# Arctangent of negative infinity.

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### Chris Schumacher

Oct 22, 2005, 6:11:54 PM10/22/05
to
What is the arctan of negative infinity? Is it pi/2, the same as infinity?
I've been looking at the graph and it seems that it's both going to infinity
and coming from negative infinity at the points where it's undefined.

-==Kensu==-
(btw, I'm not being a smartass, this came up on a improper integral question)

### David W. Cantrell

Oct 22, 2005, 6:40:48 PM10/22/05
to
Chris Schumacher <ken...@hotmail.com> wrote:
> What is the arctan of negative infinity? Is it pi/2, the same as
> infinity?

No, rather Arctan(-oo) = -pi/2.

David

### ticbol

Oct 22, 2005, 7:31:30 PM10/22/05
to
Yes, arctan (-infinity) = pi/2

So you are looking at the graph of y = tan(theta). And you see that at
theta = pi/2, the graph doesn't touch the vertical line
theta=pi/2----because that vertical line is an asymptote of the graph.

It means:
-----as theta approaches pi/2 from the left, tan(theta) approaches
positive infinity.
-----as theta approaches pi/2 from the right, tan(theta) approaches
negative infinity.

That further means:
tan(pi/2) = infinity, or negative infinity

That is why tan(pi/2) is actually indeterminate.
Tangent here being [(oppsite side} / (adjacent side)], referring to the
reference triangle of theta.
At pi/2, adjacent side is zero, while opposite side has a positive
length or height, say, "y".
So, tan(pi/2) = y / 0
That is indeterminate.
For convenience, in trig, it is assumed as infinity.
Then, due to the characteristics of the graph of y = tan(theta), one
of which is tan(pi/2 +0.0000005) = a negative number, the y=tan(theta)
is negative right after theta=pi/2 until theta=pi. That is why
tan(pi/2) is almost negative infinity right after theta=pi/2.
So, if tan(theta) = negative infinty,
Then, arctan(negative infinity) = theta = pi/2

That is only if 0 <= theta <= pi.
Due to the periodicity of y = tan(theta), where one period is one pi,
then in the interval [0,2pi],
arctan(negative infinity) = pi/2, or 3pi/2.

### David W. Cantrell

Oct 22, 2005, 8:09:05 PM10/22/05
to
"ticbol" <tic...@yahoo.com> wrote:
> Yes, arctan (-infinity) = pi/2

No. If arctan() has a single value, then we are taking about
the _principal-valued_ inverse tangent function. And that means that we do
not look at the complete "graph of y = tan(theta)", as you suggested below.
Rather, we look only at that part of the graph for which |theta| < pi/2. On
that part of the graph there is no ambiguity at all: If tan(theta)
approaches -infinity, theta must approach -pi/2.
Thus, arctan(-infinity) = -pi/2.

David

### ticbol

Oct 22, 2005, 8:35:38 PM10/22/05
to
David W. Cantrell :

"No. If arctan() has a single value, then we are taking about
the _principal-valued_ inverse tangent function. And that means that we
do
not look at the complete "graph of y = tan(theta)", as you suggested
below.
Rather, we look only at that part of the graph for which |theta| <
pi/2. On
that part of the graph there is no ambiguity at all: If tan(theta)
approaches -infinity, theta must approach -pi/2.
Thus, arctan(-infinity) = -pi/2.

David "

No, David. You know trig functions, like tan(theta) are periodic. So
arctan(...) has many values. If it was mentioned that the theta has a
domain or interval, then the values of arctan(...) are the number of
theta values whose tan is the value given in the arctan(...).
The seeker did not specify the interval of the angle in his posted
question. So, if, as popular, the interval were from 0 to 2pi, then
there two possible values for arctan(-infinity) because the period of
y=tan(theta) is one pi only.

And, we always look at the graphs so that we can see/visualize the
characteristics of the graphs. Like in this y=tan(theta) graph, if you
did not look (or you did not know by memory that the graph is like
that) at the graph, then you did not see that at pi/2, tan(theta) is
negative infinity also.

