Tag: VCE

WitCH 38: A Deep Hole

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June 20, 2020

3:06 pm

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This one is due to commenter P.N., who raised it on another post, and the glaring issue has been discussed there. Still, for the record it should be WitCHed, and we’ve also decided to expand the WitCHiness slightly (and could have expanded it further).

The following questions appeared on 2019 Specialist Mathematics NHT, Exam 2 (CAS). The questions are followed by sample Mathematica solutions (screenshot corrected, to include final comment) provided by VCAA (presumably in the main for VCE students doing the Mathematica version of Methods). The examination report provides answers, identical to those in the Mathematica solutions, but indicates nothing further.

UPDATE (05/07/20)

The obvious problem here, of course, is that the answer for Part (b), in both the examination report and VCAA’s Mathematica solutions, is flat out wrong: the function fk will also fail to have a stationary point if k = -2 or k = 0. Nearly as bad, and plenty bad, the method in VCAA’s Mathematica solutions to Part (c) is fundamentally incomplete: for a (twice-differentiable) function f to have an inflection point at some a, it is necessary but not sufficient to have f’’(a) = 0.

That’s all pretty awful, but we believe there is worse here. The question is, how did the VCAA get it wrong? Errors can always occur, but why specifically did the error in Part (b) occur, and why, for a year and counting, wasn’t it caught? Why was a half-method suggested for Part (c), and why was this half-method presumably considered reasonable strategy for the exam? Partly, the explanation can go down to this being a question from NHT, about which, as far as we can tell, no one really gives a stuff. This VCAA screw-up, however, points to a deeper, systemic and much more important issue.

The first thing to note is that Mathematica got it wrong: the Solve function did not return the solution to the equation fk‘ = 0. What does that imply for using Mathematica and other CAS software? It implies the user should be aware that the machine is not necessarily doing what the user might reasonably think it is doing. Which is a very, very stupid property of a black box: if Solve doesn’t mean “solve”, then what the hell does it mean? Now, as it happens, Mathematica’s/VCAA’s screw-up could have been avoided by using the function Reduce instead of Solve.* That would have saved VCAA’s solutions from being wrong, but not from being garbage.

Ask yourself, what is missing from VCAA’s solutions? Yes, yes, correct answers, but what else? This is it: there are no functions. There are no equations. There is nothing, nothing at all but an unreliable black box. Here we have a question about the derivatives of a function, but nowhere are those derivatives computed, displayed or contemplated in even the smallest sense.

For the NHT problem above, the massive elephant not in the room is an expression for the derivative function:

    \[\color{red} \boldsymbol{f'_k(x) = -\frac{x^2 + 2(k+1)x +1}{(x^2-1)^2}}\]

What do you see? Yep, if your algebraic sense hasn’t been totally destroyed by CAS, you see immediately that the values k = 0 and k = -2 are special, and that special behaviour is likely to occur. You’re aware of the function, alert to its properties, and you’re led back to the simplification of fk for these special values. Then, either way or both, you are much, much less likely to screw up in the way the VCAA did.

And that always happens. A mathematician always gets a sense of solutions not just from the solution values, but also from the structure of the equations being solved. And all of this is invisible, is impossible, all of it is obliterated by VCAA’s nuclear weapon approach.

And that is insane. To expect, to effectively demand that students “solve” equations without ever seeing those equations, without an iota of concern for what the equations look like, what the equations might tell us, is mathematical and pedagogical insanity.

 

*) Thanks to our ex-student and friend and colleague Sai for explaining some of Mathematica’s subtleties. Readers will be learning more about Sai in the very near future.

WitCH 37: A Foolproof Argument

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May 11, 2020

12:20 am

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We’re amazed we didn’t know about this one, which was brought to our attention by commenter P.N.. It comes from the 2013 Specialist Mathematics Exam 2: The sole comment on this question in the Examination Report is:

“All students were awarded [the] mark for this question.”

Yep, the question is plain stuffed. We think, however, there is more here than the simple wrongness, which is why we’ve made it a WitCH rather than a PoSWW. Happy hunting.

