Wednesday, October 11  (Lecture #12) 

I remarked:
arcsin
The text calls this function sin^{1} and some other sources
call it asin. I find the "1" superscript awkward, and when I compute
I may confuse it with "one over ...". So I'll call the function
inverse to sine, arcsine, and abbreviate its value at x as
arcsin(x). The "arc" part of the name refers to the angle whose sine
is tangent, and, since we use radian measure here, angles are measured
by the lengths of arcs they cut in the unit circle.
What's happening in the
pictures (left to right):
The first picture is supposed to be a
portion of the graph of sine. It is 2Pi periodic, and its range is
[1,1]. The green line is the "main diagonal", y=x, which also happens
to be tangent to y=sin(x) at (0,0). This is because the slope of the
tangent line is the derivative of sine, which is cosine, and
cos(0)=1. To get the inverse function, we interchange inputs and
outputs. Geometrically we flip the graph over the main diagonal, and
get the second picture. The tangent line is still tangent, but now,
look at the red line. This demonstrates that the flipped graph is not
the graph of a function. It fails the vertical line test to be
a graph of a function. Thus we need to cut away (!) part of the
graph. The "clouds" in bluegreen (?) demonstrate what will be cut
away. And what's left is shown in the third picture. This is the
official graph of y=arcsin(x): domain [1,1] and range
[Pi/2,Pi/2]. It has arcsin(0)=0, and the tangent lines seem always to
slope up, so the derivative should be positive. And if we are very
careful, the lines tangent to sine at +/Pi/2 are horizontal, so the
lines tangent to the flipped curve will be vertical and have no
slope so there will be no derivative at +/1. The derivative of
arcsin should have domain (1,1), the interval without
endpoints.
Consider this process:
Example: The derivative of arcsin(5x^{3}e^{x}) is [1/sqrt(5x^{3}e^{x})](5·3x^{2}e^{x}) using the chain rule.
arctan
The text calls this function tan^{1} and some other sources
call it atan. I will call the function inverse to tangent, arctan, and
abbreviate its value at x as arctan(x).
What's the picture
supposed to show? The initial picture is y=tan(x). This function is
periodic with period Pi, and its domain does not include odd
multiples of Pi/2. The function is rather simple looking (!), always
tilted up, and has vertical asymptotes at odd multiples of
Pi/2. Flipping to get an attempted inverse function reveals lots of
problems (I omitted the red line here). The standard restriction is to
throw out the "branches" that don't intersect the horizontal axis, and
that's what I've attempted to suggest with the bluegreen
"clouds". Again, y=x is a tangent line to both arctan and tan at (0,0)
because the derivative of tangent is (sec(x))^{2}, and
sec(0)=1/cos(0)=1. Arctan is very useful. It "compresses" all of the
real numbers into the interval from Pi/2 to Pi/2, so if you have lots
of data and you don't know ahead of time how big (+ or ) the data
will be, composing it with arctan will at least control it a bit. Now
for the derivative.
The inverse function to exp (arcexp?)
Should this be called arcexp? Well, it isn't. The exponential
function, e^{x}, models exponential growth. It "takes" x and
multiplies e's x number of times (can you understand that
statement?). The inverse function to an exponential function is a
logarithmic function. This logarithm function is very
important. The log functions you may deal with include
log_{10} (used in the definition of pH, and for hand
calculation in "the old days") and log_{2} (used in some
computer science applications). For reasons that will appear very
soon, the log function which identifies how many powers of e appear in
a number is called the natural log. Your text abbreviates it as
ln, and that's what I will use, but note that many sources call it
just log. What's ln? It is a function whose domain is all positive
numbers and whose range is all real numbers.
This
picture shows exp, the exponential function,
e^{x}. Since this function is onetoone, its inverse will be
a function: no more bluegreen clouds! The green line is the tangent
at (0,1) which has slope=1. Then we flip it and get the graph of
ln. What about the derivative?
If y=a^{x} is an exponential function, we could take log
(really "ln") of both sides and get ln y= x ln a. Then
we could differentiate both sides, remembering the chain rule. We
would get (1/y)(dy/dx)=ln a (remember that a and therefore
ln a are both constants!). Therefore dy/dx=y·ln a so
that the derivative of a^{x} is
a^{x}·ln a.
If we were interested in the
derivative of 10^{x} we would be looking at
10^{x}(ln(10)). ln(10) is approximately 2.303 (this is enough
for most purposes I know, but if you need more digits, here they are:
2.3025850929940456840). In computer science people sometimes consider
2^{x} whose derivative is 2^{x}(ln(2)). ln(2) is
approximately .693 (again, enough information for most purposes I
know, but ln(2) is .69314718055994530942 [more or less!]). The nicest
and simplest a^{x} for calculus purposes is one where the
derivative is simplest, where ln a is 1. And that occurs when
a=e. So
Of all the exponential functions, e^{x} is best because it is its own derivative. 

A differentiation technique you need for a calculus class
I differentiated y=x^{sin(x)} by doing this:
Take log's (ln's) of both sides. The result is
ln(y)=ln(x^{sin(x)})=sin(x)ln(x) because ln(A^{B})=B ln(A).
d/dx the resulting equation carefully (implicit differentiation,
product rule). The result of this is:
(1/y)(dy/dx)=(product rule!)cos(x)ln(x)+sin(x)/x
"Solve" for dy/dx, so that
dy/dx=y[cos(x)ln(x)+sin(x)/x]=x^{sin(x)}[cos(x)ln(x)+sin(x)/x].
even more Differentiation algorithms  
Function  Derivative 
f(g(x))  f´(g(x))·g´(x) 
Implicit differentiation: given an equation relating x and y, d/dx the entire equation, and then solve for dy/dx. Be sure to use the chain rule and all of the other differentiation rules correctly!  
arcsin(x)  1/sqrt(1x^{2}) 
arctan(x)  1/(1+x^{2}) 
ln(x)  1/x 
a^{x}  a^{x}ln(a) 
Monday, October 9  (Lecture #10) 

