# Honors HS precalculus teaching idea Classic List Threaded 7 messages Open this post in threaded view
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## Honors HS precalculus teaching idea

 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 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

 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

 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

 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