We had three very nice solutions to this challenge from Josh, from CGSB in the UK, Chuyi Yang from KECHG in the UK and Cyrille from France. They were all nice, so we include all three in full . It is well worth looking at all three to see which you prefer mathematically.

Read Cyrille's writeup, which includes proof by induction Cyrille - particularly_general.pdf

Read Chuyi's writeup, which used a proof by generic example idea to justify the full cases Chuyi-Particularly general.pdf

Rather nicely, Josh included his thinking steps in the solution, paying attention to which aspects were 'informal' and which aspects were 'formal', which is precisely how such mathematics should be approached. Josh writes:

To prove identity number one, I started with the LHS of the equation and

attempted manipulations to make it appear identical to the RHS, as

follows:

$$\begin{eqnarray}\mbox{LHS} &=& (1 - x)(1 + x + x^2 + x^3)\\

&=& 1 - x + x - x^2 + x^2 - x^3 + x^3 - x^4\\

&=& 1 - x^4 = \mbox{RHS, as required}\end{eqnarray}$$

I could also see (informally) in my head why this would hold for the nth case. When we multiply by one in the first bracket, we keep all of the powers of $x$ from zero to $n$. When we multiply by $-x$, we add one to all of the powers of $x$ in the bracket and then subtract them. Where there is overlap (from $x$ and $-x$ to $x^n$ and $-x^n$) they cancel immediately, leaving only $+1$ and $- x^{n+1}$ therefore

$$(1 - x)(1 + x + x^2 ... + x^n) =1- x^{n+1}$$

I proved identity number two in a similar manner:

$$\begin{eqnarray}\mbox{LHS} &=& (1 - x)\Big((1 + x)(1 + x^2)(1 + x^4)\Big)\\

&=& (1 - x)\Big((1 + x + x^2 + x^3)(1 + x^4)\Big)\\

&=& (1 - x)(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)

&=& 1 - x^8, \mbox{ as required}\end{eqnarray}$$

Again I could see informally why this would hold for the nth case. It is easier to see when we consider line two. When we take $(1 + x + x^2... + x^{2^n - 1})$ and multiply by $(1 + x^{2^n})$, the $1$ preserves all the powers of $x$ from $1$ to $2^n - 1$ and the $x^{2^n}$ adds $2^n$ to the powers so we continue from where we left off at (1)$(x^{2^n}) = x^{2^n}$and continue in increments of $1$ all the way up to $x^{2^n - 1 + 2^n}$, but $2^n + 2^n = 2^{n+1}$ so we finish up going from $x^0 = 1$ up to $x^{2^{n+1} - 1}$, and this is the same form as what we started with and so will continue for all values of $n$. Therefore, we are left with: $$(1 - x)(1 + x + x^2 ... + x^{2^{n+1} - 1})$$ which by the first identity simplifies into $$1 - x^{2^{n+1} - 1 + 1} = 1 - x^{2^{n+1}}$$

So $$(1 - x)(1 + x)(1 + x^2)...(1 + x^{2^n}) = 1 - x^{2^{n+1}}$$

Identity 3 I proved using a reverse method. Here I began with the RHS and

manipulated it repeatedly using the sin double angle formula. Here is what

I did:

Consider sin2A = 2sinAcosA

Therefore sinA = 2sin(A/2)cos(A/2) [1]

Let the RHS denominator = y

Then y = 16sin(x/16)

ycos(x/16) = 8{2sin(x/16)cos(x/16)}

By [1], 2sin(x/16)cos(x/16) = sin(x/8)

So ycos(x/16) = 8sin(x/8)

ycos(x/16)cos(x/8) = 8sin(x/8)cos(x/8) which similarly simplifies:

ycos(x/16)cos(x/8) = 4sin(x/4)

Similarly:

ycos(x/16)cos(x/8)cos(x/4) = 2sin(x/2)

ycos(x/16)cos(x/8)cos(x/4)cos(x/2) = sin(x)

But sin(x) = RHS numerator

Therefore the RHS can be rewritten as:

RHS = {ycos(x/16)cos(x/8)cos(x/4)cos(x/2)} / y

The y's cancel leaving:

RHS = cos(x/16)cos(x/8)cos(x/4)cos(x/2) = LHS as required

Here I reasoned that if we start with any RHS with any binary power outside

and inside the sin function on the denominator we could just repeat the

steps above.

Since we would always end on a multiplication of cos(x/2) in order to

obtain 2cos(x/2)sin(x/2) = sin(x), and begin on a multiplication of

cos(x/[2^n]) where 2^n was the binary power we used in the denominator, we

would have a multiplication in cosines ranging from cos(x/2) down to

cos(x/[2^n]).

Therefore:

cos(x/2)cos(x/4)...cos(x/[2^n]) = sin(x) / [2^n]sin(x/[2^n])

The first general identity holds true for x E R

The second holds true for x E R

The third holds true for x E R, x =/= [2^n]pi (else we would have

sin(n*pi) on the denominator which is always zero so we would have an

undefined fraction).