I then put the chart below on display and we discussed it for a while. Here's a pdf copy which you can print out separately if you like.
CHIPCO INVESTMENT DOLLARS & PRODUCTION Capital Invested Chips produced Marginal chips produced $ in millions 1,000's of units 1,000's of units per millions of $'s 200 3,000 .23 300 3,040 .28 400 3,070 .42 500 3,100 .78 600 3,190 .31 CHIPCO SALES & PROFITS Chips marketed Profit gained Marginal profit 1,000's of units $'s in millions Millions of $'s per 1,000's of units 3,000 1.2 .03 3,050 2.8 .02 3,100 3.6 .05 3,150 4.9 .01 3,200 5.1 .02
I don't think many people were familiar with the economic terms used above. In particular, the word "marginal" is used in a fairly technical sense. In the first table above, it refers to the approximate amount would increase per each million dollars of increase in capital investment. Therefore, for example, if $302 million were invested, then (according to this model) chip production would be 3,040,000 (chip production at the $300 million level) plus .28(2)(1,000) chips. The 2 comes from the additional millions of dollars of capital. The 1,000 comes from the units I use for chip production. The .28 is this "marginal" quantity. In the first table, the marginal quantity is therefore the approximate amount P/C, relating the change in chip production to the change in capital investment. It is sort of a slope, or, more likely, sort of a derivative: indeed, the use of "marginal" in economics usually means a derivative. In this model, if $297 million were invested, the approximate expected chip production would be 3,040,000 (again, chip production at the $300 million level) plus .28(3)(1,000). The novelty here is the use of the minus sign, since here we are decreasing the capital investment rather than increasing it.
The second table describes a similar phenomenon, here connecting the chip amount, C, with the profit derived from these marketing and sale of these chips. For example, the profit derived from the sale of 3,000,000 chips (the first line of the second table) is $1.2 million. If we now look at the third column, the model predicts a marginal profit of .03 (in the given units). Using this, if 3,010,000 chips are marketed (that's 10 more 1,000 units of chips) the additional profit would be .03(10) million dollars, or $300,000. And if only 2,970,000 chips were marketed, then the profit would be 1.2 million+(.03)(30)(1,000)million. (I think I got all the units correct.) The third column gives P/C for various amounts of chip marketing: the change in profits compared to the change in chips marketed. Of course the validity of such models can certainly be criticized, but I really wanted to show this to explain what's in the next paragraph.
The two tables linked together describe a complicated phenomenon. First we "input" capital, M (M is for money), which produces C, a certain number of chips. Then the chips are marketed (and sold, hopefully!) to obtain a certain amount of profit, P. Here we have a composition of functions. For example, suppose we were asked how much profit there is if we put in M=$500 million. From the first table we read off C=3,100,000 chips, and from the second table we can then see that P will be 3.6 million dollars.
I hoped that this was all fairly clear. Now I asked what I thought was
a difficult question. Suppose we increase M from 500 million dollars
to, say, 503 million dollars. What does the model predict the profit
will be? We can trace this if we are sufficiently alert. The first
marginal quantity we need to consider is
C/M. For M=$500 million, this is .78. So the new
chip production is old chip production + increase in chip production,
and this will be 3,100,000+(.78)(3)(1,000 chips). Now let us consider
the chip/profit table. With C=3,100,000, we see that profit is
supposed to be 3.6 million. But we are changing C by adding on the
(relatively small) amount of .78(3)(in units of 1,000's of chips). The
relevant marginal quantity here is P/C on the
row where C is 3,100(,000). The marginal amount here is .05, so
that the new profit will be the old profit (3.6 million) plus
(.05)(.78)(3) million dollars. The 3 comes from
perturbing the capital investment. The really interesting stuff is (.05)(.78): indeed, this
represents the marginal profit as capital invested changes, when the
capital investment is 500 million dollars. Symbolically, it might make
sense written this way:
P/M=P/C·C/M.
So the C's just seem to cancel out. Of course, this is more
complicated than just multiplying fractions, since the fractions (the
marginal stuff, the derivatives) need to be "evaluated" on the
appropriate rows of the tables.
What I've done here is tried to present heuristic evidence that would allow us to believe the chain rule.
/heuristic/ adj. 1. allowing or assisting to discover. 2. [Computing] proceeding to a solution by trial and error.The Chain Rule Suppose that f and g are differentiable functions. The F(x)=fog(x)=f(g(x)) is differentiable, and F´(x)=f´(g(x))·g´(x).
The balance of the lecture was devoted to exploiting the chain
rule. There is a correct proof of the chain rule in the book. My first
example was something like this (about as simple as I could
imagine):
If F(x)=(x^{2}+7)^{300}, what will F´(x) be? I
don't need the Chain Rule, not really (?), to compute
F´(x) because, after all, F(x) is "just" a polynomial (although
here F(x) is a polynomial of degree 600, and this polynomial is not
presented in standard fashion). Success (rapid, accurate computation)
here probably will result from recognizing that the chain rule
applies.
If F(x)=f(g(x)), then g(x) is x^{2}+7 so
g´(x)=2x, and f(x) is x^{300} so
f´(x)=300x^{299}. Thus
F´(x)=f´(g(x))g´(x)=f´(x^{2}+7)(2x)=300(x^{2}+7)^{299}(2x).
Whew!
But now comes the realistic comment. Hardly ever does anyone bother writing down all of these intermediate steps. That is, in practice very few f's and g's are actually identified. What happens is that people see and differentiate the outside most function (f above), put in the inner function (g) in that derivative, and then multiply by g´. For example, consider sin(e^{x}+x^{2}). What is its derivative? The outside function is sine, whose derivative is cosine. So I begin by writing cos(what's inside)·the derivative of what's inside. The result is cos(e^{x}+x^{2})·(e^{x}+2x). This expression is a formula for the derivative of sin(e^{x}+x^{2}). Again, I urge you to consider the significance and necessity (!) of appropriate parentheses in these expressions. The "argument" of cosine is e^{x}+x^{2} and the cosine expression is then multiplied by the expression (e^{x}+2x).
The chain rule itself can be repeated. So here, for example, we can try to differentiate cos(e^{3x2}). Its derivative is sin(e^{3x2})·(e^{3x2})·(6x). I hope that you can pick apart the layers of the functions and their compositions. One poor metaphor for using the chain rule is that it is like peeling an onion very very carefully, layer by layer, and taking care always of the outside most layer first. Confusion is certainly possible, and that's an understatement. The "chain" in the chain rule is, I think, a reference to the links (via composition) in a typical algebraically described function.
Here is an interesting application of the chain rule. Suppose we want to differentiate y=sqrt(x). Well (one student in the audience immediately and correctly said, (1/2)x^{1/2}) here is a way to do it. Square the equation to get y^{2}=x and then differentiate the resulting equation. I will switch to what is called Leibniz notation now. Although this notation is not my favorite, somehow it fits with this sort of computation. In Leibniz notation, the derivative of y with respect to x is dy/dx, and what I want to do is d/dx (differentiate!) the equation y^{2}=x. The righthand side is easy: its derivative is 1. The lefthand side needs the chain rule: it has an "outside function", squaring, and an inside function, the "unknown function" y (unknown at least as far as its derivative is concerned). The result of the chain rule is 2y(dy/dx). Since this should be equal to 1, the derivative of the righthand side, we can solve for dy/dx, and we get dy/dx=1/(2y). But y=sqrt(x), so we may recognize that the derivative of sqrt(x) is indeed (1/2)x^{1/2}.
This line of approach may be extended. For example, if y=x^{15/4} we then know that y^{4}=x^{15}. Then d/dx the result to get 4y^{3}(dy/dx)=15x^{14} (do not forget the dy/dx which the chain rule forces upon you!). Then solve for dy/dx. It is (15x^{14})/(4y^{3}). But since y=x^{15/4} we know that (15x^{14})/(4y^{3})=(15x^{14})/(4(x^{15/4})^{3})=(15x^{14})/(4x^{(15/4)·3})=(15/4)x^{14(45/4)}=(15/4)x^{11/4}. Wow! A discovery: the power rule apparently holds for rational exponents. So the derivative of sin(x^{4/7}) must be cos(x^{4/7})·(4/7)x^{3/7}.
The trick of having an equation involving x and y, then d/dx'ing the equation and solving for dy/dx is sometimes very useful. It is called Implicit Differentiation. There are times when it is difficult or impossible to get an explicit representation for y as a function of x. Then having an implicit expression is enough, if what is wanted is some information about dy/dx. I chose a rather strange example. Suppose we look at the equation x^{3}+y^{2}x=0. The picture shown here is a result of the Maple command implicitplot which allows one to plot implicitly defined functions. Of course I was waiting for students to ask why the heck anyone would ever want to plot such a curve, and I picked the curve with some intent: this is an example of an elliptic curve, and cryptographic protocols related to such curves are extremely important today (Google has almost 3,000,000 references to elliptic curves!). If x=2 in the equation x^{3}+y^{2}x=0 we see that 8+2+y^{2}=0, so that y must be +/sqrt(6). What is the slope of the line tangent to this curve at the point (2,sqrt(6))? We take the equation x^{3}+y^{2}x=0 and d/dx both sides. The righthand side gives me 0 (I always prefer to differentiate constants!) and the lefthand side is 3x^{2}+2y(dy/dx)1 which must therefore equal 0 (the chain rule forces the appearance of dy/dx). Inserting 2 for x and sqrt(6) for y, we get 12+2sqrt(6)(dy/dx)1=0 or dy/dx=11/(2sqrt(6)). Indeed, if you consider what the line tangent to the elliptic curve shown must look like, it does indeed slope "down".
Wednesday, October 4  (Lecture #9) 

L
The quiz was graded in a similar fashion to how the exam will be
graded. Students should look at their results and be advised about the
studying they need to do for the exam.
E
There will be an exam in 10 days. Both of the instructors will have
review sessions the weekend before the exam. There will be sample
material for students to see, also.
tenacity!
A former student (in several courses with me, about ten years ago)
recommended that I quote this word to students. Well, tenacity
means doggedness: persistent determination.
Reciprocals
If F(x)=1/g(x), then F(x+h)=1/g(x+h), and
(F(x+h)F(x))/h=[1/g(x+h)1/g(x)]/h (this is a compound fraction, and
I want to write it as a simple fraction, a quotient of two
expressions) =[g(x)g(x+h)]/[g(x)·g(x+h)·h]. Wow. With
effort we can recognize the pieces. First,
[g(x)g(x+h)]/h>g´(x) as h>0. And the "other stuff">
1/g(x)^{2} because g(x), since it is differentiable, must be
continuous. This is a sort of reciprocal rule. Lots
of work.
I remarked that if g(x) is getting bigger and bigger, I would expect the derivative to be positive. But then 1/g(x) would be getting smaller. The minus sign in the reciprocal rule algebraically shows this reversal.
I then computed the derivative of something like 1/x^{207}. The result was 207x^{206}/(x^{207})^{2}. In fact some algebra can be productively performed on this, and we get 207x^{2062(207)}=207x^{208}. Since the original function is 1/x^{207}=x^{207}, we can see that the power rule also holds for negative integers. Recognizing patterns is an important part of mathematics, and a very important part of being usefully lazy.
I noted that the reciprocal rule allows the power rule to be extended to negative integer powers, so that the derivative of 1/x^{33}=x^{33} is (33)x^{34} or, equivalently, 33/x^{34}.
Also we deduced the quotient rule: if F(x)=f(x)/g(x) where f and g are differentiable functions, then we can write F(x)=f(x)·(1/g(x)) so that (using the product rule and the reciprocal rule), F´(x)=f´(x)·1/(g(x)+f(x)·g´(x)/[g(x)]^{2}. This is the quotient rule. The result, F´(x), is usually written as [f´(x)·g(x)g´(x)·f(x)]/[g(x)]^{2}. I did a some "simple" examples:
Moving right along (!) I discussed the trig functions again.
Even more, we will need a kinetic view of the trig functions.
Sine: yet another sketch
I tried to draw an accurate picture of sine and then
discussed what properties the derivative of sine would have.
The derivative of sine is cosine.
In a calculus textbook, something like the following is done when f(x)=sin(x):
sin(x+h)sin(x) sin(x)cos(h)+sin(h)cos(x)sin(x)  =  = PIECE #1 + PIECE #2 h hwhere
sin(h) PIECE #1 = cos(x)  hand as h>0 this > cos(x)·1 because we arranged it this way when we decided to use radian measure! Also,
cos(h)1 PIECE #2 = sin(x)  hIf we multiply this top and bottom by cos(h)+1, the result on top is [cos(h)]^{2}1 which is [sin(h)]^{2}. Then
[sin(h)]^{2} sin(h) 1 PIECE #2 = sin(x)  = sin(x)  sin(h)  h [cos(h)+1] h [cos(h)+1]Now as h>0, I claim:
sin(h) 1 sin(x)>sin(x);  >1; sin(h)>0;  > 1/2. Nothing happens! h [cos(h)+1]So the result is sin(x)·1·0·1/2=0.
Verification that
lim_{h>0}[sin(h)/h]=1
This is sometimes a handy thing to know. Also it is a limit which is
often "requested" on calc 1 exams. Let me give you what I think is a
fairly convincing discussion. I will look at the accompanying picture
to the right. This picture shows a very small angle h, inside the unit
circle (all of the radii are equal to 1).
Smallest area  Middlest area  Largest area 

