Calculus II
Class Notes, 3/12/99
We use the Taylor polynomial of
degree 4 to indicate that the limit in the first line of the figure below is correct.
- The Taylor polynomial for the
cosine function is as indicated in the second line.
- In the third and forth lines we
substitute the Taylor polynomial for the cosine function in the expression for the limit,
and simplify the result.
- The final expression in the forth
line consists of a constant number 1/2 and a quadratic term -x^2 / 24, which approaches 0
in the limit as x -> 0.
- Had we used a Taylor polynomial of
higher order, we would have had polynomial terms of higher degrees, which also would have
approached 0 as the limit.
- We thus see that the desired limit
is indeed 1/2.
Video File #1
The DERIVE command for finding a Taylor polynomial is as indicated in the figure below.
- cos x represents the function to be expanded, x tells DERIVE what the variable is (very
handy if we have more than one variable in the expression), 0 tells DERIVE to expand about
a = 0, and 6 tells DERIVE to do the degree six polynomial.
We investigate the radius of convergence for the Taylor series of e^x.
- The nth coefficient of the expansion is 1 / n!.
- The radius convergence is the limit of | Cn | / | C(n+1) |.
- The values of | Cn | / | C(n+1) | are 1, 2, 3, 4, ..., n+1, ... .
- As n -> infinity, these values clearly approach infinity (i.e., lim(n -> inf)
(n+1) = inf).
Video File #2
We next investigate the radius of convergence of the series given in the figure below.
- We see that the nth term will have power x^n, will have a coefficient of (n+1)! in the
numerator and (n!)^2 in the denominator.
- We therefore have Cn = (n+1)! x^n / n!.
- Our expression for the radius of convergence | Cn / C(n+1) | therefore simplifies to the
expression indicated in the last line (see second figure below if the present figure is
unclear), which approaches infinity as n -> infinity.
- We therefore conclude that the radius of convergence of the sequence is infinite.
The details of the calculation are more easily seen in the figure below.
Video File #3
Video File #4
In the figure below the numerator of the nth term has a factor (2n!), which is a lot
bigger than n!. However the denominator has a factor (n!) ^2, which is also a lot bigger
than n!.
- If the numerator grows too fast compared to the denominator, the coefficients will
either get larger or will fail to get smaller fast enough for the series to converge
anywhere.
- If the denominator grows really fast the coefficients will get smaller very fast and the
series will converge everywhere.
- Which of these scenarios will dominate, and to what extent, determines whether the
sequence converges anywhere, whether it converges over some finite domain, or whether it
converges everywhere.
- Working out the details (see 2d picture below for a better view), we see that Cn /
C(n+1) simplifies to (n+1)^2 /[ (2n+2)(2n+1)].
- We easily see that as n approaches infinity, this expression approaches 1/4, so the
radius of convergence is 1/4 and the series converges for x within 1/4 of a = 0.
The figure below shows how ( 2 (n+1) )! is simplified to obtain (2n+2) (2n+1) (2n)!.
Video File #5