Monday, May 4  (Lecture #26) 

A collection of examples
r=3+sin(θ) Let's consider r=3+sin(θ). Since the values of sine are all between –1 and 1, r will be between 2 and 4. Any points on this curve will have distance to the origin between 2 and 4 (the green and red circles on the accompanying graph). When θ=0 (the positive xaxis) r is 3. As θ increases in a counterclockwise fashion, the value of r increases to 4 in the first quadrant. In the second quadrant, r decreases from 4 to 3. In the third quadrant, corresponding the sine's behavior (decrease from 0 to –1) r decreases from 3 to 2. In all of this {inde}crease discussion, the geometric effect is that the distance to the origin changes. We're in a situation where the central orientation is what matters, not up or down or left or right. Finally, in the fourth quadrant r increases from 2 to 3, and since sine is periodic with period 2Π, the curve joins its earlier points. The picture to the right shows the curve in black. I'd describe the curve as a slightly flattened circle. The flattening is barely apparent to the eye, but if you examine the numbers, the up/down diameter of the curve is 6, and the left/right diameter is 6.4.  
Converting to rectangular coordinates
A naive person might think, "Well, I could convert the equation r=3+sin(θ) to rectangular coordinates and maybe understand it better." Except under rare circumstances (I'll show you one below), the converted equation is very irritating and difficult to understand. For example, Let's start with r=3+sin(θ) and multiply by r. The result is r^{2}=3r+r·sin(θ). I multiplied by r so that I would get some stuff I'd recognize from the polar/rectangular conversion equations. r^{2} is x^{2}+y^{2} and r·sin(θ) is y. So I have x^{2}+y^{2}=3r+y, or x^{2}+y^{2}–y=3r. I would rather avoid square roots so I will square this, and get (x^{2}+y^{2}–y)^{2}=9r^{2}=9(x^{2}+y^{2}). This is a polynomial equation in x and y of highest degree 4, defining this curve implicitly. I don't get much insight from this.  
r=2+sin(θ) Now consider r=2+sin(θ). Again, the values of sine are all between –1 and 1, so r will be between 1 and 3. Any points on this curve will have distance to the origin between 1 and 3. We can begin (?) the curve at θ=0 when r=2, and spin around counterclockwise. The distance to the origin increases to r=3 at θ=Π/2 (the positive yaxis). The distance to the origin decreases back to r=2 when θ=Π (the negative xaxis). The curve gets closest to the origin when θ=3Π/2 (the negative yaxis) when r=1. Finally, r increases (as θ increases in the counterclockwise fashion) to r=3 again when θ=2Π. Here the "deviation" from circularity in the curve is certainly visible. The bottom seems especially dented.  
r=1+sin(θ) We decrease the constant a bit more, and look at r=1+sin(θ). The values of sine are all between –1 and 1, so r will be between 0 and 2. The (red) inner circle has shrunk to a point. This curve will be inside a circle of radius 2 centered at the origin. We begin our sweep of the curve at 0, when r is 1. Then r increases to 2, and the curve goes through the point (0,2). In the θ interval from Π/2 to Π, sin(θ) decreases from 1 to 0, and the curves moves closer to the origin as r decreases from 2 to 1. Something rather interesting now happens as θ travels from Π to 3Π/2 and then from 3Π/2 to 2Π. The rectangular graph of 1+sine, shown here, decreases down to 0 and then increases to +1. The polar graph dips to 0 and then goes back up to 1. The dip to 0 in polar form is geometrically a sharp point! I used "!" here because I don't believe this behavior is easily anticipated. The technical name for the behavior when r=3Π/2 is cusp. This curve is called a cardioid from the Latin for "heart" because if it is turned upside down, and if you squint a bit, maybe it sort of looks like the symbolic representation of a heart. Maybe.  
r=1/2+sin(θ) Let's consider r=1/2+sin(θ). The values of sine are all between –1 and 1, so r will be between –1/2 and 3/2. The (red) inner circle actually had "radius" –1/2, and it consists, of course, of points whose distance to the pole, (0,0), is 1/2. When θ is 0, r is 1/2. In the first two quadrants, 1/2+sin(θ) increases from 1/2 to 3/2 and then backs down to 1/2. In the second two quadrants, when θ is between Π and 2Π, more interesting things happen. The rectangular graph on the interval [0,2Π] of sine moved up by 1/2 shows that this function is 0 at two values, and is negative between two values. The values are where 1/2+sin(θ)=0 or sin(θ)=–1/2. The values of θ satisfying that equation in the interval of interest are Π+Π/6 and 2Π–Π/6. The curves goes down to 0 distance from the origin at Π+Π/6, and then r is negative until 2Π–Π/6. The natural continuation of the curve does allow negative r's, and the curve moves "behind" the pole, making a little loop inside the big loop. Finally, at 2Π–Π/6, the values of r become positive, and the curve links up to the start of the big loop. This curve is called a limaçon. The blue lines are lines with θ=Π+Π/2 and θ=2Π–Π/6. These lines, for the θ values which cross the pole, are actually tangent to the curve at the crossing points.  
r=0+sin(θ) Let's try a last curve in this family, with the constant equal to 0. What does r=sin(θ) look like? A graph is shown to the right. There are several interesting features of this graph. First, this is a polar curve which does have a nice rectangular (xy) description. If we multiply r=sin(θ) by r, we get r^{2}=r·sin(θ), so that x^{2}+y^{2}=y. This is x^{2}+y^{2}–y=0 or, completing the square, x^{2}+y^{2}–2(1/2)y+(1/2)^{2}–(1/2)^{2}=0 so that (x–0)^{2}+(y–1/2)^{2}=(1/2)^{2}. This is a circle of radius 1/2 and center (0,1/2), exactly as it looks. The moving "picture" of this curve is quite different. Between 0 and π it spins once around the circle but then from π to 2π it goes around the circle another time! So this is really somehow two circles, even though it looks like only one geometrically. 
More information about these curves is available here
Length of polar curves
The formula is
∫_{θ=α}^{θ=β}sqrt(r^{2}+(dr/dθ)^{2})dθ. This formula is gotten from the parametric
curve formula on p.652 of the textbook. I used it to find the length of the
cardioid (the double angle formula from trig is needed). Then I
used it to find the length of a circle (!), but here the novelty is
that we actually trace the circle r=cos(θ) twice from 0
to 2π, so some care is needed if we only wanted to find the length
of one circumference.
"Sketching" roses
Here are dynamic pictures of two roses. The first is the one I
sketched in class r=cos(3θ). It is covered twice and has 3
"petals". The second is r=cos(4θ). It is only covered once, and
it has 8 petals! Wow, polar coordinates can be annoying!
 

Area inside one petal of r=cos(3θ) Well, cos(3θ) "first" (going from 0 to 2π) is 0 when 3θ=π/2. So we get half a petal by integrating from 0 to π/6. The formula is ∫_{α}^{β}(1/2)r^{2}dθ for area in polar coordinates (see the discussion on p.649 of the textbook), so this becomes (for the whole petal, we need to double): 2·(1/2)∫_{0}^{π/6} cos(3θ)^{2}dθ. This can be computed using a trig identity. 
