57 Replies to “WitCH 53: Parting Ways”

  1. I have a few minor gripes with the text (and many other texts) with this material, but my big gripe is the total lack of logic (in fact, it is a contradiction, the way it is written):

    Since [equation] is true for all real numbers *except* a and b, it *must be true* for x = a and x = b…

    Total rubbish!

    OK, some form of “cheating” is required for partial fractions work, but surely we can admit that we are cheating… I feel this has been discussed before at some length!

    1. I have trouble seeing why the statement quoted by Red Five deserves this condemnation (assuming that it refers to the glossary of Example 13, Method 2): It is a sensible statement to exclude the two values x=-3 and x=1 after having multiplied away the denominators, and to note “afterwards” that the resulting linear equation is additionally valid for those two values. At least this makes sense if it is assumed to be known to the students that (i) the partial fraction decomposition does exist and (ii) lines which share two points (here, the points with abscissa values -3 and 1) must necessarily coincide.

      1. I kind of think you’re both right. But, given the style of this blog, maybe I’ll declare you both wrong.

        I agree RF is being too harsh, and yes the plugging in business can be justified (for two factors) in the manner Christian suggests. But I think it’s fair to say that the vast majority of Methods students (and more than a few Methods teachers) would not be able to make heads or tails of the text’s “explanation”.

        1. Re: Substitution of x = -3 and x = 1. I’ll usually argue something like:

          For the two expressions to be equal it’s required that the numerators 3x + 5 = A(x + 3) + B(x-1). 3x + 5 is defined for all x \in R therefore A(x + 3) + B(x-1) must be defined for all x \in R. In particular it’s convenient to consider equality of each numerator when x = -3 and x = 1.

          And then I’ll note that the expressions themselves are not defined for x = -3 and x = 1.

          1. Hi, JF. I’m not sure your approach is valid, since it is only required that the numerators be defined and equal for the non-evil values of x. I think if one is going to justify the substituting, it has to be something along the lines Christian suggests.

            I’m not overly fussed if teachers worry too much about justifying the plugging in approach, although it’s not so hard to give a sense of it. Much worse is for teachers to use their their nervous and understandable lack of understanding as an excuse to stick to the simultaneous equations nonsense. That used to be pretty standard, in texts and classes, and I imagine it’s still common, but I think (and hope) it’s dying out.

            But then, the question is, how do you present and handle the plugging in method, both when it works nicely and when it works less nicely? Hence, the WitCH.

            1. I see what you’re saying, my argument is that the left side numerator doesn’t care about the evil
              values, it’s fine for all values. So the right side numerator is required to be ok with evil values too, otherwise equality of the numerators cannot be claimed. I don’t like the “afterwards” bit, what’s being said afterwards is the same as what’s being said up front before hand.

              1. Hi, JF. It still seems to me there is something missing. Suppose you have two functions, f and g, defined for all values of x, and that we know f(x) = g(x) except for a few evil x. How do we then know f = g for those evil x as well?

                1. We know that the *numerator* on the left is defined for all values of x including evil values. So the numerator on the right has to be defined for all values of x including the evil values. Once the numerators are equated the problem is to find the values of A, B, C etc. from that sole equation, and that equation is defined for all values of x including the evil values.

                    1. If I gave you 3x + 5 = A(x + 3) + B(x-1) and asked you to solve for the values of A and B, would you worry about where the equation came from and whether x = 3 and x = -1 might be evil, or would you go ahead and substitute those values, comfortable in the knowledge that the left hand side is defined for all values of x and so the right hand side must also be defined?

                      In other words, once we have 3x + 5 = A(x + 3) + B(x-1), is the idea of evil values relevant anymore?

                    2. OK, here’s what I’m trying to say:

                      When you get to the stage

                      3x+5=A(x+3)+B(x  - 1)

                      all you’re trying to do is determine a polynomial identity, that is, what A and B will make A(x+3)+B(x - 1) identical to 3x+5 for all values of x including x=−3 and x = 1.

                      Once you’ve got those values you divide both sides of the polynomial identity by (x+3)(x - 1) with the added restriction that x \neq -3, 1. So this gives you

                      \displaystyle \frac{3x+5}{(x+3)(x - 1)} = \frac{A(x+3)+B(x-1)}{(x+3)(x - 1)} = \frac{A}{x-1}+\frac{B}{x+3}

                      which holds for all x \neq -3, 1 (which is what lets you cancel on the RHS).

