Review of CH 8: #12 #14 #16 #22 #24 #28 #30 See a few paragraphs below for the answers.
For some other problems: try CH 5 review #28; write s for sin(theta) and c for cos(theta).
Then int of s^4 c^3 d theta is int s^4 (1-s^2) ds and now the problem is trivial.
6.2 #36 (this one was listed by mistake. It's a nasty volume problem which you should skip,
but if you got 128/3sqrt(3) give yourself a pat on the back!)
Ch 6 Review #14a V=pi/3, #24 f_average = 26/9;
7.3 #12 y = sqrt(2 - sqrt(x^2 + 1)) the initial condition gives the constant c=2 and also
the choice of sign for the outer square root
#14 y = ln 2 - ln(3 - t^2) for |t| < sqrt(3) - the restriction on domain ensures ln is defined.
#26 y = + or - sqrt(2 (x + c)) (these parabolas are defined only for x >= -c and are symmetric
about the x-axis).
Problems below were reviewed last night.
5.8 #20; indef integral of x^2 arctan x = (IBP with u=arctan x) =
(x^3/3)arctan x - x^2/6 + (1/6)ln(1 + x^2)
5.10 #30; improper integral from 0 to 4 of dx/(x^2 + x - 6)
(hint: factoring denom shows integrand undefined at x = 2 so need to
evaluate two different improper integrals ending at 2 or show they
don't both converge) *** neither one converges since (using PF) 1/(x+3)(x-2) = A/(x+3) + B/(x-2),
where A= -1/5 and B= 1/5. But lim a goes to 2 from above of ln(a-2) is -oo, etc.
Ch 5 review problem #32 IBP (u=arctan x)
Ch 6 6.1 #36 A=1/12, 6.2 #8 3pi/4,#38 (16/3)r^3,#44 (shells) V=2pi, 6.3 #8 31/48
#22 skip it (too complicated and requires use of the table of integrals)
CH 8 Review problems - answers:
#12 conv by AST
#14 div by T for Div
#16 conv by Int Test - since integral (from 2 to oo) of 1/x (ln x)^2 dx conv.
#22 rearrange the n-th term to make it [(-x/4)^n]/n! so sum is e^(-x/4)
#24 Geometric series which converges if |ln x| < 1
#28 (a) conv by RT; (b) hence lim (n --> oo) a_n = 0
#30 I of conv is [-5,5]
The exam is scheduled for Saturday
at 9am in ICC 108.
See you Thurs. May 8 at 9pm in the usual room for a review session. But don't wait till then to begin reviewing. I'll update these problems as I can since I'm at a conference until Thurs. Meanwhile, the best review consists of the problems we've already discussed which (unless you have an eidetic memory ;-) are good to warm up on. After thinking for a while, you may remember what we did in class to solve them. And as I have mentioned before, the odd-numbered problems and examples also make good review - and a source for exam problems!
We began with integration theory so here is a problem to try out.
R.1a Show that
1
int (from 2 to oo) ---------------- dx < oo
q
x * (ln x)
that is, that the improper integral converges when q is
greater than 1 but diverges when q is less than or equal
to 1, even when q is zero or negative.
This requires that you think through the problem into several
cases. Then carry out the integration.
Having established what happens for certain values
of q, one can see that the behavior for other values is forced by
the integral comparison theorem.
Ok here's a second slightly easier though somewhat similar problem:
R.1b Show that
(1) int (from 0 to 1) x ln x dx < oo
(Recall that although x ln x isn't defined at x = 0 the integral can
be defined as a limit of the ordinary integrals from a to 1, as a
approaches 0 from above.) Evaluating the integrals requires you to
first find the corresponding antiderivatives and this is a task for
Integration by Parts. To integrate x ln x dx we can take u = x or
u = ln x; try it both ways as practice - one way requires an extra
step.
Now show that the convergence of the improper integral
(2) int (from 0 to 1) x^3 ln x dx < oo
is a consequence of (1).
R.2 Show that exactly the same rule (as in R.1a) applies to series
1
sum (n from 2 to oo) -----------------
q
n * (ln n)
that is, they converge if and only if q is greater than 1.
Other problems: If I have time, I'll write up some problems and leave them out in
the math department lounge (Reiss 256) - but NOT BEFORE THURSDAY mid-PM. Using
problems from our text with the possible addition of problems given out on Thurs.
(or from the above as well), we'll review Thursday night at 9. I'll also try
to have still more problems to give out on Thursday night ;-)
For Monday night's midterm, as I mentioned below, we'll include a few problems from sections 8.1 through 8.4 in addition to the material on differential equations which we've covered in class. Emphasis will be on Chapter 7 but I'll include a few from the first part of Chapter 8 to begin the reviewing process - as we'll need to go over everything before the final. Here are some problems to work on: Chapter 8 review (p. 641) #1-6 and 9-19.
Selected answers to Ch 8 (p641) above: #2 lim is zero, #4 diverges, #6 conv, #10 div by LCT with sum 1/n, #12 conv by AST, #14 div by Test for Div., #16 conv by ICT since the integral of 1/(x (ln x)^2) from 2 to oo is convergent - the antiderivative is -1/ln(x).
For Friday, as I said in class, we'll have a quiz - focus on the basic method of separation of variables applied to simple examples, including mixing problems and orthogonal trajectories. Also, go over the section on the exponential model - for population growth and radioactive decay. You should certainly know that when y' = ky, the solution is y(t) = y(0)e^{kt}, and how to use this model to find k when the half-life is given or how to find the time to doubling your money when k is given. Would you know how to handle the case when we ask for _tripling_?
For review, I suggest the following problems from the chapter review
(pp. 558-559): #6,8,10,12,16. Selected answers: #16 y = 10(1 - e^{-t/10}),
so y(6) = 10(1 - e^{-3/5}), about 4.5 kg. #6 By factoring,
1 - t + x - tx = (1 + x)(1 - t),
so separating variables, integrating and solving, x = -1 + K e^{(t-t^2)/2},
where K is an arbitrary constant.