UPDATE (11/05) Steve C’s comment below has inspired an addition:

Update (20/05/20)

The third greatest issue with the exam question is that it is wrong: none of the available answers is correct. The second greatest issue is that the wrongness is obvious: if z^3 lies in a sector then the natural guess is that z will lie in one of three equally spaced sectors of a third the width, so God knows why the alarm bells weren’t ringing. The greatest issue is that VCAA didn’t have the guts or the basic integrity to fess up: not a single word of responsibility or remorse. Assholes.

Those are the elephants stomping through the room but, as commenters as have noted, there is plenty more awfulness in this question:

  • “Letting” z = a + bi is sloppy, confusing and pointless;
  • The term “quadrant” is undefined;
  • The use of “principal” is unnecessary;
  • “argument” is better thought as the measure of an angle not the angle itself;
  • Given z is a single complex number, “the complete set of values for Arg(z)” will consist of a single number.
  • The grammar isn’t.

SACs of Shit

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May 3, 2020

11:04 pm

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SACs may not be the greatest problem with VCE mathematics, but they’re right up there. SACs are torture for teachers and torture for students. They teach nothing. As assessment, they are unnecessary, unreliable and phenomenally inefficient. They are a license for VCAA’s unaswerable auditors applying Kafkaesque rules to act either as favour-givers or as little Hitlers, as the mood takes them. These problems are currently amplified to eleven by VCAA’s “We’ll give you some kind of guidance in, oh, a little while” plan for the plague year.

For all of the awfulness of the above, that’s not the worst of it. The worst is that the majority of SACs are monumentally stupid. Literally. A SAC has the imposing presence of a monument, its towering stupidity casting a shadow over everything.

How are SACs so bad? Many contain errors, often subtle although too often not, but, as irritating as that is, that is not the main problem. The main problem is that they are mathematical nonsense. Typically they will present the student with a ridiculous model of a contrived problem, which is then all redone in greater, brain-bludgeoning generality by throwing in a needless parameter in a randomly chosen location. All of this is undertaken, of course, in the nihilistic world of CAS. Finally, somewhere near the end, the poor beleaguered student, who by this stage just wants to escape with their life, will be required to “comment on the model”, to which the usual response is “It’s really nice, please let me go” and to which the only reasonable response is “It’s fucking insane”.

How do we know SACs are this bad? Because we see them. We see the commercial SACs, and the sample SACs, and the past SACs, and the current SACs. Are they all as bad as we suggest? No, of course not. Specialist SACs are typically nowhere near as bad as Methods SACs, and even many Methods SACs will fall short of truly idiotic torture, rising only to the level of being dumb and painful. Then there are the rare few SACs we see that are good, resulting in an exchange:

“This actually makes sense. Who’s your teacher?”

“Oh, it’s Mr. ….”

“Ah. Yes.”

So, yes, the quality and worth of SACs varies widely, but the average is squarely in the neighbourhood of monumental, tortuous stupidity. Which bring us to the “why”. Why are SACs in general so awful? There are two reasons.

The first and fundamental reason is the VCAA and their view of what they imagine is a curriculum. VCE mathematics subjects are so shallow and so lacking in a foundation of solid reason, that almost any attempt at depth and substance in a SAC is destined to be farce. The VCAA has replaced foundation and depth with CAS, which reaches peak awfulness in SACs. The VCAA promotes the fantasy that CAS magically transforms students into mathematical explorers, clever little Lewises and Clarks skilfully navigating the conceptual wilderness. The reality, of course, is much less Lewis and Clark than it is Burke and Wills. To top it off, SACs must follow guidelines that Terry Gilliam would be proud of, giving us Burke and Wills’ Bogus Brazilian Journey. Or, just Eraserhead. Something like that.

The second reason is the teachers. Sort of. Even if the subjects were coherent, even if they were unpoisoned by CAS and were unconstrained by vague and ridiculous conditions, even then writing a good SAC would be a very difficult and massively time-consuming task. Most teachers just don’t have the mathematical background, or the literary skill, to write a coherent, correct and mathematically rich SAC; many cannot even recognise one. And, that’s writing a good SAC for this imaginary good subject; writing a good SAC for these fundamentally flawed subjects with their ridiculous constraints is close to impossible, even for a strong teacher. And which teachers, particularly weaker teachers, have the time to compose such a good SAC? Why bother trying? And so, with the greatest common sense, most teachers do not. Most teachers stick to the audit-proof and meaningless formulaic SAC bullshit that the VCAA expects and effectively demands.

