Particularly general

7962 /www/nrich/html/content/id/7962/ 1 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Consider these three algebraic identities $$(1-x)(1+x+x^2+x^3)\equiv (1-x^4)$$ and $$(1-x)\big((1+x)(1+x^2)(1+x^4) \big)\equiv (1-x^8)$$ and $$\cos\left(\frac{x}{2}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{8}\right)\cos\left(\frac{x}{16}\right)\equiv\frac{\sin(x)}{16\sin\left(\frac{x}{16}\right)}$$ Prove that they are true for any real number $x$ in the first two cases and any real numbers except multiples of $16\pi$ in the third case. Use the ideas in your proofs to write down general forms $$(1-x)(1+x+x^2+x^3+\dots+x^n)\equiv \quad ???$$ and $$(1-x)\big((1+x)(1+x^2)(1+x^4)\dots(1+x^{2^n}) \big)\equiv \quad ???$$ and $$\cos\left(\frac{x}{2}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{8}\right)\cos\left(\frac{x}{16}\right)\dots \cos\left(\frac{x}{2^n}\right)=\quad ???$$ In each case it is possible to write down 'infinite $n$' limiting identities. What are these, and for which values of $x$ are they valid?</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">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 <a href="/content/id/7962/Cyrille%20-%20particularly_general.pdf">Cyrille - particularly_general.pdf</a> Read Chuyi's writeup, which used a proof by generic example idea to justify the full cases <a href="/content/id/7962/Chuyi-Particularly%20general.pdf">Chuyi-Particularly general.pdf</a> 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). </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Why do this problem?</h3> This problem gives practice and insight into manipulating and constructing algebraic identities. It is based upon the idea of 'proof by generic example': the proof method of a particular example encapsulates in a clear way all of the key points of a proof of a more general case. Manipulation of algebraic identities is always good fun and will give very solid general preparation for the more challenging end of school mathematics examinations, such as Further Mathematics and STEP. It can also be used to help students to understand the differences in the uses of direct proof and proof by induction. <h3>Possible approach</h3> <div>This problem is well suited to individual solving and will require a reasonable fluency in algebraic manipulation. It would most naturally accompany work on proof by induction.</div> <div> </div> <div>A brief discussion on proof could follow attempts at this question if you feel comfortable with the topic. Some discussion points of interest are:</div> <div>o Proof by induction can often be used to prove identities involving a variable $n$, but frequently provide no insight as to why the identities make sense, or where they came from in the first place.</div> <div>o These algebraic proofs by generic example indicate clearly WHY the results are true, but for large $n$ the actual algebraic proofs would be very long sequences of symbolic manipulations. Do we need to write these down?</div> <div>o In some sense, the generalisations are justified by understanding that the manipulations made will, in some way, continue. How do students feel about this lack of formality?</div> <div>o Remind students that working through the algebra in specific instances still constitutes a 'direct' proof!</div> <div> </div> <h3>Key questions</h3> What is the difference between an identity and an equation? What method of proof can you use for the specific examples given? How can you expand the brackets or reduce the terms in a step by step way? Can you see how this procedure might generalise? <h3>Possible extension</h3> Using the ideas gained from solving this problem, can students construct some new identities of their own? <h3>Possible support</h3> Suggest the use of difference of two squares and a trigonometric double angle formula. </mdoxml> <mdoxml version="1.0">Expand out the brackets in a systematic, organised manner. You will need to use a double angle trig formula and difference of two squares. Considering $n=1$ might be useful. There is perhaps no need to provide a proof for your identities if you are very clear in your mind how such a proof would be constructed, although you might wish to produce a proof if interested.</mdoxml> 2 4 0 0 0 0 1 Particularly general By proving these particular identities, prove the existence of general cases. Algebra Manipulating algebraic expressions/formulae Algebra Difference of two squares Algebra Identities Using, Applying and Reasoning about Mathematics Mathematical induction Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof Admin FP mapping