For Tues. April 22, not for collection but I might ask something related on a quiz: Do problem #34 of section 7.3. Hint: Read the section on ``Mixing Problems'' on pp. 526--527. We talked about this problem at the end of the class just before break. I pointed out that the rate at which the "new" money is entering is going to zero asymptotically; after almost all the money has been converted to the new currency, the derivative must be small (and positive). Conversely, the initial rate will be the highest. Economics is an interesting area for mathematical applications, and for some strange reason, it tends to pay much better than math ;-) Don't let this problem spoil your Easter break - but it is a nice exercise in differential equations.
Well, it's Monday night so break is over. Here are the problems will talk about in class tomorrow and for Wed. (On Friday and next Monday, I'll go over a few additional items, including review of older material, which will be covered in our last midterm on Monday evening.) In section 7.3, I mentioned above the section on mixing problems and also you should look at the section on "orthogonal trajectories". The latter concept is discussed below. Try exercises 23-26, 32, and 36 of 7.3. Warning: 32 and 36 aren't for the faint-hearted!
Here's problem #34 in a nutshell: You have $10B (billion) in circulation, of which $50M (million) goes through bank clearance each day. When money goes through the banks, old currency is replaced with new. Let x(t) be the amount of new currency at time t (using the terminology suggested in the text. A convenient scale is to use $B = "billion dollar" units. Note that the daily fraction accessible to replacement is .005 times the total currency supply. The assigned exercise (#34) asks for (a) formulation of a model for the "initial value problem" that represents the flow of new currency into circulation. Then (b) solve the initial value problem and (c) find out how long it will take until 90 percent of the currency has been converted. (One can imagine lots of other processes described by such models ...) In other words, (a) asks for a differential equation including some initial condition and perhaps other relevant parameters necessary to describe the situation and representing the given information. Then (b) requires a solution to the differential equation and initial condition(s). Finally, (c) is an evaluation of t_1 for which x(t_1) = 9$B. For instance, x(0)=0 and clearly x(t) approaches 10$B as t approaches infinity. Moreover, the derivative x' of x is a positive, strictly decreasing function of time since there is less and less of the old money in circulation to be converted by the replacement process. Hopefully, this is enough to help you find the model. If you can't manage it, try to ask one of your classmates before class. But try hard on your own as it is quite a nice feeling when you do see how the differential equation has to work. Oh, for (c), the answer should be 200 ln(10) days (about 461 days).
Speaking of orthogonal trajectories, here's a quick sketch:
Imagine two performing families of fleas, les rouges et le noire (red and black), which wander about on the plane tracing out curves. The red guys sweep out a parallel family of curves - e.g., the horizontal lines with even spacing, and the black ones sweep out another parallel family of curves, but whenever the red and black curves cross, they do so at right angles, as would be the case with vertical black curves crossing horizontal red ones. Another extreme case, which is a bit more interesting, is to have the red curves all radiate out from one central point (where it gets crowded, even for fleas) while the black ones are circles centered at the same point. Now in general, we give you the equations for the red curves and you calculate the slope of a tangent line. Now the line L' orthogonal to a given line L has slope m' = -1/m if L has slope m, and this will give a differential equation for the second family. The constant of integration coming from solving the differential equation will select one of the various orthogonal trajectories. For instance, in #23, the first family y = kx^2, gives a family of parabolas, and the orthogonal trajectories will be a corresponding family of ellipses, so the drawing looks something like a video test pattern. Do your best to make some sketches _WITHOUT_ using the ubiquitous graphing calculator.
Collected on Tues. April 15 , problem #2 below (do it two different ways, via power series and by finding two independent solutions (as sin and cos are independent sol'ns to #1). This will count as 2 of the 5 problems. For the rest, do 7.3:#8 but with the additional condition that z is 1 when t=0. What can you say about the domain of the function z(t)? Also, 7.3:#14, and 7.4:#14.
We looked at section 7.3 today and I assigned the following problems:
(note slight change here to reflect what is written below; there
was a misprint which Katherine pointed out)
#1. Find f such that f'' = -f with f(0)=1 and f'(0)=1. (orig f(0)=1 and f(1)=1 but
there might not be such an f;
also, what is written below
uses f'(0)=1)
#2. Find f such that f''= f with f(0)=1 and f'(0)=1.
Here is the solution of #1 using three different methods I mentioned:
Knowledge of particular functions: We know that deriv of sin is cos and
deriv of cos is -sin, so sin and cos both satisfy the differential eqn.
Hence, any linear combination does so the ``generic'' solution to the diff'l
eq'n is f(x) = A*sin(x) + B*cos(x). Now we can use the given conditions to
find A and B: f(0) = B*cos(0) = B so B=1. Also, f'(x) = -B*sin(x) + A*cos(x),
so 1 = f'(0) = -B*sin(0) + A*cos(0) = A, so A=1. Hence, f(x) = sin(x)+cos(x)
is the unique solution.
Warning: The following paragraph uses notation i = sqrt(-1). Otherwise,
all operations act on i just as though it were another number. For
example, 1/i = -i since 1 = -(i^2). Any number written in the form
z = a + bi, where a and b are real, is called a complex number.
For z and z', z+z' = (a+a') + (b+b')i, and z*z' = (aa'-bb') + (ab'+a'b)i.
Define the conjugate of z, written z-bar, as a - bi (reverse sign of
the imaginary component.
Exercise: z times its conjugate is a^2 + b^2 (usually called the norm
or absolute value of z, written |z|).
Playing with exponentials. We know that the second derivative of e^{kx}
is k^2 * e^{kx} - just use the chain rule each time. If k^2 = -1, then
the resulting function e^{kx} is a solution to the differential eq'n.
Thus, the generic solution is f(x) = Ae^{ix} + Be^e{-ix} since +/-i
are the two square roots of -1. To calculate the coefficients for this
``basis'', we note that f(0) = A + B, so by the initial conditions
(1) A + B = 1
while f'(x) = Ai*e^{ix} - Bi*e^{-ix}, so 1 = f'(0) = i(A - B). Hence,
(2) A - B = -i
Adding the two equations, we get A = (1/2)(1-i) = (1/2) - (i/2), which
is a complex number (as you'd expect since we have the imaginary number
i in the exponent of e. Now B = 1 - A = (1/2) + (i/2). Two complex
numbers which have the relationship of this A and B - real parts are the
same but imaginary parts have opposite sign, are called conjugates.
