analysis – BAD MATHEMATICS

WitCH 8: Oblique Reasoning

February 13, 2019

5:58 pm

79 Comments on WitCH 8: Oblique Reasoning

analysis, asymptotes, Specialist Mathematics, textbooks, WitCH

A reminder, WitCH 2, WitCH 3 and WitCH 7 are also open for business. Our new WitCH comes courtesy of John the Merciless. Once again, it is from Cambridge’s text Specialist Mathematics VCE Units 3 & 4 (2019). The text provides a general definition and some instruction, followed by a number of examples, one of which we have included below. Have fun.

Update

With John the Impatient’s permission, I’ve removed John’s comments for now, to create a clean slate. It’s up for other readers to do the work here, and (the royal) we are prepared to wait (as is the continuing case for WitCh 2 and Witch 3).

This WitCH is probably difficult for a Specialist teacher (and much more so for other teachers). But it is also important: the instruction and the example, and the subsequent exercises, are deeply flawed. (If anybody can confirm that exercise 6G 17(f) exists in a current electronic or hard copy version, please indicate so in the comments.)

Update (05/02/20)

It is obviously long, long past time to sort out this godawful mess. We apologise to all those industrious commenters, who nailed the essential wrongness, and whose hard work was left hanging. This will also be a very long update; you should pour yourself a stiff drink, grab the bottle and get comfy. For the benefit of the Twitter addicts, however, who now find it difficult to concentrate for more than two paragraphs, here’s the punchline:

Cambridge‘s notion of “non-vertical asymptote” is so vague, falling so short of a proper definition, it is close to meaningless and it is pointless. This leads to the claim $\color{red}\boldsymbol{y \to \frac{x}{\sqrt{x}}}}$ in Example 31 being flawed in three distinct ways. In particular, it follows from Cambridge that the “curve” $\color{red}\boldsymbol{y = \frac{x+1}{\sqrt{x-1}}}$ is a “non-vertical asymptote” to the function $\color{red}\boldsymbol{F(x) = \frac{x+1}{\sqrt{x-1}}}$ .

The source of Cambridge‘s confusion is also easy to state:

More complicated asymptotes cannot be defined, interpreted or computed in the manner possible for simpler asymptotes.

Unfortunately, the discussion in Cambridge is so sparse and so far from coherent that directly critiquing the excerpt above would probably be incomprehensible. So, we’ll first try to make very careful sense of “non-vertical asymptotes”, taking some minor whacks at Cambridge along the way. Then, with that sense as foundation, it will be easy work to hammer the excerpt above. To simplify the discussion, we’ll only consider the asymptotes of a function $\boldsymbol{F(x)}$ as $\boldsymbol{x \to \infty}$ . Obviously, the case where $\boldsymbol{x \to -\infty}$ can be handled in like manner, and vertical asymptotes are pretty straightforward.

OK, take a swig and let’s go.

1. Horizontal Asymptotes

We’ll begin with the familiar and demon-free case, for asymptotes as horizontal lines. Consider, for example, the function

$\color{blue}\boldsymbol{g(x) = \frac{2x + 1}{x-3}\,.}$

We can write $\boldsymbol{g(x) = 2 \ + \ 7/(x-3)}$

, noting the second term is tiny when $\boldsymbol{x}$

huge. So, we declare that $\boldsymbol{g(x)}$

has the horizontal asymptote $\boldsymbol{y = 2}$

Formally, the asymptote in this example is captured with limits. The underlying functional behaviour is $\boldsymbol{g(x)\to 2}$ or, more officially, $\boldsymbol{\lim\limits_{x\to\infty}g(x) = 2}$ . The limit formalism is of no benefit here, however, and is merely likely to confuse. The informal manner of thinking about and writing limits, as is standard in schools, suffices for the understanding of and computation of horizontal asymptotes.