Triangle ABC  Sector ABD  Triangle ABE 
(1/2)sin(h)  (1/2)h  (1/2)tan(h) 
Comments
The function sin(h)/h occurs quite a bit when folks study vibrations
of various sorts (vibrations in a beam or vibrations in an electric
circuit or ...). Also the limit is sometimes really useful to
know For example, on my "calculator", I just asked for
sin(.0123) and got 0.0002146755. WHAT!!!??? Isn't this wrong? Isn't this way off?
Well, no. I actually asked the calculator the wrong
question. The calculator was set for degrees, not for radians. If
you insist that your trig functions be all functions of degrees, then
the derivatives will be all fouled up. In fact, the true value of sine
of .0123 is actually 0.01229969, which is pretty darn close. So sine
of h radians is darn close to h when h is small. And we can use
this to compute other strange limits, if we have to.
Since lim_{h>0}[sin(h)/h]=1 I know that lim_{h>0}[sin(h)/(5h)]=5. And I know that lim_{h>0}[sin(3h)/h]=lim_{h>0}[3sin(3h)/(3h)]=3 since lim_{h>0}[sin(3h)/3h]=1 (3h gets small along with h after all!)
I remarked that (looking again at the shape of the graphs) we can also see that the derivative of cos(x) is sin(x) (there is a shift of Pi/2 in both graphs). Thus we get two more lines in the table of derivatives.
My final specific example of the quotient rule was to compute the derivative of f(x)=sin(x)/cos(x). Here top=sin(x) and bottom=cos(x), and the derivative will be (top'·bottombottom'·top)/(bottom)^{2}. Notice that because of the minus sign, there is an asymmetry in the result, and this can lead to errors. f´(x)=[cos(x)·cos(x)(sin(x))·sin(x)]/[cos(x)]^{2}. Here is an answer which demands simplification. The top is [cos(x)]^{2}+[sin(x)]^{2} which is 1. The whole result is therefore 1/[cos(x)]^{2} or [sec(x)]^{2}. Of course, this f(x) is tan(x), so we now know that the derivative of tan(x) is [sec(x)]^{2}. Please note the parentheses. I almost always use lots of parentheses, because their use is almost required to make understandable the results of derivative algorithms.
My final gasp was to draw a quick picture, the way we all do, of sine
and cosine. I distorted the bumps as almost everyone does. The bumps
are actually rather flat. But "clearly" the curves intersect, and,
look, look!, it seems that they intersect almost perpendicularly.
Can we check this? Well, sin(x)=cos(x) for x between 0 and Pi/2 when
x=Pi/4 (that's the isosceles right triangle all the way up). Two lines
will be perpendicular when the product of their slopes is 1 (or when
"their slopes are negative reciprocals"). The slope of the line
tangent to sine when x=Pi/4 is cos(Pi/4)=1/sqrt(2). The slope of the
line tangent to cosine when x=Pi/4 is sin(Pi/4)=1/sqrt(2). The
product of these two slopes is 1/2, not 1, so the curves do
not intersect perpendicularly.
more Differentiation algorithms  
Function  Derivative 
1/g(x)  g´(x)/[g(x)]^{2} 
f(x)/g(x)  [f´(x)·g(x)g´(x)·f(x)]/[g(x)]^{2} 
sin(x)  cos(x) 
cos(x)  sin(x) 
tan(x)  1/[cos(x)]^{2}=[sec(x)]^{2} 
EVEN MORE TO COME!!! 
Monday, October 2  (Lecture #8) 

Every differentiable function is continuous.
Physically
this says that if a point moves smoothly, it must move without any
jumps or breaks.
The converse of this statement is not true. So there are continuous functions which are not differentiable (such as absolute value). Physically, there can be motion without any jumps but which is not smooth  the particle can move with kinks and jerks.
The major topic for the next few lectures are the "classical" differentiation algorithms.
/algorithm/ n. 1. [Math.] a process or set of rules used for calculation or problemsolving, esp. with a computer. /alligator/ n 1. a large reptile of the crocodile family native to S. America and China, with upper teeth that lie outside the lower teeth and a head broader and shorter than that of the crocodile. /allegory/ n. 1. a story, play, poem, picture, etc., in which the meaning or message is represented symbolically.
It turns out that if a function is given by a formula involving the standard functions (powers, exponentials, trig functions, and their inverses) then usually the function will be differentiable, and the derivative can be written in terms of the standard functions. This is very nice. The methods, which I will show you, are mostly straightforward and have been implemented in numerous computer programs.
I started discussion of the differentiation algorithms. It turns out that for functions defined by generally simple formulas, there are a series of "rules" or algorithms which allow formulas for the derivatives to be written fairly easily. We will always start with the formal definition although it is comforting to recall such intuition as "f´(x_{0}) is the slope of the line tangent to the graph of y=f(x) at x=x_{0}" and that f' is also instantaneous velocity, etc.
I should have begun with the very simplest sorts of functions. If f(x)=15 for all x, then surely f(x+h)=15 also, so that f(x+h)f(x)=0 and dividing this by h also results in 0, so the derivative is always 0. This result is true for any constant function. (Also the graph is a horizontal line, and is its "own" tangent line, with slope=0.) So we are done.
Now consider f(x)=x^{n}. We need f(x+h) using the formal
definition. f(x+h)=(x+h)^{n} and we could use the Binomial
Theorem to see exactly what the "expanded" version of f(x+h) looks
like (see also information about Pascal's
Triangle). But we actually don't need such precise
information. For example, let's look at n=4. Here (x+h)^{4} is
(x+h)·(x+h)·(x+h)·(x+h). We could multiply and
expand everything or we could look at the structure of things a
bit. There is exactly one product which is all x's, and that product
has degree 4: x^{4}. If we knock out exactly one x and take an
h instead, we would have hx^{3}. How many such terms can we
get? Well there are 4 possible h's to choose, and we only want one of
them, so there is exactly 4hx^{3} in the expansion. Every
other term has at least two h's in it. We could "collect" all those
and label them h^{2}JUNK (junk because we won't have
any need here of its precise nature). So:
(x+h)^{4}=x^{4} + 4hx^{3} + h^{2}JUNK
All x's Terms with 1 h all other terms
We can do this more generally: (x+h)^{n}=x^{n}+nhx^{n1}+ h^{2}JUNK
This is exactly a restatement, by the way, of the
[REMARKABLE] equation I mentioned last time:
f(x+h)=f(x)+f´(x)h+Err·h where you can "see" the higherorder
error terms.
But now what? We consider lim_{h>0}(f(x+h)f(x))/h= lim_{h>0}[(x+h)^{n}x^{n}]/h= lim_{h>0}[x^{n}+nhx^{n1}+h^{2}JUNKx^{n}]h= lim_{h>0}h[nx^{n1}+hJUNK]/h= lim_{h>0}nx^{n1}+hJUNK=nx^{n1} which is in the table.
Your textbook next studies the exponential functions a^{x}. If a>1, this represents exponential growth. Consider the graph of an "average" exponential growth function and the slope of the tangent lines to this curve. As the point of tangency travels from left to right, the slope, which is always positive, just increases. If we could image a graph of the slope function (which is just a graph of the function y=f´(x)) it might look a great deal like a^{x} itself. And that is the truth. A proof of this takes some effort, and could be given now, but we are supposed to run as fast as we can. So I will just write out some suggestive reasoning, following more or less what is in your text. I'll also include some pictures, mostly because pictures help me believe more.
If f(x)=a^{x}, then f(x+h)=a^{x+h}=a^{x}a^{h}. The difference quotient (f(x+h)f(x))/h becomes (a^{h}1)/h multiplying a^{x}. What is the number (a^{h}1)/h as h>0? This number (the limit, if it exists, which it does) is sort of a "fudge factor" that's needed to make the derivatives come out right.
Investigation of the "fudge factor" occurring in derivatives of exponentials  

a=2 Below are pictures of [2^{h}1]/h. The limit seems to exist, and its value seems to be
about .693. Thanks to M. Tsimaras for contributing a calculator graph in class. What's here is somewhat more accurate.  
1<h<1  .1<h<.1  .01<h<.01 
a=2 Below are pictures of
[3^{h}1]/h. The limit seems to exist, and its value
seems to be about 1.09. Thanks to E. Yi for contributing a calculator graph in class. What's here is somewhat more accurate.  
.1<h<.1  .01<h<.01  
n  (1+{1/n})^{n} 