### David W. Cantrell

Oct 22, 2005, 9:24:02 PM10/22/05
to
"ticbol" <tic...@yahoo.com> wrote:
> David W. Cantrell :
> "No. If arctan() has a single value, then we are taking about
> the _principal-valued_ inverse tangent function. And that means that we
> do
> not look at the complete "graph of y = tan(theta)", as you suggested
> below.
> Rather, we look only at that part of the graph for which |theta| <
> pi/2. On
> that part of the graph there is no ambiguity at all: If tan(theta)
> approaches -infinity, theta must approach -pi/2.
> Thus, arctan(-infinity) = -pi/2.
>
> David "
>
> No, David. You know trig functions, like tan(theta) are periodic.

Of course, they're periodic. And that's why if we are to have
a (single-valued) inverse tangent function, we conventionally
invert tan(x) where x is _restricted_ to being bewteen -pi/2 and pi/2.
Doing so gives the principal-valued inverse tangent function, which is
what I was talking about above.

> So arctan(...) has many values.

True, there is also a multivalued inverse tangent relation. For it, we have

arctan(+/-infinity) = {(2n + 1)pi/2 | n integer}

Of course, both -pi/2 and pi/2, as well as infinitely many other values,
are in that set. As such, the OP's question wouldn't have made sense. He
was asking for _one_ specific value. Hence, he seemed to have the
single-valued inverse tangent function in mind, rather than the multivalued
relation.

David

### ticbol

Oct 22, 2005, 10:35:40 PM10/22/05
to
David W. Cantrell:

"---He was asking for _one_ specific value. Hence, he seemed to have

the
single-valued inverse tangent function in mind, rather than the
multivalued
relation. "

No, he was not. He was asking for the arctan of negative infinity.. Nof
for a one specific value.

And he was looking at the graph, at the pi/2 position of the angle. So
his question made sense, meaning, he wanted to know if the negative
infinity is also tan(pi/2). Because he's seeing positive infinity and
negative infinity at pi/2.

Most likely, the graph he was looking at was the graph of tan(theta)
for 0 <= theta <= 2pi only.
Most likely, he was not looking at a graph that included the curce at
theta = negative pi/2.
Most likely, he was concerened on the positive pi/2 angle only..

Although answers or angles are sometimes easier to find if they are
counted "negatively" or clockwisely (?), still, it is customary to
present or convert the negative angles answers into their
corrresponding positive-angle values or as if they were measured the
usual counterclockwise direction. Your negative pi/2 is positive 3pi/2
correspondingly, so negative pi/2 is also an answer, but the seeker was
arctan(-infinity)--- is it pi/2 also, when in the graph he was looking
at a positive infinity also for tan(pi/2).

----------------
" Chris Schumacher Oct 23, 8:11 am show options

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Date: Sat, 22 Oct 2005 22:11:54 GMT
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### Stan Brown

Oct 23, 2005, 1:06:10 AM10/23/05
to
Sat, 22 Oct 2005 22:11:54 GMT from Chris Schumacher
<ken...@hotmail.com>:

> What is the arctan of negative infinity? Is it pi/2, the same as infinity?
> I've been looking at the graph and it seems that it's both going to infinity
> and coming from negative infinity at the points where it's undefined.

Sounds like you're looking at a graph of the tangent function, not
the arctangent function. Remember that the arctan function is the
inversion of only one part of that graph, the part between -pi/2 and
+pi/2. So what happens to the left of that asymptote of tan(x) at x=-
pi/2 isn't relevant to the arctan function.

"Negative infinity" isn't a number, so it's not in the domain of
arctan. Strictly speaking, your answer is "does not exist in the real
number system."