QotD
Has the QotD been helpful to you? This is the last QotD and
won't be returned. I would be interested in your opinion, really.
Another kind of spiral I didn't talk about Exponentials and snails, darn it! Curves of the form r=a e^{bθ} are spirals of a different kind than what we've drawn (for example, different from the spiral drawn for the QotD last time). All of these spirals have a strange and wonderful geometric property. If a ray is drawn from the pole (the origin) then the angle the ray makes with the tangent line of the spiral at any intersection is the same. I attempted to illustrate this with the first picture to the right. It turns out that this silly geometric property has natural consequences in terms of the energy efficiency of its construction. A snail shell when considered transversal (perpendicular) to the axis of symmetry usually is one of these curves. You can read a wikipedia article about these spirals, which have a number of different names (of course!). To the right is a machinedrawn picture of r=e^{.25θ} as θ goes from –Π to 5Π. The .25 was put in to make the exponential not grow too fast so the picture would be tolerable. 
Wednesday, April 29  (Lecture #25) 

Slope
dy dy/dt  =  dx dx/dt
Speed
Speed=sqrt(f´(t)^{2}+g´(t)^{2}) or
___________ /(dx)^{2} (dy)^{2} Speed= / ()+ () / (dt) (dt)As you'll see if you take Math 251, this is the sum of the squares of the horizontal and vertical components of the velocity vector: it is, in fact, the magnitude of the velocity vector.
Length of a curve
Well, suppose we move along a parametric curve given by x=f(t) and
y=g(t) from t=START to t=END. If we believe that the
speed is sqrt(f´(t)^{2}+g´(t)^{2}), then we
know that this speed can vary. In a sort time interval (dt long!) the
distance traveled is Speed·Time, or
sqrt(f´(t)^{2}+g´(t)^{2})dt. We can add up
all these distances from t=START to t=END using the
integral idea. So the distance traveled along the curve from
t=START to t=END will be given by
∫_{t=START}^{t=END}sqrt(f´(t)^{2}+g´(t)^{2})dt.
(We are integrating the magnitude of the velocity vector). Is this a
reasonable formula?
Almost no speed functions have nice, neat, simple antiderivatives. In the real world, you'll need to use numerical approximation. However, Math 152 is not the real world.
A textbook problem
Section 11.3, problems 3 through 15, are all "Find the length of the
path over the given interval" with some rather sillylooking functions
specified. Accidentally (exactly not accidentally, actually!)
all of the problems can be computed exactly with antiderivatives and
values of standard functions. There is a fair amount of ingenuity
involved in constructing such examples. I urged students to practice
with several of them. Here is problem #7:
Find the length of the path described by
(3t^{2},4t^{3}), 1≤t≤4.
The solution
Here x=3t^{2} and y=4t^{3}. We will compute the speed
and then attempt to "integrate" (actually using FTC, so we'll need to
find the antiderivative). Now dx/dt=6t and
dy/dt=12t^{2}. Therefore the speed is sqrt(6t)^{2}+(12t^{2})^{2}). This is
sqrt(36t^{2}+144t^{4}). We'll need to integrate this,
and maybe I will "simplify" first. Indeed, since we consider t in an
interval where t≥0, sqrt(t^{2})=t (otherwise we would need
to worry about t or –t etc.). But
36t^{2}+144t^{4}=36t^{2}(1+4t^{2}) so
that the square root is 6t·sqrt(1+4t^{2}). Therefore
the distance traveled along the curve is an integral,
∫_{t=1}^{t=4}6t·sqrt(1+4t^{2})dt.
Several students immediately suggested various substitutions. Here is
one which does the job efficiently. So:
If u=1+4t^{2}, du=8t dt, so (1/8)du=dt.
∫6t·sqrt(1+4t^{2})dt=(6/8)∫sqrt(u)du.
Now
(6/8)∫sqrt(u)du=(6/8)(2/3)u^{3/2}+C=(1/2)u+C=(1/2)(1+4t^{2})^{3/2}+C.
So
∫_{t=1}^{t=4}6t·sqrt(1+4t^{2})dt=(1/2)(1+4t^{2})^{3/2}_{1}^{4}=(1/2)(65)^{3/2}–(1/2)(5)^{3/2}.
A tangent line to the evolute Let me find a tangent line to the evolute of a circle (the path of the bug, discussed last time) when t=3Π/4. For this, x(t)=cos(t)+t sin(t) and y(t)=sin(t)–t cos(t). When t=3Π/4, x=–[1/sqrt(2)]+3Π/sqrt(2) (approximately 0.95897) and y=[1/sqrt(2)]+3Π/sqrt(2) (approximately 2.37319). So we know a point the line goes through. How about the slope? dy/dx=(dy/dt)/(dx/dt)=[cos(t)–cos(t)+t sin(t)]/[–sin(t)+sin(t)+t cos(t)]=tan(t) (amazing that it is so relatively simple). At t=3Π/4, this is –1. So the line is y–[1/sqrt(2)]+3Π/sqrt(2)=(–1)(x–{–[1/sqrt(2)]+3Π/sqrt(2)}). The line and the curve are shown to the right. 
Tangent lines I'll find the tangent lines at the selfintersection point of my favorite curve. The point involved is (1/2,0) and the two values of t are +1 and –1. Since I have the point, the other information I need to find the tangent lines is the slope. Well, slope is dy/dx=(dy/dt)/(dx/dt). Since x=1/(1+t^{2}), dx/dt=(–2t)/(1+t^{2})^{2}. I only fouled this up two or three times in class, but when t=–1, dx/dt=1/2, and when t=+1, dx/dt=–1/2. Since y=t^{3}–t, dy/dt=3t^{2}–1. When t=–1 or when t=+1, dy/dt=3–1=2.
Therefore when t=–1, dy/dx=(2)/(1/2)=4. The line goes through (1/2,0),
so an equation for it is y=4(x–1/2).
I had Maple graph the parametric curve and the two lines just found (in green and blue). I then asked what the angle between the lines (the angle which encloses the xaxis). The angle is 151.9 degrees. Hey, please remember that the slope of a line is the tangent of the angle that the line makes with the positive xaxis. Here the angle between the line y=4(x–1/2) and the positive xaxis has angle equal to arctan(4), approximately 1.326 radians. Double this is about 151.9 degrees. Hey: we have the machines. Please use them.
Uniform speed?
Nonuniform speed
Now this does resemble what I know about the motion of a planet in
orbit. When it is far away from the center of the orbit, the planet
will have large potential energy and relatively small kinetic energy
(it will move slowly). This is near t=0 and t=Π. When it is close
to the center of the orbit, the planet moves faster, and the kinetic
energy is larger, while the potential energy, measured by the work
needed to move closer/farther from the center, decreases.
