = 1, we proceed
is in the figure below.
- The first three derivatives of f(x) are evaluated at a = 1, yielding coefficients -1/2,
3/4 and -15 / 8.
- The resulting polynomial in shown in the last line of the figure below.
We substitute x = 1 - z^2 to obtain a Taylor expansion of 1 / `sqrt( 1 - z^2 ) about z
= 0, as shown below.
To find the Taylor series of f(t) = t^2 / (1 + t) about a = 0, we can first find the
series for 1 / (1 + t) and then multiply the result by t^2.
- We will begin by expanding 1/x about x = 1, then we will substitute x = 1 + t.
- This will give us an expansion of 1 / (1 + t) about t = 0.
- g(x) in the last two lines is the Taylor series for 1/x about x = 1.
Video File #1
When we substitute x = 1 + t, x - 1 will become t, and we obtain the expansion in the
first line below. Note that the polynomial is in powers of (t - 0) = t, so the
expansion is about t = 0.
- Multiplying this expansion by t^2 we obtain the expansion of t^2 / (1 + t) about t = 0,
shown in the last line below.
Video File #2