Informally, draw the graph of the arctan function. It sure looks like
the left branch approaches -pi/2 asymptotically as you go further and
further left. While we can't say that arctan(-inf) "is" -pi/2, we can
say that as x gets more and more negative, arctan(x) gets closer and
closer to -pi/2, and you can get the arctan as close as you like to
that -pi/2 (without ever reaching it) just by going far enough left.
Informally, you might say arctan(x) = -pi/2, though if your teacher
is good he won't accept that. :-)

In calculus you'll find that this idea gets formalized with the
concept of a limit. You'll learn to say "the limit of arctan(x) as x
approaches negative infinity is -pi/2".

--
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
"I try not to fly in the face of public opinion."

### Guess who

Oct 23, 2005, 1:04:55 PM10/23/05
to
On Sun, 23 Oct 2005 01:06:10 -0400, Stan Brown
<the_sta...@fastmail.fm> wrote:

>you might say arctan(x) = -pi/2, though if your teacher
>is good he won't accept that. :-)

Well, not precisely as stated, but not so different from saying
loudly and clearly that sin(pi/2) = 1, which is a limiting case of an
arm of a right triangle approaching the same length as the
hypoteneuse.

It's like the old story of the difference between the engineer and the
mathematician. At a school dance, they each approached a girl on the
opposite side of the room. To do so they strolled half the distance
then half that ... the same old story, except that the mathematician
realised he would never get to the girl and gave up, but the engineer
figured that to all intents and purposes he would, and did. [Now,
let's keep it clean, boys and girls.]

### David W. Cantrell

Oct 23, 2005, 11:21:57 AM10/23/05
to
Stan Brown <the_sta...@fastmail.fm> wrote:
> Sat, 22 Oct 2005 22:11:54 GMT from Chris Schumacher
> <ken...@hotmail.com>:
> > What is the arctan of negative infinity? Is it pi/2, the same as
> > infinity? I've been looking at the graph and it seems that it's both
> > going to infinity and coming from negative infinity at the points where
> > it's undefined.
>
> Sounds like you're looking at a graph of the tangent function, not
> the arctangent function. Remember that the arctan function is the
> inversion of only one part of that graph, the part between -pi/2 and
> +pi/2.

Thanks, Stan, for again stating what is correct here. Whether "ticbol" is
well intentioned or not, I don't know; but his contribution did nothing
but "muddy the waters", as far as I can tell.

> So what happens to the left of that asymptote of tan(x) at x=-pi/2 isn't

> relevant to the arctan function.

And similarly, what happens to the right of the asymptote of tan(x) at
x = +pi/2 isn't relevant to the arctan function either.

> "Negative infinity" isn't a number,

Well, at least it's not a _real_ number, which I think is what you had in

> so it's not in the domain of
> arctan. Strictly speaking, your answer is "does not exist in the real
> number system."
>
> Informally, draw the graph of the arctan function. It sure looks like
> the left branch approaches -pi/2 asymptotically as you go further and
> further left. While we can't say that arctan(-inf) "is" -pi/2, we can
> say that as x gets more and more negative, arctan(x) gets closer and
> closer to -pi/2, and you can get the arctan as close as you like to
> that -pi/2 (without ever reaching it) just by going far enough left.
> Informally, you might say arctan(x) = -pi/2, though if your teacher
> is good he won't accept that. :-)

What you say in the above paragraph is perfect _if_ we are forced to
restrict ourselves to the real numbers. But of course, if Arctan(-oo) is,
literally, to make any sense, then we must deal with a system having -oo
as an element, and so we cannot be dealing with just the real number
system. Rather, we would need to deal with the two-point extension of the
reals, [-oo, +oo]. In that system, we do, rigorously, have

Arctan(-oo) = -pi/2.

This is perfectly standard. Look, for example, at equation (2) of
<http://mathworld.wolfram.com/InverseTangent.html>.

Regards,
David Cantrell

### Stan Brown

Oct 23, 2005, 7:57:05 PM10/23/05
to
23 Oct 2005 15:21:57 GMT from David W. Cantrell
<DWCan...@sigmaxi.org>:

> Thanks, Stan, for again stating what is correct here. Whether "ticbol" is
> well intentioned or not, I don't know; but his contribution did nothing
> but "muddy the waters", as far as I can tell.