(We don't use this fact here but it is a nice example for future reference.)
So the unique solution with respect to the basis e^{ix} and e^{-ix}
which satisfies the initial conditions is
f(x) = (1/2 - i/2)e^{ix} + (1/2 + i/2)e^{-ix}.
Although this looks different from the last solution, they are in fact
exactly the same function since there is a _unique_ solution to such a
differential equation with initial conditions. In fact, if you write
down the power series for e^x and replace x by ix (or -ix) and multiply
by the two given conjugate complex numbers above, you get something
simpler and all of the complex numbers drop out! (This is a nice
exercise if you want to practice complex arithmetic ... ;-)
So what about the third solution using power series to solve a diff eq?
Suppose a solution f(x) has a power series representation
f(x) = Sum_(n from 0 to oo) c_n x^n. Then differentiating term by term,
f'(x) = Sum_(n from 1 to oo) n*c_n x^{n-1}
= Sum_(n from 0 to oo) (n+1)*c_{n+1} x^n
and
f''(x) = Sum_(n from 2 to oo) n(n-1)*c_n x^{n-2}
= Sum_(n from 0 to oo) (n+2)(n+1)*c_{n+2} x^n
Using the differential equation, we have
f'' = -f so for all n, c_n = -(n+2)(n+1)*c_{n+2}, or c_{n+2} = -(c_n)/(n+1)(n+2),
since two power series are equal if and only if all pairs of corresponding
coefficients are equal.
Now c_0 = f(0) so c_0 = 1; c_2 = -c_0/2 = -1/2, c_4 = -(-1/2)(1/12) = 1/24,
etc. Similarly, c_1 = 1, using the initial condition that f'(0)=1, so the
coefficients are c_3 =-1/6, c_5 = -(-1/6)(1/20), etc. With a bit of practice,
you recognize that for all n, |c_n| = (1/n!), whether n even or n odd. Or one
can prove it using induction (for the odd and even cases separately) since
c_0 = 1/0! and c_1 = 1/1! so that if |c_n| = 1/n! by the inductive hypothesis,
then |c_{n+2}| = |c_n|/(n+1)(n+2) = (1/n!)/(n+1)(n+2) = 1/(n+2)! as required.
However, this is not the series Sum_(n from 0 to oo) (-1)^n (1/n!) * x^n
which is the power series for e^{-x). This is good since g(x) = e^{-x}
is not a solution to diff'l eq'n 1. above; rather, g'' = g.
If you look more carefully, you'll see that we actually have the sum of two
power series (the odd terms and the even ones), and these are just the power
series for sin(x) and cos(x), resp., so again the unique solution f to the
differential equation is sin(x) + cos(x). More specifically, the power series
has the first two terms positive, then the next two are negative, then the
next two are positive again, etc. But for e^{-x}, the power series is
changing sign at every term. (This is an example which shows that the
condition in the alternating series test is sufficient but not necessary
for convergence.)
Now see if you can do this for differential equation #2 above: f'' = f.
As mentioned in class, f(x) = e^x is certainly a solution, and it isn't
Too difficult to find another similar one. Hence, you can carry out the
first method. The second method is essentially the same here. For the
third method, use power series and see what happens. I will include
this problem on the homework which will be due on Tuesday. FYI, Wed.
we do have class which will be primarily review of the new material.
For section 7.3, which we will discuss in class on Monday April 14, try
problems 2,4,8,10,16. On problem 6, covered in class, I left out the +/-
sign preceding the square root on the RHS. Let me summarize:
Given y' = xy/2*ln(y), we separate variables and integrate to get
integral (2*ln(y)/y) dy = integral x dx,
so (ln y)^2 = x^2/2 + C. Hence, taking square roots,
ln y = +/- sqrt(x^2 + C),
so that, as Melanie described, when we apply the exponential function
to both sides of the last equation, we get y = e^{+/- sqrt(x^2 + C)},
which is the solution.
Please read section 7.4 (up to the part about "orthogonal trajectories") and try problems 4,8,10,12,14,18 for Wed. Here are some of the answers. 7.4: #4(a) y(0)=120, (b) y(t) = 120 * 5^(t/2), (c) y(5) = 120 * 5^(5/2) (this is sufficient for the answer but clearly 5^(5/2) = 25*sqrt(5) so y(5) = 120*25*sqrt(5) = 3000 sqrt(5) = 6708 (a bit more than 6000 as sqrt(5) is a bit more than 2). I actually prefer that you only show the abstract form of the answer, e.g., 120 * 5^(5/2, since it is easier to see from it whether you've done the calculations correctly and also since it eliminates a major source of error. On a test or quiz in class, I don't allow use of calculators which is another reason to keep the answer as simple and clear as possible. However, in many cases, one can estimate the answer approximately - e.g., as above. That is the best way to answer these problems as far as I am concerned.
Here is a basic exercise, similar to what I asked in class but a bit more general. I won't ask for it on Friday - but I might ask you to do it on a quiz! Show that log_a (b) = ln(b)/ln(a). Hint: Multiply through by ln(a) and then exponentiate both sides of the resulting equation; i.e., if u=v, then e^u = e^v.
Another idea for a quiz: Prove the "rule of 69" which is the correct form (not 70 or 72), using the fact that ln(2.0) is 0.693147, according to Mathematica. The rule of 69 says that if you are earning a rate of R percent per year interest and the interest is being compounded continuously, then the principle will double in approximately 69/R. Probably, in practice, interest is only given, say, quarterly, so that one would have a slightly smaller rate of increase, which could reflected in a slightly longer time to doubling: Hence the possibility of a traditional "rule of 72" being correct after all when the interest is not compounded continuously.