Note also that horizontal asymptotes can easily be spotted with essentially no calculation. Consider the function

$\color{blue}\boldsymbol{h(x) = \frac{4x^2 + 2}{x^2 + 1}\,,}$

which is Example 29 in Section 6G of Cambridge. Here, it is only the highest powers, the $\boldsymbol{x^2}$

terms in the numerator and denominator, that matter; the $\boldsymbol{x^2}$

s cancel, and so the function has horizontal asymptote $\boldsymbol{y=4}$

. This simplification also applies to suitable rooty (algebraic) functions. The function

$\color{blue}\boldsymbol{k(x) = \frac{4x + 2}{\sqrt{x^2 + 1}}\,,}$

for example, also has horizontal asymptote $\boldsymbol{y=4}$

, for the same reason.

2. Linear Asymptotes

We’ll now consider general linear asymptotes, which are still reasonably straight-forward (and which include horizontal asymptotes as a special case). There are, however, two demons lurking, one conceptual and one computational.

Cambridge begins with the example

$\color{blue}\boldsymbol{f(x) = \frac{8x^2 -3x+2}{x}\,.}$

The function can be rewritten as $\boldsymbol{f(x) = 8x -3 \ + \ 2/x}$

, and it seems simple enough: a similar “when $\boldsymbol{x}$

is huge, the leftovers are tiny” intuition leads us to declare that $\boldsymbol{f(x)}$

has the linear asymptote $\boldsymbol{y = 8x -3}$

. The trouble is that the example, and the discussion that follows, are too simple, so that the demons remain hidden.

3. A Conceptual Demon

How do we formally (or semi-formally) express that a function has a linear asymptote? For the example above, Cambridge writes “f(x) will approach the line y = 8x – 3”. This is natural, intuitive and sufficiently clear. It is considering the asymptote to be a geometric object, as the line, which is fine at the school level.

What, however, if we want to think of the asymptote as the function $\boldsymbol{g(x) = 8x -3}$ ? The problem is that both $\boldsymbol{f(x)}$ and $\boldsymbol{8x-3}$ are zooming off to infinity, which means that writing $\boldsymbol{f(x) \to 8x -3}$ , or anything similar, is essentially meaningless. Those arrows are shorthand for limits and, fundamentally, the limit of a function must be a number, not another function. (For vertical asymptotes we consider infinity to be an honorary number, which may seem dodgy, but which can be justified.)

The obvious way around this problem is simply to ignore it. As long as we stick to asymptotes of reasonably simple functions – rational functions and carefully chosen others – Cambridge’s intuitive approach is fine. But in the, um, unlikely event that our situation is not so simple, then we have to carefully consider the limiting behaviour of functions. So, we’ll continue.

Suppose that the function $\boldsymbol{F(x)}$ appears to have the linear function $\boldsymbol{L(x)}$ as asymptote. To capture this idea, the simple trick is to consider the difference of the two functions. If $\boldsymbol{x}$ is huge then this difference should be tiny, and so $\boldsymbol{L(x)}$ being the asymptote to $\boldsymbol{F(x)}$ is captured by the limit statement

$\color{blue}\boxed{\boldsymbol{F(x)- L(x)\ \longrightarrow \ 0}}$

Again, none of this is really necessary for capturing linear asymptotes of simple functions. But it is necessary, at least as underlying guidance, if we wish to consider less simple functions and/or more general asymptotes.

WARNING: It may seem as if our boxed definition of linear asymptote $\boldsymbol{L(x)}$ would work just as well for $\boldsymbol{L(x)}$ non-linear, but there is a trap. There is another, third demon to deal with. Before that, however, our second demon.

