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Title: Geometric Sequence Post by Sir Col on Jul 15th, 2003, 2:16pm I stumbled on an interesting problem the other day and I'd like to know what you make of it... Four real numbers are in a geometric progression. Their sum is 13 and the sum of their squares is 1261. Find the four numbers. What if the sum is a and the sum of the squares is b? |
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Title: Re: Geometric Sequence Post by towr on Jul 16th, 2003, 1:29am [hide] we have x+rx+r^2x+r^3x = 13 and x^2+r^2x^2+r^4x^2+r^6x^2 = 1261 so x = 13/(1+r+r^2+r^3) fill this into second equation solve for r the only real solutions are -3/2 and -2/3 filling this into equation 1 we get r=-2/3,x=27 or r=-3/2,x=-8 in both cases we get [-8, 12, -18, 27] [/hide] |
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Title: Re: Geometric Sequence Post by Sir Col on Jul 16th, 2003, 12:10pm I'm intrigued as to know how you got to the solutions, Towr? I got as far as the quartic: [hide]42r4+97r3+97r2+97r+42=0[/hide], but factoring it into [hide]+-(2r+3)(3r+2)(7r2+r+7)=0[/hide], is hardly straight forward; I can't see how a high school student, aged 16-18 (the target group), is supposed to make that leap. Have I missed some major simplification? This was why I said it was an 'interesting problem' and asked about the generalisation for a (the sum of the terms) and b (the sum of the squares of the terms). This problem is supposed to be solved without the aid of any electronic devices. For reference, this is how I arrived at the quartic: [hide]From Towr's substitution into equation 2, we get (13/(1+r+r2+r3))2(1+r2+r4+r6)=1261. Then writing 1+r2+r4+r6=(r8–1)/(r2–1)=( (r4–1) (r4+1))/((r–1)(r+1)) and (1+r+r2+r3)2=((r4–1)/(r–1))2=((r4–1)(r–1)(1+r+r2+r3))/(r–1)(r–1))=((r4–1)(1+r+r2+r3))/(r–1). After tidying up we get the quartic given above. [/hide] |
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Title: Re: Geometric Sequence Post by towr on Jul 16th, 2003, 2:36pm I did use some computational aid, but I have an idea to solve it without one. That will however have to wait till after I've finished watching TV :P [e]I think I'm a bit closer, but I need sleep now..[/e] |
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Title: Re: Geometric Sequence Post by Icarus on Jul 16th, 2003, 3:53pm Since the solutions are rational, the rational root theorem would have provided them. |
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Title: Re: Geometric Sequence Post by Sir Col on Jul 16th, 2003, 4:52pm It would still require a lot of work, as there are a considerable number of rational values obtained from the factors of 42: +-42/1, 21/1, 21/2, 14/1, 14/3, 7/1, 7/3, 7/6, 7/12, 6/1, 6/7, 3/1, 3/2, 3/7, 3/14, 2/1, 2/3, 2/7, 2/21, 1/2, 1/3, 1/6, 1/7, 1/14, 1/21, 1/42. If we were allowed to use a calculator I'd probably use the Newton-Raphson method anyway, as a couple of iterations would locate the rational roots. Of course, an even easier approach would be to use a graphical calculator; substitution into the function would confirm the two rational roots. It may be the case that students were expected to use electronic devices, but I suspect there is an elegant solution at hand. Could anyone find an easier way to get to the quartic, or is it possible to derive a quadratic (with the two rational roots) by eliminating the imaginary solutions in some way? [e] Spaces added! Sorry about that; that's the problem with having a display 1600 pixels wide. [/e] |
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Title: Re: Geometric Sequence Post by towr on Jul 17th, 2003, 1:51am starting from a2·(r4 + 1) - b·((r + 1)2·(r2 + 1)) = 0 == r4·(a2 - b) - 2·b·r3 - 2·b·r2 - 2·b·r + (a2 - b) = 0 The first thing to notice from the original problem is that the reciprocal of a solution is also a solution. so we want to reform the equation to p(r-z)(r-1/z)(r-f)(r-1/f) = 0 obviously p = a2-b, so we don't have to find that variable any more, and there are now only 2 others left. so we're left to matching r4 + 2·b·r3/(b - a2) + 2·b·r2/(b - a2) + 2·b·r/(b - a2) + 1 = 0 to (r-z)(r-1/z)(r-f)(r-1/f) = 0, setting (z2+1)/z = s and (f2+1)/f = t this is equal to (r2 - s r + 1)(r2 - t r + 1) = 0 == r4 - r3·(s + t) + r2·(s·t + 2) - r·(s + t) + 1 so we get - (s+t) = 2b/(b - a2) (s·t+2) = 2b/(b - a2), let's say (b - a2) = u and solve for s and t you get two solution for each, but since you can swap t and s there are really just two in total, just pick one for t and one for s, it won't make a difference for the solution. s = - (sqrt(b2 - 2·b·u + 2·u2) + b)/u and t = (sqrt(b2 - 2·b·u + 2·u2) - b)/u now solve (r2 - s r + 1) = 0 and (r2 - t r + 1) = 0 again you only need one solution from each let's pick r = (sqrt(s2 - 4) + s)/2 or r = (sqrt(t2 - 4) + t)/2 Take whichever of these is the real solution. now fill that into the original equation 1, x (1+r+r2+r3) = a and solve for x. No let's see if it works on our problem.. a = 13 b = 1261 u = 1261 - 13*13 = 1092 s = -13/6 t = -1/7 now we get (from s) r = -2/3, which is what we were looking for. as they say, w00t! |
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Title: Re: Geometric Sequence Post by towr on Jul 17th, 2003, 2:04am btw Sir Col, could you put some spaces in that line with fractions in your last post? it kinda messes up the layout on my computer (the line is too long to not have breaks) |
