01
(a)
(b)
02
(a) Notice the matching powers. Collect powers and then observe the geometric series pattern:
(b)
03
Notice that
We have this series for
Therefore:
Now we use the “Next Term Bound” rule. Calculate terms until we find a term less than
So we take the following partial sum approximation:
04
(a)
(b)
05
Calculate derivatives. Use
| 0 | |||
| 1 | |||
| 2 | |||
| 3 | |||
| 4 | |||
| 5 | |||
| 6 |
Each new derivative takes down the next power as a factor, and switches the sign. The accumulation of powers follows a factorial pattern, and these factorials cancel those added to the denominator to make
So we have:
Alternate method:
We can derive this Taylor series using some algebraic tricks with the standard geometric series:
06
(a)
(b)
(c)
07
(a)
(b)
(c)
(d)
08
Use the formula
Therefore:
09
(a)
Recall the series for
This matches our series if we set
(b)
10
The Coefficient-Derivative Identity says that
Solve:
So, for the coefficient
11
Write the alternating series for
This is an alternating series, so we can apply the “Next Term Bound” rule. Calculate some terms:
(Without a calculator, we can see that
So we add up the prior terms:
12
Write the series of the integrand:
Integrate:
Now apply the “Next Term Bound” and look for the first term below
So we simply add the first two terms: