Honors HS precalculus teaching idea

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Honors HS precalculus teaching idea

Dave L. Renfro
I have a few minutes free before I go home, so I
thought I'd toss out something that lurking high
school teachers might want to challenge their
better students with.

Recall the closed form sum of an arithmetic series
with n terms, a common difference of d, and a
first term of 'a':

a + (a + d) + (a + 2d) + ... + [a + (n-1)d]

= (a + a + ... + a) + d(1 + 2 + ... + n-1)

= na + d(n-1)n/2

= 2a(n/2) + d(n-1)(n/2)

= (n/2)[2a + (n-1)d].

Suppose we didn't know that n was supposed to
be a positive integer. For example, suppose we
look at this formula for n = 5/3.

(5/3)/2 * [2a + (5/3 - 1)d]

= 5/2 * [(1/3)(2a) + (1/3)(2/3)d]

= 5/2 * [2(a/3) + (3-1)(d/9)]

Note that this is the same as the sum of an
arithmetic series with first term a/3, n = 3,
and a common difference of d/9.

More generally, if n = p/q where p and q are
integers, then we get

(p/q)/2 * [2a + (p/q - 1)d]

= p/2 * [(1/q)(2a) + (1/q)((p-q)/q)d]

= p/2 * [2(a/q) + (p-q)(d/q^2)],

which is the same as the sum of an arithmetic
series with first term a/q, n-1 = p-q, and
common difference of d/q^2.

With this interpretation we can make sense of an
arithmetic series with a fractional number of terms.

Questions for the student: (1) Can you think of any
applications for this idea of having an arithmetic
series with a fractional number of terms? (2) Can the
same idea be used to define, in a formal way, the
idea of a geometric series with a fractional number
of terms?

Incidentally, I saw this idea (but not the questions)
in a short paper in The Mathematical Gazette, from
around 1942 or 1943. I don't have the exact reference
with me now, but I believe the short note didn't go
beyond the n = p/q computation I gave above.

Dave L. Renfro
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Re: Honors HS precalculus teaching idea

Dave L. Renfro
Dave L. Renfro wrote (in part):

http://mathforum.org/kb/thread.jspa?messageID=6123229

> More generally, if n = p/q where p and q are
> integers, then we get
>
> (p/q)/2 * [2a + (p/q - 1)d]
>
> = p/2 * [(1/q)(2a) + (1/q)((p-q)/q)d]
>
> = p/2 * [2(a/q) + (p-q)(d/q^2)],
>
> which is the same as the sum of an arithmetic
> series with first term a/q, n-1 = p-q, and
> common difference of d/q^2.
>
> With this interpretation we can make sense of an
> arithmetic series with a fractional number of terms.

I looked up the article and it wasn't from 1942 or 1943,
which I said I thought it was.

Here's the full citation:

G. Osborn, "Fractional and negative values of n in
arithmetical progressions", Mathematical Notes #411,
The Mathematical Gazette 7 #109 (January 1914), p. 248.

The article is about 2/3 of a page in length,
with the second half devoted to considerations
where n is a negative number. Note that I
overlooked in the above the possibility that
p < q. (It occurred to me when I was driving
home, a few minutes after I made the post.)

Osborn gives a slightly different interpretation
than I gave above, even for p > q:

first term = a/q - d/(2q) + d/(2q^2)

difference = d/q^2

number of terms = p

Osborn considers the case in which n is
a negative integer by putting q = -1, and
gives the following example:

How many terms of the (finite) arithmetic series
3 + 5 + 7 + ... are needed to obtain a sum of 8?

If we use the arithmetic series sum formula,
we get the equation

(n/2)*[2(3) + (n-1)(2)] = 8

n(3 + n - 1) = 8

n^2 + 2n - 8 = 0

(n + 4)(n - 2) = 0

n = -4, 2

Of course, n = 2 corresponds to 3 + 5.

What about n = -4?

Using Osborn's formula for n = -4 = p/q
(taking p = 4 and q = -1), we get

first term = a/q - d/(2q) + d/(2q^2) = -3 + 1 + 1 = -1

difference = d/q^2 = 2

number of terms = p = 4

Thus, n = -4 corresponds to -1 + 1 + 3 + 5.

**********************************************

Here's another "honors precalculus" item I'll toss
out for those who are interested.

Find a way to show the following is an even function,
that is f(x) = f(-x) for all real numbers x, without
tediously expanding everything out to see that f(x)
is a polynomial in x^2. The letters 'a' and 'b'
represent real number constants.

f(x) = x + { [(a+b+x)^2 + ab]*(x-a)^2*(x-b)^2 / 2ab(a+b)^3 }

Dave L. Renfro
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Re: Honors HS precalculus teaching idea

Dave L. Renfro
In reply to this post by Dave L. Renfro
Here's yet another "honors precalculus" item I'll toss
out for those who are interested.