1  2.000000000 
2  2.250000000 
3  2.370370369 
4  2.441406250 
5  2.488320000 
10  2.593742460 
100  2.704813829 
1,000  2.716923932 
10,000  2.718145926 
100,000  2.718268237 
1,000,000  2.718280469 
I did not verify that there is such a number (e, that is). But the following "argument" provides one way to approximate it.
This is, by the way, 2.704813829. Well, you can see it in the accompanying table. This method of "computing" or approximating e is actually very very slow. The first several million digits of e are online, if you need them.
Then I worked on building new functions. If F(x)=f(x)+g(x), and the
derivatives of f and g exist, what can one predict about the existence
and value of the derivative of F?
Since F(x)=f(x)+g(x) we know that F(x+h)=f(x+h)+g(x+h), and the difference quotient for F can be written this way:
(F(x+h)F(x))/h=[(f(x+h)+g(x+h))(f(x)+g(x))]/h=[f(x+h)f(x)+g(x+h)g(x)]/h=[f(x+h)f(x)]/h+[g(x+h)g(x)]/h.
And now let h>0, and we see that the derivative of the sum is the sum of the derivatives.
Now I did a hard problem from the textbook (I think #50 of section 3.1), to show that we have gotten already to some level of achievement. The problem asks us to find the equations of the lines tangent to the parabola y=x^{2}+x which also go through the point (2,3). Note that although the problem statement does not request it, I would almost always begin the solution by making a sketch.
If P=(x,y) is the point of tangency on the parabola, we can solve the problem by realizing that the slope of the tangent line at P, m_{TAN}, can be written in two different ways. First, since the tangent line goes through P and (2,3), its slope is (y(3))/(x2), which is (x^{2}+x+3)/(x2). But m_{TAN} is also f´(x) if f(x)=x^{2}+x. So m_{TAN}=2x+1. Therefore 2x+1=(x^{2}+x+3)/(x2), and (2x+1)(x2)=x^{2}+x+3. Then 2x^{2}3x2=x^{2}+x+3 so that moving everything to one side, we get x^{2}4x5=0. Since this is a problem in a textbook, the quadratic factors into (x5)(x+1)=0. If x=5, then y=5^{2}+5=30 and the derivative is 2(5)+1=11. So the tangent line is y30=11(x5). We can make a cheap check: does this line go through (2,3)? Well, 330=33 and 11(25)=33, so the answer is "Yes." The point of tangency is (5,30) which explains why we can't see it in the picture. You can find the equation of the other line yourself.
Now we began to discuss what is called the product rule. The statement of the product rule begins "The derivative of the product is ..." There is an expectation of simplicity and symmetry here, which should be eliminated as soon as possible. Consider x^{2} which is also, of course, x·x. The derivative of x is 1, and 1·1=1, but the derivative of x^{2} is 2x, so the product of the derivatives is not the formula we want.
If F(x)=f(x)·g(x), then F(x+h)=f(x+h)·g(x+h), so that
(F(x+h)F(x))/h=(f(x+h)·g(x+h)f(x)·g(x))/h. Now the
game is to somehow write this fraction in terms of the difference
quotient of f and the difference quotient of g. Here the picture may
help. It tries to show a sort of decomposition of
f(x+h)·g(x+h)f(x)·g(x). The suggestion is that
f(x+h)·g(x+h)f(x)·g(x)=(f(x+h)f(x))·g(x)+f(x)·(g(x+h)g(x))+(f(x+h)f(x))·(g(x+h)g(x)). If we now divide
by h and let h>0, then:
[(f(x+h)f(x))·g(x)]/h>f´(x)·g(x) and
[f(x)·(g(x+h)g(x))]/h>f(x)·g´(x).
The blue rectangle in the corner is a curiosity. It is algebraically
(f(x+h)f(x))·(g(x+h)g(x)) (then divided by h)
In the [REMARKABLE] equation I mentioned last time:
F(x+h)=F(x)+F´(x)h+Err·h the blue rectangle belongs to the Error term. So what happens is this:
[(f(x+h)f(x))·(g(x+h)g(x))]/h=[(f(x+h)f(x))/h]·(g(x+h)g(x)). The first term >f´(x) but the second term: g(x+h)>g(x as
h>0 since g is continuous (because differentiable functions are continuous).
Therefore the blue rectangle divided by h >f´(x)·0 which is 0: it contributes nothing to the limit. This is a bit elaborate, but I'd like to be honest
when I can be (!?). So now we know that the derivative of f(x)·g(x) is
f´(x)·g(x)+f(x)·g(x). This is called the product rule or, sometimes, the Leibniz rule, memorializing one of the inventors of calculus.
Examples: If f(x)=x and g(x)=x, the product rule gives us 1·x+x·1=2x, the correct answer. We can also differentiate something like e^{x}·x^{178}. And we can differentiate 37·x^{23} with f(x)=37 (a constant function) and g(x)=x^{23}. The product rule predicts that the derivative is 0·x^{37}+37·23x^{22}. Frequently people use this special case of the product rule without thinking about it: the derivative of a constant times a function is a constant times the derivative of the function.
Differentiation algorithms  
Function  Derivative 
f(x)  lim_{h>0}(f(x+h)f(x))/h This is the formal definition 
Constant  0 
x^{n} when n is a positive integer  nx^{n1} 
e^{x} (Here e is approx. 2.71828)  e^{x} 
f(x)+g(x)  f´(x)+g´(x) 
f(x)·g(x)  f´(x)·g(x)+f(x)·g´(x) 
(Const)f(x)  (Const)f´(x) 
MORE TO COME!!! 
Wednesday, September 27  (Lecture #7) 

One example
I'm going to look at g(x)f(x). This is sqrt(x^{2}+1)x. I
want to investigate
lim_{x>infinity} sqrt(x^{2}+1)x.
"Clearly" (always use that word when you can't explain where the idea
comes from) multiply this by a fraction which has the same thing on
the top and the bottom (so the fraction is a fancy way of writing
"1".) Here the following fraction was suggested:
sqrt(x^{2}+1)+x  sqrt(x^{2}+1)+xI was told that the top and bottom were each the conjugate of the formula for g(x)f(x). Then
(sqrt(x^{2}+1)x)(sqrt(x^{2}+1)+x) g(x)f(x) = sqrt(x^{2}+1)x =  = sqrt(x^{2}+1)+x) (x^{2}+1)x^{2} 1  =  sqrt(x^{2}+1)+x) sqrt(x^{2}+1)+x(Here we are using (AB)(A+B)=A^{2}B^{2} with A=sqrt(x^{2}+1) and B=x. The final fraction shows us that the result >0 since the bottom grows to "infinity". Therefore lim_{x>infinity}g(x)f(x)=0.
Another example
Now let me look at h(x)f(x). This is sqrt(x^{2}+x)x. I'd
like to consider
lim_{x>infinity} sqrt(x^{2}+x)x.
Now the appropriate conjugate is (sqrt(x^{2}+x)+x).
(sqrt(x^{2}+x)x)(sqrt(x^{2}+x)+x) h(x)f(x) = sqrt(x^{2}+1)x =  = sqrt(x^{2}+x)+x) (x^{2}+x)x^{2} x  =  sqrt(x^{2}+x)+x) sqrt(x^{2}+x)+xHere things are bit more complicated.
Now look: sqrt(x^{2}+x)=sqrt(x^{2}[1+{1/x}])=sqrt(x^{2})sqrt(1+{1/x}). We've seen sqrt(x^{2}) before, and what is it? If x is positive, then sqrt(x^{2})=x. (If x were negative, sqrt(x^{2} is x.) Here we go:
x x x 1  =  =  =  sqrt(x^{2}+x)+x sqrt(x^{2})sqrt(1+{1/x})+x x[sqrt(1+{1/x})+1] [sqrt(1+{1/x})+1]Take a look at the last bit of "mess". The only appearance of x is in 1/x, and certainly as x>infinity, this should>0. All of the other pieces of the expression don't change. If you carefully examine them, you will see that the result is 1/2. Therefore lim_{x>infinity}h(x)f(x)=1/2.
Numerical evidence?
I haven't talked enough about numerical evidence for limits in class,
mostly because I am scared of using a calculator or computer in front
of people. But I certainly use such things on my own. So here is some
numerical information.
x=10  x=10^{2}  x=10^{3}  x=10^{4}  x=10^{5}  

sqrt(x^{2}+1)x  0.0498756211  0.0049998750  0.0004999998  0.0000499999  0.0000049999 
sqrt(x^{2}+x)x  0.4880884817  0.4987562112  0.4998750624  0.4999875006  0.4999987500 
I needed 25 digit accuracy to get the last few entries in the first row to 10 digit accuracy. Computations with small numbers which are almost equal can be imprecise. But the numbers should help you accept the previous results which were obtained with algebraic manipulation.
If you prefer graphical evidence then maybe the picture to the right can help. It shows graphs of f(x)=x and g(x)=sqrt(x^{2}+1) h(x)=sqrt(x^{2}+x) on the interval [0,10]. Although 10 is not very large, I think the asymptotic relationships between the functions are already apparent.
Short cuts are good but ...
So I know lim_{x>infinity}x=infinity and
lim_{x>infinity}sqrt(x^{2}+1)=infinity,
but lim_{x>infinity}sqrt(x^{2}+1)x=0.
And I also know lim_{x>infinity}x=infinity and
lim_{x>infinity}sqrt(x^{2}+x)=infinity,
but lim_{x>infinity}sqrt(x^{2}+x)x=1/2.
Please notice that 1/2 and 0 are not
equal.
Therefore for this kind of limit, simple algebraic manipulations
are not guaranteed to give valid results. We all like
computational shortcuts, but limits involving infinity can be
difficult to manipulate. Such limits should be treated more as
descriptions of geometric situations rather that anything else.
Vocabulary?
On a wellknown web page (concerning the Fungi
of Australia, the word conjugate is defined to mean
"copulation, especially isogamic copulation". I think I am too scared
to look up "isogamic" since it might mean something illegal.
How to evaluate limits
As was remarked in the previous lecture, if the limiting function were
defined with familiar formulas, I'd first try to "plug in" if there is
any chance this strategy can be applied. If this simplest method can't
be used, I'd try to "massage" the function (algebraically) and get to
some algebraically equivalent restatement where plugging in can be
used.
The derivative
Here is the most important single use of limit in Math 151.
Suppose f(x) is a function. Then we write f is
differentiable at x=a if
lim_{h>0}[f(a+h)f(a)]/h exists. If the limit exists, it is
called the derivative of f at a and
the notation f´(a) is used for
the value of the limit.
Comments There are other notations for the derivative (almost
every chunk of applied science and engineering has its favorite
notation!). We'll see some of them. Please notice that the quotient
involved in the definition of the derivative is exactly of the
form which prevents direct "plugging in". If we insert h=0 in
[f(a+h)f(a)]/h we get [f(a+0)f(a)]/0 and that's 0/0, a meaningless
arithmetic expression.
One nice example
Suppose f(x)=1/x^{2}. We already looked at this function in
the lecture on September 13 (whose diary entry is not done, which I
regret). Let's look at [f(a+h)f(a)]/h:
1 1    1/(a+h)^{2}1/a^{2} (a+h)^{2} a^{2}  =  h hMy goal is to understand what happens to this as h>0. If I "plug in" h=0 now I get no information. I will use part of my brain to manipulate this mess algebraically, and hope that I will eventually get to an equivalent algebraic form whose behavior as h>0 will be apparent (by plugging in: that would be the easiest thing). Now what we have here is a compound fraction, and I will convert it to a simple fraction (with one "division"). Experience tells me that's easier to understand.
1 1 a^{2}(a+h)^{2}     (a+h)^{2} a^{2} (a+h)^{2}a^{2} a^{2}(a+h)^{2}  =  =  h h h(a+h)^{2}a^{2}The major "transition" here is done by multiplying the top and bottom each by 1/h, and the resulting fraction has 1 in the bottom and is therefore a "simple" fraction. Now we expand part of the top, cancel a^{2} additively and h multiplicatively:
a^{2}(a+h)^{2} a^{2}[a^{2}+2ah+h^{2}] 2ahh^{2} h(2ah) 2ah  =  =  =  =  h(a+h)^{2}a^{2} h(a+h)^{2}a^{2} h(a+h)^{2}a^{2} h(a+h)^{2}a^{2} (a+h)^{2}a^{2}Now finally I can "plug in" h=0. More properly for this course, I can see what happens as h>0. The limit does exist, and its value is 2a/{a^{2}a^{2}]. Usually people write this as 2/a^{3}.
Comments There are many opportunities to make algebraic errors in what's done above. One amazing thing is that we will develop very easy ways to compute derivatives for almost all familiar functions combined in interesting ways (including algebraic combinations and composition and "inversing"). Most of next week will be devoted to stating and understanding these results. Although you should (and will, darn it!) practice these rules (algorithms!), learning them is not the peak of the course. There are very nice programs which can compute such derivatives. A major purpose of the course is to understand why derivatives are interesting to people. We need to know how to use them. And maybe machines are not yet up to that level of cognition.
Not a nice example
Consider f(x)=x. I would like to see if this f is differentiable at
a=0. Therefore I must consider:
f(0+h)f(0) 0+h0 h  =  =  h h hI need to understand what happens to h/h as h>0. Any computations with absolute value will go better if they are split into two parts, one from each side.
h<0 so we are considering h>0^{}  h>0 so we are considering h>0^{+} 