4. A Computational Demon

It is very import to understand that the simple, “highest powers” trick we indicated above for spotting horizontal asymptotes does not work for general linear asymptotes. Cambridge‘s introductory example $\boldsymbol{f(x)}$ illustrates the issue, albeit poorly and with no subsequent examples. For a better illustration, we have Exercise 12 in Section 6G, which presents us with the function

$\color{blue}\boldsymbol{m(x) = \frac{4x^2 + 8}{2x +1}\,.}$

Here, it would be invalid to divide the $\boldsymbol{4x^2}$

by the $\boldsymbol{2x}$

and then declare the linear asymptote to be $\boldsymbol{y =2x}$

. The problem is the $\boldsymbol{+1}$

in the denominator has an effect, and we are forced to perform long division or something similar to determine that effect. So, $\boldsymbol{m(x) = 2x -1 \ + \ 10/\left(2x +1\right)}$

, and then it is clear that the linear asymptote is $\boldsymbol{y =2x -1}$

All that is fine, as far as it goes. Cambridge routinely performs long division to correctly determine linear asymptotes (including horizontal asymptotes, for which the simpler highest powers trick would have sufficed). What, however, if the function is not rational? Consider, for example, Example 31 above. Can we safely ignore the $\boldsymbol{-1}$ in the denominator of the function, as Cambridge has done? And, if so, why?

In order to stay in the linear world for now, we’ll leave Example 31 and instead consider two other examples:

$\color{blue}\boldsymbol{p(x) = \frac{x^2}{\sqrt{x^2 -1}} \qquad\qquad q(x) = \frac{x^2}{\sqrt{x^2 -x}}\,.}$

So, are we permitted to ignore the $\boldsymbol{-1}$

and the $\boldsymbol{-x}$

in the denominators of $\boldsymbol{p(x)}$

and $\boldsymbol{q(x)}$

? It turns out that the answers are “Yes” and “No”: the function $\boldsymbol{p(x)}$

has linear asymptote $\boldsymbol{y =x}$

, but $\boldsymbol{q(x)}$

has linear asymptote $\boldsymbol{y =x +\frac12}$

The behaviour of $\boldsymbol{p(x)}$ and $\boldsymbol{q(x)}$ is, of course, anything but obvious. In particular, the “highest powers” and long division tricks are of no assistance here. Moreover, similar difficulties arise with the analysis of non-horizontal asymptotes of pretty much all rooty functions. It is essentially impossible for a Victorian school text to cover non-horizontal asymptotes of non-rational functions. The text must either work very hard, or it must cheat very hard; Cambridge takes the latter approach.

5. A Nonlinear Demon

Almost there. We have non-linear “asymptotes” left to consider, which are fundamentally demonic. This is demonstrated by Cambridge‘s very first example, Example 26, which considers the function

$\color{blue}\boldsymbol{r(x) = \frac{x^4 +2}{x^2}\,.}$

Dividing through by $\boldsymbol{x^2}$

gives $\boldsymbol{r(x) = x^2 + 2/{x^2}}$

, after which Cambridge baldly declares “The non-vertical asymptote has equation $\boldsymbol{y = x^2}$

“. But why, exactly? All we have to go on is Cambridge‘s intuitive approach to $\boldsymbol{f(x)}$

above. Intuitively, the function $\boldsymbol{r(x)}$

approaches the parabola $\boldsymbol{y = x^2}$

. That’s fine, and we can formalise it by setting $\boldsymbol{Q(x)=x^2}$

. Then guided by our boxed definition above, we can write $\boldsymbol{r(x) - Q(x) \ \to \ 0}$

, and we seem to have our asymptote.

The problem is that there are a zillion functions that will fit into that blue box. It is also true, for example, that $\boldsymbol{r(x) - r(x) \ \to \ 0}$ . This means that $\boldsymbol{r(x)}$ itself just as good an asymptote as $\boldsymbol{Q(x)}$ . So, yes, one can reasonably declare that $\boldsymbol{r(x)}$ is asymptotic to $\boldsymbol{Q(x)}$ , but we cannot declare $\boldsymbol{Q(x)}$ to be THE asymptote to $\boldsymbol{r(x)}$ .

