WitCH 19: A Powerful Solvent

The following WitCH is from VCE Mathematical Methods Exam 2, 2009. (Yeah, it’s a bit old, but the question was raised recently in a tutorial, so it’s obviously not too old.) It is a multiple choice question: The Examiners’ Report indicates that just over half of the students gave the correct answer of B. The Report also gives a brief indication of how the problem was to be approached:

    \[\mbox{\bf Solve } \boldsymbol{\frac{1}{k-0} \int\limits_0^k \left(\frac1{2x+1}\right)dx = \frac16\log_e(7) \mbox{ \bf for $\boldsymbol k$}.\ k = 3.}\]

Have fun.

Update (02/09/19)

Though undeniably weird and clunky, this question clearly annoys commenters less than me. And, it’s true that I am probably more annoyed by what the question symbolises than the question itself. In any case, the discussion below, and John’s final comment/question in particular, clarified things for me somewhat. So, as a rounding off of the post, here is an extended answer to John’s question.

Underlying my concern with the exam question is the use of “solve” to describe guessing/buttoning the solution to the (transcendental) equation \mathbf {\frac1{2k}{\boldsymbol \log} (2k+1) = \frac16{\boldsymbol \log} 7}.  John then questions whether I would similarly object to the “solving” of a quintic equation that happens to have nice roots. It is a very good question.

First of all, to strengthen John’s point, the same argument can also be made for the school “solving” of cubic and quartic equations. Yes, there are formulae for these (as the Evil Mathologer covered in his latest video), but school students never use these formulae and typically don’t know they exist. So, the existence of these formulae is irrelevant for the issue at hand.

I’m not a fan of polynomial guessing games, but I accept that such games are standard and that  “solve” is used to describe such games. Underlying these games, however, are the integer/rational root theorems (which the EM has also covered), which promise that an integer/rational coefficient polynomial has only finitely many candidate roots, and that these roots are easily enumerated. (Yes, these theorems may be a less or more explicit part of the game, but they are there and they affect the game, if only semi-consciously.) By contrast, there is typically no expectation that a transcendental equation will have somehow simple solutions, nor is there typically any method of determining candidate solutions.

I find something generally unnerving about the exam question and, in particular, the Report. It exemplifies a dilution of language which is at least confusing, and I’d suggest is actively destructive. At its weakest, “solve” means “find the solutions to”, and anything is fair game. This usage, however, loses any connotation of “solve” meaning to somehow figure out the way the equation works, to determine why the solutions are what they are. This is a huge loss.

True, the investigation of equations can continue independent of the cheapening of a particular word, but the reality is that it does not. Of course, in this manner the Solve button on CAS is the nuclear bomb that wipes out all intelligent life. The end result is a double-barrelled destruction of the way students are taught to approach an equation. First, students are taught that all that matters about an equation are the solutions.  They are trained to give the barest lip service to analysing an equation, to investigating if the equation can be attacked in a meaningful mathematical manner. Secondly, the students are taught that that there is no distinction between a precise solution and an approximation, a bunch of meaningless decimals spat out by a machine.

So, yes, the exam question above can be considered just another poorly constructed question. But the weird and “What the Hell” incorporation of a transcendental equation with an exact solution that students were supposedly meant to “solve” is emblematic of a an impoverishment of language and of mathematics that the CAS-infatuated VCAA has turned into an art form.

51 Replies to “WitCH 19: A Powerful Solvent”

  1. Side note before delving into the problem more deeply – this concept is taught by many as a formula, rather than a concept which is only emphasised by the examiners’ comments.

    1. Number 8, granted it’s off the point, but I’m not sure what you mean. You mean the formula for the average is commonly taught without justification for the formula? (Which, yes, is all the question requires.)

  2. Correct. Many students, in my experience are taught that average value has a formula and to learn it. A methods paper in 2010(ish) may have tested the idea of a rectangle having the same area as an integral and asked for the height, but beyond this, it seems to be very commonly assessed as just a formula.

