Math is Math! Why would they change Math?!

[Brother Oni]

I know exactly how Mr Incredible feels now. (https://www.youtube.com/watch?v=adjS3H8eYJ0) :smallsigh:

So my daughter has a maths question: "Show every term in this sequence is positive by completing the square or otherwise: n2 - 6n + 14"

I don't know completing the square, so I'm thinking 'quadratic equation formula, plug and chug':

(-b ± SQRT(b2 -4ac)) / 2a

(- (-6) ± SQRT(-62 - (4x1x14) ) ) / (2x1)

(6 ± SQRT(36 - 56) ) / 2

(6 ± SQRT(-20) ) / 2

(6 ± 4.46i / 2)

Which means the answers are (6 + 4.46i)/2 and (6 - 4.46i)/2.

TLDR: I get two imaginary roots, which is probably not the answers they're looking for.

I've tried various maths resources online to explain completing the square, and I can't wrap my head around the answer I get from that ( (n - 3)2 + 5 = 0 ) as proof that all the terms are positive.

Help please? :smallfrown:

[Ninja_Prawn]

I think I remember this from school...

Completing the square means factorising it, and then seeing what is left over. n2 - 6n + 14 = (n - 3)(n - 3) + 5

Since the constant on the end is positive, all values in the sequence must be? :smallconfused: I can't remember the proof for that, but it must be true, because the question implies it is. According to this (https://www.mathsisfun.com/algebra/completing-square.html), if a quadratic equation is expressed as (x + a)2 = b, the x/y co-ordinates of the vertex are (-a,-b). So if b is negative the curve is all above 0.

That said, your approach with the imaginary roots should also be valid, since the question allows "or otherwise".

[Jay R]

First, this is not an equation. The quadratic formula applies to an equation of the form ax2 + bx + c = 0.

This is an expression, not an equation. There's no "= 0" part.

The expression n2 – 6n + 14 is the same as n2 – 6n + 9 + 5.

Since (n – 3)2 = n2 – 6n + 9, this is the same as (n – 3)2 + 5

Any square number is 0 or positive. Add 5 to the square number at you get something that is 5 or more, and therefore positive.

And, finally, they didn't change math. Completing the square is a technique that's been around at least as long as the quadratic formula, and was in fact the tool used to derive and prove the quadratic formula.

[Unavenger]

It's a sequence, not an equation - they're saying "For any value of n, this expression is positive. Prove this by completing the square or otherwise". You could also do so by induction, say.

(Actually, I'm not sure it's possible by induction, and even if it is, you'd only get rational values no matter how meticulous you were about the induction. But that's not the point.)

[Knaight]

First things first - two imaginary roots in the quadratic equation means that the system is either positive at all points or negative at all points, so setting the expression equal to 0 to perform the quadratic equation works fine; you just also need to solve for any individual point (or the limit at positive or negative infinity).

Next up, completing the square is just writing it down differently. Consider the expression ax2+bx+c, where a, b, and c are constants. Now consider that (sqrt(a)x+b/(2*sqrt(a)))2 can, through foiling be found to be equal to ax2+bx+b2/4a. Now consider that you can always add 0 to an expression without changing it, and that (b2/4a)-(b2/4a) is definitely 0. So, we add (b2/4a)-(b2/4a) to the original expression of ax2+bx+c and change that to (sqrt(a)x+b/(2*sqrt(a)))2-(b2/4a)+c. The portion that gets squared is obviously at least 0 for all normal numbers, so if -(b2/4a)+c is positive, clearly the whole expression is.

That -(b2/4a)+c term probably looks instinctively familiar to you, so let's do some things that won't change whether it is positive or negative at all. First off, multiply it by 4a/a, for (-b2+4ac)/a. Then, we can multiply it by -1 while now knowing that a negative result means a positive result, and we get (b2-4ac)/a, which is probably looking really familiar. Now we change the understanding slightly. That a in the denominator changes the sign if it's a negative value, but that's about all it can do to a sign, so we can take it out if we reframe getting a negative result as the result being the same sign as a (in graph terms this means the "bottom" of the curve is the same side of 0 as the "top", using non-technical terms), and we get (b2-4ac). That's the very term from the quadratic equation that you took the square root of to get imaginary numbers.

This forum needs an equation editor.