Here h is negative, and h=h. Therefore h/h is h/h and this is 1: there's no appearance of h in this result. I think that the limit as h>0^{} of h/h must be 1.  Here h is positive, and h=h. Therefore h/h is h/h and this is +1: there's no appearance of h in this result. I think that the limit as h>0^{+} of h/h must be +1. 
But the limits from the two sides (+/) don't agree. Therefore the
limit lim_{h>0}[f(0+h)f(0)]/h does not
exist.
We conclude that this function is not differentiable
at x=0.
Interpretations
Here are a collection of simple interpretations of what we are
discussing. Other interpretations will be shown in you in virtually
every technical course you take from now on.
Math 151  Geometry in the plane  Simple physical motion 

We're given some function f, and want to understand how it "changes".  The object studied is the graph of f, which is the collection of points in the plane with coordinates (x,y) which satisfy y=f(x).  Here we study rectilinear motion, where the position of a point on the coordinate line is given by f(x) (usually the variable is called t, not x, because this would help people remember that f(x) is the position at time "x"). 
The change in f over an interval from a to a+h is just f(a+h)f(a).  f(a+h)f(a) denotes the difference in the heights of the function at the xvalues a and a+h.  f(a+h)f(a) is the difference in position at the two times indicated. This is also called displacement. In general, this is not the distance that the point travels between those two times, because the moving point could wiggle back and forth. 
[f(a+h)f(a)]/h is called the average rate of change of f over the interval.  [f(a+h)f(a)]/h is the slope of the secant line through the points (a,f(a)) and (a+h,f(a+h)).  [f(a+h)f(a)]/h is called the average velocity of the point over the time interval. 
If, as h>0, the average rate of change of f approaches a limit, this limit is called the derivative of f at a and written f´(a).  If, as h>0, the slope of the secant line approaches a limit, this limit is called the slope of the tangent line to y=f(x) at x=a.  If, as h>0, the average velocity of the point over the time interval approaches a limit, this limit is called the instantaneous velocity of the point at the time x=a. The limit in this case is frequently written ds/dt (difference in s, a Latin abbreviation for distance, divided by a difference in t, meaning time. 
I will usually abbreviate the word "instantaneous" while discussing velocity, because I will almost never again refer to average velocity.
O.k.: a tangent line
Well, I made a mistake in class. I will not repeat the mistake
here. Here I will use a different function. I will try hard not to
make another mistake.
Suppose f(x) is the function defined by the formula
sqrt(17+x^{3}). Then it turns out that this function is
differentiable (if you are "naive" this is not at all obvious and
would be quite difficult to verify directly from the definition!) and
f´(x)=[3x^{2}]/[2sqrt(17+x^{3})]. Please: we will
very soon see how this formula gets computed! Suppose I ask for an
equation of the line tangent to the graph of this f(x) when x=2? We need some
information.
A point on the line Well, (2,f(2)) is a point on the line, and
f(2)=sqrt(17+2^{3})=sqrt(17+8)=sqrt(25)=5.
Slope of the line Well, f´(2) is supposed to be the slope
of the tangent line, and this is
f´(2)=[3·2^{2}]/[2sqrt(17+2^{3})]=3·4/[2·5]=6/5.
Therefore an
equation of the line tangent to the graph of this f(x) when x=2
is (y5)=[6/5](x2). I am lazy and I will not "simplify".
Some numerical "work"
Very very very few of you will need to write the equations of tangent
lines after getting through a first calculus course. But you will
almost certainly need to consider many, many derivatives and make
judgements based on these derivatives. What the heck is going on?
Well, let us think about the example above. Since f´(2)=6/5, I
know that
lim_{h>0}[f(2+h)f(2)]/h=6/5.
If I omit the "lim_{h>0}" phrasing, then the equality is no
longer ture. But the limit definition implies that what is true is a
statement with an error term, and this error term is small when h is
small. That is:
[f(2+h)f(2)]/h=6/5+{ERROR [small when
h is small]}. I can multiply by h and then add f(2)
and get the following equation:
f(2+h)=f(2)+(6/5)h+{ERROR [small when
h is small]}h
This equation is really why people study derivatives. The important qualitative aspect is that the Error is multiplied by h, and when h is small the Error is small. Products of "smalls" are even smaller, and the effect on the output of changing the input to f by h for small h is almost entirely determined by the multiplier, (6/5)h. Here are some numbers.
 

Reinterpreting the definition
The function f is differentiable at a if for any very small
{perturbationkickchange} to the input, h, the output will be
approximately f(a) (the old output) plus a constant multiplying the
perturbation. That constant is the derivative of f at a.
The strategies available now for computing limits are:
Intermediate Value Theorem, restated
Suppose that the function f is
defined and continuous on the interval [a,b]. Then the equation f(x)=y
has at least one solution for every y which is between f(a) and f(b).
QotD, version 2 (fill in the blanks)
The QotD last time was not suitable. Here, maybe is something I should
have asked, an alternative QotD for the last lecture, with a
fillintheblanks format.
Suppose
f(x)=x^{3}+cos(7x^{2}+5)+sin(3x^{4}8)+2.
Then
f(2) is # 1
because (2)^{3}=8 and the remainder of the formula
describing the function f(x) is at most # 2 at x=2.
And
f(+2) is # 3 because 2^{3}=8
and the most negative the remainder of the formula describing
the function f(x) can be at x=2 is # 4 .
Since f(x) is # 5 in the interval [2,2]
and the signs of f(x) at the endpoints differ, f(x) must have
# 6 root in [2,2].
So the QotD would have been to fill in the blanks. Here we go:
Horizontal and vertical asymptotes: geometric and algebraic
vocabulary
The limit idea was fairly successful, and people over years decided to
use it to cover more and more situations. Sometimes the extensions
were not as simple as the original setting. One extension that is
useful describes certain geometric behavior of graphs called vertical
and horizontal asymptotes.
Near x=3
To the right is the graph of a function, f(x), whose
domain includes all x except for 3. The "arrows" at the top of the
curve are supposed to indicate that as x gets closer and closer to 3
(on either the right or the left side) then f(x) gets large. But
there's more, really. It doesn't just "get" large in some uncontrolled
way. The function is
supposed to
Examples
1/(x3)? 1/x2?? 1/(x3)^{2}???
1/(x3)^{EVEN POS INT}???
A rational function
x^{2}/[(x+2)^{3}(x+4)^{6}]
Horizontal asymptotic behavior
Be a bit careful ...
sin(x)/x
Not damned oscillation, just damped oscillation
x/sqrt(x^{2}+3)
Review of the definition of limit
Discussion, criticism, comments about the definition of limit
Framework of ideas supporting the definition of limit
Algebra
Order (1)
Wiggling
Squeezing the wiggling
Order (2)
Plugging in
An important official word: continuity
The Garden State Parkway, from Cape May to Montvale
and my friend Francine ...
The
Garden State Parkway runs most of the length of New Jersey. Mile
0 is at Cape May, while the other end, mile 172, seems to be close to
Montvale. Suppose that my friend Francine leaves Cape May at 7 AM one
morning, and drives north on the Garden State Parkway. Further,
suppose she arrives at mile 172, the northern end, at, say, 10
AM. Must Francine at some time be at mile 135 (fairly near Busch
campus)? The parkway seal here was "borrowed" from a State of New
Jersey webpage.
We discussed various curves which could represent the position of Francine on the parkway in terms of miles from the start of the parkway at time t, in terms of hours elapsed from 7 AM. I tried to show that our everyday intuition lead to the graph being increasing (as you travel from left to right, the points on the graph go up). The graph can have level spots, where Francine pulls over for a rest stop. Legally Francine isn't supposed to drive backwards, though.
If we believe that motion is continuous (so Francine does not have a Star Trek transporter or other device) then the graph of Francine's position goes from (7 AM, 0 miles) to (10 AM, 172 miles) and therefore the graph must have on it at least one point with coordinate description (*,135). All of this, by the way, rests on some complicated assumptions, some of them philosophical (why should motion be continuous?). Today, though, I believe that motion is continuous, and therefore at sometime Francine must be at Mile 135. By the way, I will retain this information for later, when we analyze the rate of change of position (velocity) so that we can see whether Francine deserves a speeding ticket.
The Intermediate Value Theorem
Suppose that the function f is
defined and continuous on the interval [a,b]. Then the equation f(x)=y
has at least one solution for every y which is between f(a) and f(b).
In mathematics, the word theorem is applied to results that are deduced from basic principles, and usually the term is used for more important conclusions in the subject. In this case, the Intermediate Value Theorem follows from basic principles governing the real numbers. A particular basic principle which is used in the proof of the theorem is the "least upper bound" property of the real numbers. This essentially declares that there are "no holes" in the real numbers. A precise statement is fairly delicate, and this property essentially shows that the reals and the rationals are distinct. Several upperlevel math courses spend quite a bit of time exploring the statement. You can read about it in Wikipedia but I do add that detailed knowledge of such foundational material is not needed for success in Math 151 (or, for that matter, for successful careers in almost all of science and engineering!).
The square root of 2
If we were desperate to compute
sqrt(2) (that is, really, desperate to approximate sqrt(2)),
for example, we could look at f(x)=x^{2}2 on the interval
[0,2]. This f(x) is certainly continuous. (We already observed that we
could "plug in" values to evaluate limits for polynomials. I know that
f(0)=2<0 and f(2)=+2>0. Therefore according to the Intermediate
Value Theorem there will be at least one x inside the interval [0,2]
so that f(x)=0: x^{2}2=0. This is a positive number whose
square is 2, which we call sqrt(2). Now we have "trapped" sqrt(2)
inside the interval [0,2]. If we compute f(1)=1^{2}2=1, we
know that a root must be inside [1,2] since the signs of f(x) at the
two endpoints differ. We can continue this "game", each time halving
the interval, and chosing a halfsubinterval so that the signs of f(x)
differ on the endpoints. The graph of f(x) on the first few
subintervals is shown below.
Interval: [0,2]  Interval: [1,2]  Interval: [1,1.5]  Interval: [1.25,1.5]  Interval: [1.375,1.5] 