Returning to the answers for #4 (d) y(t) = 120 * 5^(t/2) = 120 * e^((t/2)ln 5) - the latter equality is just the "swiss army knife" trick, or equivalently ignoring the simplification that rewrote e^((t/2)ln5) as 5^(t/2). Hence, y'(t) = 5^(t/2) * 120 * ln(5)/2 (using the Chain Rule) and so y'(5) = 25*sqrt(5)*60*ln(5) (best form) = approximately 5398 bacteria/hr. (e) t = 2*ln(5000/3)/ln(5) or t = 2*log_5(5000/3). Since 5^4 = 625 and 5^5 = 3125 while 5000/3 = 1667 (approx), t is between 8 and 10. The answer you get by explicit calculation using a computer is about 9.2 hours.
Problem 8 was done in class. For 7.4: #10, (a) t_h = - 3*ln(2)/ln(0.58). Since 0.58 is close to 1/2, ln(0.58) is nearly equal to -ln(2) so t_h should be near 3. Actual numeric answer: 3.82 days approxly. For #14 (a), y(t) = 5 + 15*e^(kt), where k=ln(7/15). Hence, y(t) = 5 + (7/15)^t. This is reasonable since the second term is approaching zero from above. In fact, as 7/15 < 1/2, y(10) is within 1/1000 of a degree of 5 - i.e., the coffee has cooled from room temp. to nearly the ambient outdoor temp. This sort of physical fidelity should be expected in differential equation problems.
For Friday, read section 7.3 (on separable differential equations) and do problems 7.3: #1-6. Those who got a pass today should be ready! I'll also post some easier (numerically) problems involving exponentials.
For previous Monday, started with problems 1-18 of 8.7.
Homework due on Tuesday April 1 8.6:#6,10,14,22,24 (collected).
For Fri. April 4, please do as many as you can of the problems 1 through 18 of section 8.7, and also review those from section 8.6 which we haven't yet been over. Can you easily recognize when a power series has a finite or an infinite rad of conv? Can you tell in the case of a finite R when one, both or none of the endpoints are included? Check by then doing the problem via the Ratio Test and specific endpoint tests - but see if you can usually guess it just from appearance!
Remember we have our 3rd midterm next Monday evening and it will be on Chap. 8, sections 1 through 7.
For Tues. March 25 to be collected 8.5: 4,8,10,16,18. Note that when the power series involves, e.g., (x-a)^n, then the Interval of Convergence is centered at a, not at 0. Also, remember that the Ratio Test will settle these problems, as far as the radius of convergence is concerned, then you need to check endpoints using various other tests.
For Wed. 26th, review sections 8.3 through 8.5; there will be a quiz! Also, no class on Friday (as announced in class). To review, go over the homework problems and similar odd numbered problems.
Homework due on Tuesday April 1 will be selected from the following set of problems which will be discussed on Monday 31 March: 8.6:#4-16,21-24 (all, of course the odd ones have the answers).
For Friday March 7, read sections 8.3 and 8.4 and try 8.3:#12--24 (inclusive) and 8.4:#3--8,19-26. For Tuesday March 18, collected 8.3:#14,16,22; 8.4:#8,24. For Wed., we reviewed earlier material with a quiz and I suggested you try a couple of exercises for Friday March 21. For instance,
Show that sum_(n from 1 to oo) P(n)/c^n converges absolutely provided that P(n) is any polynomial and c > 1 is a constant. Also, check that the derivative f'(x) is negative when x is large enough if f(x) = (ln x)/x and similarly for g'(x) if g(x) = x/(x^2 + 1). (These last two are just a review of your ability to calculate derivatives and to do a bit of algebra.)
Also, for Fri. the 21st, do 8.5 #4-12 (inclusive) and, more challenging, #24(a). I didn't go over these in class today so I'll give you an example below:
8.5:#6 Find the radius of convergence and interval of convergence of
the series sum_(n from 1 to oo) (x^n)/(n^2).
These power series problems require the use of the Ratio Test to find
the interval of convergence. Then a separate argument will determine
whether or not the endpoints of the interval are included. In this case,
n+1 2
|x| n n 2
lim -------- ------ = lim ( --- ) |x| = |x| which
n-->oo 2 n n-->oo n+1
(n+1) |x|
is less than 1 if and only if |x| < 1. (This last statement is trivial
here but in other examples, when you calculate the limit of the absolute
value of the n+1-th term divided by the absolute value of the n-th term,
you don't always get |x|.) Thus, the Radius of Convergence for this problem
is 1. (See problem 9' below for another example.) The Interval of
Convergence is -1 to 1 but we have to check whether or not to include
the endpoints. Plugging in x = +1 into the series, you get a p-series
with p=2 which is convergent so +1 is included. Plugging in x = -1 gives
the corresponding alternating series which also must converge since it
converges absolutely. Hence, the I of C is [-1,1].
8.5:#9' (modified from text) sum_(n from 0 to oo) 3^n x^n / (n+1)
(I've just dropped the square in the denominator.) Now the ratio test
limit is lim_(n from 0 to oo) 3|x| n/(n+1) = 3|x| which is less than 1
if and only if |x| < 1/3 so the Ratio of Convergence is 1/3. To check
the endpoints, plug in x = +/- 1/3 and see what happens to the series.
It will become a harmonic series for x=1/3 and an alternating harmonic
series for x=-1/3, so the I of C is [-1/3, 1/3); that is, the lower
endpoint is included but the upper endpoint is not.
I went over 8.3 #24 in class on 3/14. In case you missed it, the
problem asks us to find out whether or not the series
Sum_(n from 1 to oo) a_n
converges, where
n + 5
a_n = ------------------
(n^7 + n^2)^(1/3)
The approach on this and similar problems is to recognize that the Limit
Comparison Test shows that the series Sum a_n is equivalent with respect
to convergence to the series Sum b_n, where
n
b_n = --------
n^(7/3)
which is clearly a p-series with p = 4/3 and so is convergent by the p-test.
To show that the two series are LCT-equivalent, we need to check that the
conditions on p. 588 hold. This is true since both a_n and b_n are positive,
and, by rearranging slightly, lim_(n --> oo) a_n/b_n =
n + 5 n^(7/3)
lim_(n --> oo) ----- . ----------------- = 1,
n (n^7 + n^2)^(7/3)
so the limit of the ratio is 1 which belongs to the interval (0,oo).
(A p-series just means a series of the form Sum_(n = 1 to oo) 1/n^p.)
