Yesterday, I introduced J and K to recurrence relations. I set a few problems for homework. This post contains descriptions of one student's (J) response to three of these problems. Here is the first one:
J (unsuccessfully) tried to solve part (a) using trial and error. This is a common approach when unsure what to do. There is some value in trial and error approaches for getting a sense of a problem, but often a shift towards an algebraic approach is required. How might J learn how and when to make this shift?
It is something that many learners find difficult to do. I think there are a number of reasons for this. To solve this problem you can substitute the information about the terms into the recurrence relation to create two simultaneous equations, that can then be solved. How might a learner realise this? Recognising the 'type' of question (due to familiarity) might be useful, but this will not always be enough: it is necessary to prepare for variations in what is possible (see below).
Something like a leap of faith is required. It might be necessary to form an equation without knowing if or how it is going to be useful. Often, it will not be clear how the equations may be helpful (or whether they will be solvable!) until they are formed. Then, solving the equation that we have formed requires working on a level of abstraction at least once removed from the original problem.
Here's a second problem we worked on:
Once J had formed the quadratic equation, it was very easy for him to know what to do - solve it! I think our work on the previous problem had prepared him well. Flexibility is required, something along the lines of: it might be useful to use the information to form some kind of equation, and then solve whatever is formed, whilst bearing in mind its relevance to the original problem. Had he memorised the previous question as a type - perhaps 'form a pair of simultaneous equations and solve them' - he may have been puzzled.
Here's a third - rather challenging - problem we worked on:
It was not clear to J what to do here, not least because the coefficients in the recurrence relation are themselves variable! It takes a quite large degree of faith to substitute the expression 1/2.sin(x) into the recurrence relation in order to form a trigonometric equation in x that can then be solved to find the variable x! J was able to solve this problem, but only after all of this work on forming equations, and after a discussion about how we 'normally' form an equation when trying to find the limit.
Here are J's responses to the three questions; the equations formed are boxed (click to enlarge):