Of course we want asymptotes to be unique, so what do we do? There are two ways out of this mess, the first solution being to restrict the type of function that we’ll permit to be an asymptote. That’s intrinsically what we did when considering horizontal and linear asymptotes, and it is exactly why the multiple asymptote problem didn’t arise in those contexts: linear functions only had to compete with other linear functions. Now, for general rational functions, we must broaden the notion of asymptote, but it’s critical to not overdo it: we now permit polynomial asymptotes, but nothing more general. Then, every rational function will have a unique polynomial (possibly linear, possibly horizontal) asymptote.

But now, what do we do for non-rational functions? These functions may naturally have rooty asymptotes, as illustrated by Example 31 above. Such functions can also be dealt with, by suitably and carefully generalising “polynomial asymptote”, but it’s all starting to get fussy and annoying. It is making a second solution much more attractive.

That second way out of this mess is to just give up on unique nonlinear “asymptotes”. Then, our blue box becomes a definition of two functions being “asymptotically equivalent”. We don’t get unique asymptotes, but we do get a clear way to think about the asymptotic behaviour of functions. (You can’t always get what you want, But if you try sometimes, you just might find …)

One final comment. The computational demon we mentioned above is of course still around to bedevil non-linear asymptotes. This is well illustrated by Exercise 17, which considers the function

$\color{blue}\boldsymbol{s(x) = \frac{x^2 + x + 7}{\sqrt{2x+1}}\,.}$

As discussed by commenters below, this exercise used to have a part (f), asking for the vertical asymptote. That part was deleted in later editions, and it is easy to guess why; the $\boldsymbol{+1}$

in the denominator of $\boldsymbol{s(x)}$

turns out to matter, so that the “highest powers” technique that Cambridge invalidly employs on such examples gives the wrong answer.

6. The Devil in Cambridge‘s detail

Finally. We can now deal with the Cambridge excerpt above. There is no need to write at length here, since everything follows from the discussion above, as indicated. We begin with a clarification of the punchline:

The claim $\color{red}y \to \frac{x}{\sqrt{x}}}$ in Example 31 is triply flawed: it is meaningless (conceptual demon); the calculation is invalid (computational demon); and the conclusion is wrong (nonlinear demon).

To be fair, it turns out that the -1 in the denominator of the function in Example 31 can be ignored, but not because of the highest powers trick that Cambridge appears to employ. This also means y = √x will be the non-linear asymptote to y = (x + 1)/√(x-1) if we suitably generalise the notion of “polynomial asymptote”.

That’s plenty wrong, of course. But to round off, here are the other problems with the excerpt:

The “graph of y = f(x)” doesn’t approach anything. It is the function that approaches the asymptote.
Underlying the punchline, the definition of “non-vertical asymptote” is hopelessly vague and does not remotely mean what Cambridge thinks it means.
We do not “require √(x – 1) > 0; we require x – 1 > 0.

And, we’re done. Thank Christ.

Numberphile and the Cult of Collegiality

by marty

February 25, 2018

4:34 pm

4 Comments on Numberphile and the Cult of Collegiality

media

analysis, infinite series, mathematics education, Numberphile, Riemann zeta function

Mathologer recently posted a long video addressing the “proof” by Numberphile of the “astounding result” that 1 + 2 + 3 + … = -1/12. As well as carefully explaining the underlying mathematical truth, Mathologer tore into Numberphile for their video. Mathologer’s video has been very popular (17K thumbs up), and very unpopular (1K thumbs down).

Many who objected to Mathologer’s video were Numberphile fans or semi-literate physicists who were incapable of contemplating the idea that Numberphile could have gotten it wrong. Many others, however, while begrudgingly accepting there were issues with the Numberphile video, strongly objected to the tone of Mathologer’s critique. And it’s true, Mathologer’s video might have been improved without the snarky jokes from that annoying cameraman. (Although, awarding Numberfile a score of -1/12 for their video is pretty funny.) But whining about Mathologer’s tone was mostly a cheap distraction from the main point. Fundamentally, the objections were to Mathologer’s engaging in strong and public criticism, to his lack of collegiality, and these objections were ridiculous. Mathologer had every right to hammer Numberphile hard.