            2. Re: “Much worse is for teachers to use their their nervous and understandable lack of understanding as an excuse to stick to the simultaneous equations nonsense.”

              I think you’ll find that a lot of teachers use the ‘plugging in approach’ and just don’t care about justifying the use of evil values. The only justification they give is that it works so just do it …

              1. Hi, JF.

                Of course I’m happier if teachers are justifying formulas and statements and methods. Or even just trying to justify them. Or even just acknowledging that a justification is required. Many teachers regularly fall short of even the last one, a consequence of the textbooks regularly falling short, in turn a consequence of the VCAA not giving a flying fuck. But the justification of plugging for partial fractions, although poorly done in the text above, was not my motivation for this WitCH.

                With hindsight, I can see why RF caught on that aspect of the text excerpt, it’s well worth discussing, and anyway I’m happy for the discussion go wherever people want it to go. (Within bounds: praising of ScoMoFo is verboten.) There are other aspects of the excerpt, however, that bother me way, way more.

      2. I’m not saying the conclusion is wrong. It is the logic used to reach the conclusion I have a problem with.

  2. In example 13, it might be better to argue as follows.

    If we can find constants A and B such that (2) is true for all x, then (1) will be true for all x other than 1 and -3.

    1. Why do we need to mention (1) not being true for 1 and -3 at all?

      In this process, it doesn’t seem to help at all.

      1. I don’t think it’s a matter of helping, but a matter of a hurdle to overcome. We only know (1) is true for non-evil x, and so we only know the consequence (2) for non-evil x. But then we want to plug in evil x.

      1. If you are referring to my comment, I do not see my comment as a sleight of hand. As far as I can see this is a valid argument, even though students will not meet this style of argument often.

  3. OK… this may sound *really* stupid (more so than some of my other recent statements…)

    BUT

    Could we get around the issue by considering the limit as x approaches a instead of substituting x = a?

    I haven’t yet had the head space to consider whether or not this actually works.

    1. Hi RF, using limits looks good to me. The only snag is that partial fraction decomposition is taught (in Specialist Unit 1) well before limits are first taught (in Methods Unit 2). But I really like the idea of using limits in Specialist Unit 3.

  4. So… (and this is perhaps the critical point) IF we use limits (but perhaps don’t tell students at first that is what we are doing), is the rest of the process mathematically valid?

    I’ve done a few scribbles and I *think* it works.

    This doesn’t mean there isn’t a better way though!

    The Specialist Unit 1 students I teach these days seem to be 90% Methods Unit 3 at the same time, so should remember limits (I know it is “in the course” for Unit 2 Methods, but that doesn’t always mean they are taught/remembered)

    1. Hi, RF. The limits thing is a good idea (modulo that it’s in a way different part of the curriculum). But, you can’t always take limits of functions. Why can you take limits here?

      1. Because the limit is the same if you approach from both sides. Thought that was obvious?

        Or… have I missed something?

        1. No problem with limits, RF. Once the numerators are equated, you’re dealing with polynomials and the limit x \rightarrow a (a \in R) of a polynomial always exists.

          1. JF, I think this is the key point. You need to note that the functions on both sides of (2) are polynomials or are continuous or something.

            There’s a further issue in justifying the plugging in method, but I’ll leave it there for now.

            1. My issue with this is that limits are disjoint from this treatment. You can’t really get around using them in some form, but you can get around using the student-meaningless terms “algebra of continuous functions” and/or “continuity of polynomials” by just subbing in successive approximations. You do need to still take a limit, but having students do this hands-on with all the terms makes it better IMO for their learning.

            2. I agree. When I wrote “(ii) lines which share two points (here, the points with abscissa values -3 and 1) must necessarily coincide.”, I was first thinking about writing “polynomials of degree one”, which is what the two sides of (2) are. I admit that this would have been a bit bombastic, especially for a classroom.

              But the message that is in action here – we fill a gap in a restricted domain of definition through some knowledge of the functions at hand, or we extend an equality known initially for some x to all x – is one which is certainly good to know at least to teachers, whatever small ounce of this they may be able to get across to students. I am not familiar with the curriculum but this may come handy when examining power series, for example. Outside of secondary school, complex analysis (where x is replaced by z) contains such reasoning in great abundance.