The bisection method for rootfinding
I discussed this too rapidly and then asked the QotD about it. I
regret the hurry. In particular, I mentioned the word "algorithm". Let
me give some further information about this word in the form of quotes
from
The Art of Computer Programming by D. E. Knuth:
The modern meaning for algorithm is quite similar to that of recipe, process, method, technique, procedure, routine, except that the word "algorithm" connotes something just a little different. Besides merely being a finite set of rules which gives a sequence of operations for solving a specific type of problem, an algorithm has five important features:Knuth continues on the same page to contrast his definition of algorithm with what could be found in a cookbook:
 Finiteness An algorithm must always terminate after a finite number of steps.
 Definiteness Each step of an algorithm must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case.
 Input An algorithm has zero or more inputs, i.e., quantities which are given to it initially before the algorithm begins. These inputs are taken from specified sets of objects.
 Output An algorithm has one or more outputs, i.e., quantities which have a specified relation to the inputs.
 Effectiveness An algorithm is also generally expected to be effective. This means that all of the operations to be performed in the algorithm must be sufficiently basic that they can in principle be done exactly and in a finite length of time \e
Let us try to compare the concept of an algorithm with that of a cookbook recipe: A recipe presumably has the qualities of finiteness (although it is said that a watched pot never boils), input (eggs, flour, etc.) and output (TV dinner, etc.) but notoriously lacks definiteness. There are frequently cases in which the definiteness is missing, e.g., "Add a dash of salt." A "dash" is defined as "less than 1/8 teaspoon"; salt is perhaps well enough defined; but where should the salt be added (on top, side, etc.)?
... a computer programmer can learn much by studying a good recipe book
Discussion of the bisection algorithm  Specification of the bisection algorithm 

The bisection algorithm is one of the simplest and neatest algorithms. It approximates roots very nicely. The entry conditions are a continuous function defined on an interval which includes the interval's endpoints. Also the sign of the function at the endpoints differs: the function's value is positive at one endpoint and negative at the other. Then, according to the Intermediate Value Theorem, there must be at least one root (f(x)=0) inside the interval. Another entry condition is a positive number, here called epsilon, which serves as the error tolerance for the root. 
Entry conditions A continuous function f(x) defined on an interval [a,b], with f(a)·f(b)<0; a positive tolerance epsilon for the error. 
Here we check if the interval we're looking at already fulfills the error tolerance condition. In later steps, we will be altering the interval, and shrinking it.  Exit condition If ba<epsilon, report the interval [a,b] as the answer. 
Compute the value of f(x) at the middle of the interval.  Computation Let c=(1/2)(a+b). Compute f(c). 
Here's the heart (?) of the algorithm. If f(a) and f(c) have different signs, then the root desired is in [a,c], the left half of the interval. We then change the interval (shrinking it) and see if the length of the interval is small enough. If the signs of f(a) and f(c) are not different, the root we're looking for (whose presence is guaranteed by the Intermediate Value Theorem!) is in the right half of the interval. So we redefine [a,b] as the right half interval and check if the exit condition is satisfied.  Decision If f(a)·f(c)<0, then change b to c, and return to check the Exit condition. Otherwise, change a to c, and return to check the Exit condition. 
Workshop work ...
Workshops are important, as I've said previously. And they will be
graded (by the recitation instructor and by me) equally for both
mathematical content and technical exposition. Please try to do both,
and please ask for advice about writeups from either of us.
Textbook homework assignments
The textbook homework assignments to be handed in on a Thursday should
be available each week on Monday. I hope this will help you schedule
your work.
Slope of a tangent line
Let's look at the curve y=1/x^{2} near x=3. The point (3,1/9)
is on the curve, since 1/3^{2}=1/9. We can try to find the
equation of a line tangent to the curve at (3,1/9). Since we know a
point this line goes through, we will be able to find an equation for
the line if we know a slope. I'll call the slope of the tangent line
m_{tan}. The traditional calculus way to find this slope is to
approximate it with m_{sec}, the slope of a secant line. This
will be a line which goes through the point (3,1/9) and
(3+h,1/(3+h)^{2}) when h is small. We can get this slope by
writing it as the difference in the second coordinates divided by the
diffeence in the first coordinates:
1 1    (3+h)^{2}  9  (3+h)3I'd like to see what happens when h gets very very small. In the geometric picture the secant line is held down at (3,1/9) and as h gets small, the secant line will sort of revolve into the tangent line, and m_{sec} will get close to m_{tan}. Algebraically, if I inspect the quotient I've just written and try to quickly see the behavior as h gets small by replacing h by 0, the result will be 0/0, and we can't assign any value to this quotient. But let's try some simple algebra on the quotient, first replacing the compound fraction by an equal simple fraction:
1 1 9(3+h)^{2}     (3+h)^{2}  9 9(3+h)^{2} 9(3+h)^{2}  =  =  (3+h)3 h h9(3+h)^{2}Now let's "expand" the top and cancel the 9's.
9(3+h)^{2} 9(9+6h+h^{2}) 6hh^{2}  =  =  h9(3+h)^{2} h9(3+h)^{2} h9(3+h)^{2}But we can cancel factors of h from the bottom and from the top (we must cancel an h from both terms on the top).
6hh^{2} 6h  =  h9(3+h)^{2} 9(3+h)^{2}Now the "asymptotic" nature of the fraction shouldn't be too difficult to see: when h gets close to 0, the top of the fraction gets close to 6 and the bottom gets close to 9·9=81. So I think that m_{tan} should be 6/81, which is 2/27.
Is m_{tan}=2/27 "reasonable"? Well, if we look at the graph we can see that the y values on the curve are getting smaller as the x values increase (near x=3). So the tangent line should tilt down, which makes the minus sign of the answer more agreeable. The magnitude of the slope, 2/27, is small, and, in fact, if you really look at the curve, the tilt is quite small. Some graphs are shown below. Of course 0 is not in the domain of 1/x^{2}.
y=1/x^{2}; x between 5 and 5  y=1/x^{2}; x between 2 and 4  y=1/x^{2}; x between 2.9 and 3.1 

Possibly the slope of another tangent line
Suppose now that we define f(x) in a piecewise fashion:
11 2x if x<3 f(x)= x^{2}4 if x≥3Now the analysis of the slopes of the approximating secant lines is more complicated. A graph of the function (with lots of other "stuff") is shown to the right. The point (3,5) is on the graph, and the point (3,5) also satisfies y=112x and y=x^{2}4. There is no jump or break in the graph at (3,5). Please note, though, that the vertical and horizontal scales differ. Let's analyze m_{sec} if h>0:
f(3+h)f(3) (3+h)^{2}45 9+6h+h^{2}9  =  =  = 6+h h h hTherefore as h gets small (with h positive) m_{sec} seems to get close to 6. So 6 seems to be our candidate for m_{tan}.
m_{sec} if h<0:
f(3+h)f(3) 112(3+h)5 1162h5  =  =  = 2 h h hTherefore m_{sec} seems to be 2.
There does not seem to be one value of m_{tan} here. So I guess there may not be a unique satisfactory tangent line. Some pictures may help.
y=1/x^{2}; x between 2 and 6  y=1/x^{2}; x between 2 and 4  y=1/x^{2}; x between 2.9 and 3.1 