(Actually, I'm not sure it's possible by induction, and even if it is, you'd only get rational values no matter how meticulous you were about the induction. But that's not the point.)
You could. Moving from the vertex in either direction lets you prove that it is monotonically increasing by induction, a term above 0 that is monotonically increasing can't go below 0, this starts above 0, QED it is always above 0. It's a needlessly painful way of doing things though.

Personally my first instinct would be to note that this is technically a series but that the series will be entirely on a continuous function with the same equation, take the first derivative, solve for 0, solve the equation for that solution, then take the second derivative, note that it's a constant positive, and call it a day. That said if this problem is being put forth I'm pretty sure the kid isn't in a calc class.


I've tried various maths resources online to explain completing the square, and I can't wrap my head around the answer I get from that ( (n - 3)2 + 5 = 0 ) as proof that all the terms are positive.
That's because that's not the answer. It's just a change in expression, so there's no =0, just (n - 3)2 + 5. (n-3)2 is always 0 or positive so the expression is always above 5, and thus positive.

[eggynack]

You could. Moving from the vertex in either direction lets you prove that it is monotonically increasing by induction, a term above 0 that is monotonically increasing can't go below 0, this starts above 0, QED it is always above 0. It's a needlessly painful way of doing things though.
Does this qualify as a proof by induction? You're not drawing any direct relationship between an arbitrary element and a specific subsequent element so much as you are just identifying a quality of the function in a broad sense. Unless I'm missing something, this is taking place in the real numbers. Standard induction requires that you be working with something that can line up with the natural numbers, and even transfinite induction requires that the set be well ordered.

[Unavenger]

You could. Moving from the vertex in either direction lets you prove that it is monotonically increasing by induction, a term above 0 that is monotonically increasing can't go below 0, this starts above 0, QED it is always above 0. It's a needlessly painful way of doing things though.

Ah. I'm used to inductive arguments in the form "This is true for x=1, and also true for x=k+1 wherever it's true for x=k, therefore it is true for x in the natural numbers" and such, so I didn't consider doing it like that.

[DavidSh]

Yes, induction here would be more like: "This is true for n=3. If it is true for n=k, where k >= 3, then it is true for n=k+1 (do the calculation). If it is true for n=k, where k <= 3, then it is true for n=k-1 (again, do the calculation). Therefore it is true for all integer n."

[Caerulea]

Yes, induction here would be more like: "This is true for n=3. If it is true for n=k, where k >= 3, then it is true for n=k+1 (do the calculation). If it is true for n=k, where k <= 3, then it is true for n=k-1 (again, do the calculation). Therefore it is true for all integer n."
But you are not asked to prove it for all n∈Z, you are asked for n∈R. Thus, incrementing by one doesn't work.

Given f(x) = (n-3)2 + 5
Knaights' point was that you could start at n=3 (the vertex of the function), and notice that it evaluates to 5. Then, take some epsilon ε > 0 and notice that, for any n = 3 + ε, f(n) > 5, so the function is greater than 5. Also, for any ε, all ζ > ε cause f(3 + ζ) > f(3 + ε), so the function is strictly increasing after n=3.

[Radar]

But you are not asked to prove it for all n∈Z, you are asked for n∈R. Thus, incrementing by one doesn't work.

Given f(x) = (n-3)2 + 5
Knaights' point was that you could start at n=3 (the vertex of the function), and notice that it evaluates to 5. Then, take some epsilon ε > 0 and notice that, for any n = 3 + ε, f(n) > 5, so the function is greater than 5. Also, for any ε, all ζ > ε cause f(3 + ζ) > f(3 + ε), so the function is strictly increasing after n=3.
Actually, the question is about a sequence, so it is implied that n∈N. Nevertheless, there are many ways to show that every element of said sequence is positive and every one works.

[Caerulea]

Actually, the question is about a sequence, so it is implied that n∈N. Nevertheless, there are many ways to show that every element of said sequence is positive and every one works.
True. My mistake.

This whole discussion reminds me of a Tom Leher song, titled (appropriately) New Math (https://www.youtube.com/watch?v=UIKGV2cTgqA).

Hooray for ... New Math! New math! It won't do you a bit of good to review math!