Thank _you_. As for ticbol, I won't speculate on his intentions, but
I haven't seen his article because I killfiled him a few weeks ago,
when he boasted of deliberately being obnoxious and posting wrong
things. Seems he hasn't changed.

> Stan Brown <the_sta...@fastmail.fm> wrote:
> > "Negative infinity" isn't a number,
>
> Well, at least it's not a _real_ number, which I think is what you had in
> mind (considering your next sentence).

Yes, that's exactly what I was thinking. Glad it turned out clear.

> What you say in the above paragraph is perfect _if_ we are forced to
> restrict ourselves to the real numbers. But of course, if Arctan(-oo) is,
> literally, to make any sense, then we must deal with a system having -oo
> as an element, and so we cannot be dealing with just the real number
> system. Rather, we would need to deal with the two-point extension of the
> reals, [-oo, +oo]. In that system, we do, rigorously, have
> Arctan(-oo) = -pi/2.

Agreed, but there are two reasons why I didn't say so:

(1) As you know, when students write something non-standard, usually
they actually mean something more elementary but didn't (bother to)/
(know how to) write it correctly. Not too long ago we had someone
needing to solve y = 2e^x+1 for x, and nearly all the responders
interpreted it literally, only to find out later that the student
meant y = 2e^(x+1). That's par for the course, as you know.

(2) I don't know much about the two-point extension, only what I've

> This is perfectly standard. Look, for example, at equation (2) of
> <http://mathworld.wolfram.com/InverseTangent.html>.

I don't deny that it's a standard possibility, but AFAIK students are
first taught to deal only with real numbers.

Can other teachers weigh in? In pre-calculus and first-year calculus,
what's the answer to sec(pi/2)? "undefined" or "infinity"?

### Stan Brown

Oct 23, 2005, 8:01:55 PM10/23/05
to
Sun, 23 Oct 2005 10:04:55 -0700 from Guess who
<notreal...@here.com>:

> On Sun, 23 Oct 2005 01:06:10 -0400, Stan Brown
> <the_sta...@fastmail.fm> wrote:
>
> >you might say arctan(x) = -pi/2, though if your teacher
> >is good he won't accept that. :-)
>
> Well, not precisely as stated, but not so different from saying
> loudly and clearly that sin(pi/2) = 1, which is a limiting case of an
> arm of a right triangle approaching the same length as the
> hypoteneuse.

I don't think we need to invoke any limiting case for sin(pi/2). The
co-ordinates of a point on the unit circle pi/2 radians
counterclockwise from the x axis are (0,1); therefore sin(pi/2)=1.

It's true that you can't have a right triangle with a second right
angle, but (1) the law of sines works just exactly, not as a limiting
case, on a 90° angle as on any other >0 and <180°, and (2)
historically the sine function was defined by Ptolemy in terms of
chords on a circle. While we normally teach right-0angle trigonometry
first, that's not a logical necessity.

### David W. Cantrell

Oct 23, 2005, 8:49:41 PM10/23/05
to
Stan Brown <the_sta...@fastmail.fm> wrote:
> 23 Oct 2005 15:21:57 GMT from David W. Cantrell
[snip]
Referring to Arctan(-oo) = -pi/2:

> > This is perfectly standard. Look, for example, at equation (2) of
> > <http://mathworld.wolfram.com/InverseTangent.html>.
>
> I don't deny that it's a standard possibility, but AFAIK students are
> first taught to deal only with real numbers.

I believe you're right about how they're first taught. However, since the
OP asked about Arctan(-oo), it seems -- at least if we are to
take "Arctan(-oo)" literally -- that he was not dealing only with real
numbers.

> Can other teachers weigh in? In pre-calculus and first-year calculus,
> what's the answer to sec(pi/2)? "undefined" or "infinity"?

For better or worse, normally, those courses are restricted to the real
and complex number systems, and so the answer is "undefined".

