Wild About Math! Making Math fun and accessible

26Jan/082

Some amazing comments posted on this blog

I know that this blog is doing really well when I'm attracting some very high caliber comments, many of which deserve their own posts to address.

Here are some comments worthy of mention:

1. Efrique made the connection between nomograms and the odd way a parabola can be used to perform multiplication regarding the post A clever use of a parabola to perform multiplication. Efriuqe has, in fact, two helpful comments about the post. I know only a little about nomograms but will learn more and post about them. My sense is that nomograms were forgotten when the slide rule was replaced by the calculator. You can read about nomograms at Wikipedia.

2. Clueless posted a comment to Seven old Wild About Math! posts that I wish had been more popular noting that if you modify the graph of the parabola referenced in item #1 above to y=|x| then the same approach you used with the parabola yields a line that crosses the y axis at a value that is the harmonic mean of the y-values of the two points you drew the line through. Great observation and a great exploration for students. If the harmonic mean is new to you then you can get a background at this Wikpiedia article.

3. Eric has a really interesting way to square large numbers. It's not the approach I am going to document in my followup to How to square large numbers quickly (part 1) but Eric's approach is quite interesting. Can you figure out what is going on in the example Eric provided looking at just his example and not at his other comments? Can you prove why his approach works?

4. The Mad Hatter commented on A fun little counting puzzle to start 2008. He proposes a closed-form formula for the sum of the digits of consecutive integers starting from 1 and going up to (and including) abcd (a 4-digit number). I've not made the time to explore this proposed formula but I'm very curious as to how this was derived. Look at the formula. It's intense! The Mad Hatter is obviously mad about Math!

5. Mathmom took the time to document a solution to A fun little counting puzzle to start 2008 in a great amount of detail.

Comments (2) Trackbacks (1)
  1. Ooh! Ooh!

    “Clueless … noting that if you modify the graph of the parabola referenced in item #1 above to y=|x| then the same approach you used with the parabola yields a line that crosses the y axis at a value that is the harmonic mean of the y-values of the two points you drew the line through.”

    A much more well-known nomogram – this one is sometimes called the “resistance nomogram” or the “focal-length nomogram” (because of the fact that if you draw a scale on the y-axis that has values of y/2, but leave the other scales untouched, you get something that does resistance and focal length calculations).
    Take a look at http://en.wikipedia.org/wiki/Nomogram#Parallel-resistance.2Fthin-lens_nomogram

    See that image on the right? Mentally rotate it anti-clockwise 45 degrees. Now note the formula on the left. That’s just half the harmonic mean.

    Incidentally, the third of the posts by Ron Doerfler on Dead Reckonings (http://myreckonings.com/wordpress/2008/01/09/the-art-of-nomography-iii-transformations/)
    illustrates the projective transformations between conic sections I mentioned in the other comment (I hadn’t read the third one in detail when I posted before, so I didn’t realize it was all explained there).

    I recently located some (free) software for nomograms. I have yet to try it out, but it looks like it will be very good.

  2. The relationship between the points on y=x^2 and the relationship between the points on y=|x| seem to stem from the same general relationship… In general, I think, (hope) if you take the absolute value of any power function, y=|x^n| and pick points (-a, a^n) and (b, b^n) the line connecting them will cross the y-axis at the point (a^n b + b^n a)/ (a+b).

    For the absolute value case (n=1) we get ab+ba in the numerator and the harmonic mean pops out..

    for the square case, (a^2 b + b^2 a) factors to ab(a+b) and the a+b cancels out the denominator giving us the product ab…

    Of course all of these can be rewritten as
    (ab)[a^(n-1) + b^(n-1)] / (a+b) and we see in an instant that when n=2, the second term is just a+b and cancels out the denominator.

    So for instance if you did the same thing with y = x^10, the y-intercept would be (ab)(a^9 +b^9)/(a+b)

    Hope I didn’t mess that up too badly..
    Pat


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