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Title: Re: Geometric Sequence Post by Sir Col on Jul 17th, 2003, 11:16am on 07/17/03 at 01:51:31, towr wrote:
I should say, too; a truly impressive solution, spotting the reciprocal symmetry and the ingenius substitution. Bravo, Towr! Okay, what about, "Five real numbers are in a geometric progression..."? :P |
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Title: Re: Geometric Sequence Post by towr on Jul 17th, 2003, 11:45am I found that for any geometric sequence of n numbers there should be n div 2 solutions plus their reciprocal. So for n=5 case first the same trick can be used again, but then I think you have to get a cube-root, which I can't generally do. I have thought about it a little, and the general case this afternoon, but I'm not sure if there is a general solution (or how hard the n=5 case is). [e](just forget there was anything here at all ;))[/e] I'll work on it a little more. And see if I can find a way to easily crack n=5. |
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Title: Re: Geometric Sequence Post by Sir Col on Jul 17th, 2003, 11:58am Actually, I was just kidding about n=5; I suspect, like you, that it cannot be solved in the general case. Although the polynomial stucture is not random and we are dealing with a special relationship between coefficients, I suspect that, ultimately, we are asking the question of solving a general quintic. Before tackling n=5, it may be worth considering n=3 and comparing it with the solution of a general cubic; I think it is equivalent. In which case, the most ABEL mathemtician has already denied general solutions for n>4. I am fascinated by your suggestion that, Quote:
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Title: Re: Geometric Sequence Post by towr on Jul 17th, 2003, 12:09pm [e]hmm.. I thought I made a mistake but I was mistaken it seems.. must be getting late..[/e] if you have a geometric sequence you have some x and x*r and x*r^2 etc take y = x*r^n, then the same sequence is begotten in reverse order by y, y/r, y/r^2 etc so both r and 1/r must be solutions. The only exceptions is when y = x, or for real r's r=1 I don't think it generally matters much if the solutions are complex or only real. |
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Title: Re: Geometric Sequence Post by towr on Jul 17th, 2003, 1:50pm If I'm done screwing up, I think I have the solution to the n=5 case starting with r^4 + r^3·(a^2 + b)/(b - a^2) + r^2 + r·(a^2 + b)/(b - a^2) + 1 = 0 u := (a^2 + b)/(b - a^2) -(s+t) = u and (st+2) = 1 s := (sqrt(u^2 + 4) - u)/2 t := - (sqrt(u^2 + 4) + u)/2 r = (sqrt(s^2 - 4) + s)/2 or r = (sqrt(t^2 - 4) + t)/2 x = a/(1+r+r^2+r^3+r^4) I've checked with a=211 b=11605 (which should, and do yield r=3/2 and x=16) I think it should also be solvable for n=6,7 |
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Title: Re: Geometric Sequence Post by towr on Jul 17th, 2003, 2:17pm I've just solved an n=7 problem as well :P with o = (a^2 + b)/(b - a^2) and finding a cuberoot for - (s + t + u) = o AND s·(t + u) + t·u + 3 = 1 AND - (s·(t·u + 2) + 2·(t + u)) = o (taking just any set s,t,u that solves it, the other sets are again just permutations) You really only need the one with the largest magnitude, since that one leads to the real solution. Again using the same solution+reciprocal trick. |
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Title: Re: Geometric Sequence Post by towr on Jul 17th, 2003, 3:05pm from the general formula a^2*(r - 1)*(r^n + 1)/((r + 1)·(r^n - 1)) = b you can derive r^n - 2b/(a^2 - b) * sum(r^i, i, 1, n - 1) + 1 = 0 which works find for all even numbers, and gives n/2 pairs of roots but for odd n, somehow you can get another equation sum(r^(2i), i, 0, (n - 1)/2) - (a^2 + b)/(a^2 - b) * sum(r^(2*i+1), i, 0, (n - 3)/2) =0 which gives (n-1)/2 pairs of roots, which is one root less than you might expect (but as I said, I think there should generally only be pairs, so it makes sense there's either one more or one less). I'm not yet sure what the 'somehow' is though. Getting the equation by following the pattern is easy enough, but I haven't derived it from the original equation yet. And it will probably have to wait toll after the weekend unless someone else does it.. |
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Title: Re: Geometric Sequence Post by Sir Col on Jul 17th, 2003, 4:48pm That's some impressive work, Towr. I had a play with the general case and derived the following result. Given n terms in a geometric sequence, the sum of the terms is a and the sum of the squares is b, then r, the common ratio, is the root of the equation: (b–a2)rn+1 + (b+a2)rn – (b+a2)r – (b–a2) = 0 In the original problem, for which n=4, a=13 and b=1261, we get the equation: 42r5 + 55r4 – 55r – 42 = 0, which, upon checking, satisfies the expected roots. However, I've not been able to do much else with this yet. Do you think there is any mileage in it? |
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Title: Re: Geometric Sequence Post by towr on Jul 18th, 2003, 12:43am It works for n=5 and n=7 as well.. I think I actually had that formula at some point, but dismissed it because the polynomial is of a higher degree than I wanted/expected (and at the time I was making mistakes all over the place, so I thought it was just plain wrong) The computer seems to have no problem at all finding roots for it, so maybe we can too.. |
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