Limits
The relationship between the approximating m_{sec} and
m_{tan} needs to be investigated more precisely. There's a
name to the process: LIMIT. Limits are
fundamental to precise statements about asymptotic relationships of
all kinds.
Exactness and reality
"Real" functions are not very exact. A chemical engineer might have
(I'm simplifying hugely) some process which produces the correct kind
of plastic if, say, a certain amount of benzene is used in the
mix. (Hey: benzene, with chemical formula
X_{3}Y_{7}Z_{11} (nah), was suggested by a
student. So a certain percent of benzene, say 3%, may produce a
plastic which has the desired amount of, say, translucence. But in
reality, even very precise measurements may not create an input with
exactly 3% benzene. Maybe some days we get 3.5, or 4.2. And maybe the
desired output measurement is not necessary  we can deal with some
error in the output. Real measurements imply that we need to
contemplate error, not as a moral defect (at least in this case), but
as part of our mathematical model which we must deal with.
Output tolerance as controlled by input tolerance
So maybe we need to understand the following idea: we want a
certain output tolerance: that is what we can "live with" in the
material created. Is there an input tolerance describing inputs in an
interval around the ideal input which makes the corresponding outputs
close enough to the desired output? This is too darn abstract. Let me
consider a numerical example, with a very simple function.
x^{2} near x=3
If the input to f(x)=x^{2} is 3, the output will be 9. Suppose we are willing to live with a +/ error of 1 in the output. That is, we can tolerate f(x)9<1. Is there some simple specification of an interval of inputs near 3 which will guarantee this? In this case, almost everyone thinks that the simplest kind of input specification would be an interval of x's centered at 3 (your feelings may differ here, but this is what is usually done). So I ask if there is some convenient number ("CN") so that if x3<CN then x^{2}9<9. Most people don't want an exact, precise, most perfect (?) value of CN. It may be difficult to get something like that. They are willing to settle for something convenient. In this case, the graph to the right shows the needed output tolerance using horizontal red lines.  
The next graph, shown here to the right, adds part of the vertical
lines corresponding to x=2.9 and x=3.1. The part of f(x)=x^{2}
shown which is between the vertical blue lines is "clearly" forced to
be between the red horizontal lines. This geometry reflects the
following algebraic statement: If x3<.1 then x^{2}9<1. So an input tolerance of .1 around 3 will guarantee that the output tolerance of 1 around 9 is true. 
Change the output tolerance to .2
What if we changed the desired output tolerance to .2? That is, we
want to get x^{2}9<.2 by controlling the size of x3.
Here's the picture of f(x)=x^{2} with the two horizontal red lines indicating the output tolerance needed.  
Now here is the "old" input tolerance lines which we just used. Please note that this input tolerance won't work: there is part of the graph of x^{2} between 2.9 and 3.1 which is not between y=9.2 and y=9+.2.1.  
Here is a display showing a satisfactory input tolerance. I used .1 as
the input tolerance (yes, I experimented a bit). I repeat that in
reality people rarely care about the "best possible" input
tolerances. They usually just want to get some number that works.
To me this graph doesn't show very much. So let's try a different scale.  
I hope that you can see here a graphical "verification" of the
algebraic implication:
If x3<.01 then x^{2}9<.2 An input tolerance of .01 satisfies the output tolerance of .2. 
Suppose I want the outputs to be within 1 of the ideal output. That
is, I want x^{2}9<1. The horizontal lines shown indicate this restriction.
Warning!  
The vertical lines shown are x=9.9 and x=10.1, which would be the geometric side of the input restriction x10<.1, the input tolerance we used in "If x3<.1 then x^{2}9<1" The same input tolerance won't work about different "ideal inputs". If you really believe in the graph, this isn't too surprising since the graph is much more tilted around x=10 than it was around x=3.  
The vertical segments shown are parts of the lines x=9.97 and
x=10.03. I hope this is pictorial evidence that the following
implication is correct: If x10<.03 then x^{2}100<1 An input tolerance of .03 satisfies the output tolerance of 1 near the ideal input/output pair (10,100). 
Complicated? Yes, it is.
These pictures should convince you that the limit business when
addressed "officially" is indeed complicated. In your courses and
careers, there may rarely be times when you'll need to cope with these
implications. But I believe you should have some idea of what's going
on.
Graphs don't really work too well for complicated functions or
irritating input/output pairs. People have therefore developed
extremely intricate strategies relating the inequalities
needed. Serious investigation of this algebra is not the aim of the
course but the equipment is there when/if you need it.
The official definition
The statement lim_{x>a}f(x)=L means:
Given any (positive) output tolerance, there is a (positive) input tolerance so thatComments
If 0<xa<the input tolerance, then f(x)L<the output tolerance.
The piecewise linear function
The inverse(s)
Example
Some trig functions and their inverses
Some exponential functions and their inverses
Dwarf apple trees
Growth rate
Ultimate height
Chapter 2 and its fictions
Slope of a tangent line
QotD
The diary
I mentioned the diary and remarked that it served several
purposes. First, even very enthusiastic students may sometimes miss a
class. The diary can help them find out what was covered. Another
reason for the diary's existence is so that I can fix errors which I
will certainly make, so while the diary will be a representation of
what went on in class it won't be a totally accurate one: improvements
may occur! Also, even if I don't make a mistake in the lecture,
sometimes I may not explain things optimally, so the diary may give
students another chance to understand the presentation. I really don't
think that a diary entirely takes over the teaching/learning aspect of
the lectures. There is still a good amount to be said about the
interaction between the instructor and the student, provided that
students are willing to ask questions and (sigh!) the instructor is
really willing to listen and answer. I will try to do the latter (also
with errors, of course, since I plead the defects of humanity!).
QotD
Most days I will ask students who attend the lecture a question or
two, and receive their answers in writing. Students will get full
credit for any answer. This "Question of the Day" has several
purposes. First and most minor, it is a way of taking
attendance. More important is that it serves as an additional method
of communicating between the lecturer and the audience. If I ask some
question which I think is "simple" and, hey, almost no one answers
the question correctly, well ... something's wrong. I will "grade" the
question very roughly, only retaining for records whether or not an
answer was made. For the student, this gives an opportunity to see if
some content in the lecture has been learned satisfactorily, and, if
not, maybe that also signals that something's wrong. I'll try
to get the graded QotD's back to you in recitation.
Study groups
I asked students in the class to tell me their email addresses and
likely majors, and, with appropriate permission, this information is
now displayed on a web page. Now at the
beginning of this course, there may be little that is new to
students. I caution you, though, that even if you have seen much of
this material before, the course will move swiftly. It is a good idea
to form study groups meeting at appropriate times and durations to go
over homework, etc. I hope that the student list will help people get
in touch with each other.
Calculus [Advertisement]
Calculus originated as a way of describing certain methods used in
understanding and solving problems originating in geometry and
physics. The language of the subject shows this ancestry. Calculus has
been very successful. Calculus also "happens" to be suitable to
describe and study many problems in biology and economics. Any
problems which involve rates of change (how soluble is substance A in
substance B as the concentration of substance C varies?) or
accumulation of changes (what is the present value under
certain interest rate assumptions of a stream of mortgage payments to
be made monthly over the next 20 years?) is a problem which likely can
be described using calculus. Frequently this description will suggest
certain ways of solving the problems using the methods of
calculus. That's why the subject is required in so many majors.
Functions
The word function is used in a technical sense in calculus, and
is one of the most important vocabulary words. It is the logical
setting for how things are transformed to other things. In the case of
Math 151, the "things" are numbers. So functions will change numbers
to numbers. I decided to begin with what I hoped was a known example
described verbally.
Hooke's law
Think of a spring with a rest or "equilibrium" position. Then Hooke's
Law states that the deviation from the rest position is in direct
proportion to the impressed force. There are many assumptions involved
in Hooke's Law, but to a considerable extent it is a widely valid
shorthand for a great variety of real phenomena. If I further add something like: the spring gets stretched 10 inches by a weight of 300 pounds, then I can formulate an equation relating the impressed force and the deviation from the equilibrium position:
D(x)=(1/30)x
Here D(x) is the deviation in inches from the
equilibrium position (with positive meaning the spring is forced to be
longer) and x is the impressed force in poinds (I'm thinking maybe of
a thick mechanical spring or something like that). x is called the
independent variable and D(x) is called the dependent variable. Notice
that sigh matters. If x is negative, then we could be compressing the
spring, and the end of the spring is higher than the equilibrium
position in our model.
Notice that as in all mathematical models, you need to be alert a bit about how "correct" this is. Uhhh ... if we stretch a rubber band with a force of six trillion tons, I don't necessarily think it will end up having a length of one lightyear.
Functions
One metaphor I will use (and probably overuse!) in this course is the
idea of a function as a machine with an input and a unique output
associated to each input. The collection of all valid inputs to the
machine, those inputs which don't cause the machine to break, is
called the domain. The collection of all of the outputs for
these valid inputs is called the range.
Functions via formulas
Functions can be described simply with formulas. Actually, most of the
functions we will study in this course will have such
descriptions. Warning: students interested in obviously experimental
disciplines (engineering, physics, chem, etc.) will mostly encounter
functions which are described by data sets, and must work with them
using the techniques of this course which will mostly be presented
using functions defined by formulas. This can be ... irritating.
The domain of a function defined by a formula is usually assumed to be
what's called the natural domain: this is the collection of all
inputs of real numbers which make sense in the formula. But
there will be exceptions (see below!).
Equalsized squares are cut from the corners of an 8 inch by 11 inch rectangular piece of paper. The flaps in the resulting piece of paper are folder up. Write a formula for the resulting volume which the paper object encloses, using the edge length of the squares as the independent variable. What is a useful domain for this formula?I hope that the accompanying illustrations are helpful. I'll call the edge length x. Then the "solid" brick will have volume, V(x), which will be the product of the area of the base multiplied by the height. The height is x, and the base has sides with lengths 82x and 112x. So V(x)=x(112x)(82x). This formula is "just" a polynomial of degree 3. But please notice that x=50 or x=200 makes no sense in this problem. The physically reasonable domain, the x's that can actually be used, are just those x's in the interval [0,4]. So already here is an example of a function given by a formula where the applicable domain for the mathematical model is much smaller than the natural domain. You may think that this is unreasonable, and I'm being much too picky. But the restriction is essential.
Piecewise defined functions
Here is a silly example:
x^{2} if x<1 f(x)= x+2 if x≥1Such examples will be useful to consider when we look at limits. Here f(2)=(2)^{2}=4 and f(5)=5+2=7.
Part of the graph of this function is shown to the right. Please note the "empty circle" on the parabolic arc, because of the restriction x<1 (x is not allowed to be equal to 1). There's a "filled circle" on the other piece, a half line.
A more elaborate "real" example
Now here is a somewhat more complicated example:
f(x)= .1x if 0≤x≤15,100 1,510+.15(x15100) if 15,100<x≤61,300 8,440+.25(x61300) if 61300 <x≤123,700 24,040+.28(x123700) if 123,700<x≤188,450 42,170+.33(x188450) if 188450<x≤336,550 91,043+.35(x336,550) if 336,550<xThis is the U.S. individual tax rate schedule (form 1040) for 2005. The function has many weird numbers in it. This is an example of a piecewise linear function because each of the pieces is defined by a polynomial of degree 1. The slopes of the pieces increase as the x's increase because this is a progressive income tax. But the other strange numbers are caused by a desire to avoid jumps and breaks in the graph of the tax function, so that strange situations don't arise (examples: yes, you earn $1 more and therefore will pay $100 less in taxes, etc.).
Graph of a function
The graph of a function is the collection of all points in the plane with coordinates (x,f(x)) if x is in the domain of the function. To the right is the graph of a function. A collection of points in the plane is the graph of a function if every vertical line which touches the graph intersects it exactly once. This is called the Vertical Line Test, and corresponds to the fact that we want exactly one output determined by every "legal" input. If we have the graph, the domain is all x's on the xaxis for which the vertical line through that x hits the graph. Similarly the range is all y's on the yaxis for which the horizontal line through that y hits the graph. Sometimes if you only have the graph and no other information, precise descriptions of the domain and range can be difficult.
An inverse function Consider the function defined by f(x)=2x^{3}+7. Part of the graph of this function is shown to the right. It was produced by my silicon friend, Maple. (Available on eden, just type maple). The graph is a bit deceptive. I asked that the function be shown for x's in the interval from 3 to 3. The program automatically "autoscales" and squeezes so that the image is in a square. This can be confusing at times if you don't realize it. In this case, the vertical range seems to be between 45 and 60.  
The part of the graph I want to emphasize here is the qualitative
aspect: certainly every vertical line hits the graph once (the
Vertical Line Test) but so does every horizontal line. That is, every
output corresponds to a unique input. For this function, this
uniqueness claim can be verified algebraicaly quite easily: if w is an
output, we can write a formula for the unique x it came from. w=2x^{3}+7 > w7=2x^{3} > (w7)/2=x^{3} >[(w7)/2]^{1/3}=x A function which has unique outputs for each input is called 1to1. Such a function has an inverse. The function g(w)=[(w7)/2]^{1/3} is inverse to f. It undoes f, so that f(g(w))=w and g(f(x))=x. The graph of g is shown to the right. It is gotten algebraically by interchanging the two coordinates of each point. Geometrically the graph of the inverse function is gotten by picking up the original graph of f and "flipping it over" the main diagonal, y=x. 
A piecewise linear function to play with
I defined a function by sketching its graph, as shown to the right. I
declared that the function, f(x), was piecewise linear. One part was a
half infinite line (a ray) passing through (3,0) and ending at
(0,2). Then there was a line segment from (0,2) to (3,1), followed by
a line segment from (3,1) to (5,4). The domain of this function was
(infinity,5] and its range was (infinity,4]. My real intention was
to use this function to study the problems involved in specifying
inverses, but ...
QotD
Suppose H(x)=f(x^{2}) and K(x)=f(x)^{2}. I asked
students to tell me the domain and range of H and K.
Below, by the way, are graphs of f and H and K, as Maple "sees" them. Notice again that vertical and
horizontal scales differ.
Graph of part of f  Graph of all of H  Graph of part of K 