Basically, one just ignores the lower order parts of the term and looks
to see what simpler form the term approaches asymptotically.
I also went over 8.4 #4. Here we can't use the Alternating Series Test
(AST) since the series -(1/3) + (2/4) - (3/5) + (4/6) - (5/7) + - ...
has the limit of its sequence of terms approaching 1, not 0. But by
the Test for Divergence (TD), that means that it cannot be convergent.
Note, however, that if the series were, for example,
-(1/3^2) + (2/4^2) - (3/5^2) + (4/6^2) - + ...
then it would converge by AST since the terms alternate in sign,
each is smaller than the preceding term, and the necessary condition
that the limit of the sequence of terms is zero does hold. For this
second version of the series, we could write the general term c_n as
c_n = (-1)^n . n/(n+2)^2,
and the series is now Sum_(n = 1 to oo) c_n. To see that c_(n+1) < c_n,
as I claimed above, you can either show that f(x) = x/(x+2)^2 has its
derivative negative for x any sufficiently large positive integer or
one can use algebra. The term n/(n+2)^2 is larger than (n+1)/(n+3)^2
if and only if n(n+3)^2 > (n+1)(n+2)^2. But this is equivalent to
n^3 + 6n^2 + 9n > n^3 + 5n^2 + 8n + 4. Subtracting, we get n^2 + n > 4,
and this does hold once n is at least two. If you look at the series,
-1/9 + 2/16 + 3/25 - 4/36 + - ..., you can see that after the first
term, successive terms are getting smaller in absolute value. Also,
the limit of the sequence of terms is clearly zero.
Ok, I hope these hints are helpful.
For Tuesday March 4, please do 8.1:#22,40; 8.2:#12,34,36.
Of course, you all know that we have our second midterm on Tuesday
night at 8:15 in Reiss 283.
Other general information, including for other courses,
on the index (classroom) page .
Meanwhile, in terms of which sections and what specific material to be
studying for the test, I suggest the following as a list of topics which
is representative:
Sections 6.1 through 6.4, 6.7, 8.1 and 8.2.
From 6.1 (on areas between curves): Especially examples 1,2,5.
6.2 (on volumes of various solids): Especially examples 2,3,4,6,7,9;
see p. 459 for the basic formulas for ``disks'' and ``washers''
and for example 9, the formula is also the integral of areas
but this time each of them has the form A(x) = 2pi * r(x) * h(x)
so the shells method integrates the areas of cylinders.
6.3 (on arclength of curves): Esp. examples 1,3,4. For these examples,
put your emphasis on the ability to set up the integrals;
that is, focus on how to use calculus to calculate the lengths.
6.4 (average value): Esp. examples 1 and 2, and the def. on p. 473,
and the mean value theorem for integrals p.474.
6.7 (probability): Esp. examples 1,2,3,4, p. 494 (def of mean value).
8.1 (sequences): Esp. examples 1,3,4,5,6,8,9,10,11, Thm. 2 p.565, Limit
Law and Squeeze Thm. p566, sequence increasing or decreasing p.568,
bounded p.569, monotone sequence thm. p.570.
8.2 (series): Esp. examples 1,2,3,5,7,8. You should know the definition
of series and convergence of series in terms of the two associated
sequences: the sequence of terms and the sequence of partial sums
(p.574), and the consequences of the definition: Thm 6 and the
Test for Divergence (p.578); also you should know the arithmetic
of convergent sequences (p.579) and its consequences.
Please see the background page for calculus
for information regarding the course mechanics and grading, and also for
the schedule of four midterm exams, given in the _evening_ to allow for
more time.
In general, your review of a section should always include doing some
of the odd number problems which have answers in the back of the text.
Also, examples from the sections provide additional practice - if you
try them on your own and then compare your solutions to those given.
For Fri. 21st, we did: 6.1:#16,6.2:#32,46;6.3:#18;6.4:#4. You should
do 6.4:#14,16 and 6.7:#4,6 which we will go over in class on Friday;
be prepared to do them at the board.
On Monday 2/24 I'll tell you about the normal distribution. Meanwhile,
just work on problems 4 and 6. Here are a few more from the earlier
parts of the chapter (sections 1 through 4): pp. 500-501 #2,4,6,12,14.
Some of these problems will be selected for collection on Tuesday.
Collected on Tuesday 25 Feb.: 6.7:#6 (a)(ii) and (b), and pp. 500-501 #2,4,6,12.
For Wed. please read sections 8.1 and 8.2; we'll go back to Chapter 7 after
we have finished some material in Chap. 8. The second midterm is coming up
next week and it will include the sections from Chapter 8 that we have time
to cover. For Friday (or Monday if the snow is heavy), please do:
8.1:#4-24 (even),38,40; 8.2:#10-22(even),26,28,40. In doing these problems,
you will need to use such techniques as partial fractions and l'Hospital's
rule for a couple of them. Also, recall that we have some results on when
improper integrals converge (based on the squeeze theorem) and they can be
applied to give easy proofs that some of the series converge - however, if
one wants to find the particular value to which a series converges, then
other techniques may be needed. For instance, the series
1 + 1/4 + 1/9 + 1/16 + ... = sum_(n = 1 to oo) (1/n^2) must converge by
the p-test for integrals. However, finding the value to which it converges
is harder (we won't do it here but the actual value is pi^2/6).
I'll be here on Sunday afternoon (3/2), and I'll give a review for those who
can use some extra interaction at 3pm in Reiss 281,283 or 284 (depending
on availability).
Here are some hints for 6.7, #4: a function f is a
pdf (probability density function)
if int_(-oo,oo) f(t) dt = 1 and f(t) >= 0; that is, if the function
is always nonnegative and has definite integral 1 on the real line.
(Note that the last integral is an improper integral but for many
choices of pdf the function is zero except on a finite interval,
so in those cases, the integral is just an ordinary integral.)
The function given for problem #4 is clearly nonnegative,
and it is easy to check that its integral is 1 using geometry. You should
say just how this is done in your answer. One gets the answer to 4b(i)
also by calculating an area using geometry. As a corresponding example,
note that P(8 < X < 10) = P(X >8) = area of triangle = (1/2)*2*0.1 = 1/10.