[Douglas]

So my daughter has a maths question: "Show every term in this sequence is positive by completing the square or otherwise: n2 - 6n + 14"

I don't know completing the square, so I'm thinking 'quadratic equation formula, plug and chug':
Completing the square goes like this:
Given an expression of the form an2 + bn + c, first divide it all by a. In this case a is 1, so you can skip this step.
Divide b by 2 and then square it. I'll call the result d.
Form the expression n2 + bn + d, and tack on the addition of the value of c - d at the end, making it equivalent to the original expression.
Because of how you computed d, n2 + bn + d = (n + b/2)2 will always be true, so replace the first with the second.
You now have an expression consisting of a number squared (which will always be non-negative if you're working in the real numbers) and a constant. The value of this expression obviously cannot be less than the constant.


For your example:
n2 - 6n + 14
n2 - 6n + 9 + 5
(n - 3)2 + 5
This expression is always at least 5, so it's always positive.


TLDR: I get two imaginary roots, which is probably not the answers they're looking for.
That's what happens when the expression is never 0 for real number inputs. That plus evaluating the expression at any arbitrary spot to check whether it's positive or negative is enough to prove that the expression is positive (or negative) for all real number inputs.

[Knaight]

Does this qualify as a proof by induction? You're not drawing any direct relationship between an arbitrary element and a specific subsequent element so much as you are just identifying a quality of the function in a broad sense. Unless I'm missing something, this is taking place in the real numbers. Standard induction requires that you be working with something that can line up with the natural numbers, and even transfinite induction requires that the set be well ordered.

You can draw the direct relationships just fine - you're just splitting it into two sets, and I didn't cover how to actually do it at a detail level.

We'll call the vertex k=0. k=1 would be 1 away in either direction (you might want a quick symmetry proof), k=2 would be 2 away in either direction, so on and so forth.

We can see that the kth term is always k2+5. We also know that the (k+1)th term is k2+2k+1+5, which is the kth term +2k+1. The k=0 term is positive, 2k+1 is always positive, thus the k+1 term is greater than the k term, done.

[gomipile]

I don't understand the title. They taught us this in Algebra 2 at my high school back in the 1990s. I don't see how this is a change to math?

Maybe she's younger than high school, but that would still only imply a reshuffling of curriculum, not a change to its contents?

[eggynack]

You can draw the direct relationships just fine - you're just splitting it into two sets, and I didn't cover how to actually do it at a detail level.

We'll call the vertex k=0. k=1 would be 1 away in either direction (you might want a quick symmetry proof), k=2 would be 2 away in either direction, so on and so forth.

We can see that the kth term is always k2+5. We also know that the (k+1)th term is k2+2k+1+5, which is the kth term +2k+1. The k=0 term is positive, 2k+1 is always positive, thus the k+1 term is greater than the k term, done.
But isn't there also k=.5? Or k=pi? If k can take on non-integer values, then using only integer values isn't sufficient to prove the function is monotonically increasing.

[factotum]

I don't understand the title. They taught us this in Algebra 2 at my high school back in the 1990s. I don't see how this is a change to math?

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).

[Knaight]

But isn't there also k=.5? Or k=pi? If k can take on non-integer values, then using only integer values isn't sufficient to prove the function is monotonically increasing.

This was explicitly defined as a "series", which generally means integer values. That said, k doesn't actually need to be an integer for that to work in any way, so it would work just fine in a continuous case, with the notable exception of needing to prove that the entire k=0 to k=1 range is positive before induction really kicks in in the continuous sense (which is why induction is generally used for discrete math to begin with).

Granted, by the time you're even in position to prove by induction you can already just point to a term that gets squared added to a positive term and call it a day.

[jayem]

There's a proof by contradiction for well ordered sets that can be used more or less the opposite way to induction. You assume you have the least element that doesn't fit the bill and then create ANY lower one. This is probably easier to make watertight.
Classical proof by induction you know you get each number, if your elements are even the rationals that goes. I mean it's 'clearly' the case [basically what Knaight says]

This is a case where knowing and understanding the theory is different from just using it. By stopping the factorizing half way, you get a square of a real number (positive) plus a positive which is positive (as you did).
Alternatively if you do go straight for QF you could know that the roots are where the graph cuts the axis, and again it follows that it must be one side or another (assumming continuity and assorted things).

Personally I like completing the square as my first resort for finding the roots anyway, it means I don't need to do a temporary substitution when I may already have used x,a,b,c.