I should also mention that the answer should not be either of the signed
infinities of the two-point extension of the reals. Rather, if one wishes
to use a system in which sec(pi/2) is defined, one should use the one-point
extension: R U {oo}, where oo is _unsigned_ infinity. Then

sec(pi/2) = oo.

(I suppose you're aware of this, Stan, but I'll mention it for others who

That's very nice in several ways. For example, if we restrict ourselves
to reals numbers, then the identity

cos(x) = 1/sec(x) is not valid for x = pi/2 + N pi.

But if we use the one-point extension of the reals for codomains of
functions such as secant, then we have

cos(x) = 1/sec(x) for _all_ real x.

David Cantrell

### Guess who

Oct 24, 2005, 12:16:35 AM10/24/05
to
On Sun, 23 Oct 2005 20:01:55 -0400, Stan Brown
<the_sta...@fastmail.fm> wrote:

>I don't think we need to invoke any limiting case for sin(pi/2). The
>co-ordinates of a point on the unit circle pi/2 radians
>counterclockwise from the x axis are (0,1); therefore sin(pi/2)=1.

How do you define the sine function in a coordinate system?
Certainly, the point on the circle is (0,1), but the function is not
defined by the coordinates of a point. It is defined by y/r, and *in
the limit* as x approaches zero, y approaches r, so defining the value
for the sine of pi/2. As it is then familiar to state that sin(pi/2)
= 1 [as a limiting value], then it is as feasible to talk similarly

### Guess who

Oct 24, 2005, 12:30:55 AM10/24/05
to
On Sun, 23 Oct 2005 19:57:05 -0400, Stan Brown
<the_sta...@fastmail.fm> wrote:

>Can other teachers weigh in? In pre-calculus and first-year calculus,
>what's the answer to sec(pi/2)? "undefined" or "infinity"?

In my book [an expression, not a text] "undefined" should be taken
literally. I do hope there is not an endless discussion as has
occurred in the distant past about 0/0, but there we have something
*really* indefinite [undefined.] It can be 0, 1, or oo depending upon
viewpoint. We all know, of course that it's "value" will depend upon
circumstances leading to consideration of 0/0 in limiting cases. In
any event, its value is not specific, or definite. again, in my book,
a definition [hence "definite"] has only one meaning. That is its
purpose. The definition might vary slightly in wording, but there is
still only one meaning, or significance.

"Infinity" is a definite statement, and meaning simply "not of this
number system." It can be said many ways including "WAY out there!".
Either way, it has only one significance. A chair or a seat, it's the
same single concept, and definitely something to sit on. So, to say
that something has an "infinite value" is simply yet another way of
saying large beyond bound, or any other word-combination with the same
meaning. However it's said, there is only one meaning, not three.
Pour moi, "Equals infinity" gives me the message needed that something
has increased without bound, and does so without quandry on my part.
It certainly simplifies notation, as the notation is used in limit
problems.

### Stan Brown

Oct 23, 2005, 9:42:17 PM10/23/05
to
Sun, 23 Oct 2005 21:16:35 -0700 from Guess who
<notreal...@here.com>:

> On Sun, 23 Oct 2005 20:01:55 -0400, Stan Brown
> <the_sta...@fastmail.fm> wrote:
>
> >I don't think we need to invoke any limiting case for sin(pi/2). The
> >co-ordinates of a point on the unit circle pi/2 radians
> >counterclockwise from the x axis are (0,1); therefore sin(pi/2)=1.
>
> How do you define the sine function in a coordinate system?
> Certainly, the point on the circle is (0,1), but the function is not
> defined by the coordinates of a point. It is defined by y/r,

It's always dangerous to claim a function "is defined" in a certain
way -- I've made that mistake myself. You definition is never the
only one.

One definition -- IMHO the easiest to work with though it's not the
only possible definition -- is to say that cos(theta) and sin(theta)
are the x and y co-ordinates of a point on the unit circle at an
angle "theta" counterclockwise from the x axis, and tan(theta) is
y/x. (If theta is negative then it's clockwise by minus theta.)