    The word average is fraught with all sorts of strange problems in high school mathematics (and to real estate agents but that is another thing altogether)

    1. Within the context of the Methods course, the average value of a function is given by a very specific formula and no justification is explicitly required. So, given that the question is within *this* context, the crap is not in the meaning of average value itself or the formula.

      By calculating the (simple) integral as ln(2k+1)/(2k) and comparing this with ln(7)/6, the question is obviously cooked so that the (only) solution is k = 3 by inspection. So for me, making it a CAS-active question (and multiple choice to boot!) is the crap here. Because it’s just another brainless push the buttons question (which is what the Report encourages).

      I think it’s worth remarking that in the limit k –> 0, the average value approaches 1, which is what you’d expect.

    1. Yeah, I was wondering what mistakes could you possibly make that would lead to *any* of the wrong options. The wrong options are definitely crap, but that’s because it’s a really dumb question to have in Section A of Exam 2. At the risk of over-looking some even more obvious crap, the question would be much more appropriate in Exam 1.

  3. Thank you, RF and JF, for your comments. There is, as you have both noted, the stark clunkiness of the question. But there is something else there, and I think this is another one which, in that other aspect, would annoy the shit out of mathematicians but not teachers.

    Ignoring the clunkiness of the question, teachers would presumably tend to think the question is difficult for students to get wrong (although 48% did …), or at least be unfairly misled, and so it’s at worst a no harm no foul thing. Mathematicians, however, and rightly or wrongly, would more likely be fixated on the foul.

  4. Well, k = 3 is the only answer, so no troubles with the implication in the question that there’s only one value.

    The function is well-defined, so no trouble there.

    Is it a matter of definition: Mathematicans define the average value to be over the domain, not a subset of the domain …? I wouldn’t have thought so. So it’s OK to have an average value over the given interval (the function is continuous over the interval), so no trouble there.

    The average value is clearly defined in Maths Methods, so there can’t be a quibble based on using that formula.

    The unique answer is obvious by inspection, so no troubles with solutions beyond the scope of the course (and a CAS gives an approximate answer of exactly 3 anyway, so no need to even know how to integrate or solve something by inspection ….)

    Square brackets are OK.

    The wrong options are weak, but that’s not a mathematical error (in fact, the trouble is that none of the wrong options seem to arise from a good error!).

    I see no obvious foul, the only thing that annoys me is that the question is on Exam 2 and is a multiple choice question.

    1. JF, I agree with all you write. As I indicated, this one is like the error in WitCH 10. It annoys me, and I’m willing to bet it would annoy mathematicians in general, and I *wish* that it annoyed teachers, but I’m not surprised that it doesn’t.

  5. OK… be gentle with me everyone… I’m going to ask why the question defines the function over a different domain to that which the average value is based on. Could they not have said f: [0,k]->R, f(x)= and thus avoided a lot of issues?

        1. Marty, I put my comment in the trash can after realising I’d misread an earlier comment. No apology needed – please return it to the bin where it belongs.

  6. Is there a difference here between log and Log? My (growing distant) memories of Complex Analysis are coming back looking at this – my recollection is that when spelt with a capital it is a single valued function, but spelt lowercase, the complex arguments come into play.

  7. To think about this in a (hopefully) less VCAA examiner way… is the issue in the question or in the suggested solution? I have an idea of what you might be thinking of if the issue is in the VCAA intended solution.

    1. Hi RF. The issue is in the question. As I suggested, it’s similar in mathematician-annoyingness to WitCH 10. Not as bad as that one, but similar.

    1. Better and worse. There is a unique solution, so “could be” would be a little weird. But I think you’re getting to it.

  8. OK, I’ll bet it’s the potential ambiguity:

    “… f(x) = 1/(2x+1) over the interval [0, k] …”

    instead of the better worded

    Over the interval [0, k], the average value of the function …..