Back to my favorite curve 
A tilted ellipse To the right is a graph of the parametric curve x=sqrt(3)sin(t)+(1/2)cos(t) y=–sin(t)+(sqrt(3)/2)cos(t) for t between 0 and 2Π. As you might expect, this is a prepared (nonrandom!) example. It is a rotated ellipse with 2to1 ellipticity. Those of you who take linear algebra (Math 250) will be able to construct examples like this easily.) One thing which turns out to be important in computer typesetting (typography) and, more generally, computer graphics, is the idea of a bounding box: the smallest box with vertical and horizontal sides containing the figure. I would like to find the bounding box for this curve. I will admit that in the real world, if I had just one curve, I would probably just guess numerically  I would "eyeball" it. But I would like to show you a more systematic way to obtain the bounding box.  
The top border To find the top border, we must look for a place where the tangent line is horizontal. This means that the derivative should be 0. But we now know that dy/dx=(dy/dt)/(dx/dt). This will be 0 if the top is 0 (hey, I am neglecting the possibility that the bottom is 0 also but that doesn't happen here and rarely should happen in practice). But y=–sin(t)+(sqrt(3)/2)cos(t) so dy/dt=–cos(t)–(sqrt(3)/2)sin(t) so this is 0 when tan(t)=2/sqrt(3). There isn't a simpler answer (!!!), sorry. Even in such an easy example, the numerics get complicated. The approximate numerical value of t is.85707. The point corresponding to this value of t is (–(3/14)sqrt(21),(1/6)sqrt(63)) or approximately (–.98189,1.32288). This is where the red line touches the ellipse in the picture to the right.  
The right border How about the right border? How could we find the point which is most to the right on this tilted ellipse? We could look for a vertical tangent. Since the slope of the tangent line is still given by dy/dx=(dy/dt)/(dx/dt) the points which will be candidates for vertical tangent lines will have dx/dt=0. (You see, looked at the correct way, the "tilt" or direction of the tangent line is not quite the same as m in y=mx+b. There's a better way to think about this, and this better way will be shown to people in Math 251. Well, x=sqrt(3)sin(t)+(1/2)cos(t) so dx/dt=sqrt(3)cos(t)–(1/2)sin(t) and this is 0 if tan(t)=2sqrt(3) so that t=arctan(2sqrt(3)). There isn't a simpler answer (!!!), sorry. Even in such an easy example, the numerics get complicated. The approximate numerical value of t is 1.28976. The point corresponding to this value of t is exactly (sqrt(13)/2,–(3/26)sqrt(39)) or approximately (1.80278,–.72058). And that pair of numbers designates the blue point in the picture to the right here.  
The bounding box The numbers for the two blue points shown allow us to deduce the vertices of the bounding box for this tilted ellipse. For example, the top side of the box is y=the second coordinate of the first point found. The right side is x=the first coordinate of the second point found. By symmetry we can get the other sides (I centered the ellipse at the origin) and the coordinates of the corners can be deduced fairly easily. 
The idea of polar coordinates
You have found a treasure map supposedly giving directions to the
burial spot of a chest full of gold, jewels, mortgages, etc., stolen
by the Dread Pirate Penelope. The information you have is that
A buried treasure is located 30 feet from the
old dead tree, in a
NorthNorthWest direction.
So there you are, on the island. Perhaps the Old dead tree is
still visible. You could mentally draw a circle 30 feet in radius
around the Old Dead Tree. Then you find the North
direction. π/4=45^{o} to the West is NW (Northwest) and
then NNW (Northnorthwest) is π/8 towards North (anyway, you decide
on the direction). Where that direction intersects the circle is
probably where to dig, unless Penelope is tricky, etc.
The whole idea of located a point in a 2dimensional setting using distance from a fixed point and angle with respect to a fixed direction is called polar coordinates.
"Standard issue" polar coordinates Fix a point (usually called "the center" or sometimes "the pole" and in most common situations, the origin of the xycoordinate system. Also fix a direction  if needed this might be called "the initial ray". Almost always this is the positive xaxis in an xycoordinate system. Then locate another point in the plane by giving its distance from the center (called r) and by drawing the line segment between the center and the point you are locating. Measure the angle between that and the initial ray (note: counterclockwise is a positive angle!): this is called θ. r and θ are the polar coordinates of the point. 
An example and the problem with polar coordinates Well, make the standard choices for "the pole" and "the initial ray". Let's get polar coordinates (the values of r and θ) for the point whose rectangular coordinates are x=sqrt(3) and y=1. Of course this is not a random point (sigh). So we consider the picture, and decide that the hypotenuse (r) should be 2 units long, and the acute angle (θ) should be π/6. Fine. But suppose that the point (sqrt(3),1) is operating in a sort of dynamic way. Maybe it is the end of a robot arm, or something, and suppose that the arm is swinging around the pole, its angle increasing. It might be true that we somehow are computing various angles, and since the arm is moving continuously (still no teleporting robot arms!) the angles which are θ's should change continuously. If the arm swings completely around the pole, and comes back to the same geometric location, it would make more sense to report its polar coordinates as r=2 and θ=13π/6 (which is better understood as 2π+π/6). 
Some valid polar coordinates for the point whose rectangular
coordinates are x=sqrt(3) and y=1:
r=2 and θ=π/6 r=2 and θ=13π/6
r=2 and θ=25π/6 ETC.
But the "robot arm" could also swing around backwards, so other
possible polar coordinates for the same geometric point include
r=2 and θ=–11π/6 r=2 and θ=–23π/6
ETC.
Generally, r=2 and θ=π/6+2π(multiplied by any integer): the integer
could be 0 or positive or negative.
The irritation ("It's not a bug, it's a feature ...") is that there
are further "reasonable" polar coordinate pairs for the same point!
For example, go around to π/6+π. If you position your robot arm
there, and then tell the arm to move backwards 2 units, the arm will
be positioned at (sqrt(3),1). Sigh. So here are some more polar
coordinates for the same point:
r=–2 and θ=7π/6 and
r=–2 and θ=–19π/6
and r=–2 and θ=–31π/6 ETC.
but we are not done yet,
because there are also (going backwards in the angle and the
length)
r=–2 and θ=–5π/6 and
r=–2 and θ=–17π/6 and
r=–2 and θ=–29π/6 ETC.
Generally, r=–2 and θ=7π/6+2π(multiplied by any integer): the integer
could be 0 or positive or negative.
Common restrictions on polar coordinates and the problems they have
This is irritating. Any point in the plane has infinitely many valid
"polar coordinate addresses". In simple applications, people
frequently try to reduce the difficulty. Much of the time, we expect
r>0 always. And maybe we also make θ more calm. The
restriction 0≤θ<2π is used, except when it isn't, so
the restriction –π<θ≤π is used in other
circumstances. I am not trying to be even more incomprehensible than
usual. I am merely reporting what different people do. As we will see,
this is all very nice, except that there are natural circumstances,
both in physical modeling (the robot arm I mentioned) and in the
mathematical treatment, where it will make sense to ignore the
artificial restrictions, even if this makes life more
difficult. You'll see a few of these circumstances.
Conversion formulas
If you consider the picture to the right, I hope that you can fairly
easily "read off" how to go from r and θ to x and y:
x=r cos(θ)
y=r sin(θ)
Going from x and y to r is easy enough:
r=sqrt(x^{2}+y^{2}). If we divide the y equation above
by the x equation, the r's drop out and we get y/x=tan(θ) so
that θ=arctan(y/x). Please note that there are infinitely
many valid r and θ pairs for every point, so this method
will only give you one such pair! Be careful in real applications,
please.