The VCAA’s SAC system is a crime against mathematical humanity.

UPDATE (15/5)

We received the following from a student acquaintance (who hadn’t read this post):

Hi Marty, given the upcoming math SACs approaching soon, the pressure is on to practice and practice. Attached below is last year’s Methods SAC1 (Unit 3/4) for [the student’s school]. I remember many talented friends of mine who were stumped, and didn’t do very well on this SAC. Personally, I thought this SAC was horrifying. In contrast to Specialist, (I actually quite enjoy Specialist!), Methods seems to be a huge prick because of frustrating, ambiguous SACs containing questions seemingly cooked from the pits of hell itself. Are these sort of SACs common across the state?

The student is, of course, correct. The SAC, which comes from a highly respected school, is a nightmare in all of the ways canvassed above. From start to end it is idiotic CAS-driven pseudo-modelling, complete with Magritte nonsense and a pointlessly prissy grading scheme. And, yes, the SAC contains an error.

Of course we won’t reveal the school, much less any teachers involved, which means that we are also unable to critique the SAC in detail. But that is one of the insidious aspects of the SAC system; an entirely proper concern for privacy means that SAC nonsense, although endemic, fails to be exposed to the public critique that is so very much needed.

MitPY 6: Integration by Substitution

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April 17, 2020

9:26 pm

73 Comments on MitPY 6: Integration by Substitution

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From frequent commenter, SRK:

A question for commenters: how to explain / teach integration by substitution? To organise discussion, consider the simple case

    \[\boldsymbol{ \int \frac{2x}{1+x^2}dx \,.}\]

Here are some options.

1) Let u = 1 + x^2. This gives \frac{du}{dx} = 2x, hence dx = \frac{du}{2x}. So our integral becomes \int \frac{2x}{u}\times \frac{du}{2x} = \int \frac{1}{u}du. Benefits: the abuse of notation here helps students get their integral in the correct form. Worry: I am uncomfortable with this because students generally just look at this and think “ok, so dy/dx is a fraction cancel top and bottom hey ho away we go”. I’m also unclear on whether, or the extent to which, I should penalise students for using this method in their work.

2)  Let u = 1 + x^2. This gives \frac{du}{dx} = 2x. So our integral becomes \int \frac{1}{u}\times \frac{du}{dx}dx = \int \frac{1}{u}du. Benefit. This last equality can be justified using chain rule. Worry: students find it more difficult to get their integral in the correct form.

3) \frac{2x}{1+x^2} has the form f'(g(x))g'(x) where g(x)=1+x^2 and f'(x) = \frac{1}{x}. Hence, the antiderivative is f(g(x)) = \log (1+x^2). This is just the antidifferentiation version of chain rule.  Benefit. I find this method crystal clear, and – at least conceptually – so do the students. Worry. Students often aren’t able to recognise the correct structure of the functions to make this work.

So I’m curious how other commenters approach this, what they’ve found has been effective / successful, and what other pros / cons there are with various methods.

UPDATE (21/04)

Following on from David’s comment below, and at the risk of splitting the discussion in two, we’ve posted a companion WitCH.

The Methods Intelligence Test

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April 13, 2020

11:03 pm

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Psychologist Daniel Kahneman dedicates his book Thinking Fast and Slow  to the memory of Amos Tversky, his long-time collaborator. Tversky was considered so brilliant by his colleagues that they came up with the Tversky Intelligence Test:

The faster you realise that Amos Tversky is smarter than you, the smarter you are.

It has occurred to us that there is a similar Methods Intelligence Test:

The slower you realise that Methods is stupider than you, the stupider you are.

MAV’s Dangerous Inflection

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February 24, 2020

11:02 am

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This post concerns a question on the 2019 VCE Specialist Mathematics Exam 2 and, in particular, the solution and commentary for that question available through the Mathematical Association of Victoria. As we document below, a significant part of what MAV has written on this question is confused, self-contradictory and tendentious. Thus, noting the semi-official status of MAV solutions, that these solutions play a significant role in MAV’s Meet the Assessors events, and are quite possibly written by VCE assessors, there are some troubling implications. Question 3, Section B on Exam 2 is a differential equations problem, with two independent parts. Part (a) is a routine (and pretty nice) question on exponential growth and decay.* Part (b), which is our concern, considers the differential equation

    \[\boldsymbol{\color{blue}\frac{{\rm d}Q}{{\rm d}t\ } = e^{t-Q}}\,,\]

for t ≥ 0, along with the initial condition

    \[\boldsymbol{\color{blue}Q(0) =1}\,.\]