QotD answers
Let's see: H(x)=f(x^{2}). The inputs to f are
(infinity,5]. But squaring makes things nonnegative, so the only
numbers I can push into x so x^{2} feeds into f are numbers
which won't square to bigger than 5. Therefore I should put in only
numbers between sqrt(5) and +sqrt(5). So the domain will be
[sqrt(5),+sqrt(5)]. The range will be the outputs of f when the
inputs of f are in the interval [0,5], and those outputs are [1,4].
K(x)=f(x)^{2}. Now K has the same domain as f: (infinity,5]
because the domain of squaring is every possible number  there are
no restrictions to worry about. The outputs, however, are squared. The
collection of outputs of f(x) are (infinity,16]. When these are
squared we will get every nonnegative number.
Here is more detail. For x>0, the outputs of f(x)^{2}
go from 0 to 16. On the left, though, the outputs of f itself
go down to infinity, so squaring means the results are all
nonnegative numbers. The range of K is [0,infinity).
Numbers
Geometry
The real numbers are usually thought of as corresponding to a specific
geometric object, the real line. I usually think of this line
as horizontal with 0 sitting "in the middle". 1 is to the right of
0. And this geometric picture brings up the idea of order. In addition
to the algebraic structure, there is order: a<b. To me this means
(in my picture of the line) that a is to the left of b. Negative
numbers are to the left of 0 and positive numbers, including all the
positive integers, are to the right of 0. Here are some interesting
aspects to note.
The decimal expansion of a real number provides an "address" to locate the number geometrically. For example, 23.4680472... means we first move right from 0 by 23 steps, each 1 long. Then look between 23 and 24, and divide that length in tenths. The number we're looking for resides (?) in the 4^{th} subinterval. Now divide that subinterval into tenths. The number is in the 6^{th} subinterval of that collection. Etc. The "Etc." of course conceals an approximation process which will locate the real number with the specified decimal expansion.
Some real numbers have two valid decimal expansions. This doesn't bother me, since, say, a house on a corner of a grid system of streets could also possibly have two valid addresses (if it is on the corner of Cedar Avenue and Third Street, the house could have the address 12 Cedar Avenue and also 42 Third Street). So, for example, the real number with decimal address 23.45999... (with the 9's repeating) also have the decimal address 23.46, a terminating expansion (there are a string of 0's which are usually not written.
Distance including a discussion of  
The distance between two points is a nonnegative real number whose
size expresses how far apart the numbers are. This will be important
when we study approximation schemes. We'd like to know that the
approximation gets "close" to the correct answer, and the closeness
will be measure by the distance. Algebraically, if the points
correspond to the real numbers a and b, the distance between them is
ab and this is the same as ba, so that distance has some
symmtery. But I just used absolute value, and here is the
piecewise definition of absolute value:
x=x if x≥0 and x=x if x<0
Therefore absolute value is always a nonnegative number. The absolute
value of a number is 0 only when the number itself is 0. And absolute
value of a product is the product of the absolute values (this
actually is not totally obvious, and needs a bit of thought, I
believe).
Intervals
Suppose we want to discover what numbers x are closer to 9 than a
distance of 4. Algebraically this requirement translates to
x9<4. We can sort of "unroll" the inequality. The absolute value
will be less than 4 if the number itself is both less than 4 and
greater than 4. The two inequalities can be compactly written as
follows:
4<x9<4 which implies
5<x<13
This is an interval and an interval which does not contain
either endpoint is called an open interval. The notation for
this interval is (5,13). Intervals which
contain both endpoints are called closed. An examples of such
an interval is [4,6], which means the numbers x satisfying
4≤x≤6. There are also
halfopen intervals, unbounded intervals (with notation using + or 
infinity), etc. Please see the textbook.
Warning!
If you wanted to "solve" (better: understand!) the inequality
x9>4 you can't just "unroll" it to 4<x9>4. This
inequality has no solutions. There is no number
which is simultaneously less than 4 and greater than 4. You
can't write this so compactly and using such implications represents
an invalid (wrong!) method of solution.
A valid method of solution would involve separately solving
the inequalities:
4<x9 orx9>4 which gives
5<x or x>13. This is actually, therefore, two
intervals:
(infinity,5) and (13,infinity). So the inequality
x9>4 has a solution set which is two intervals.
The plane
The conventional way to describe the plane algebraically is to drop
down two lines perpendicular to each other: coordinate axes. A point
in the plane will then be described by an ordered pair of real
numbers. The first coordinate will usually be called the xcoordinate
and the second, the ycoordinate. This pair describes ordered
distances from the horizontal (the xaxis) and vertical (the yaxis)
lines. Please see the text for more about this.
The embarrassment of all this, especially with "new" students, is that (3,8) could describe both a point in R^{2}, the plane, and could also describe an open interval (with [missing!] endpoints 3 and 8). The context is supposed to help, but still the notational confusion is possible, and this is lousy.
Distance in the plane, R^{2}
Well, I went through the usual diagram to try to motivate the
algebraic definition of distance in R^{2}. Look, please, at
the diagram to the right. In the plane, points correspond to ordered
pairs of numbers. So a point p might correspond to an ordered pair,
(x_{1},y_{1}), and q might correspond to (x_{2},y_{2}). Then the point (x_{1},y_{2}) is
the vertex of a right triangle whose hypotenuse is a line segment
connecting p and q. One leg of the right triangle is on a line where
all the first coordinates are x_{1}, and the length of that leg is given
by the one dimensional formula, y_{1}y_{2}. The other leg is on the line
where all the second coordinates are y_{2}, and the length of that leg is
x_{1}x_{2}. Then by Pythagoras, the hypotenuse has length
sqrt(x_{1}x_{2}^{2}+y_{1}y_{2}^{2}). And usually the
absolute values are discarded since we are squaring the
quantities. Therefore we officially define:
dist(p,q)=sqrt((x_{1}x_{2})^{2}+(y_{1}y_{2})^{2}) if p
has coordinates (x_{1},y_{1}) and q has coordinates (x_{2},y_{2}).
Example
The distance between between (3,2) and (6,4) is
sqrt([36]^{2}+[4(2)]^{2})=sqrt(3^{2}+2^{2})=sqrt(13).
A circle and its algebraic description
Suppose we wanted to describe algebraically the collection of all
points which have distance sqrt(13) from (6,4). This is, of course, a
circle of radius sqrt(13) and center (6,4). Well, if (x,y) is such a
point, then sqrt([x6]^{2}+[y4]^{2})=sqrt(13). This
does describe the circle. Of course, if you must, all sorts of
algebraic things could be done to the equation. But I am the laziest
person in the room, and therefore ...
A (nonvertical) line and its algebraic description
Suppose we wanted an algebraic description of points with coordinates
(x,y) which lie on the straight line which goes through (4,3) and
(8,13). If (x,y) is such a point, then look at the picture: two right
triangles indicated are similar, so the corresponding sides have the
same ratios:
13y 133  =  8x 84and the point (x,y) is on the line exactly when y13=([133]/[84])(x8). The quantity ([133]/[84]) is called the slope, and multiplies changes in x to give changes in y. It is frequently designated with the letter m.
QotD
I gave two points and asked for an equation for the line containing
the two points. Most people were able to do this.
Maple
Much intricate algebraic computation can now be done by computer
programs. A course like Math 151 can therefore comcentrate more on
understanding rather than elaborate computational skills. On
the other hand, knowing what results should look like (if you
calculator declares that 56.007 is the sum of 1,345 and 14,891, do
you believe it?) is important, so a secure command of various
algebraic algorithms is still important. At Rutgers, the program Maple is widely available since there is a
university license for it. Students may also get copies under certain
circumstances.
Basic information about the program is here
and here
are some local help pages.
Read the text!
Please read the first few sections of chapter 1, as on the Math 151
syllabus.
Maintained by greenfie@math.rutgers.edu and last modified 9/7/2006.