Specifying regions in the plane in polar fashion
It is useful to try to get used to thinking in polar fashion, because
then you will be able to see problems (usually physical or geometric
problems with lots of central symmetry) where this coordinate system
can be used to really simplify computations. So here are some simple
examples of regions which can be easily specified with polar
inequalities.
 
We will study the equations and graphs of some polar curves next time, and we will do a bit of calculus (arc length and area). That will conclude the course lectures.
Last QotD I can have returned
Please sketch as well as possible all points in the plane which have
r=θ using any (!) determination of polar coordinates (no
restrictions on r and θ). The result is weird until you get used
to it (as usual!). I will try to have the results returned on
Thursday, April 29.
To the right is Maple's graph of r=θ when –12≤θ≤12. Think hard about what the graph would look like for θ small positive (in the first quadrant) and θ small negative (although the "robot arm" is pointing in the fourth quadrant, the corresponding r value is negative, so the curve appears in the second quadrant, near the negative xaxis). The curve is smooth with no corners or cusps, and has infinitely many selfintersections on the yaxis.
Most people got the θ>0 part of the curve right but the part with θ<0 was incorrect, because it is difficult to remember to go backwards when r is negative (this happens because the curve is r=θ). I drew something like the small picture to the left on some people's papers. I thought maybe that might help them understand.
I don't think negative r's will occur very often in practice so maybe this wasn't a great choice as QotD. This curve is an example of an Archimedean spiral (although usually r>0 in the description of that spiral).
A curious collection of facts
The most intriguing (strange, weird?) entries in the table below occur
as a result of Euler's Formula. If you are willing to
accept that there is a number i whose square is –1, then something
strange happens as you consider e^{ix}. Notice that the powers
of i have this behavior:
Then because sine is odd and cosine is even, e^{–ix}=cos(x)–i sin(x). This gets the entries in the formula column for sine and cosine (add and subtract the two equations).
There are other hyperbolic functions defined in a fashion which is
similar to trig functions. For example, the hyperbolic tangent, tanh
(pronounced ludicrously as "tanch") is sinh/cosh. As I previously
wrote, you may look in your textbook. But here also are some links: 
Monday, April 27  (Lecture #24) 

We continue the schizophrenic (?) progression of this course by spending three lectures on parametric curves and polar coordinates. These topics would be better covered after other applications of the definite integral, and I will use this order if I teach this course again. Also, the material will make more sense from the point of view of Math 251. Sigh. So, onwards.
Parametric curves
We begin our rather abbreviated study of parametric curves. These
curves are a rather clever way of displaying a great deal of
information. Here both x and y are functions of a parameter. The
parameter in your text is almost always called t. The simplest
physical interpretation is that the equations describe the location of
a point at time t, and therefore the equations describe the motion of
a point as time changes. I hope the examples will make this more
clear. The t here is usually described for beginners as time, but in
applications things can get a great deal more complicated. Parametric
curves could be used to display lots of information. I mentioned that
some steels contain chromium. Maybe the properties of the steel such
as ductility (a real word: "The ability to permit change of shape
without fracture.") and density, might depend on the percentage of
chromium. So the t could be that and the x and y could be measurements
of some physical properties of the steel. Here x=f(t) and y=g(t), as
in the text. Now a series of examples.
Example 1
Suppose x(t)=cos(t) and y(t)=sin(t). I hope that you recognize almost
immediately that x and y must satisfy the equation
x^{2}+y^{2}=1, the standard unit circle, radius 1,
center (0,0). But that's not all the information in the equations.
The point (x(t),y(t)) is on the unit circle. At "time t" (when the parameter is that specific value) the point has traveled a length of t on the unit circle's curve. The t value is also equal to the radian angular measurement of the arc. This is uniform circular motion. The point, as t goes from –∞ to +∞, travels endlessly around the circle, at unit speed, in a positive, counterclockwise direction.
Example 2
Here is a sequence of (looks easy!) examples which I hope showed
students that there is important dynamic (kinetic?) information in the
parametric curve equations which should not be ignored.
 
 
 

Example 3
A bug drawing out a thread ...
Thread is wound around the unit circle centered at the origin. A bug
starts at (1,0) and is attached to an end of the thread. The bug
attempts to "escape" from the circle. The bug moves at unit speed.
I would like to find an expression for the coordinates of the bug at time t. Look at the diagram. The triangle ABC is a right triangle, and the acute angle at the origin has radian measure t. The hypotenuse has length 1, and therefore the "legs" are cos(t) (horizontal leg, AB) and sin(t) (vertical leg, BC). Since the line segment CE is the bug pulling away (!) from the circle, the line segment CE is tangent to the circle at C. But lines tangent to a circle are perpendicular to radial lines. So the angle ECA is a right angle. That means the angle ECD also has radian measure t. But the hypoteneuse of the triangle ECD has length t (yes, t appears as both angle measure and length measure!) so that the length of DE is t sin(t) and the length of CD is t cos(t).
The coordinates of E can be gotten from the coordinates of C and the lengths of CD and DE. The xcoordinates add (look at the picture) and the ycoordinates are subtracted (look at the picture). Therefore the bug's path is given by x(t)=cos(t)+t sin(t) and y(t)=sin(t)–t cos(t).
t between 0 and 1  t between 0 and 10 Note that the scale is changed! 

Finally to the right is an animated picture of the bug moving. Maybe
you can understand this picture better: maybe (!!). This curve is more typical of parametric curves. I don't know any easy way to "eliminate" mention of the parameter. This seems to be an authentically (!) complicated parametric curve, similar to many curves which arise in physical and geometric problems. It has an official name. It is called the evolute of the circle. 
Calculus?
Finally, very late in the lecture, I attempted some calculus. Here's
what I said.
Suppose we want to analyze what happens when the parameter changes just a little bit, from t to t+Δt. Well, the point starts at (f(t),g(t)). What can we say happens at t+Δt? Well, f(t+Δt)≈f(t)+f´(t)Δt. Why is this true? You can think of this either 151 style as linear approximation, or from our more sophisticated 152 approach, this is the constant and linear terms in the Taylor series for f(t). Similarly for g(t) we know g(t+Δt)≈g(t)+g´(t)Δt. Therefore the point in the interval [t,t+Δt] moves from (f(t),g(t)) to (approximately!) (f(t)+f´(t)Δt,g(t)+g´(t)Δt). What is the slope of the line segment connecting these points?
Slope
Take the difference in second coordinates divided by the difference in
the first coordinates. The result (there is a lot of cancellation) is
g´(t)/f´(t). If this were an xy curve, this would be noted
as dy/dx, the slope of the tangent line. In fact, people usually
remember the result in the following way:
dy dy/dt  =  dx dx/dtand this can be used to get tangent lines (which I will do next time!).
Speed
Since Distance=Rate·Time, and in the time
interval [t,t+Δt] we move from (f(t),g(t)) to (approximately!)