But even if you want to use y/r and not necessarily the unit circle,
r has to be positive (or it's not a circle). Thus at pi/2 you have
y=r=/=0 so that y/r = 1, exactly and not only as a limit.

> and *in
> the limit* as x approaches zero, y approaches r, so defining the value
> for the sine of pi/2. As it is then familiar to state that sin(pi/2)
> = 1 [as a limiting value], then it is as feasible to talk similarly

Sine sin is a continuous function on the reals, it is true that
LIM[sin @, @->pi/2] = 1
But it sounds like you think that is somehow true _only_ in the
limit. That's not accurate: the sine function is defined and
continuous at pi/2.

### Stan Brown

Oct 23, 2005, 9:46:58 PM10/23/05
to
24 Oct 2005 00:49:41 GMT from David W. Cantrell
<DWCan...@sigmaxi.org>:

> since the
> OP asked about Arctan(-oo), it seems -- at least if we are to
> take "Arctan(-oo)" literally -- that he was not dealing only with real
> numbers.

It's always hard to know when to answer what students actually ask
and when to answer what we think they probably meant to ask. With
algebraic equations I generally answer what the student said, even if
I'm fairly sure s/he meant something else. This time I made a value
judgment and guessed at what I thought the student probably meant.
(The original questioner has been strangely silent, unless I missed
his/her follow-up.)

Literally, as you say, arctan(-inf) presupposes that "infinity" is a
number. I guess I'm jaded, because 99 times out of a hundred when I
hear a student use "infinity" s/he should have said "undefined".

### Darrell

Oct 23, 2005, 11:49:06 PM10/23/05
to

"Guess who" <notreal...@here.com> wrote in message
news:njnol1d4sleon9hpf...@4ax.com...

> On Sun, 23 Oct 2005 20:01:55 -0400, Stan Brown
> <the_sta...@fastmail.fm> wrote:
>
>>I don't think we need to invoke any limiting case for sin(pi/2). The
>>co-ordinates of a point on the unit circle pi/2 radians
>>counterclockwise from the x axis are (0,1); therefore sin(pi/2)=1.
>
> How do you define the sine function in a coordinate system?
> Certainly, the point on the circle is (0,1), but the function is not
> defined by the coordinates of a point.<...>

As Stan stated it depends on what definitions we're using. The trig
functions are, most certainly, defined in terms of points in a coordinate
system.

> It is defined by y/r,

And is not "y" a member of the Cartesian coordinate system, as is not
r=sqrt(x^2+y^2) also a member? We're talking x's and y's here...

I don't know about everyone else but when I first learned sin(pi/2)=1 I
hadn't the foggiest idea what a "limit" was, since I had never encountered
the word in a mathematical context. Trig class came before calc class then,
and probably still does :-).

--
Darrell

### Darrell

Oct 24, 2005, 12:22:49 AM10/24/05
to
"Guess who" <notreal...@here.com> wrote in message
news:r1ool1dhbbb7b608u...@4ax.com...

> On Sun, 23 Oct 2005 19:57:05 -0400, Stan Brown
> <the_sta...@fastmail.fm> wrote:
>
>>Can other teachers weigh in? In pre-calculus and first-year calculus,
>>what's the answer to sec(pi/2)? "undefined" or "infinity"?
>
> In my book [an expression, not a text] "undefined" should be taken
> literally. I do hope there is not an endless discussion as has
> occurred in the distant past about 0/0, but there we have something
> *really* indefinite [undefined.]

Just by mentioning it you are inviting discussion ;-).

<...>

> again, in my book,
> a definition [hence "definite"] has only one meaning. That is its
> purpose. The definition might vary slightly in wording, but there is
> still only one meaning, or significance.
>
> "Infinity" is a definite statement, and meaning simply "not of this
> number system."

My cat is also "not of this number system" but her name is not Infinity.

> It can be said many ways including "WAY out there!".
> Either way, it has only one significance.

http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html

http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html

> A chair or a seat, it's the
> same single concept, and definitely something to sit on.