[Radar]

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).
Well, I for example didn't at first understand the phrase "completing the squares" either, but this was due to language difference. Back in my school days however we were tought all three forms to write down a quadratic expression and we did use all three to solve specific problems.

[Lord Torath]

Completing the square goes like this:
Given an expression of the form an2 + bn + c, first divide it all by a. In this case a is 1, so you can skip this step.
Divide b by 2 and then square it. I'll call the result d.
Form the expression n2 + bn + d, and tack on the addition of the value of c - d at the end, making it equivalent to the original expression.
Because of how you computed d, n2 + bn + d = (n + b/2)2 will always be true, so replace the first with the second.
You now have an expression consisting of a number squared (which will always be non-negative if you're working in the real numbers) and a constant. The value of this expression obviously cannot be less than the constant.
I was taught the Completing the Square method (7th grade - age 12-13), but I don't remember the algorithm ever being explicitly laid out like that by my teacher. Thanks, Douglas!

[eggynack]

This was explicitly defined as a "series", which generally means integer values.
Maybe, but I don't think there's anything in the expression being evaluated that only takes on integer values.


That said, k doesn't actually need to be an integer for that to work in any way, so it would work just fine in a continuous case, with the notable exception of needing to prove that the entire k=0 to k=1 range is positive before induction really kicks in in the continuous sense (which is why induction is generally used for discrete math to begin with).
I mean, yeah, I guess you're fine if you can prove this for that interval. That scenario is super bizarre though, where you can prove something for an uncountable infinity of cases without having already proved that thing for the higher numbers where it's also true. I wonder if that's ever occurred in all of math.

[Unavenger]

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).

I did A level maths (it's no longer a choice to do A levels [unless you want to do some other kind of further education], although you can very well decide not to do maths) and learned about completing the square. IIRC, it's the stuff of C1. My mother's a maths tutor, who has been teaching this stuff a long time, and she never mentioned completing the square as being a new thing in any sense.

[jayem]

There was a nice video on youtube that have completing the square done graphically for the case x*x+b*x=(-c) [where x, b, -c is positive]

You have your square + your (same widthed) rectangle.
Divide the rectangle in half (lengthways) and attach it to your square
Complete the square and add the same area to the other side
Reassemble the other side into a square somehow (if necessary by using explicit arithmetic, but if you can go via Pythagoras it shows it's more generally true)
Identify the two sides of the constructed squares as being the same (as the squares are the same area and square) and hence take off the known length to find 'x'

Which makes you identify with the Babylonians, which is kind of cool.
I think deriving a similar picture for negative bx (when b<x) is also just about intuitively understandable. When b>x I didn't like it so much

To me converting the two (eight?) pictorial cases to algebra seems to give a justification for the basic rules of negative numbers and it being much easier to treat them as genuine numbers.

[Brother Oni]

I don't understand the title. They taught us this in Algebra 2 at my high school back in the 1990s. I don't see how this is a change to math?

Because the exact curriculum and methods taught vary from era to era, country to country, school to school and even teacher to teacher.

I was never taught completing the square, but I went to public school in the UK in the 90s. Your experiences may vary.

Anyway, it wasn't a serious title, just that scene from The Incredibles 2 seemed apt.


Nevertheless, thank you all for the replies! I'll try and explain it to my daughter when she stops giggling at her old man running to the forums to complain. :smallsigh:

[Knaight]

Maybe, but I don't think there's anything in the expression being evaluated that only takes on integer values.
As it's explicitly defined as a sequence you can assume this holds for x, or more generally any domain term (technically you should explicitly define this, in practice that tends not to happen, similarly whether the first term is n=0 or n=1 should have been specified, though it doesn't actually matter). The very term "sequence" is a discrete math term.

The range here is also all integers as a result, but that's not a requirement at all. It could be octonions with irrational numbers for every term, and that would still be fine, though the very idea of trying proof by induction on an octonion is pretty terrible.


I mean, yeah, I guess you're fine if you can prove this for that interval. That scenario is super bizarre though, where you can prove something for an uncountable infinity of cases without having already proved that thing for the higher numbers where it's also true. I wonder if that's ever occurred in all of math.
It's a pretty textbook example of why proof by induction is a discrete math process generally. If you treat it as a discrete process (and again, sequences) you just need to explicitly show the k = 0 case and the rest takes care of itself.