(f(t)+f´(t)Δt,g(t)+g´(t)Δt), we can get the speed
(the Rate) by taking the distance between these points and dividing by
Δt. There is more cancellation here, and the result is
Speed=sqrt(f´(t)^{2}+g´(t)^{2}) or
___________ /(dx)^{2} (dy)^{2} Speed= / ()+ () / (dt) (dt)As you'll see if you take Math 251, this is the sum of the squares of the horizontal and vertical components of the velocity vector: it is, in fact, the magnitude of the velocity vector.
Please: I will show you a few simple examples of this Wednesday, and then go on to Polar Coordinates.
The second exam
was returned. More information is available.
Wednesday, April 22  (Lecture #23) 

As I mentioned in class, I am not allowed to tell you why these series resemble each other because your heads might explode. e^{ix}=cos(x)+isin(x) The notaccidential resemblance will be discussed in your differential equation course.Book problem: 10.7, #21
Book problem: 10.7, #14
Find the Maclaurin series of cos(sqrt(x)). Since
cos(x)=1–[x^{2}/2!]+[x^{4}/4!]–[x^{6}/6!]+[x^{8}/8!]–[x^{10}/10!]...=∑_{n=0}^{∞}(–1)^{n+1}x^{2n}/(2n)! I know that
cos(sqrt(x))=1–[(sqrt(x))^{2}/2!]+[(sqrt(x))^{4}/4!]–[(sqrt(x))^{6}/6!]+[(sqrt(x))^{8}/8!]–[(sqrt(x))^{10}/10!]...=∑_{n=0}^{∞}(–1)^{n+1}(sqrt(x))^{2n}/(2n)!
and so
cos(sqrt(x))=1–[x/2!]+[x^{2}/4!]–[x^{3}/6!]+[x^{4}/8!]–[x^{5}/10!]...=∑_{n=0}^{∞}(–1)^{n+1}x^{n}/(2n)!,
and please be LAZY.
Book problem: 10.7, #19 Find the Maclaurin series of (1–cos(x^{2})/x. Since cos(x)=1–[x^{2}/2!]+[x^{4}/4!]–[x^{6}/6!]+... I know that cos(x^{2})=1–[x^{4}/2!]+[x^{8}/4!]–[x^{12}/6!]+... and 1–cos(x^{2})=[x^{4}/2!]–[x^{8}/4!]+[x^{12}/6!]+... so that (1–cos(x^{2})/x=[x^{3}/2!]–[x^{7}/4!]+[x^{11}/6!]+... 
An integral
The function e^{–x2} is extremely important in
probability. Its integral is called the error
function.
Suppose we want to compute
∫_{x=0}^{.5}e^{–x2}dx. It can
be proved that e^{–x2} has no
antiderivative which can be written in terms in familiar
functions. How could we then compute this definite integral?
Its value, a pal of mine tells me, is approximately 0.46128. Well, I
could uses the Trapezoid Rule or Simpson's Rule or ... Look at this:
e^{x}=∑_{n=0}^{∞}x^{n}/n!
Substitute –x^{2} for x.
e^{–x2}=∑_{n=0}^{∞}(–x^{2})^{n}/n!=∑_{n=0}^{∞}(–1)^{n}x^{2n})/n!
Integrate.
∫_{x=0}^{x=.5}e^{–x2}dx=∫_{x=0}^{x=.5}∑_{n=0}^{∞}(–1)^{n}x^{2n})/n! dx=∑_{n=0}^{∞}∫_{x=0}^{x=.5}(–1)^{n}x^{2n})/n!=∑_{n=0}^{∞}(–1)^{n}x^{2n+1}/([2n+1]n!)_{x=0}^{x=.5}
Evaluate.
The integral is∑_{n=0}^{∞}(–1)^{n}(1/2)^{2n+1})/([2n+1]n!)
This series is alternating, and satisfies all the hypotheses of the Alternating Series Test. Any partial sum is within the accuracy of the first omitted term (the last workshop!). Well, if I want 5 digit accuracy, I just need to find n so that (1/2)^{2n+1})/([2n+1]n!) is less than .00001, which is 1/(100,000).
If n=4, then (1/2)^{9}/[9·24] is {1/(512)}·[1/(216)] which is 110,592. The sum from n=0 to 4, that is, S_{4}, is 0.46128 (it is actually accurate one digit beyond that).
A familiar series
We know 1/(1–x). It is the sum of a geometric series with first term 1
and ratio equal to x. So
1/(1–x)=∑_{n=0}^{∞}x^{n}. This
equation is valid when x<1, or –1<x<1. Some remarkable
things can be done with this series.
Logarithm
What the Maclaurin series of ln(x)? This is a trick question because
y=ln(x) looks like this and the
limit of ln(x) as x→0^{+} is –∞ so ln(x)
can't have a Taylor series centered at 0. What's a good place
to consider? Since ln(1)=0, we could center the series at 1. Most
people would still like to compute near 0, though, so usually the
function is moved instead! That is, consider ln(1+x) whose graph now
behaves nicely at 0 so we can
analyze it there.
If f(x)=ln(1+x), I want to "find" ∑_{n=0}^{&infin}[f^{(n)}(0)/n!]x^{n}. Well, f(0)=ln(1+0)=ln(1)=0, so we know the first term. Now f´(x)=1/(1+x) so that ... wait, wait: remember to try to be LAZY.
Look at
1  1+xThis sort of resembles the sum of a geometric series. We have two "parameters" to play with, c, which is the first term, and r, which is the ratio between successive terms. The sum is c/(1–r). If we take c=1 and r=–x then 1/(1+x)=1/(1–{–x}) is the sum of a geometric series. So
Computing with ln
What if x=–1/2 in the previous equation? Then ln(1–1/2)=ln(1/2)=–ln(2)
and this is approximately –69314. A friend of mine has just computed
&sum_{n=1}^{10}[(–1)^{n+1}(–.5)^{n}/n]
and this turns out to be –.69306. We only get 3 decimal places of
accuracy. It turns out that this series converges relatively slowly
compared to the others we've already seen, which have the advantage of
factorials in the denominator. So this series is usually not directly
used for numerical computation, but other series related to it are used.
Book problem: 10.7, #9 Find the Maclaurin series of ln(1–x^{2}). Be LAZY. We know that ln(1+x)=x–[x^{2}/2]+[x^{3}/3]–[x^{4}/4]+[x^{5}/5]+...=&sum_{n=1}^{∞}[(–1)^{n+1}x^{n}/n]. This so we can substitute –x^{2} for x and get ln(1–x^{2})=–x^{2}–[(–x^{2})^{2}/2]+[(–x^{2})^{3}/3]–[(–x^{2})^{4}/4]+[(–x^{2})^{5}/5]+...=&sum_{n=1}^{∞}[(–1)^{n+1}(–x^{2})^{n}/n] and further ln(1–x^{2})=–x^{2}–[x^{4}/2]–[x^{6}/3]–[x^{8}/4]–[x^{10}/5]+...=–&sum_{n=1}^{∞}[(–1)^{n+1}x^{2n}/n] (valid for x<1).