    (If it is, it doesn’t annoy me greatly. Not nearly as much as the mere presence of the question on Exam 2)

      1. First I would look for a simple integer solution, starting by testing 0 or 1 (not for this question, but for questions of that nature). Failing that, I would try some form of numerical approximation, unless a calculator was available to do this for me.

          1. “Solve” to me has (in a highschool context) always meant “find the value of the pronumeral which makes the statement true” however once literal equations are encountered, it is perhaps more appropriate to say, “transpose” rather than “solve” although perhaps the two are used interchangeably?

            In VCAA paper 2 exams, “Solve” is a function on the calculator more often than not.

  9. Do the integration: 1/(2k) Ln(2k+1).

    Equate to the given mean: 1/(2k) Ln(2k+1) = 1/6 Ln(7).

    Simplify: 1/k Ln(2k+1) = 1/3 Ln(7).

    Require simultaneous solution to 1/k = 1/3 and 2k+1 = 7. By inspection the solution is clearly and trivially k = 3.

    No powerful solvent required here.

    But a powerful solvent means strong dissolving …. And by putting this question on Exam 2 it is certainly a strong dis to solving.

  10. If your question was in Exam 2, both I and students would press some buttons (or write some Mathematica code) and get 3.92884 (correct to five decimal places). The wording of the question would need to change:

    “The value of k is *closest* to”

    My complaint would still be that the question is pointless in Exam 2 because CAS reduces it to triviality. At least average value can be used in Exam 1 as a context for testing integration skills.

    The VCAA question has clearly been cooked so that the solution is readily obtained by inspection in the way that I posted. Although this method fails for your example, I don’t think that failure invalidates its use for the ‘cooked’ VCAA question (although I don’t see the point of ‘cooking’ questions for Exam 2). And if the VCAA question is solved using a CAS then my method is moot.

    (As an aside, there is undoubtedly some fancy use of the Lambert W-function that would give exact solutions).

    1. Thanks, JF. Yes, the solution can readily be obtained by inspection. And, it can only be obtained by inspection. You’re not bothered by that, I guess because of the “readily”: students are not likely to go down a rabbit hole, and there’s no evidence that students did. I’m much less comfortable with the question, and the Report.

      1. The word “Solve” in the Examiners Report should be decoded as “Use a CAS to solve the following equation”, which is what most if not all students would have done in one form or another (some may have evaluated the integral first, with the CAS, and then solved, with the CAS. In which case the only thing wrong with the question is that, mathematically, it’s utterly pointless and the crap here is the very presence of the question.

        The fact that the exact solution can only be obtained by inspection doesn’t bother me, except in the sense that I don’t see the point in cooking the question so that it *can* be done by inspection (after all, it’s a CAS exam, for better or for worse).

  11. Thanks, JF and RF. The word “solve” appears to make me more uncomfortable than you guys. For me, giving a fundamentally unsolvable (transcendental) equation that happens to have an exact solution, is weird and disorienting. Anyway, there’s not more there (except the general clunkiness), and I’ll update soon.

    1. OK, my turn to ask a question, Marty. A quintic equation is generally unsolvable, so would you have the same unease towards the following question:

      Solve x^5 – x^4 + x^3 – x^2 + 2x – 2 = 0.

      By the way, although I (and others) have been using the word ‘solve’, the question doesn’t explicitly use this word. It just asks for an option that gives the value of k. So I’m guessing it’s the comment in the Examiners Report that causes your unease. But again, for better or for worse it’s a CAS exam – the assumption is that students are going to ‘solve’ the equation by pressing buttons, that is, essentially get an answer using a numerical method that happens to be exact rather than a decimal approximation. I can better understand your unease if it was a ‘Solve’ question in Exam 1, but then again, would that unease extend to the above question?