The differential equation is separable, and parts (i) and (ii) of the question, worth a total of 3 marks, asks to set up the separation and use this to show the solution of the initial value problem is

    \[\boldsymbol{\color{blue}Q =\log_e\hspace{-1pt} \left(e^t + e -1\right)}\,.\]

Part (iii), worth 2 marks, then asks to show that “the graph of Q as a function of t” has no inflection points.** Question 3(b) is contrived and bitsy and hand-holding, but not incoherent or wrong. So, pretty good by VCE standards. Unfortunately, the MAV solution and commentary to this problem is deeply problematic. The first MAV misstep, in (i), is to invert the derivative, giving

    \[\boldsymbol{\color{red}\frac{{\rm d}t\ }{{\rm d}Q } = e^{Q-t}}\,,\]

prior to separating variables. This is a very weird extra step to include since, not only is the step not required here, it is never required or helpful in solving separable equations. Its appearance here suggests a weak understanding of this standard technique. Worse is to come in (iii). Before considering MAV’s solution, however, it is perhaps worth indicating an approach to (iii) that may be unfamiliar to many teachers and students and, possibly, the assessors. If we are interested in the inflection points of Q,*** then we are interested in the second derivative of Q. The thing to note is we can naturally obtain an expression for Q” directly from the differential equation: we differentiate the equation using the chain rule, giving

    \[\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - Q'\right)}\,.\]

Now, the exponential is never zero, and so if we can show Q’ < 1 then we’d have Q” > 0, ruling out inflection points. Such conclusions can sometimes be read off easily from the differential equation, but it does not seem to be the case here. However, an easy differentiation of the expression for Q derived in part (ii) gives

    \[\boldsymbol{\color{magenta}Q' =\frac{e^t}{e^t + e -1}}\,.\]

The numerator is clearly smaller than the denominator, proving that Q’ < 1, and we’re done. For a similar but distinct proof, one can use the differential equation to replace the Q’ in the expression for Q”, giving

    \[\boldsymbol{\color{magenta}Q'' = e^{t-Q}\left(1 - e^{t-Q}\right)}\,.\]

Again we want to show the second factor is positive, which amounts to showing Q > t. But that is easy to see from the expression for Q above (because the stuff in the log is greater than \boldsymbol{e^t}), and again we can conclude that Q has no inflection points. One might reasonably consider the details in the above proofs to be overly subtle for many or most VCE students. Nonetheless the approaches are natural, are typically more efficient (and are CAS-free), and any comprehensive solutions to the problem should at least mention the possibility. The MAV solutions make no mention of any such approach, simply making a CAS-driven beeline for Q” as an explicit function of t. Here are the contents of the MAV solution:

Part 1: A restatement of the equation for Q from part (ii), which is then followed by 

.˙.  \boldsymbol{ \color{red}\ \frac{{\rm d}^2Q }{{\rm d}t^2\ } = \frac{e^{t+1} -e^t}{\left(e^t + e -1\right)^2} } 

Part 2: A screenshot of the CAS input-output used to obtain the conclusion of Part 1.

Part 3: The statement   

Solving  .˙.  \boldsymbol{\color{red} \ \frac{{\rm d}^2Q }{{\rm d}t^2\ } = 0} gives no solution  

Part 4: A screenshot of the CAS input-output used to obtain the conclusion of Part 3.

Part 5: The half-sentence

We can see that \boldsymbol{\color{red}\frac{{\rm d}^2Q }{{\rm d}t^2\ } > 0} for all t,

Part 6: A labelled screenshot of a CAS-produced graph of Q”.