Also, P(3 < X < 8) = P(0 < X < 8) - P(X < 3) - P(X > 8). To calculate
the mean of the pdf f, one determines int_(-oo,oo) t f(t) dt. Here the
function f is given in graphical form so you must write it in the form
f_1(t) 0 < t < 6
f(t) = {
f_2(t) 6 < t < 8
and f(t) = 0 elsewhere. In fact, f_1 and f_2 here are linear functions,
though not the same one since f_1 has positive slope while f_2's is negative.
For any integrand f(t) which is continuous on the real line,
int_(-oo,oo) = int_(-oo,0) + int_(0,6) + int_(6,8) + int_(8,oo).
For this particular f, the first summand is zero since f(t) is zero for
t negative, and similarly the fourth summand is zero. The second summand
is int_(0,6) f_1(t) dt, etc.
For #6 in this section, use the exponential pdf (pp.493-4) and the result
of example 3 (p.495), which tells you how to determine the parameter from
information about the mean. For example, you should get 1 - e^(-1/5) for
part a(i).
==============================================================
I defined the notion of probability in class as follows:
For a set S (nonempty) we define a number P(A) for every subset A of S
subject to the following conditions:
(i) 0 <= P(A) <=1 (probabilities are between 0 and 1),
(ii) P(S) = 1 and P(0) = 0 (probability of S is 1 and of 0 is 0),
(iii) P(A + B) = P(A) + P(B) provided AB = 0
(probability is added when the subsets are disjoint).
(The notation "0" means either zero or the empty set depending on context.
Similarly, "+" means ordinary addition for numbers but "union" for subsets,
where the union of subsets A and B is the set of all elements which belong
to _either_ A or B, or both, while "AB" means the "intersection" of the
subsets A and B, which is the set of all elements common to _both_ of them.)
If you want a nice exercise, show that the second part of condtion (ii)
above, P(0)=0, is redundant. It follows from (iii). I sketched in class
another consequence of the axioms above: If U is a subset of V, then
P(U) <= P(V) since V = U + (V-U), where V-U is the set of all elements in
V which are not elements in U. Since the intersection of U and (V-U) is
the empty set, P(V) = P(U) + P(V-U) by (iii) with A = U and B = V-U.
Example: Take a finite set S and for any subset A of S, let P(A) be the
ratio of the number of elements in A divided by the number of elements in S.
Since we require S to be nonempty, the denominator is not zero so the
quotient makes sense. This corresponds to the usual notion of probability
when all elements are "equally likely" and you can check that it does
satisfy the three conditions above. For instance, the probablilty that
a subset with one element from a standard deck of cards is the king of
diamonds is the number of elements in the subset divided by 52 - i.e., 1/52.
The notion of probability can be applied to measurements as follows:
If one makes a measurement X of some ``random variable'' (r.v.) x, there
is a probability that X is between a and b in value - this is denoted
P(a < X < b). The set S corresponds to the ensemble of all possible
measurements. It is not at all obvious that such a concept is logically
rigorous - how can one say what "all possible" measurements are? - but
it works quite well in practice. I'm telling you a bit more than you
need to know (and less than I would need to tell you if we were going
to do a course in probability, e.g., figuring out the probablity of
getting four of a kind in poker). But hopefully you will find it of
some interest.
To actually carry out the problems from section 6.7,
all you need to understand is the notion of pdf given above and the
examples described in the section. For Monday, problems 4 and 6 only
need the first part of the section - up to the subsection on normal
distributions.
Two points to please note: The use of "x" and "X" in probability theory
is different from that we've been following. The lower case means the
r.v., while the upper case means a particular value of the r.v. Also,
for probability of continuous random variables (the only ones we are
studying), there is _no_ distinction between "<" and "<=", whereas for
some problems in analysis, such as the existence of a maximum value,
it makes a very big difference if an interval is (a,b) or [a,b].
The reason is that the probability that a temperature measurement,
for instance, is between 10 and 20 degrees F. is the same whether we
allow 10 and 20 to be in the interval or not since the chance that a
measurement would be _exactly_ 10 or 20 degrees is zero. That is,
the kind of numbers one considers for a r.v. are by their nature
not exactly determined.
For Monday Feb. 10, please read sections 6.1 and 6.2 and do the following
6.1:#4,6,8,10,12,16; 6.2:#2,4,22 (hint: use the formula for a cone which
we proved today in class: Vol of cone = pi*(radius of base)^2 * height,
and apply it by describing the figure given in the problem as what is
left over after cutting off a smaller cone on top from a larger cone;
you'll need to use similar triangles to find the heights of the two cones).
I didn't (yet) assign any problems using the "method of washers" so these
can all be done using the "stacked coins" (i.e., disks).
For Wed. (to be done at the board by students ;-) 6.2:28,44; 6.3:4,18,24
For Friday, here are a few more: 6.4: 4,6,10,14,16. These problems
involve average value of a function on an interval, which is the integral
divided by the length of the interval, where we only consider finite length
intervals. What is the average value of an odd function on a symmetric
interval (i.e., an interval of the form [-a,a], for a > 0)? The average
value of sin x on the interval [0,pi/2] is the integral (which is
cos 0 - cos (pi/2) = 1) divided by pi/2, so 2/pi is the average.
What is the lim_(b --> oo) avg value of arctan x for x in the interval
[0,b]? (Hint: You don't have to evaluate the integral to see the
answer ;-)
(Collected on Tuesday 2/11/03 ) 6.1#4,8,10; 6.2:#8,10
Find least value of n such that the error in
approximating integral of (1/x) on [1,4] by M_3 is not more than 1/10.
ans. n=5.
The integral of x^(-p)
from 0 to oo converges if and only if p > 1. For p <=1, the integral
diverges. This is also covered in the book. The idea of improper
integrals also applies to situations where both of the endpoints of the
integral is finite but the integrand isn't defined at one or both of them.
For example, the integral of ln(x) from 0 to 1 converges since it is
defined to be the limit, as a --> 0+, of the integral of ln(x) from a to 1.