      I also note that there are many questions (eg. set in the context of drug dosages) with models of the form x = t e^t and questions such as how long before x < …. are asked. These equations are also fundamentally unsolvable but the expectation is that a decimal approximation is found using a CAS, and I don't think anyone has too much unease about this. (Such questions can be solved using the Lambert W-function, but I guess it opens up a new can of worms if equations are solved exactly by defining special functions to do the job ….)

      1. Thanks, John. It’s a very good question. I was writing a reply, but it seemed to end up sufficiently long to make it the rounding-off update to the post, which I’ll do now.

        Briefly on using the W-function or whatever to “solve” the equation, it is standard in mathematics to invent a new function so as to somehow capture the “solutions” of an equation. It is unclear, however, whether and when one wants to call this solving. For example, in what sense can the equation x3 = 2 be solved? (Feel free to continue the discussion, but I’ll try to put up a dedicated post to this question.)

  12. Decrappers,

    Clearly as JF mentions you don’t need a CAS enabled root extractor to “guess” a solution for k in this example.

    So I tried plugging in sin(kx) = x for k close to 1 into the solve X root finding algorithm on the CX
    And it came back with X= 0 and the helpful? hint that there may be other solutions.
    Wolfram Alpha was better giving the other solutions without having to provide ranges

    That said I was taught to use Newton’s iterative method for ‘guessing ‘ roots of polynomials etc which mainly converges quickly provided initial guess is close and is intuitive .

    https://en.m.wikipedia.org/wiki/Newton%27s_method

    Steve R

    1. Thanks, Steve. That seems pretty weird and unsolvy. Just to clarify, what does the CX do if you try to get it to solve sin(9x/10) = x? What about sin(11x/10) = x?

  13. A simple graph also helps:

    For sin(kx) = x and 0 < k < 1, there is clearly only one intersection point of the line y = x and the curve y = sin(kx) (since the gradient of y = x is 1 and the gradient m of y = sin(kx) is 0 < m < k < 1 over -pi/2 < x < pi/2). And that intersection point is clearly (0, 0).

    So the unique solution to sin(kx) = x, where 0 < k 1, there are clearly more solutions than just x = 0 (and the number of solutions depends on the value of k).

    1. Not sure what happened to the above comment – bits and pieces are missing (and the edit option seems to be playing up – using inequality symbols seem to be causing the trouble so I’ve substituted lt and gt for them). The original comment (with the substitution) was:

      A simple graph also helps:

      For sin(kx) = x and 0 lt k lt 1, there is clearly only one intersection point of the line y = x and the curve y = sin(kx) (since the gradient of y = x is 1 and the gradient m of y = sin(kx) is 0 lt m < k lt 1 over -pi/2 lt x lt pi/2). And that intersection point is clearly (0, 0). So the unique solution to sin(kx) = x, where 0 lt k lt 1 is x = 0. (And *ahem* does solving this equation make anyone uncomfortable?).

      For sin(kx) = x and k gt 1, there are clearly more solutions than just x = 0 (and the number of solutions depends on the value of k).

      1. JF,

        Yes ! graphing/sketching the ‘ squashed sin ‘ function on your tablet ,whiteboard,calculator ,envelope … against y=x quickly Shows there should be 3 intersections for k> 1 and only one for 0 <= k<= 1.

        I picked the equation to test the functionality of solve X on the CX (and WA )which appears to be root finding Algorithm in Mathematica. I was wondering if it used a Taylor's series to approximate the solution close to 1 (similar to the approximation mentioned in link below) or more likely an iterative approach using Newton's method or …?

        https://en.m.wikipedia.org/wiki/Transcendental_equation

        Steve R

  14. Hi,

    For k =1.1 then sin(1.1x) = x into WA gives roots 0 & +- 0.680897 and for 0< k < 1 root will be 0

    On my 4 year old CX solve( sin(1.1x)=X,X) only gives 0 with the warning of other solutions

    Steve R

      1. Ahh yes. That was the other thing that disappeared from my earlier comment: Does the CX still give the warning that there may be other solutions when 0 lt k lt 1 ….? The TI-89* does (and it gives the same warning for both Arcsin(x) = x and Arcsin(x) = 1.3x).

