MATH 232 – Scientific Calculus II
Homework Supplement S11
Text Exercises:
Section 8.2: 1, 2, 3, 4, 11, 17, 21, 35, 36
More About Geometric Series:
A geometric series is important for understanding infinite series, because it is one of only a very few
series for which we have an explicit formula for S n , the nth partial sum of the series. This, in turn, gives
us an explicit formula for the limit (when it exists) of the series itself. Specifically, we learned in class
that

1
rk 
, for | r | 1 .

1 r
k 0

What if the index on a geometric series starts with a value different from 0, e.g. what does
r
k
k 1
converge to? Here’s an easy way to figure it out:



k 1
k 0
k 0
 r k   r k  r 0   r k 1 

So,
rk 
k 1

1
1
1 r 11 r
r
1 



.
1 r
1 r 1 r
1 r
1 r
r
. In a similar fashion, you can compute the limit of series like
1 r

Exercise A1. Compute the limits of the geometric series
 
k 1
5 k
6

r
, etc.
k 2

and
k
 
k 2
1 k
2
.
Some Unfinished Work from Probability:
Now that we know about geometric series, we can go back to something we skipped in our probability
unit: Recall that we use a geometric random variable to model waiting time1 for a binomial experiment.
We derived a formula for probabilities of a geometric RV like this: For repeated independent Bernoulli
trials (i.e. a binomial experiment), with Pr( S )  p for each trial, we let Y = # trials until the first success.
Then, Pr(Y  k )  (1  p) k 1 p . This formula allows us to make a (partial) pdf table for Y. (It is “partial”,
because Y has an infinite number of possible values, so the complete table would be infinitely long.) One
of the things we never verified (but certainly must be true) is that, for a geometric RV, the sum of all the
probabilities in the complete table must equal 1 (because it includes all possibilities). In this case, our
probability formula gives
1
Recall that waiting time refers to the number of Bernoulli trials until the first success occurs.
Pr(Y  1)  Pr(Y  2)  Pr(Y  3)  

  Pr(Y  k )
k 1

  (1  p) k 1 p
k 1

p 
 (1  p) k
1  p k 1
So, we end up with a geometric series (starting at k  1 ), with r  1  p . This is supposed to converge to
1, as we said before.

Exercise A2. Let Y be the waiting time random variable for a binomial experiment with Pr( s)  16 .
Use the formula above, along with what we know about geometric series, to confirm that the sum
of probabilities for all possible values of Y equals 1.
What Can Tangent Polynomials Do for Me?
Tangent polynomials provide another method (in addition to tools like Simpson’s Rule) for estimating the
value of a definite integral. The idea is simple: To estimate the definite integral

b
a
f ( x)dx , we replace f in
the integral with one of its tangent polynomials, then integrate the polynomial. In symbols, we have

b
a
b
f ( x)dx   pn ( x)dx .
a
As always, it matters where you build the tangent polynomial. Unless you have a good reason for doing
otherwise, the usual practice is to build the tangent polynomial at the midpoint of the interval of
integration. For example, if we were integrating over the interval 1  x  2 , we would build the tangent
polynomial at x  32 .
This approximation method will give a decent estimate of the value of the definite integral, provided the
tangent polynomial fits the original integrand at each x-value in the interval of integration. (We will make
this more precise, with an error bound, one of these days.)

Exercise A3. Use a degree-2 tangent polynomial to estimate the value of the definite integral
1

1
2
1
2
e
 x2
dx . (Irrelevant note: You should recognize this integrand as the pdf of a normal
random variable with   0 and   1.)
Integrals with parameters: For an integral like the one in the previous exercise, Simpson’s Rule will
usually give a more-accurate estimate, and is preferred over this tangent polynomial approach. As in
many instances, the importance of tangent polynomial approximations lies in the fact that, unlike Simpson
and friends, they can provide estimates in terms of parameters. This is illustrated in the following
exercise.

Exercise A4.
1
a. Repeat Exercise A3, but this time for the integral

1
1
2 
e

x2
2 2
dx
(another
normal RV, but, now, without a pre-selected value for the standard deviation  ).
b. Assuming this approximation is valid, what should we expect to happen, eventually, to the
value of this definite integral if we make  very large?