Numberphile’s video is mathematical crap and it continues to do great damage. The video has been viewed over six million times, with the vast majority of viewers having absolutely no clue that they’ve been sold mathematical snake oil. Numberphile made a bad mistake in posting that video, and they’re making a much worse mistake in not admitting it, apologising for it and taking it down.

The underlying issue, a misguided concern for collegiality, extends far beyond one stupid video. There is so much godawful crap around and there are plenty of people who know it, but not nearly enough people willing to say it.

Which brings us to Australian mathematics education.

There is no shortage of people happy to acknowledge privately their frustration with or contempt for the Australian Curriculum, NAPLAN, VCE, AMSI, AAMT, MAV, teacher training, textbooks, and on and on. Rarely are these people willing to formally or publicly express any such opinions, even if they have a natural platform for doing so. Why?

Many feel that any objection is pointless, that there is no hope that they will be listened to. That may well be true, though it may also be self-fulfilling prophecy. If all those who were pissed off spoke up it would be pretty noisy and pretty difficult to ignore.

More than a few teachers have indicated to us that they are fearful of speaking out. They do not trust the VCAA, for example, to not be vindictive. To us, this seems far-fetched. The VCAA has always struck us as petty and inept and devoid of empathy and plain dumb, but not vengeful. The fear, however, is clearly genuine. Such fear is an argument, though not a clinching argument, for remaining silent.

It is also clear, however, that many teachers and academics believe that complaining, either formally or publicly, is simply not nice, not collegial. This is ridiculous. Collegiality is valuable, and it is obviously rude, pointless and damaging to nitpick over every minor disagreement. But collegiality should be a principle, not a fetish.

At a time when educational authorities and prominent “experts” are arrogantly and systemically screwing things up there is a professional obligation for those with a voice to use it. There is an obligation for professional organisations to encourage dissenting voices, and of course it is reprehensible for such organisations to attempt to diminish or outright censor such voices. (Yes, MAV, we’re talking about you, and not only you.)

If there is ever a time to be quietly respectful of educational authority, it is not now.

There’s Madness in the Methods

by marty

November 15, 2017

2:11 pm

3 Comments on There’s Madness in the Methods

VCE

analysis, Australia, differentiation, exams, limits, Mathematical Methods, school mathematics, VCAA, VCE

Yes, we’ve used that title before, but it’s a damn good title. And there is so much madness in Mathematical Methods to cover. And not only Methods. Victoria’s VCE exams are coming to an end, the maths exams are done, and there is all manner of new and astonishing nonsense to consider. This year, the Victorian Curriculum and Assessment Authority have outdone themselves.

Over the next week we’ll put up a series of posts on significant errors in the 2017 Methods, Specialist Maths and Further Maths exams, including in the mid-year Northern Hemisphere exams. By “significant error” we mean more than just a pointless exercise in button-pushing, or tone-deaf wording, or idiotic pseudomodelling, or aimless pedantry, all of which is endemic in VCE maths exams. A “significant error” in an exam question refers to a fundamental mathematical flaw with the phrasing, or with the intended answer, or with the (presumed or stated) method that students were supposed to use. Not all the errors that we shall discuss are large, but they are all definite errors, they are errors that would have (or at least should have) misled some students, and none of these errors should have occurred. (It is courtesy of diligent (and very annoyed) maths teachers that I learned of most of these questions.) Once we’ve documented the errors, we’ll post on the reasons that the errors are so prevalent, on the pedagogical and administrative climate that permits and encourages them.

