tag:blogger.com,1999:blog-8231784566931768362.post29210936021108230..comments2021-06-16T05:56:38.112-04:00Comments on MathNotations: A Different 'View' of Sums of Cubes? An Algebraic "Proof Without Words!"Dave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-8231784566931768362.post-67035359779141011892010-07-02T10:32:40.081-04:002010-07-02T10:32:40.081-04:00Thaumkid -- thanks!!
ZeroSum--
One method of proo...Thaumkid -- thanks!!<br /><br />ZeroSum--<br />One method of proof uses math induction but that may be above your student's level. There is a general strategy for deriving the sum of powers of integers. However, I left it in the margins of my notebook and I misplaced it!!<br /><br />Seriously, when I have time I will do a video of this cool method which is related to the Taylor expansion of polynomials.Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-59132619299541983292010-06-27T22:10:24.618-04:002010-06-27T22:10:24.618-04:00If I posted a comment before, forget it. I made a...If I posted a comment before, forget it. I made a major mistake. I took this proof to mean that the sum of ANY two consecutive integers is equal to the square of the sum of the integers. I failed to see that starting from 1 is necessary.<br /><br />Can anyone show a way to prove this algebraically? I'd love to use it in class, but I'm not the best at proofs, especially ones where there is a set number (here, 1).ScaffoldedMathhttps://www.blogger.com/profile/12991099683629425350noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-87117125428323660812009-05-13T21:56:00.000-04:002009-05-13T21:56:00.000-04:00Just an aside, this is called Nicomachus's theorem...Just an aside, this is called Nicomachus's theorem. I found that at http://mathworld.wolfram.com/PowerSum.html<br /><br />I thought that site had some pretty and "visual" demonstrations... =) Happy May!thaumkidhttps://www.blogger.com/profile/11319338489107643359noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-61238664940948827882008-12-30T13:06:00.000-05:002008-12-30T13:06:00.000-05:00You may be right Jonathan but I wasn't really cons...You may be right Jonathan but I wasn't really considering the 'cube of a sum', rather I was constructing a 'sum of cubes.' I was actually trying to set up a visualization of the inductive step. Thus (1+2+3+4)^2 = ((1+2+3) + 4)^2 = <BR/>(1+2+3)^2 +(2)(4)(1+2+3) + 4^2. Assuming we have already represented (1+2+3)^2 as a sum of cubes, we need to show that the remaining terms can be re-formed to represent 4^3. This can be done numerically but the fun is to do this visually by actually assembling the pieces to form another cube. I have no idea what value this has. I was just playing around with it and thought it would be "fun" but then again my idea of fun is...Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-56358382221626229352008-12-30T08:53:00.000-05:002008-12-30T08:53:00.000-05:00Dave,I think (but do not know!) that the diagram w...Dave,<BR/><BR/>I think (but do not know!) that the diagram would be more effective if you were trying to show that:<BR/>(n+1)^3 = n^3 + 3n^2 + 3n + 1,<BR/>in other words, if n was relatively large compared to 1, the eye would focus on the "frosted parts" and the single vertex.<BR/><BR/>I like your use of color.<BR/><BR/>JonathanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-17797556375281092672008-12-30T06:41:00.000-05:002008-12-30T06:41:00.000-05:00That's beautiful, Eric. There's something elegant ...That's beautiful, Eric. There's something elegant about representing n^3 as n(n^2) visually! This reminds me of the elegant but simpler 'proof' of the formula for the nth triangular number which I should also demonstrate visually. As you know, this also uses the idea of stacking up a triangular number (L-shaped) and interlocking it with a congruent shape but upside down. The two pieces form a rectangle leading to the famous formula. There must be many websites that show this (I'm sure it's in MathWorld and Cut-the-Knot). It would be great if there were a compendium of such Proofs Without Words so that we could look these up just as they have an encyclopedia of number sequences.<BR/><BR/>The method I used is probably out there already although I haven't yet found it. I was just playing around with it and it seemed to lend itself to an inductive argument both arithmetically and geometrically.<BR/><BR/>Thank you for this idea and Happy New Year!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-63380447817763687812008-12-30T00:27:00.000-05:002008-12-30T00:27:00.000-05:00I remember a proof without words of this that is p...I remember a proof without words of this that is purely planar, though I don't remember where I saw it. The case n = 2 goes like this:<BR/><BR/>First, draw a square with sides (1+2) and (1+2), marking the various subrectangles.<BR/><BR/>Second, draw a 1 by 1 square in the lower-left of the figure. Then, draw a 2 by 2 square immediately to its right, and another immediately above it. In chess terms, first you have a1, and then b1, c1, b2, c2, and then a2, b2, a3, b3. These three squares almost fill the 3 by 3 square but leave a 1 by 1 hole in the upper right (c3), and then a 1 by 1 square (b2) is the overlap in the middle. Those two regions have the same size, and they cancel, leading to (1+2)^2 = 1 1^2 + 2 2^2,<BR/><BR/>For the n = 3 case, place 3 3 by 3 square around the figure in an L shape. (Well, an upside-down L.)<BR/><BR/>For the n = 4 case, place a vertical stack of n/2 (= 2) 4 by 4 squares on the right of the n = 3 figure. Place another horizontal stack of (n/2) 4 by 4 squares above the n = 3 figure. The two stacks overlap in a 2 by 2 square, but they don't quite fill the 10 by 10 square, leaving a 2 by 2 hole.Anonymousnoreply@blogger.com