*Which is what I should have said when I said series; it's just the sequence points on the parabola, not their sum.

[Algeh]

I don't know how this varies from state to state in the US, but I know I never learned anything about "completing the square" in my algebra lessons here in the UK, and I did maths up to and including A levels (the exams that are set for people who choose to stay on at school for two years after 16 and which are used to determine university entrance requirements).

In much, but not all, of the USA, it's currently in the high school Common Core standards as:



A.SSE.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.

b. Complete the square in a quadratic expression to reveal the maximum or minimum value of the function it defines.




A.REI.4 Solve quadratic equations in one variable.

a. Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x – p)² = q that has the same solutions. Derive the quadratic formula from this form.

and



F.IF.8 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function.

a. Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context.


Which is, generally speaking, taught in Algebra 1 near the end of the school year (if you actually get through the whole curriculum), touched on again the next year in Geometry since it's relevant to the equation of a circle:



G.GPE.1 Derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation.


(which is taught near the end of the year if you actually have time to get through the whole curriculum)

and then quickly gone over again by an aggravated teacher in Algebra 2 early in the year when it becomes clear that most of the class still doesn't know how to do it...

Or maybe that's just the case in districts that I've taught in.

(I currently teach all 3 of those classes, and we do indeed get to the end of the curriculum each year, but completing the square is still one of those things that most of my students don't seem to retain from year to year...the time I most wish they understood it is the geometry case, really, since that's the one I find hardest to end-run with other ways to deal with quadratics that they do understand and remember.)

[snowblizz]

Because the exact curriculum and methods taught vary from era to era, country to country, school to school and even teacher to teacher.

I was never taught completing the square, but I went to public school in the UK in the 90s. Your experiences may vary.

Anyway, it wasn't a serious title, just that scene from The Incredibles 2 seemed apt.


Nevertheless, thank you all for the replies! I'll try and explain it to my daughter when she stops giggling at her old man running to the forums to complain. :smallsigh:

I'm told you are by definition old when you try to do your kids homework and realise they've changed all the rules.


My mother was a teacher and every 15 years or so she'd complain how they "changed math". Some of the stuff she showed me to complain before she retired also made me go "huh? why would they try and explain it like that???".

At least it's not the 1970s "New Math".

Which ironically would be superbly adapted for today's computer age.

[Radar]

I can't beleive I only now remembered this (https://www.youtube.com/watch?v=UIKGV2cTgqA).

Enjoy. :smallsmile:

[Rockphed]

I can't beleive I only now remembered this (https://www.youtube.com/watch?v=UIKGV2cTgqA).

Enjoy. :smallsmile:

I was about to post that exact video. Tom Leher is great.

[OutOfThyme]

Because the exact curriculum and methods taught vary from era to era, country to country, school to school and even teacher to teacher.

I was never taught completing the square, but I went to public school in the UK in the 90s. Your experiences may vary.

Anyway, it wasn't a serious title, just that scene from The Incredibles 2 seemed apt.

It's kind of ironic, because the person responsible for coming up with the idea of completing the square (at least, IIRC, he was the first person to publish about it) is Al-Khwarizmi, who wrote the book that the word "algebra" is derived from.

I don't recall if that text is the one where he discusses completing the square.

[Grey_Wolf_c]

I can't beleive I only now remembered this (https://www.youtube.com/watch?v=UIKGV2cTgqA).

Enjoy. :smallsmile:

Wait, so this "new math" American keep complaining about is the borrow method of subtraction I was taught in a british-style school decades ago?

Bloody hell, America really is stuck in the 50s and refusing to be dragged kicking and screaming into the century of the fruitbat, isn't it?

I'm told below that it is from the 60s, so that makes sense.

Grey Wolf

[Khedrac]

Wait, so this "new math" American keep complaining about is the borrow method of subtraction I was taught in a british-style school decades ago?

Bloody hell, America really is stuck in the 50s and refusing to be dragged kicking and screaming into the century of the fruitbat, isn't it?

Grey Wolf

That being a Tom Lehrer video I would guess that it was recorded in the 60s (70s at the latest).

[Grey_Wolf_c]

That being a Tom Lehrer video I would guess that it was recorded in the 60s (70s at the latest).