Computing a value of a derivative That's if you desperately wanted to know the value of the derivative. An alternate strategy would be to compute the 8^{th} derivative and evaluate it at x=0. Here is that derivative: 6 8 4 10080 (28 x + x + 70 x + 28 x + 1)   2 8 (1 + x ) 
arctan
Let me try to find a Taylor series centered at 0 (a Maclaurin series)
for arctan. Well, the general Maclaurin series is
∑_{n=0}^{∞}[f^{(n)}(0)/n!]x^{n}
so we can just try to compute some derivatives and evaluate them at
0. Let's see:
n=0 f(x)=arctan(x) so f(0)=arctan(0)=0.
n=1 f´(x)=1/(1+x^{2}) so
f´(0)=Stop this right now! Why?
Because this way is madness. Here is the 7^{th} derivative of
arctan(x):
6 4 2 720 (7 x  35 x + 21 x  1)  2 7 (1 + x )Does this look like something you want to compute?
Instead look at
1  1+x^{2}This sort of resembles the sum of a geometric series. We have two "parameters" to play with, c, which is the first term, and r, which is the ratio between successive terms. The sum is c/(1–r). If c=1 and r = –x^{2} then 1/(1+x^{2})=1/(1–{–x^{2}}) is the sum of a geometric series. So
Computing π This series has been used to compute decimal approximations of π. For example, if x=1, arctan(1)=π/4, so this must be 1–1/3+1/5–1/7+... but the series converges very slowly (for example, the 1000^{th} partial sum multiplied by 4 gives the approximation 3.1406 for π which is not so good for all that arithmetic!) . Here is a history of some of the classical efforts to compute decimal digits of π. You can search some of the known decimal digits of π here. There are more than a trillion (I think that is 10^{12}) digits of π's decimal expansion known. Onward! The methods used for such computations are much more elaborate than what we have discussed. 
In physics they say ...
Many of the force "laws" stated in physics are quadratic (second
degree) and therefore it is not surprising that squares and square
roots need to be computed frequently. What does sqrt(1+x) "look like"
near 0? Well, in this case I will try a direct computation. If
f(x)=sqrt(1+x) then ...
Function  Value at x=0 

f(x)=(1+x)^{1/2}  1 
f´(x)=(1/2)(1+x)^{–1/2}  1/2 
f´´(x)=(1/2)(–1/2)(1+x)^{–3/2}  (1/2)(–1/2) 
f^{(3)}(x)=(1/2)(–1/2)(–3/2)(1+x)^{–5/2}  (1/2)(–1/2)(–3/2) 
I forget but look at the pattern.  (1/2)(–1/2)(–3/2)(–5/2) 
So this Taylor series looks like
(1+x)^{1/2}=1+(1/2)x+[(1/2)(–1/2)/2]x^{2}+[(1/2)(–1/2)(–3/2)/6]x^{3}+[(1/2)(–1/2)(–3/2)(–5/2)/24]x^{4}+...
where those other strange numbers come from the factorials, of
course. Well, how might this be used in physics. Suppose you are
trying to analyze sqrt(1+w). If w is very small, well then I bet
that sqrt(1+w) is quite close top sqrt(1+0) which is 1. But the value
will not equal 1 if w is not 0. What sort of "first order" estimate
would I believe in? I bet that sqrt(1+w) is approximately 1+(1/2)w for
small w. I also believe that the error will be (roughly) proportional
to the size of w^{2} (that's the Error Bound again). For many
applications, knowing this is enough. But what if I wanted more
accuracy, and I wanted an estimate which was correct to "second order
terms". I bet this would take sqrt(1+w) and then the estimate woul dbe
1+(1/2)w–(1/8)w^{2}, with an error which would be (roughly)
proportional to w^{3}. Depending on the application you were
interested in, the estimate would be a bigger and bigger partial sum
of the Taylor series.
Why do I insist on writing the coefficients of the series in the silly
way done above? Why not just multiply out and write that? Well, if I
do the result may be rather deceptive. It would be
1+(1/2)w–[1/8]w^{2}+[1/16]w^{3}–[5/64]w^{4}+...
so if I accidentally saw only the first 4 terms I might think there is
some obvious pattern to the series. Actually, the pattern is more
complicated, as the coefficient of w^{4} showns. There is an
abbreviation which is used (binomial coefficients) but the numbers are
complicated.
QotD
What are the terms up to order 10 (x^{10}) in the Maclaurin
series for cos(5x^{3}) (this is pretty close to a trick
question). The answer is short and easy, and lots of the terms
have their coefficient equal to 0.
An answer We know that cos(x)=1–x^{2}/2+x^{4}/24+... and we should be LAZY. So substitute 5x^{3} for x in that equation. This is substitution or composition, it is not multiplication! For example, the x^{2} term becomes (5x^{3})^{2} which is 25x^{6}. In fact, we get cos(5x^{3})=1–(25/2)x^{6}+(625/24)x^{12}/24+... and the actual answer to this particular question is only the terms 1–(25/2)x^{6}.
Here is what I was able to do last year, without the darn "blizzard"
causing a class to be canceled. Maybe you might find it helpful:
Binomial series with m=1/3
1+(1/3)x+[(1/3)(–2/3)/2]x^{2}+[(1/3)(–2/3)(–5/3)/2·3]x^{3}+[(1/3)(–2/3)(–5/3)(–8/3)/2·3·4]x^{4}+... I'll come back to the general ideas later, but let's see how to use this in various ways.
Naive numerical use What if we wanted to improve this estimate? Well, we can try another term. By this I mean use 1+(1/3)(.05)+[(1/3)(–2/3)/2](.05)^{2} as an estimate of (1.05)^{1/3}. How good is this estimate? Again, we use the Error Bound: Error≤[K/(n+1)!]x–a^{n+1}. Now n=2 and a=0 and x=.05, and K comes from considering f^{(3)}(x)=(1/3)(–2/3)(–5/3)(1+x)^{–8/3}. We need to look at (10/27)(1+x)^{–8/3} on [0,.05]. The exponent is again negative (what an accident not  these methods are actually used and things should be fairly simple!) and therefore the function is again decreasing and an overestimate is gotten by looking at the value when x=0, so (10/27)(1+x)^{–8/3} becomes (10/27)(1+0)^{–8/3}=(10/27). Hey, [K/(n+1)!]x–a^{n+1} in turn becomes [(10/27)/3!](.05)^{3}, about .000008, even better.
Approximating a function on an interval
Improving the approximation
Binomial series in general

The list ...
Wednesday, April 15  (Lecture #22) 

Power series ...
I recalled what had already been said about power series. Here we
go:
Definition A power series centered at x_{0} (a fixed number) is an infinite series of the form ∑_{n=0}^{∞}a_{n}(xx_{0})^{n} where the x is a variable and the a_{n} are some collection of coefficients. It resembles an infinite degree polynomial. Usually I (and most people) take x_{0}=0 because this just makes thinking easier. 
