Eric Nicholas K's answer to LNG's Junior College 2 H2 Maths Singapore question.
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Not 100 percent sure, but here are my workings for this question. For the third column of my workings (with the characteristic equation), it is noteworthy to know that the golden ratio for the Fibonacci sequence (1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...) with recurrence relation a (n + 2) = a (n + 1) + a n is obtained using very similar workings.
Date Posted:
4 years ago
I have previously verified that my formula works for at least the first six terms (a1 to a6) in the sequence, but I have yet to verify this for future expressions.
Also, I don't recall this question being taught in JC during my JC days more than 10 years ago (and in fact I already forgot most of my JC and uni stuffs!), so I had to use my uni knowledge (with a little modification by introducing lg on both sides first) to do this question.
Also, I don't recall this question being taught in JC during my JC days more than 10 years ago (and in fact I already forgot most of my JC and uni stuffs!), so I had to use my uni knowledge (with a little modification by introducing lg on both sides first) to do this question.
It’s quite a complex idea. The introduction of the “characteristic equation” is very very similar in idea to the techniques of differential equations. I simply used those ideas to conclude that both values of r are accepted.
I am not so sure what you have learnt in differential equations so far, because I only learned this technique during university (my A Level syllabus ten years ago had no such level of differential equations).
Below I have provided a general idea to solve the recurrence relations. It’s not apparently clear how they found out such general solutions, because I would have never guessed it. I suspect it was due to years of analysis of equations that the older generations of mathematicians undertook.
I am not so sure what you have learnt in differential equations so far, because I only learned this technique during university (my A Level syllabus ten years ago had no such level of differential equations).
Below I have provided a general idea to solve the recurrence relations. It’s not apparently clear how they found out such general solutions, because I would have never guessed it. I suspect it was due to years of analysis of equations that the older generations of mathematicians undertook.
http://discrete.openmathbooks.org/dmoi3/sec_recurrence.html
They key idea in how I got general expression = a * (r1)^n + b * (r2)^n is that the powers of 10 appear to be geometric (in the sense that they have a “multiplier”), but their geometric ratios (“multipliers”) are not really fixed.
In fact, I noticed that the numerators follow a pattern, which is simply the previous term multiplied by 4 and then either a plus 1 or -1, where the +-1 is done alternately. Hence, it cannot take up a single ratio r1, but because there are two repeating but related patterns, it could be made of two ratios r1 and r2.
Of course, to talk about the powers only rather than the entire value of 10^power, I used lg to kill off the base 10.
In fact, I noticed that the numerators follow a pattern, which is simply the previous term multiplied by 4 and then either a plus 1 or -1, where the +-1 is done alternately. Hence, it cannot take up a single ratio r1, but because there are two repeating but related patterns, it could be made of two ratios r1 and r2.
Of course, to talk about the powers only rather than the entire value of 10^power, I used lg to kill off the base 10.
Many thanks for the explanation, Sir. Still trying to wrap my head around the derivation of that general expression = a * (r1)^n + b * (r2)^n.
This question was given as a "question of the week" by my teacher !
This question was given as a "question of the week" by my teacher !
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