              1. Right. (There are many others too!) Not only that, but well before, when defining limits (hopefully in high school) the limit point does not need to be in the domain. It’s a punctured neighbourhood. That’s built into the definition of a limit and this would be good practice at seeing why the definition is that way in action.

            1. Thanks, JF. They’re nice notes. Knill is always worthwhile, if only because of his great collection of maths movie clips. It’s a little odd in the presentation, since the [l’]Hopital part is hidden. I’m also not sure that presented as a method for the given example (rather than sort of a justification for plugging in), it’s an improvement.

  5. Change of tack… One of my other gripes with the text here concerns the repeated linear factors and the irreducible quadratic factors. Mostly the former.

    I feel the text doesn’t do enough to justify *why* when there is a repeated linear factor there are two partial fraction parts, one with the linear factor and one with the square of the linear factor (and, in this case, the numerator is a constant, not a linear as is the case in other quadratic denominators).

    That said… I’ve never gone into much detail about the *why* in any of my classes on this point, hence it being a minor gripe.

    1. I get my class to try and figure out the partial fraction decomposition for any other ‘intuitively reasonable’ form.

    1. Thanks TM – I would never require students to use limits. However, trying to get my head around the bigger picture of why and how the process *really* works, I think they are a nice approach.

    2. Terry, your argument as far as it goes is simpler. But I’m not sure it’s simpler to see your argument is simpler.

      1. In a totally pragmatic sense, I do like Terry’s argument.

        When textbooks try to “show off” as it were, they more often than not make things worse (and often unreadable, hence 50+ WiTCHes), so I think simpler (if valid, which in this case it is) is best.

  6. Slight side nitpick:
    The bit “cannot be reduced to linear factors” on the topic of irreducible quadratic factors. Shouldn’t that read “reduced to real linear factors”?

    1. Thanks, Sai. Yes, that’s a nitpick, and definitely not motivation for a WitCH, but it’s definitely an important point (in the general scheme of things).

  7. OK, no one has mentioned this, so I will:

    In the statement \frac{3x+5}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3} I would instead write \frac{3x+5}{(x-1)(x+3)} \equiv \frac{A}{x-1} + \frac{B}{x+3}.

    From which, I would then write 3x+5 \equiv A(x+3) + B(x-1) and since I have used the \equiv symbol, I have no issue in then saying let x = 1 to determine the value of A

    As with a lot of these WiTCHes, I feel I’ve missed the point though…

  8. RF: Some older, and venerable, books, use your approach. However, we don’t need \equiv. Let me elaborate on my point above.

    We want to find constants A, B such that

    \frac{3x+5}{(x-1)(x+3)} = \frac{A}{x-1} + \frac{B}{x+3} if x \neq 1,-3.

    So, we want constants A, B such that

    3x+5 = A(x+3) + B(x-1) if x \neq 1,-3.

    We can go further, and find constants A, B such that 3x+5 = A(x+3) + B(x-1) for all x, including 1,-3. They will suffice.

      1. I have always believed (perhaps wrongly) that equivalence was actually quite different to equality in that you solve an equation but you use an identity.

        Willing, as always, to be proven wrong by any of the superior minds who frequent this site. It is appreciated.

    1. OK… the textbook(s) I learned from in the 1990s used an equivalence sign whenever they wanted to say “for all x“. Were they wrong or just unnecessary? Genuinely curious.

      They were British books if that makes any difference.

      1. At least unnecessary and arguably wrong. I’ve been wanting to post on this for a long time. I’ll put it on the short list (so look back again in June).

      2. RF: Accepting your definition of the equivalence sign as meaning “for all x“, I suggest that we could not write

        \frac{3x+5}{(x-1)(x+3)} \equiv ...

        because the left-hand side is not defined for all x.

    2. The problem I have with this is that students may think it is a general principle when trying to simplify say 1/f(x)g(x). What is making this work is continuity and not mentioning it at all I think is a mistake.

      It’s giving a justification that works in this case, but isn’t useful when they need to do similar things later. Even if they aren’t caught by it, I kind of wonder if it is any better than just saying “yes, it looks dodgy. It’s actually ok.”.

      1. I’m not sure whether or not it’s better, but I’m pretty fine with it. My concerns with the text passage are way more pedestrian.

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