Calculus and power series Hypothesis Suppose the power series ∑_{n=0}^{∞}a_{n}(x–x_{0})^{n} has some positive radius of convergence, R, and suppose that f(x) is the sum of this series inside its radius of convergence. Differentiation The series ∑_{n=0}^{∞}n(a_{n})(x–x_{0})^{n–1} has radius of convergence R, and for the x's where that series converges, the function f(x) can be differentiated, and f´(x) is equal to the sum of that series. Integration The series ∑_{n=0}^{∞}[a_{n}/(n+1)](x–x_{0})^{n+1} has radius of convergence R, and for the x's where that series converges, the sum of that series is equal to an indefinite integral of f(x), that is ∫f(x)dx. 
Convergence and divergence A power series centered at x_{0} always has an interval of convergence with the center of that interval equal to x_{0}. Inside this interval of convergence, the series converges absolutely and therefore converges. Outside the interval, the series diverges. The power series may or may not converge on the boundary of the interval. The interval may have any length between 0 and ∞. Half the length of the interval is called the radius of convergence. 
If a function has a power series then ...
Suppose I know that f(x) is equal to a sum like
A+B(x–x_{0})+C(x–x_{0})^{2}+D(x–x_{0})^{3}+E(x–x_{0})^{4}+...
and I would like to understand how the coefficients A and B and C and
D etc. relate to f(x). Here is what we can do.
Step 0 Since
f(x)=A+B(x–x_{0})+C(x–x_{0})^{2}+D(x–x_{0})^{3}+E(x–x_{0})^{4}+...
if we change x to x_{0} we get f(x_{0})=A. All the
other terms, which have powers of x–x_{0}, are 0.
Step 1a Differentiate (or, as I said in class, d/dx) the
previous equation which has x's, not x_{0}'s. Then we have
f´(x)=0+B+2C(x–x_{0})^{1}+3D(x–x_{0})^{2}+4E(x–x_{0})^{3}+...
Step 1b Plug in x_{0} for x and get
f´(x_{0})=B. All the
other terms, which have powers of x–x_{0}, are 0.
Step 2a Differentiate (or, as I said in class, d/dx) the
previous equation which has x's, not x_{0}'s. Then we have
f´´(x)=0+0+2C+3·2D(x–x_{0})^{1}+4·3E(x–x_{0})^{2}+...
Step 2b Plug in x_{0} for x and get
f´´(x_{0})=2C, so
C=[1/2!]f^{(2)}(x_{0}). All the other terms,
which have powers of x–x_{0}, are 0.
Step 3a Differentiate (or, as I said in class, d/dx) the
previous equation which has x's, not x_{0}'s. Then we have
f^{(3)}(x)=0+0+0+3·2·1D+4·3·2E(x–x_{0})^{1}+...
Step 3b Plug in x_{0} for x and get
f^{(3)})(x_{0})=3·2·1D=3!C
so D=[1/3!]f^{(3)}(x_{0}). All the other terms,
which have powers of x–x_{0}, are 0.
ETC. Continue as long as you like. What we
get is the following fact: if
f(x)=∑_{n=0}^{∞}a_{n}(x–x_{0})^{n}
then a_{n}=[f^{(n)}(x_{0})/n!]. This is
valid for all nonnegative integers, n. Actually, this formula is one
of the reasons that 0! is 1 and the zeroth derivative of f is f
itself. With these understandings, the formula works for n=0.
What this means is
If a function is equal to a power series, that power series must
be the Taylor series of the function.
I hope you notice, please please please, that the partial sums of
the Taylor series are just the Taylor polynomials, which we studied earlier.
Usually I'll take x_{0}=0, as I mentioned so that (xx_{0})^{n} becomes just x^{n}. Then the textbook and some other sources call the series the Maclaurin series but I am too lazy to remember another name. A useful consequence of this result (it will seem sort of silly!) is that if a function has a power series expansion, then it has exactly one power series expansion (because any two such series expansions are both equal to the Taylor series, so they must be equal). This means if we can get a series expansion using any sort of tricks, then that series expansion is the "correct one"  there is only one series expansion. I'll show you some tricks, but in this class I think I will just try some standard examples which will work relatively easily.
e^{x}
I'll take x_{0}=0. Then all of the derivatives of
e^{x} are e^{x}, and the values of these at 0 are all
1. So the coefficients of the Taylor series, a_{n}, are
[f^{(n)}(x_{0})/n!]=1/n!. The Taylor series for
e^{x} is therefore
∑_{n=0}^{∞}[1/n!]x^{n}.
e^{–.3}
Let's consider the Taylor series for e^{x} when
x=–.3. This is
∑_{n=0}^{∞}[1/n!](–.3)^{n}. What
can I tell you about this? Well, for example, my "pal" could compute a
partial sum, say
∑_{n=0}^{10}[1/n!](–.3)^{n}. The
result is 0.7408182206. That's nice. But what else do we know? Well,
this partial sum is T_{10}(–.3), the tenth Taylor
polynomial for e^{x} centered at x_{0}=0, and
evaluated at –.3. The Error Bound gives an estimation of
T_{10}(–.3)–e^{–.3}. This Error
Bound asserts that this difference is at most
[K–.3–0^{11}/11!], where K is some overestimate
of the 11^{th} derivative of e^{x} on the interval
between –.3 and 0. Well, that 11^{th} derivative is also
e^{x}. And we know that e^{x} is increasing
(exponential growth after all!) so that for x's in the interval
[–.3,0], e^{x} is at most e^{0}=1, and we can
take that for K. So the Error Bound is
1–.3–0^{11}/11!. Now let's look at some numbers:
–.3^{11}=0.00000177147 and 11!=39,916,800, and their
quotient is about 4·10^{–14}.
This means that e^{–.3} and T_{10}(–.3) agree at least
to 13 decimal places. Indeed, to 10 decimal places, e^{–.3} is
reported as 0.7408182206, the same number we had before. Wow? Wow!
Let's change 10 to n and 11 to n+1. Then T_{n}(–.3)–e^{–.3} is bounded by K–.3–0^{n+1}/(n+1)!. Here K=1 again because all of the derivatives are the same, e^{x}. Since 1–.3–0^{n+1}/(n+1)!→0 as n→∞ what do we know?
I think that the sequence {T_{n}(–.3)} converges, and
its limit is e^{–.3}. Since this sequence of Taylor polynomial
values is also the sequence of partial sums of the series
∑_{n=0}^{∞}[1/n!](–.3)^{n}, I
think that the series converges, and its sum is
e^{–.3}. Therefore
e^{–.3}=∑_{n=0}^{∞}[1/n!](–.3)^{n}.
e^{.7}
We tried the same sequence of ideas with x=.7. We could first examine
T_{10}(.7). This is
∑_{n=0}^{10}[1/n!](.7)^{n}. To 10
decimal places, this is 2.0137527069. We have information from the
Error Bound. It declares that T_{10}(.7)–e^{.7} is
no larger than K.7–0^{11}/11!. Here K is an overestimate of
the 11^{th} derivative, which is e^{x}, on the
interval [0,.7]. The exponential function is (still!) increasing, so
the largest value is at x=.7. But I don't know e^{.7}. I do
know it is less than e^{1} which I hardly know also. I will
guess that e<3. So a nice simple K to take is 3. Let me try
that. The Error Bound is less than 3.7–0^{11}/11!. Let's do
the numbers here.