        *The TI-89 is where I drew the line in the stupid CAS calculator arms race, no TI-Insipid for me (although I had to work with it with students).

  15. This is going off on a tangent to a tangent, but since the issue of mindlessly using CAS to solve equations has been broached, I would be interested to read what others (especially users of the TI handheld) think of Question 5(c)(ii) in the 2017 Specialist Exam 2…. The examiners report states that only 15% of students answered correctly, and apparently “incorrect answers involving other locally minimum values were frequent”.

      1. My complaint is not very deep. The issue with the question is that if students use the TI to (attempt to) find the minimum distance by solving for when the derivative of the distance function is zero, the TI finds a handful of stationary points, but not all of them. And, unfortunately for students, the missing stationary point turns out to be the minimum distance!

        1. Um, it may not be very deep, but it sounds very fucked up. I try to stay out of CAS swamps, but what were students supposed to do? (Also, does the stray square root soon in the report on 5(c)(i) mean anything?)

          1. Yes, clearly the question was poorly vetted.

            Two CAS approaches to get the answer: (1) use the “fmin” command to find the time at which the distance is minimised; (2) enter the equation on a graph page and use the graph minimum tool.

            If solving for when the derivative of the distance function is zero, CAS gives a warning that additional solutions may exist. It would be consistent with VCAA’s arrogance to proclaim that students should use alternative methods to verify an answer if their technology gives them this warning.

  16. I have very mixed feelings about this.

    On the one hand, it’s clear that students have been set up for failure with this question.

    But on the other hand, I would hope (and expect) that students have been taught to “use alternative methods to verify an answer if their technology gives them this warning”. This would probably happen with experienced teachers and/or teachers highly familiar and comfortable with the CAS calculator, but I suspect it would often be over-looked in the classroom. It’s a good example of where a teacher (and students) can learn a lot from an Examination Report in terms of preparing for the exam. Sanctimonious advice can nevertheless still be valuable advice to students and teachers in later years.

    When ‘solving’ complicated equations (particularly like this one) I always try to encourage students to draw a graph (easily done with a CAS) to get an idea of the expected solution. It’s frustrating that you need a list of ‘CAS traps for young players’ with salient examples illustrating those traps as part of your teaching repertoire. It’s meant to be a *maths* examination.

      1. Indeed. But it will be a cold day in hell before that happens. Last I heard, the exams were meant to be vetted using the common types of CAS calculators so that the questions are ‘calculator neutral’. However, I don’t see much evidence of this happening. There was a question some time ago (I can’t remember when) on the Methods Exam 2 where the equation was easily ‘solved’ using the TI but the Casio froze. There was a lot of angst among students and teachers who use the Casio.

        It’s going to be very interesting this year in the Specialist exam 2 because there are several schools whose students will be using Mathematica rather than a CAS calculator. All students will sit the same exam. Having taught with both CAS calculators and Mathematica, I can confidently say that students using Mathematica will have an advantage. And students with an aptitude for coding will have a huge advantage. As far as I can read the tea-leaves, Mathematica will not be the compulsory CAS for at least another couple of years, so that advantage is going to remain (one could speculate that this is a deliberate ploy by VCAA to get all schools on board with Mathematica ….) But now we’re getting off-topic.

        1. “There was a question some time ago (I can’t remember when) on the Methods Exam 2 where the equation was easily ‘solved’ using the TI but the Casio froze. There was a lot of angst among students and teachers who use the Casio.”

          John, I believe the question you’re referring to is Q3(e) on the [2010 Methods Exam 2](https://www.vcaa.vic.edu.au/Documents/exams/mathematics/2010mmcas2-w.pdf). The CAS (whichever company) technology can sometimes get caught up in convoluted trig expressions etc, wnich causes a freeze when trying to solve an equation such as that required in the question.

          Clearly not vetted correctly…

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