Ah, fair enough. Never heard of the guy. But OK, so the new math is something else - presumably not the borrow method. WOuld make sense, that is a terrible way to do subtraction - counting up makes so much more sense when it comes to understanding what's actually going on.

Grey Wolf

[Tyndmyr]

I suggest that, as a fun way to explore problems, you just plug them into Wolfram Alpha. Now, you still need to know what you're looking for, but it handles a lot of the tedious calculation steps.

Glancing at the graph resulting from graphing the function would suffice to demonstrate if it is always positive or negative. There are shortcuts here, in that some equations are always going to have specific results. The pattern of n2 - 6n + 14 always ends up with a parabola, for instance. Therefore, if you're attempting to prove something's always positive, figuring out the minimal value would suffice.

[Grey_Wolf_c]

Therefore, if you're attempting to prove something's always positive, figuring out the minimal value would suffice.

This, however, runs into what I call the Barometer Issue (https://www.snopes.com/fact-check/the-barometer-problem/). You certainly can prove that this curve never dips under 0 by calculating its minimum... but this being homework, the purpose is to demonstrate the comprehension and applicability of the day's lesson, not that you have mastered a different lesson.

It ties in with the "change Math" thing - it is not random cruelty on the math teacher's part to insist that math problems be solved by the method they are teaching and not the way the parents are used to. There is often serious thought put into the methods, and teachers are right* to object to alternative methods being used, because the only way you can get through a curriculum when teaching multiple kids at once is for everyone to be on the same page.

Grey Wolf

*I mean, they can also be wrong, certainly, and cruel, definitely, but in this case, clearly the day's lesson is function manipulation into squares, not derivation, or whatever it was that you needed to calculate minimums and maximums.

[Caerulea]

I can't beleive I only now remembered this (https://www.youtube.com/watch?v=UIKGV2cTgqA).

Enjoy. :smallsmile:
I posted that upthread a bit ago.

This whole discussion reminds me of a Tom Leher song, titled (appropriately) New Math (https://www.youtube.com/watch?v=UIKGV2cTgqA).

Tangentially related to Math, but it annoys me so much when teachers insist on teaching an outdated model or view of the subject. Yes, you know it's wrong, I know it's wrong, but we are still going to spend a week on the Bohr model of the atom just because it is easier than explaining the quantum mechanics behind the correct model, which we are going to do anyway next week! It is infuriating.

[Radar]

I posted that upthread a bit ago.


Tangentially related to Math, but it annoys me so much when teachers insist on teaching an outdated model or view of the subject. Yes, you know it's wrong, I know it's wrong, but we are still going to spend a week on the Bohr model of the atom just because it is easier than explaining the quantum mechanics behind the correct model, which we are going to do anyway next week! It is infuriating.
Well, there are some merits to talk about Bohr's model first, since it introduces ideas important for quantum mechanics and it might be easier to understand the more accurate description of a hydrogen atom once you know Bohr's framework first. Besides, below university level any talk about quantum mechanics lacks the mathematical language needed to fully explain it, so being able to properly and fully describe an earlier empirical model has some advantage. Another reason is historical: Bohr's model made many other breakthroughs possible and after corrections by Sommerfeld was actually very accurate. As such, it would be good for any educated person to at least know abuot what Bohr did.

As a sidenote, I would like to say, that one can understand principles of quantum mechanics practically without any mathematics, since they can be derived from experiment (if you can get it somewhere, Dirac wrote a book on quantum mechanics with a fantastic introduction). Still, making any predictions with it requires university level knowledge.

[factotum]

Well, there are some merits to talk about Bohr's model first, since it introduces ideas important for quantum mechanics and it might be easier to understand the more accurate description of a hydrogen atom once you know Bohr's framework first.

Yeah. Complaining about that seems to me to be like complaining that they still teach Newtonian mechanics before (or even instead of) Einsteinian ones, and the simple answer there is that Newtonian mechanics are perfectly adequate for 99% of situations you'll ever encounter in real life. Similarly, the Bohr model of the atom answers most questions--you rarely need the quantum model.

[crayzz]

This, however, runs into what I call the Barometer Issue (https://www.snopes.com/fact-check/the-barometer-problem/). You certainly can prove that this curve never dips under 0 by calculating its minimum... but this being homework, the purpose is to demonstrate the comprehension and applicability of the day's lesson, not that you have mastered a different lesson.

