Geometric Proof

[modest]

Find a geometric proof that,

 

[math](1+2+3...+n)^2 = 1^3 + 2^3 + 3^3... +n^3[/math]

 

Proving it with induction or looking it up are relatively easy, but figuring out a geometric proof I found very satisfying.

 

~modest

[Qfwfq]

Interesting question. I came across that result algebraically, a few years ago when I worked out a complicated recipe for a polinomial giving the sum of the first n number raised to a chosen power. I noticed that case and I remember trying to find a geometric way of seeing it but I can't remember succeeding. I did read there is a geometric argument of that specific case and it's even called Somebody's Theorem. It's the umbralitic version of:

 

[math]\left(\int_0^y dx\,x\right)^2=\left(\frac12 x^2\right)^2=\int_0^y dx\,x^3=\frac14 x^4[/math]

[modest]

Interesting question. I came across that result algebraically, a few years ago when I worked out a complicated recipe for a polinomial giving the sum of the first n number raised to a chosen power. I noticed that case and I remember trying to find a geometric way of seeing it but I can't remember succeeding. I did read there is a geometric argument of that specific case and it's even called Somebody's Theorem.

 

Right There is an illustration—a visual proof—by someone in the 1500's. I'll leave the name out in case anyone is working on the riddle and not wanting to look it up. The theorem also, you are correct, it is named after a Greek mathematician.

 

It's the umbralitic version of:

 

[math]\left(\int_0^y dx\,x\right)^2=\left(\frac12 x^2\right)^2=\int_0^y dx\,x^3=\frac14 x^4[/math]

 

Interesting. I've never heard of umbralititc before. Looking it up, wiki has:

In mathematics, Faulhaber's formula, named after Johann Faulhaber, expresses the sum

 

[math]\sum_{k=1}^n k^p = 1^p + 2^p + 3^p + \cdots + n^p[/math]

 

<...>

 

In the classic umbral calculus one formally treats the indices ''j'' in a sequence ''B''

_''j''

as if they were exponents, so that, in this case we can apply the binomial theorem and say,

 

[math]\sum_{k=1}^n k^p = {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} B_j n^{p+1-j}

= {1 \over p+1} \sum_{j=0}^p {p+1 \choose j} B^j n^{p+1-j} [/math]

Faulhaber's formula

 

I've never used, nor even heard, of umbralititc calculus before, so I'm not sure about the umbral form. But, I believe you are correct that my example is a specific case of a Faulhaber polynomial.

 

It could be expressed,

 

[math]1^3 + 2^3 + 3^3 + \cdots + n^3 = \left({n^2 + n \over 2}\right)^2[/math]

 

~modest

[CraigD]

Find a geometric proof that,

 

[math](1+2+3...+n)^2 = 1^3 + 2^3 + 3^3... +n^3[/math]

Visualizing the number [imath]n^3[/imath] as a [imath]n \times n^2[/imath] rectangle, I can see that each such rectangle can be divided into a pair of [imath]n \times \left( \sum_{x=1}^{n-1} x \right)[/imath] rectangles and a single [imath]n \times n[/imath] rectangle, which can be arranged into an L-shape (chevron) that can be adjoined to a [imath]\sum_{x=1}^{n-1} x[/imath] wide square to make a [imath]\sum_{x=1}^{n} x[/imath] wide square.

 

It sketches prettily in ASCII text, for example (spoiler - click/drag/select to see the sketch):

1 2 2 3 3 3 4 4 4 4 5 5 5 5 5

2 2 2 3 3 3 4 4 4 4 5 5 5 5 5

2 2 2 3 3 3 4 4 4 4 5 5 5 5 5

3 3 3 3 3 3 4 4 4 4 5 5 5 5 5

3 3 3 3 3 3 4 4 4 4 5 5 5 5 5

3 3 3 3 3 3 4 4 4 4 5 5 5 5 5

4 4 4 4 4 4 4 4 4 4 5 5 5 5 5

4 4 4 4 4 4 4 4 4 4 5 5 5 5 5

4 4 4 4 4 4 4 4 4 4 5 5 5 5 5

4 4 4 4 4 4 4 4 4 4 5 5 5 5 5

5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

5 5 5 5 5 5 5 5 5 5 5 5 5 5 5

 

I’m so spoiled by algebra and unpracticed in construction I can scarcely recall how to describe a geometric proof, though, so can’t quickly do better than this vague description. :(

[modest]

You've got it precisely.

 

When I was presented with the problem (this problem was presented verbally by a player on Poker After Dark) I thought that the proof must be 3D. The trick, I believe, is that it is 2D.

 

The solution is here: http://users.tru.eastlink.ca/~brsears/math/oldprob.htm#s32 (the last solution on that page)

 

There is an image much like your ASCII text.

 

Very well done.

 

~modest

[Turtle]

Find a geometric proof that,

 

[math](1+2+3...+n)^2 = 1^3 + 2^3 + 3^3... +n^3[/math]

 

Proving it with induction or looking it up are relatively easy, but figuring out a geometric proof I found very satisfying.

 

~modest

 

i found no solution but wanted to point out that the first part of the equality, [math](1+2+3...+n)^2 ...[/math] is the square of the n^th triangular number. after seeing that, from then on i got stuck on stupid gnomons. not sure if anything different geometrically can be made of it from that angle or not. :shrug: (:doh:) :hyper:

[modest]

i found no solution but wanted to point out that the first part of the equality, [math](1+2+3...+n)^2 ...[/math] is the square of the n^th triangular number.

 

If anyone knows their gnomons... :)

 

In fact, the wikipedia article is entitled "Squared triangular number". The equality at the top of that page,

[math]\sum_{i=1}^{n} i^3 = \Bigl(\sum_{i=1}^{n} i\Bigr)^2.[/math]

is the equality in the OP, and to affirm what Q was saying,

This identity is sometimes called Nicomachus's theorem.

Squared triangular number

 

 

~modes

[Qfwfq]

Gosh I managed to figure it out! Without even having pencil and paper on hand. :beer:

 

Last night I had the stationery but I tackled the square of the triangle differently, translating it from 4D into 2D but I only ended up with an algebric demonstration that's essentially inductive too, thus failing the requisites.

 

Just now I instead worked the 4D figure into a sequence of prisms and figured how it equates to the sequence of cubes. Each prism shorter than the one, with two square sides, contains a smaller one with square sides and all these need to be completed to a cube having sides of that number. The largest prism can be completed with one slice from each of the shorter ones, each slice has one side of n and a shorter side, down to one. This leavs the next biggest prism with two square sides of (n - 1) and the corresponding triangular section; the others become shorter prisms with the same triangular section. Each prism contains one with square sides, which remains when extra slices are removed and is then completed with slices removed from the shorter ones. This goes on, down till the flat prism of thickness 1 is left with a single block, when the 2 by 2 by 2 cube is completed.

 

Now that I went to look at Craig's answer and selected the ASCII text:

I’m so spoiled by algebra and unpracticed in construction I can scarcely recall how to describe a geometric proof, though, so can’t quickly do better than this vague description.

Yeah, that was my problem last night; your diagram is what I had tried but I couldn't see how to geometrically equate each L shape with the right cube :doh: but skimming through your lines it hit on me that it can be done quite simply.