Part 7: The second half of the sentence,

so Q(t) has no points of inflection

This is a mess. The ordering of the information is poor and unexplained, making the unpunctuated sentences and part-sentences extremely difficult to read. Part 3 is so clumsy it’s funny. Much more important, the MAV “solution” makes little or no mathematical sense and is utterly useless as a guide to what the VCE might consider acceptable on an exam. True, the MAV solution is followed by a commentary specifically on the acceptability question. As we shall see, however, this commentary makes things worse. But before considering that commentary, let’s itemise the obvious questions raised by the MAV solution:
  • Is using CAS to calculate a second derivative on a “show that” exam question acceptable for VCE purposes?
  • Can a stated use of CAS to “show” there are no solutions to Q” = 0 suffice for VCE purposes? If not, what is the purpose of Parts 3 and 4 of the MAV solutions?
  • Does copying a CAS-produced graph of Q” suffice to “show” that Q” > 0 for VCE purposes?
  • If the answers to the above three questions differ, why do they differ?
Yes, of course these questions are primarily for the VCAA, but first things first. The MAV solution is followed by what is intended to be a clarifying comment:

Note that any reference to CAS producing ‘no solution’ to the second derivative equalling zero would NOT qualify for a mark in this ‘show that’ question. This is not sufficient. A sketch would also be required as would stating \boldsymbol{\color{red}e^t (e - 1) \neq 0} for all t.

These definitive-sounding statements are confusing and interesting, not least for their simple existence. Do these statements purport to be bankable pronouncements of VCAA assessors? If not, what is their status? In any case, given that pretty much every exam question demands that students and teachers read inscrutable VCAA tea leaves, why is it solely the solution to question 3(b) that is followed by such statements? The MAV commentary at least makes clear their answer to our second question above: quoting CAS is not sufficient to “show” that Q” = 0 has no solutions.  Unfortunately, the commentary raises more questions than it answers:
  • Parts 3 and 4 are “not sufficient”, but are they worth anything? If so, what are they worth and, in particular, what is the import of the word “also”? If not, then why not simply declare the parts irrelevant, in which case why include those parts in the solutions at all?
  • If, as claimed, it is “required” to state \boldsymbol{e^t(e-1)\neq 0} (which is indeed the key point of this approach and should be required), then why does the MAV solution not contain any such statement, nor even the factorisation that would naturally precede this statement?
  • Why is a solution “required” to include a sketch of Q”? If, in particular, a statement such as \boldsymbol{e^t(e-1)\neq 0} is “required”, or in any case is included, why would the latter not in and of itself suffice?
We wouldn’t begin to suggest answers to these questions, or our four earlier questions, and they are also not the main point here. The main point is that under no circumstances should such shoddy material be the basis of VCAA assessor presentations. If the material was also written by VCAA assessors, all the worse. Of course the underlying problem is not the quality or accuracy of solutions but, rather, the fundamental idiocy of incorporating CAS into proof questions. And for that the central villain is not the MAV but the VCAA, which has permitted their glorification of technology to completely destroy the appreciation of and the teaching of proof and reason. The MAV is not primarily responsible for this nonsense. The MAV is, however, responsible for publishing it, promoting it and profiting from it, none of which should be considered acceptable. The MAV needs to put serious thought into its unhealthily close relationship with the VCAA.   *) We might ask, however, who refers to “The growth and decay” of an exponential function? **) One might simply have referred to Q, but VCAA loves them their words. ***) Or, if preferred, the points of inflection of the graph of Q as a function of t.

Update (26/06/20)

The Examination Report is out and is basically ok; none of the nonsense and non sequiturs of the MAV solutions are included. The solution to (b)(iii) correctly focuses upon the factoring of Q”, although it needlessly worries about the sign of the denominator. There is no mention of the more natural approach to obtaining and analysing Q” but, given the question is treated by the VCAA and pretty much everyone as just another mindless exercise in pushing buttons, this is no surprise.

WitCH 35: Overly Resolute

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February 14, 2020

9:53 am

54 Comments on WitCH 35: Overly Resolute

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This WitCH (arguably a PoSWW) comes courtesy of Damien, an occasional commenter and an ex-student of ours from the nineteenth century. It is from the 2019 Specialist Mathematics Exam 2. We’ll confess, we completely overlooked the issue when going through the MAV solutions.

Update (16/02/20)

What a mess. Thanks to Damo for pointing out the problem, and thanks to the commenters for figuring out the nonsense.