It ties in with the "change Math" thing - it is not random cruelty on the math teacher's part to insist that math problems be solved by the method they are teaching and not the way the parents are used to. There is often serious thought put into the methods, and teachers are right* to object to alternative methods being used, because the only way you can get through a curriculum when teaching multiple kids at once is for everyone to be on the same page.

Grey Wolf

*I mean, they can also be wrong, certainly, and cruel, definitely, but in this case, clearly the day's lesson is function manipulation into squares, not derivation, or whatever it was that you needed to calculate minimums and maximums.

Aye, exactly.

"Everyone on the same page" is one concern; the other being that, if you can only solve a problem one particular way, you aren't very good at those kinds of problems. You're passably adequate, certainly, but if you only have one technique under your belt, you're going to be very limited at what you can do. If you want to learn a subject in depth, you should learn to approach it in a variety of ways.

In my experience, multiple methods are usually taught, and when it comes to evaluation, you'll be asked both "solve this problem" and "solve this problem using this particular method" type questions. So you'll have ample opportunity to solve problems with your preferred technique, but for perfect marks you'll need to learn a broader understanding of the subject.



Well, there are some merits to talk about Bohr's model first, since it introduces ideas important for quantum mechanics and it might be easier to understand the more accurate description of a hydrogen atom once you know Bohr's framework first. Besides, below university level any talk about quantum mechanics lacks the mathematical language needed to fully explain it, so being able to properly and fully describe an earlier empirical model has some advantage. Another reason is historical: Bohr's model made many other breakthroughs possible and after corrections by Sommerfeld was actually very accurate. As such, it would be good for any educated person to at least know abuot what Bohr did.

As a sidenote, I would like to say, that one can understand principles of quantum mechanics practically without any mathematics, since they can be derived from experiment (if you can get it somewhere, Dirac wrote a book on quantum mechanics with a fantastic introduction). Still, making any predictions with it requires university level knowledge.

The Bohr model is great because it's simple but still widely applicable; you can make a career out of chemistry and never stray out of using the Bohr model. Models are tools as much as they are theories, and not teaching the Bohr model because the quantum model is better is like not teaching someone how to use a hammer because nail guns exist.

Incidentally, the quantum models aren't "correct" in any strict sense. They're better models than the Bohr model, usually, but there's plenty of fudging and simplifying assumptions that go into those models. At the end of the day, they're just models, and you pick your model to match the level of precision you need; quite often, you just need the Bohr model. Like how you often just need a simple hammer, because hanging family photo's with a nail gun is just plain overkill.

[Radar]

Incidentally, the quantum models aren't "correct" in any strict sense. They're better models than the Bohr model, usually, but there's plenty of fudging and simplifying assumptions that go into those models. At the end of the day, they're just models, and you pick your model to match the level of precision you need; quite often, you just need the Bohr model. Like how you often just need a simple hammer, because hanging family photo's with a nail gun is just plain overkill.
The key difference is, Bohr made an empirical model without having knowledge why electrons have to obey such specific laws. Quantum mechanics is a general theory, which can be in particular used to predict behaviour and qualities of an hydrogen atom.

Also: for chemistry, you really need more then Bohr's model, since you cannot describe molecular bonds with it. Even for things like polarizability you Bohr's model does not give any clue - it was built empiricaly to do one thing and only one thing.

[Aran nu tasar]

In this case, clearly the day's lesson is function manipulation into squares, not derivation, or whatever it was that you needed to calculate minimums and maximums.

Interestingly, the two approaches are one and the same. The minimum of an upwards-pointing parabola is the vertex. You find the vertex by manipulating the equation of the parabola into the form y = a(x-h)2+k; the vertex is (h,k) when the equation is in this form. You do that manipulation by, essentially, completing the square.

If somebody starting trying to compute the minimum of the parabola using calculus, of course, that's serious overkill for the problem and not the approach the teacher wants the students to practice/demonstrate understanding of.

[Knaight]

If somebody starting trying to compute the minimum of the parabola using calculus, of course, that's serious overkill for the problem and not the approach the teacher wants the students to practice/demonstrate understanding of.