But I_a = int_(a to 1) ln x dx = - 1 - (a ln a - a) since ln 1 = 1. Hence,
lim_(a --> 0+) I_a = -1 - lim_(a --> 0+) (a ln a) = -1 because the limit
as a --> 0+ of a ln a is zero by l'H rule since
a ln a = (ln a)/(1/a)
and both numerator and denominator goes to oo as a --> 0+ (though the
algebraic signs are different).
Here are some review problems to do for Tuesday. Chapter review pp. 438-440:
T/F Quiz: 2-18 (even number problems);
Regular review exercises: 4,8(a,b,c),12,22,26,28,30,34(hint: use symmetry!),
40,54,56!,60,70.
Ans. 12. 1/10
18. make substitution u = 16-3x and get (1/9)int (from 16 to 4, note
reverse direction) of (u^(3/2) - 16*u^(1/2)) ... messy ...= 3008/135.
22. (1/3)ln 4
26 t=3x sub gives (1/9) integral t cos t dt = (1/9)(t sin t + cos t)
(using IBP with u=t) and with orig'l var. (1/3)x sin 3x + (1/9) cos 3x + C
30 sub y = x^2, ans. is (1/2) arcsin (x^2) + C
32 IBP u=arctan x, ans. is x arctan x - (1/2)ln(1+x^2) + C
34 zero (the function is an odd function)
40 ans. is -(sin x)^(5/3) (remember than (1-cos^2 x)=sin^2 x)
58 since 2-3x=0 for x = 2/3, 1/(2-3x) isn't continuous on [0,1]. Hence,
the integral of 1/(2-3x) from 0 to 1 must be considered as the sum of two
improper integrals from 0 to 2/3 and from 2/3 to 1. But by a substitution,
lim_(a --> 0+) int (from a to 2/3) of 1/(2x-3) = lim int (from a to 1) of 1/u,
and this diverges since lim ln a is not finite.
60 u = sqrt(y-2) ans. is 40/3
62 converges if and only if the parameter a is negative. The integral
turns out to equal (1/(1+a^2)) lim_(b --> oo) [e^(ab)(a cosb + sin b) - a].
For a>0, the limit blows up, while for a=0, the limit is the limit of sin b,
which oscillates between 1 and -1 so doesn't converge. For a < 0, the limit
converges by the Squeeze theorem
Earlier assignments:
I also would like you to do one improper integral
integral (from 0 to oo) x*e^(-x) dx
As I mentioned in class, you first write it as a limit of ordinary
integrals (with finite endpoints), then use integration by parts,
and finally, to evaluate what you get, you'll need to use l'Hospital's
rule.
For section 5.8, do 6,8,10; you may need to "massage" a problem
first before you can apply the table - i.e., by breaking up an
integral into a sum of integrals or by making a substitution.
Don't worry about computer-aided-algebra systems.
Then read section 5.9 first three pages (pp. 416-419) and
do the following problem for Friday.
Approximate the integral of 1/x for the interval [1,4] using n=3 with
the lefthand endpoint, midpoint, righthand endpoint and trapezoid rule;
that is, for f(x)=1/x, find L_n, M_n, R_n, and T_n. The actual value
of the integral I = ln 4 = 2 ln 2 = 1.4 (roughly). What is the error
produced by the various rules? How does it compare with the theoretical
bound given by formula 3? For the theoretical bound, you have to get
an upper bound K on the second derivative of f on the interval [1,4].
When f''(x) is continuous (as it is in this case), its absolute value
must have a maximum
value in any closed interval so we can use the max value as K. Sometimes,
however, it can be hard to find the exact max value even when we know it
exists, but it may be possible to give an argument that (whatever it is
exactly) the max value is certainly bounded above by some constant.
In this example, the max value is easily computed.
If we had a different example, where the second derivative were, say,
t^2 sin(t^3/3), we could use, for the interval [1,4], K=16, since t^2
can't exceed 16 on that interval while sin can't exceed 1 in absolute
value anywhere. The process of finding such bounds usually involves
a trade-off between smallness of the bound (hence the degree of
accuracy guaranteed) and the effort required to calculate the bound.
Similarly, as the text observes, the size of n (i.e., the number of
steps involved in carrying through the approximation algorithm) is
inversely proportional to the error - more effort guarantees a smaller
error hopefully.
For section 5.7, in addition to the problems listed below and previously
assigned (though not yet discussed in class), do 28, 30 and 32. These
are challenging but good practice! Some hints follow below. We will
go over some of the problems (other than the ones for Tues. homework)
so I'll call on some of you to try them at the board on Monday.
On Tuesday (1/28), I'll review briefly 5.8 and then we'll move on to 5.9
and 5.10 for Wed., Fri., Mon. and Tues. The first midterm, on Tues.
Feb. 4 in the evening, will cover Chapter 5 sections 5.6 - 5.10.
For Tuesday for collection 5.7:#12,14,18,24,30.
Hints: Divide first for 28, then PF
with two distinct linear factors. For 30, make a substitution of u some
function of x; in this case, you can also express x as a function of u.
When you're done, you have a rational function in u, so use PF, where
you will again have two distinct linear factors. For 32, you need to
rearrange the stuff under the radical sign. Then you can make a trig
substitution and solve the integral in terms of an angle. Last, figure
out what the expression in terms of the angle looks like in terms of x.
For the problems from 5.7 below, here are two of the answers: #8 ans. 8/15,
and for #20 ans. is ln |x(x-1)/(x+1)| + C (remember ln (a/b) = ln a - ln b,
etc.).
One more example for your benefit: The trig substitution/trig integral problem,
#10 of 5.7. Try it first! Then you can look through the (slightly tricky)
calculations.
Find I = integral of sqrt(x^2 -1)/x^4. Seeing x^2 -1, one tries
x = sec(t) since, from sin^2 t + cos^2 t = 1, we have (by dividing by cos^2)
sec^2 t = 1 + tan^2 t or equivalently tan^2 t = -1 + sec^2 t.
Since in the integral, we have sqrt(x^2 - 1), x must be at least 1 in
absolute value. If x is at least 1, then we can take t in [0, pi/2)
(see below for the description of the corresponding triangle). If x
is at most -1, the corresponding angle t is in [pi, 3pi/2).