Our first post concerns Exam 1 of Mathematical Methods. In the final question, Question 9, students consider the function $\boldsymbol{ f(x) =\sqrt{x}(1-x)}$ on the closed interval [0,1], pictured below. In part (b), students are required to show that, on the open interval (0,1), “the gradient of the tangent to the graph of f” is $(1-3x)/(2\sqrt{x})$ . A clumsy combination of calculation and interpretation, but ok. The problem comes when students then have to consider tangents to the graph.

In part (c), students take the angle θ in the picture to be 45 degrees. The pictured tangents then have slopes 1 and -1, and the students are required to find the equations of these two tangents. And therein lies the problem: it turns out that the “derivative” of f is equal to -1 at the endpoint x = 1. However, though the natural domain of the function $\sqrt{x}(1-x)}$ is [0,∞), the students are explicitly told that the domain of f is [0,1].

This is obvious and unmitigated madness.

Before we hammer the madness, however, let’s clarify the underlying mathematics.

Does the derivative/tangent of a suitably nice function exist at an endpoint? It depends upon who you ask. If the “derivative” is to exist then the standard “first principles” definition must be modified to be a one-sided limit. So, for our function f above, we would define

$f'(1) = \lim_{h\to0^-}\frac{f(1+h) - f(1)}{h}\,.$

This is clearly not too difficult to do, and with this definition we find that f'(1) = -1, as implied by the Exam question. (Note that since f naturally extends to the right of x =1, the actual limit computation can be circumvented.) However, and this is the fundamental point, not everyone does this.

At the university level it is common, though far from universal, to permit differentiability at the endpoints. (The corresponding definition of continuity on a closed interval is essentially universal, at least after first year.) At the school level, however, the waters are much muddier. The VCE curriculum and the most popular and most respected Methods textbook appear to be completely silent on the issue. (This textbook also totally garbles the related issue of derivatives of piecewise defined (“hybrid”) functions.) We suspect that the vast majority of Methods teachers are similarly silent, and that the minority of teachers who do raise the issue would not in general permit differentiability at an endpoint.

In summary, it is perfectly acceptable to permit derivatives/tangents to graphs at their endpoints, and it is perfectly acceptable to proscribe them. It is also perfectly acceptable, at least at the school level, to avoid the issue entirely, as is done in the VCE curriculum, by most teachers and, in particular, in part (b) of the Exam question above.

What is blatantly unacceptable is for the VCAA examiners to spring a completely gratuitous endpoint derivative on students when the issue has never been raised. And what is pure and unadulterated madness is to spring an endpoint derivative after carefully and explicitly avoiding it on the immediately previous part of the question.

The Victorian Curriculum and Assessment Authority has a long tradition of scoring own goals. The question above, however, is spectacular. Here, the VCAA is like a goalkeeper grasping the ball firmly in both hands, taking careful aim, and flinging the ball into his own net.

UPDATE (20/09/20)

Above, we hammered Q9(c) on the 2017 Mathematical Methods, Exam 1. We regret not having hammered also the idiotically misleading diagram, but another issue has arisen, pointed out to us by frequent commenter SRK.

In Q9(b), students were asked to show that the derivative of $\boldsymbol{ f(x) =\sqrt{x}(1-x)}$ is $\boldsymbol{ (1-3x)/(2\sqrt{x})}$ . as we noted, the question was pointlessly verbose in classic VCAA style, but no big deal; an easy 1-mark question. What could go wrong?

Well, what went wrong is that 2/3 of students scored 0/1 on this very easy question. How? The Examination Report explains:

When answering ‘show that’ questions, students should include all steps to demonstrate exactly what was done, but many students often left steps out. A common pattern was to go straight from the first line of differentiation immediately to the final line, with no indication of obtaining a common denominator.

For fuck’s sake.

The stark incompetence of VCAA is often stunning. And, the nasty, meaningless pedantry of the VCAA is often stunning. But, on a question like this, when you see the two in seamless combination, that’s when you realise that you’re in the presence of true greatness.