...but not all chairs are equal.

> So, to say
> that something has an "infinite value" is simply yet another way of
> saying large beyond bound,

Look at the 2nd paragraph from David's article at the 1st link above.

As an analogy, take the English language. There are many words with more
than one meaning. As long as the specific definition is clear in the
context in which we're communicating, there is no problem. My dog's papers
has a line where it specifies the name of her "bitch" but I ain't gonna
knock on the neighbor's door and ask (even politely) to "keep your bitch out
of my yard" lest the man of the house may assume I intend a differenet
definition.

...but that's not all it's used for.

--
Darrell

### Darrell

Oct 24, 2005, 12:26:17 AM10/24/05
to
"Darrell" <dr6...@comcastsnip.net> wrote in message
news:cNidnY4Siv8...@comcast.com...

> "Guess who" <notreal...@here.com> wrote in message

>

> http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html
>
> http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html
>
>
>> A chair or a seat, it's the
>> same single concept, and definitely something to sit on.
>
> ...but not all chairs are equal.
>
>> So, to say
>> that something has an "infinite value" is simply yet another way of
>> saying large beyond bound,
>
> Look at the 2nd paragraph from David's article at the 1st link above.

--
Darrell

### Guess who

Oct 24, 2005, 12:24:50 PM10/24/05
to
On Sun, 23 Oct 2005 22:49:06 -0500, "Darrell" <dr6...@comcastsnip.net>
wrote:

>I don't know about everyone else but when I first learned sin(pi/2)=1 I
>hadn't the foggiest idea what a "limit" was, since I had never encountered
>the word in a mathematical context. Trig class came before calc class then,
>and probably still does :-).

I hope you believe this ...It's good to see you back in the fray.

It [limits] still existed at the time. You [and I] were just not
ready for it then, but were ready for the consideration in order to
have continuity. There is a philosophy and pedagogy to teaching math,
[or any other subject], but when young we are generally not ready for
such in-depth study. Still, again, it is there.

Sorry about your cat. If I do ever get one, I'll name it accordingly.
It would make an interesting name and topic for discussion after
dinner. :-)

My point is not complicated and need not be made so. The definition
of trig functions in the coordinate system is still based upon the
definitions derived in the geometric right triangle from consideration
of similarity. As applied in physical problems, one might think of
the radius as a straight stiff rod rotating about the origin. It has
an 'elastic' line connected from its tip to the horizontal axis, which
changes length as the connection to the x-axis moves back and forth
[consider problems in simple harmonic motion.] At any point in time
the line, the said axis and the rod form a right triangle, thus
forming similar relations to those defined in ordinary geometry. I
readily accepted that at the time, since it seemed at the time to make
sense. A real study of limits came later, but a slight preview didn't
hurt a bit.

You had no idea of limit at the time, neither did I. But we could
readily accept the fact that the relationships we were learning at the
time depended heavily on what we had learned in geometry, and that
there was no apparent reason to suspect their failure in a limiting
situation where the reference triangle, with its trig and pythagorean
relationships reduced to the trivial, no longer existed.

Just incidentally, my visualisation of the motion I describe not only
made life in a world of trig and physics so much simpler, I found it
worked wonders with students struggling with these concepts. As they
say, "If it ain't broke, dont fix it."

### Stan Brown

Oct 24, 2005, 9:54:44 AM10/24/05
to
Mon, 24 Oct 2005 09:24:50 -0700 from Guess who
<notreal...@here.com>:

> My point is not complicated and need not be made so. The definition
> of trig functions in the coordinate system is still based upon the
> definitions derived in the geometric right triangle from consideration
> of similarity.

Your point is not complicated, just inaccurate. The problem, as I
said before, is that you simply cannot talk about "the" definition of
nearly anything in math. You can talk about "a" definition, and then
decide which definition is most useful in a given situation.