Assume that a, b, c are in arithmetic progression
and x, y, z are in geometric progression. Show that

(x^b)(y^c)(z^a) = (x^c)(y^a)(z^b).

Dave L. Renfro
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Re: Honors HS precalculus teaching idea

Hypatia
In reply to this post by Dave L. Renfro
Thanks so much for these. Was the last one from the same source?

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Re: Honors HS precalculus teaching idea

Dave L. Renfro
In reply to this post by Dave L. Renfro
Dave L. Renfro wrote:

http://mathforum.org/kb/message.jspa?messageID=6124591

>> Assume that a, b, c are in arithmetic progression
>> and x, y, z are in geometric progression. Show that
>>
>> (x^b)(y^c)(z^a) = (x^c)(y^a)(z^b).

vlm217 wrote:

http://mathforum.org/kb/message.jspa?messageID=6124974

> Thanks so much for these. Was the last one from
> the same source?

If you mean the problem just above, it's from the
same journal, a few years earlier:

Problem 82, Mathematical Gazette 2 #28 (July 1901),
p. 77.

The page I have copied doesn't specify an author or
a source, but my guess is that it's from one of the
late 1890s England college's examinations (but clearly
not from the math tripos exam), since almost all of
the published problems in this journal's first few
years came from college examinations.

Here's another interesting example from a different
journal, which I posted about a month ago in sci.math:

- -------------------------------------------------

"Questions D'Examen" column, Mathesis Recueil
Mathematique (4) 3 (1913), p. 54.

Solve for x, if

(a^2)(x^4) + ax^3 + bx - b^2 = 0.

Here's how they indicate this can be solved:

Begin by solving for b, using the quadratic formula.

After cleaning things up a bit, you'll get

b = [x +/- x(2ax + 1)] / 2,

which leads to the following two equations:

ax^2 + x - b = 0

ax^2 + b = 0

Now use the quadratic formula to solve for x in
each of these equations.

- -------------------------------------------------

A poster replied:

> Hmmm ...
>
> But the original polynomial factors easily by grouping:
> (a^2)(x^4) + ax^3 + bx - b^2
>
>    = ( (a^2)(x^4) - b^2 ) + (ax^3 + bx)
>
>    = (a*x^2 + b)*(a*x^2 - b) + x*(ax^2 + b)
>
>    = (a*x^2 + b)*(a*x^2 - b + x)

I then replied:

> Darn! I had not tried to work it by another way.
> I just assumed the method they outlined was easier
> than any other obvious approach would be. Still,
> the idea they gave is interesting. When I have time
> (and the desire), I think I'll see if I can come up
> with an example for which their method works and
> which is resistent to other approaches. Or, if anyone
> in sci.math has the time and interest to take up this
> challenge now, feel free to post an example if you
> can come up with one.

I spent a little time trying to come up with an
example (but not very long) and didn't find one.

Dave L. Renfro
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Re: Honors HS precalculus teaching idea

Hypatia
In reply to this post by Dave L. Renfro
An interesting problem. When using the quadratic formula originally as they suggested, the discriminant is a perfect square, which would indicate that the original problem was factorable. This was the case, as factoring by the grouping method work - and much simpler than the formula. I haven't spent a lot of time working on this, but what if the polynomial ended in plus b^2 instead of minus b^2. Wouldn't you then be forced to use their suggested method?
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Re: Honors HS precalculus teaching idea

Dave L. Renfro
In reply to this post by Dave L. Renfro
Dave L. Renfro wrote (in part):

http://mathforum.org/kb/message.jspa?messageID=6124384

> G. Osborn, "Fractional and negative values of n in
> arithmetical progressions", Mathematical Notes #411,
> The Mathematical Gazette 7 #109 (January 1914), p. 248.

This morning, by accident, I came across another paper
on this topic. I suspect there are many similiar papers
on this topic, largely forgotten and unknown because
they wouldn't have been reviewed by the standard
mathematical reviewing journals of that time
("Jahrbuch ├╝ber die Fortschritte der Mathematik" and
"Revue Semestrielle Des Publications Mathematiques"
are the two best known from this time). Thus, I'll
post (in this thread) additional citations on this
topic if/when I happen to come across them, much like
I've done in several follow-up posts to my "Acoustic
circles" and "sound-ranging" references at

http://mathforum.org/kb/message.jspa?messageID=4685296

and

http://groups.google.com/group/sci.math/msg/71a7d52f12944a47

(Regarding this last URL, it should be fairly obvious
to anyone looking up things I've written that, given
my frequent references to sci.math (and other usenet)
posts, you'd want to search google-groups in addition
to Math Forum archived groups.)

F. C. Boon, "On the sum of an A.P.", Mathematical
Notes #436, The Mathematical Gazette 8 #116
(March 1915), 45-47.

Dave L. Renfro