11!=39,916,800 (again) but .7^{11}=0.0197732674 (small, but
not as small as –.3^{11}). The Error Bound
3.7–0^{11}/11! is about 2·10^{–9}, not quite
as small.
Now e^{.7}, to 10 decimal places, is 2.0137527074 and this is
close enough to the sum value quoted before.
Again, go to n and n+1: T_{n}(.7)–e^{.7} is less
than 3.7–0^{n+1}/(n+1)!, and again, as n→∞ this
goes to 0. Our conclusion is:
The sequence {T_{n}(.7)} converges, and
its limit is e^{.7}. Since this sequence of Taylor polynomial
values is also the sequence of partial sums of the series
∑_{n=0}^{∞}[1/n!](.7)^{n}, I
think that the series converges, and its sum is
e^{.7}. Therefore
e^{.7}=∑_{n=0}^{∞}[1/n!](.7)^{n}.
e^{50}
Just one more example partly because we'll see some strange
numbers. Let's look at T_{10}(50)
which is
∑_{n=0}^{10}[1/n!]50^{n}.
This turns out to be (approximately!) 33,442,143,496.7672, a big
number. The Error Bound says that T_{10}(50)–e^{50}
is less than K50–0^{11}/11! where K is the largest
e^{x} can be on [0,50]. That largest number is e^{50}
because e^{x} is increasing. I guess e^{50} is less
than, say, 3^{50}, which is about
7·10^{23}. I'll take that for K. Now how big is that Error?
K50–0^{11}/11! still has 11! underneath but now the top is
growing also. This is approximately 9·10^{34}, a sort
of big number.
The situation for x=50 may look hopeless, but it isn't really. To
analyze T_{n}(50)–e^{50} we need to look at
K[(50)^{n+1}/(n+1)!]. Here the fraction has power growth on
the top and factorial growth on the bottom. Well, we considered this before. I called it a
"rather sophisticated example". Factorial growth is faster eventually
than power growth. So this sequence will →0 as
n→∞. A similar conclusion occurs:
e^{50}=∑_{n=0}^{∞}[1/n!](50)^{n}.
In fact, e^{50} is 5.18470552858707·10^{21} while the partial sum with n=100, ∑_{n=0}^{100}[1/n!](50)^{n} has value 5.18470552777323·10^{21}: the agreement is not too bad, relatively.
And generally for exp ...
It turns out that
∑_{n=0}^{∞}[1/n!]x^{n} converges
for all x and its sum is always e^{x}. The way
to verify this is what we just discussed. Most actual computation of
values of the exponential function relies on partial sums of this
series. There are lots of computational tricks to speed things up,
but the heart of the matter is the Taylor series for the exponential
function.
Sine
We analyzed sine's Taylor polynomials, taking advantage of the cyclic
(repetitive) nature of the derivatives of cosine:
sine→cosine→sine→cosine then back to sine. At
x_{0}=0, this gets us a cycle of numbers:
0→1→0→1→0 etc. The Taylor series for sine
centered at 0 leads off like this:
x–[x^{3}/3!]+[x^{5}/5!]–[x^{7}/7!]+[x^{9}/9!]–...
It alternates in sign, it has only terms of odd degree, and each term has the reciprocal of an "appropriate" factorial (same as the degree) as the size of its coefficient. Using summation notation, which is convenient and compact, this series is ∑_{n=0}^{∞}[(1)^{n}/(2n+1)!]x^{2n+1}.
What happens to the error bound?
This is similar to what we did with exp. There are two claims: the
series
∑_{n=0}^{∞}[(1)^{n}/(2n+1)!]x^{2n+1}
converges and the sum of the series is sin(x). Well, to see that this
is true we investigate the difference between sin(x) and
S_{N}, the N^{th} partial sum of the series. But
S_{N} is the same as T_{N}(x), the N^{th}
Taylor polynomial. And the error bound tells us that
sin(x)T_{N}(x)≤[K/(N+1)!]x0^{n+1}. Just as
before, [x^{N+1}/(N+1)!]→0 as N→∞. What about
the K's? If they misbehave (get very big) that could make the whole
estimate lousy. But in fact in this specific case, K is an
overestimate on the size of some derivative of sine. But all of the
derivatives of sine are +/sine and +/cosine, and these all are ≤1
in their absolute values. So, in fact, we're done. We have verified
that the series converges and that sin(x) is its sum.
Cosine
We could duplicate this work for cosine, or, as I mentioned in class,
be a bit cleverer. Since we know that
sin(x)=∑_{n=0}^{∞}[(1)^{n}/(2n+1)!]x^{2n+1}
(valid for all x) we could differentiate this equation. The result is
cos(x)=∑_{n=0}^{∞}[(1)^{n}/(2n+1)!](2n+1)x^{2n}.
In fact, most people realize that (2n+1)/(2n+1)! is 1/(2n)! so that we
have verified the equation
cos(x)=∑_{n=0}^{∞}[(1)^{n}/(2n)!]x^{2n}
for all x.
A numerical example: cos(1/3) How close is 1–[(1/3)^{2}/2!]+[(1/3)^{4}/4!]–[(1/3)^{6}/6!]+[(1/3)^{8}/8!]–[(1/3)^{10}/10!] to cos(1/3)? Here we sort of have two candidates because T_{10}(1/3) is the same as T_{11}(1/3) since the 11^{th} degree term is 0. Error bound, n=10 So we have K(1/3)–0^{11}/11!, where K is a bound on the size of the 11^{th} derivative of cosine. Hey: I don't care much in this example, because I know that this derivative is +/–cosine or +/–sine, so that I can take K to be 1. Now it turns out that (1/3)^{11}/11! is about 1.4·10^{–14}. This is tiny, but ... Error bound, n=11 This is even better. So we have K(1/3)–0^{12}/12!, where K can again be taken as 1 (this is easier than exp!) So (1/3)^{12}/12! is about 4·10^{–15}, even tinier. Hey, cos(1/3)=0.944956946314738 and T_{10}(1/3)=0.944956946314734.
Cosine and sine estimates
Is success guaranteed? But these are the first and nicest and simplest examples. The situation is not always so easy. We will see a few functions where things don't work out. I can even think about one of them with you now. Consider tangent. Certainly if we take x_{0} to be 0, we can differentiate tangent lots of times and "get" a Taylor series for tangent centered at 0. The reason I wrote "get" with quotes is that the coefficients of the Taylor series for tangent are wellknown in the math world to be rather, well, rather irritating and difficult to predict and understand. So already there's a problem. How about convergence? Things also don't go well there because, if you remember tangent's graph, vertical asymptotes occur at odd multiples of Π/2. You can't expect that the series will converge, for example, at +/Π/2 and, in fact, the radius of convergence turns out to be only Π/2 (this is not so obvious, actually). Most calculators I know compute values of tangent by computing values of sine and cosine and then dividing. This is easier than direct computation of tangent. 
Maintained by greenfie@math.rutgers.edu and last modified 4/16/2009.