A moment's thought should make it obvious that your definition is
unusable even for triangles. By your definition, there's no such
thing as the sine or cosine of 120°, since that angle can't exist in
a "geometric right triangle". But the law of sines works quite well
for obtuse angles, so any definition that is limited to acute angles
is, er, sub-optimal.

### Darrell

Oct 25, 2005, 1:11:49 AM10/25/05
to

"Guess who" <notreal...@here.com> wrote in message
news:la1ql1tpa7v47eeqn...@4ax.com...

> My point is not complicated and need not be made so. The definition
> of trig functions in the coordinate system is still based upon the
> definitions derived in the geometric right triangle from consideration
> of similarity.

That's _one_ definition.

> As applied in physical problems, one might think of
> the radius as a straight stiff rod rotating about the origin. It has
> an 'elastic' line connected from its tip to the horizontal axis, which
> changes length as the connection to the x-axis moves back and forth
> [consider problems in simple harmonic motion.] At any point in time
> the line, the said axis and the rod form a right triangle, thus
> forming similar relations to those defined in ordinary geometry.

No, not at "any point in time," specifically not when you're rod coincides
with a coordinate axis. No triangle. Surely you can agree this definition,
while useful as a starting point in learning trig and useful for many
applications, is limited compared to other more general definitions.

--
Darrell

### Chris Schumacher

Nov 1, 2005, 7:44:45 PM11/1/05
to
Stan Brown <the_sta...@fastmail.fm> wrote in
news:MPG.1dc60720c...@news.individual.net:

> Literally, as you say, arctan(-inf) presupposes that "infinity" is a
> number. I guess I'm jaded, because 99 times out of a hundred when I
> hear a student use "infinity" s/he should have said "undefined".

I'm the original poster, so maybe I should step in and clear up some
things.
The problem that I was solving was an improper integral. The anti-
derivative was Arctan and the limits of integration were something like 0
to negative infinity.
So, you see, I was literally looking for the arctan of negative infinity.
I had encountered something similar a few problems before this, and was
rather shocked when I was asked for the arctan of infinity. I almost wrote
infinity or zero, but that I remembered that there actually was a place
where a tangent approached infinity.
When asked for negative infinity, I wasn't sure. I mean, after all, it
couldn't be both, could it? I had forgotten about one-sided limits, which
is what lead to my confusion.

Hope that clears up some things...

-==Kensu==-

### Darrell

Nov 2, 2005, 2:28:57 AM11/2/05
to
"Chris Schumacher" <ken...@hotmail.com> wrote in message
news:Xns9701BEB07AC2...@207.115.63.158...

> Stan Brown <the_sta...@fastmail.fm> wrote in
> news:MPG.1dc60720c...@news.individual.net:
>
>> Literally, as you say, arctan(-inf) presupposes that "infinity" is a
>> number. I guess I'm jaded, because 99 times out of a hundred when I
>> hear a student use "infinity" s/he should have said "undefined".
>
> I'm the original poster, so maybe I should step in and clear up some
> things.
> The problem that I was solving was an improper integral. The anti-
> derivative was Arctan and the limits of integration were something like 0
> to negative infinity.
> So, you see, I was literally looking for the arctan of negative infinity.

No, you were not. What you should have been looking for is a limit. (Yes,
a definite integral is itself a limit but what I'm talking about is a limit
of one of the _limits of integration_.) That's quite a few related--but
different--uses of the word "limit" that I would understand if that left you
confused. Here's what it means (the definition) of such an improper
integral:

int(-oo to b) f(x)dx := lim a-->-oo int(a to b) f(x)dx

IOW, in your case antidifferentiate arctan as you normally would but when
you go to "plug in" the limits of integration, for the infinite one don't
plug in "-oo" but plug in "a." This will leave you with an evaluated
definite integral (so to speak) in terms of a. Then take the limit as
a-->-oo.

In standard calculus, when we speak of infinity we may _conceptualize_ it as
real infinity but the meahanical _mathematics_ of it implies a limit. The
"oo" notation in standard calculus, if you refer to the definitions in your
text, will involve a limit of some type.

--
Darrell

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