In general form, the (intended) scenario of the exam question is

The vector resolute of \boldsymbol{\tilde{a}} in the direction of \boldsymbol{\tilde{b}} is \boldsymbol{\tilde{c}},

which can be pictured as follows: For the exam question, we have \boldsymbol{\tilde{a}} = \boldsymbol{\tilde{i}} + \boldsymbol{\tilde{j}} - \boldsymbol{\tilde{k}}, \boldsymbol{\tilde{b}} = m\boldsymbol{\tilde{i}} + n\boldsymbol{\tilde{j}} + p\boldsymbol{\tilde{k}} and \boldsymbol{\tilde{c}} = 2\boldsymbol{\tilde{i}} - 3\boldsymbol{\tilde{j}} + \boldsymbol{\tilde{k}}.

Of course, given \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{b}} it is standard to find \boldsymbol{\tilde{c}}. After a bit of trig and unit vectors, we have (in must useful form)

\boldsymbol{\tilde{c} = \left(\dfrac{\tilde{a}\cdot \tilde{b}}{\tilde{b}\cdot \tilde{b}}\right)\tilde{b}}

The exam question, however, is different: the question is, given \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}}, how to find \boldsymbol{\tilde{b}}.

The problem with that is, unless the vectors \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} are appropriately related, the scenario simply cannot occur, meaning \boldsymbol{\tilde{b}} cannot exist. Most obviously, the length of \boldsymbol{\tilde{c}} must be no greater than the length of \boldsymbol{\tilde{a}}. This requirement is clear from the triangle pictured, and can also be proved algebraically (with the dot product formula or the Cauchy-Schwarz inequality).

This implies, of course, that the exam question is ridiculous: for the vectors in the exam we have |\boldsymbol{\tilde{c}}| > |\boldsymbol{\tilde{a}}|, and that’s the end of that. In fact, the situation is more delicate; given the pictured vectors form a right-angled triangle, we require that \boldsymbol{\tilde{a}} - \boldsymbol{\tilde{c}} be perpendicular to \boldsymbol{\tilde{c}}. Which implies, once again, that the exam question is ridiculous.

Next, suppose we lucked out and began with \boldsymbol{\tilde{a}}- \boldsymbol{\tilde{c}} perpendicular to \boldsymbol{\tilde{c}}. (Of course it is very easy to check whether we’ve lucked out.) How, then, do we find \boldsymbol{\tilde{b}}? The answer is, as is made clear by the picture, “Well, duh”. The possible vectors \boldsymbol{\tilde{b}} are simply the (non-zero) scalar multiples of \boldsymbol{\tilde{c}}, and we’re done. Which shows that the mess in the intended solution, Answer A, is ridiculous.

There is a final question, however: the exam question is clearly ridiculous, but is the question also stuffed? The equations in answer A come from the equation for \boldsymbol{\tilde{c}} above and working backwards. And, these equations correctly return no solutions. Moreover, if the relationship between \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} had been such that there were solutions, then the A equations would have found them. So, completely ridiculous but still ok?

Nope.

The question is framed from start to end around definite, existing objects: we have THE vector resolute, resulting in THE values of m, n and p. If the VCAA had worded the question to find possible values, on the basis of a possible direction for the resolution, then, at least technically, the question would be consistent, with A a valid answer. Still an utterly ridiculous question, but consistent. But the VCAA didn’t do that and so the question isn’t that. The question is stuffed.

Further Update (26/06/20)

As commenters have noted, the Examination Report has finally appeared. And, as predicted, answer A was deemed correct, with the Report noting

Option A gives the set of equations that can be used to obtain the values of m, n and p. Explicit solution would result in a null set as it is not possible for a result of a vector to be of greater magnitude than the vector itself.

Well, it’s something. Presumably “result of a vector” was intended to be “resolute of a vector”, and the set framing is weirdly New Mathy. But, it’s something. Seriously. As John Friend notes, it is at least a small step along the way to indicating the question is not all hunky-dory.

That step, however, is way too small. We’ll close with two comments, reiterating the points made above.

1. The question is wrong

Read the question again, and read the first sentence of the Report’s comment. The question and report justification are fundamentally stuffed by the definite articles, by the language of existence. All answers should have been marked correct.

2. The question is worse than wrong

Even if the vectors \boldsymbol{\tilde{a}} and \boldsymbol{\tilde{c}} had been chosen appropriately, the question is utterly devoid of mathematical sense. It suggests a long and difficult method to solve a problem that, if indeed is solvable, is trivial.