Also, for x = sec t, dx = sec t tan t dt.
Making this substitution for t either in [0, pi/2) or t in [pi, 3pi/2)
ensures that tan t is nonnegative so sqrt(-1 + x^2) = sqrt(tan^2 t) = tan t.
Hence, I = int (tan t) sec t tan t dt/sec^4 t = int tan^2 t dt/sec^3 t.
But by replacing tan and sec by sin/cos and 1/cos, respectively, the integral
I simplifies to I = int sin^2 t cos t dt = int u^2 du, for u = sin t.
Hence, I = (1/3)u^3 + C = (1/3) sin^3 t + C so all that remains is to
evaluate sin t in terms of x. Draw a right triangle with angle t; the side
adjacent to the angle has length 1 and the hypoteneuse has length x so that
sec t = x as required. But the side opposite to the angle t has length
sqrt(x^2 -1) by the Pythagorean theorem so sin t is sqrt(x^2 -1)/x.
For Wed. Jan 22, read section 5.7 and try some of the odd numbered
problems - for which, as you know, the book has answers in the back.
Also for Wed. Jan. 22, I _may_ give a quiz ;-) If so, it will be on topics
previously covered - not on the new material assigned from 5.7.
Note that the solutions manual has worked out solutions for the odd number
problems (that's why the homework is usually only among even numbers).
For Friday, Jan. 24, do 5.7: #8,10(!),12,14,16,18,20,24,26; (!) means hard!
These problems will be discussed and some of you will get to do them at the
board. Also, look at Appendix G - try the examples yourself and then look
to see how the author did them.
Review session on Sat. Jan. 25 from 1 to 2 pm in Reiss 284 (or nearby).
Those who (i) didn't take calculus in the previous semester, (ii) feel that
you can use extra review, or (iii) want to have a little more interaction
should come to the review. I won't cover brand new material but will go over
questions on older material - and even things from last semester as needed.
For the practice problems I've suggested, here are a few of the
answers. If you got the same answer, then you most likely did
the problem correctly but sometimes a wrong answer may really be
just a different form while at other times it can mean that you
were way off or merely that you made a minor algebraic or arithmetic
slip. Note that a minor slip at the beginning can damage the structure
and lead you to use inappropriate methods - compounding the mistake
by trying to force the answer to fit some mold which is wrong. When
things look strange, you may want to check your first few steps to
see if a careless error is responsible.
For section 5.3 #28 ans. is 24, for #54b ans. is 89/6 meters.
For 5.4: #10 ans. is -tan(x). For 5.5: #20 ans. is (1/2)(arctan x)^2 + C,
#40 ans. is e-1; #46 (use a fact about odd functions to get ans. = zero);
#50 (u = a^2 - x^2), #56 ans. is pi/8 (based on area of a circle).
For section 5.6 #16 is (16/3) ln 4 - (28/9), for #20, ans. is pi/4 + (1/2)ln 2.
For Tuesday Jan. 21, homework for collection 5.6: 4,10,16,20,26.
Here is an example of IBP (#6):
Let I be the integral of arcsin(x). Take u = arcsin(x) and dv = dx.
Then du = dx/sqrt(1-x^2) and v = x so, using IBP,
I = x arcsin(x) - int xdx/(sqrt(1-x^2) = x arcsin(x) + (1-x^2)^(1/2) + C
where the second integral is computed using substitution (u = 1-x^2).
For Tuesday, for collection as I described in class, please do
the following problems from section 5.2: #32,34,36 and also the
second Riemann sum problem below - i.e., the one involving the
Righthand Rule.
Not for collection but as a refresher on the Fundamental Thm of Calc,
try the following problems, which will be discussed in class
on Tuesday: 5.3: #20,22,24,28,34,46,52,54; 5.4: #8,10,14.
Also, no class on Wed. 15th (I'll be at the American
Mathematical Society meetings in Baltimore). For Friday, read
section 5.5 and try 5.5: #8,12,14,20,42,48.
I realize that a few of you have not had calculus recently.
If you need some extra time to get back up to speed, I'll be
glad to help. You can also utilize: (i) your classmates and
(ii) the Math Assistance Center.
For Monday the 13th, we are going over the definition of definite
integral as a limit of a sequence of Riemann sums - so read (or
reread) the first two sections. Then in section 5.2, do problems
18,20,22,24 (hint: use eq'n 1 and problem 18b from section 5.1 for
#22,24), 32,34,36. Also, use the Midpoint Rule with n=3 to approximate
int(from 0 to 6) of (x^2 - 7x) dx, and use the Righthand Rule with
n=2 to approximate int(from 0 to pi/2) of sin x dx. Finally, use
properties of the integral to estimate int(from 0 to 2) of e^x dx.
For Tues., a subset of these problems will be selected to be written
up for collection. I'll go over some of them on Monday, calling on
various people in class to help ;-), and then tell you which ones to
write up for collection on Tues. Also, don't forget that I sometimes
give unannounced quizzes - usually short and designed to check that
you have been at least vaguely following what's going on.
The homework for Friday Jan. 10 was collected.
The first exercise is 0.5 on the first page (not too clear after the
xerox): Show that N_+ is in one-to-one correspondence with 2*N_+
i.e., show that there is a function from the positive integers
to the even positive integers which is 1-1 and onto, as defined
on the handout. You can take the proof I gave on the handout
for the function S from N to N_+ and modify it slightly.
The second exercise, 1.3, which I mentioned
in class, is to show that lim (as n goes to infinity) of 1/n is zero,
using the definition given on the handout: i.e., show that for every
epsilon > 0 (no matter how small) there is some (sufficiently large)
positive integer N such that whenever n is bigger than N, 1/n is less
than epsilon. The goal is to write N as a function of epsilon, as I
did on the example on the handout - that lim (as n goes to infinity)
of (1/10)^n is zero. Again, just modify the proof I gave on the handout.
Other general information, including for other courses,
on the index (classroom) page .
Please remember to always use the "reload" feature to be sure
you have the current version of this page!
Homework, tests, etc., which are not picked up in class will be in the
box outside my door (Reiss 252).