It's not the approach, but I'm not sure I'd call it overkill. Taking derivatives of polynomials is arguably a lot easier than the algebra involved, and you can also do it once and call it a day. Take the generic case:
ax2+bx+c

For calculus, take a derivative and set it to 0. That's just
d/dx(ax2+bx+c)=0
2ax+b=0
x = -b/2a

The generic case of taking the squares was solved upthread, and it was a whole lot more complicated than that.

[Kato]

Re outdated models :
It is absolutely sensible to use (partially wrong) models in school or even lower university courses. Bohr is a prime case, since understanding the quantum model requires math most people will never come across. I had a pretty good grasp of chemistry in school and we touched on the existence of orbitals without anything resembling quantum physics. It needs some handwaving but...
Even if you look at the better model right after, it's sensible to see what came before to understand where it comes from.

[Algeh]

Interestingly, the two approaches are one and the same. The minimum of an upwards-pointing parabola is the vertex. You find the vertex by manipulating the equation of the parabola into the form y = a(x-h)2+k; the vertex is (h,k) when the equation is in this form. You do that manipulation by, essentially, completing the square.

If somebody starting trying to compute the minimum of the parabola using calculus, of course, that's serious overkill for the problem and not the approach the teacher wants the students to practice/demonstrate understanding of.


It's not the approach, but I'm not sure I'd call it overkill. Taking derivatives of polynomials is arguably a lot easier than the algebra involved, and you can also do it once and call it a day. Take the generic case:
ax2+bx+c

For calculus, take a derivative and set it to 0. That's just
d/dx(ax2+bx+c)=0
2ax+b=0
x = -b/2a

The generic case of taking the squares was solved upthread, and it was a whole lot more complicated than that.

Well, sure, but if a student has a completing the square problem for homework they probably haven't seen calculus yet and won't for several years (at least in the system I teach in).

It reminds me of checking geometry homework about a decade ago. This was using a "spiraling" curriculum where homework problems would be a combination of stuff from a while ago, stuff we're doing right now, and stuff that will help you build intuition about things we'll study formally later in the year.

Anyway, one of the homework problems was, and this is not the exact wording, asking about the different possible number of intersections between a line and a circle. Since we haven't hit circles yet, students wouldn't have formal ideas about things like tangent and secant lines to work from, but it's perfectly reasonable to ask a kid to sit there with a picture of a circle and try to empirically see the different ways a line might or might not touch it. (Phrased the right way, I could probably get some reasonable conclusions out of elementary students about that, since it's a visual question that only relies on definitions of common things that most students "get".) The idea was to be building their intuition for a much later lesson about tangent and secant lines because they would have at least have thought about the visual relationship before hitting a bunch of formal math stuff.

This homework problem was mostly memorable because it taught me that pretty much my whole class was copying homework answers they didn't understand from the internet, because their answers all were the same verbatim explanation involving tangent and secant lines from the website they were all using to "check their work". Sure, that answer was technically correct, but it served no pedagogical purpose for them to copy it into their homework that night rather than try and think about what circles and lines look like. Fun times.

[Radar]

This homework problem was mostly memorable because it taught me that pretty much my whole class was copying homework answers they didn't understand from the internet, because their answers all were the same verbatim explanation involving tangent and secant lines from the website they were all using to "check their work". Sure, that answer was technically correct, but it served no pedagogical purpose for them to copy it into their homework that night rather than try and think about what circles and lines look like. Fun times.
This is one of the most difficult things to eradicate. I only ever tought at university level, but it was a huge problem with any kind of projects or even lab reports. Granted, I did have many more options to make students do their work, but still.

It also contributes to a bigger issue of remembering instead of understanding, but this is a different topic.

[Caerulea]

Re outdated models :
It is absolutely sensible to use (partially wrong) models in school or even lower university courses. Bohr is a prime case, since understanding the quantum model requires math most people will never come across. I had a pretty good grasp of chemistry in school and we touched on the existence of orbitals without anything resembling quantum physics. It needs some handwaving but...
Even if you look at the better model right after, it's sensible to see what came before to understand where it comes from.
I would agree, and perhaps overstated my frustration. It came mostly from spending far too long on something that should have been a half hour review of stuff that was done two years ago. Also, the issue with the quantum mechanical model is not math—if the term "Hamiltonian" was mentioned, 3/4ths of the class would immediately fall asleep or lose all interest—we covered it this week with a very basic overview.

With regards to people copying answers of the internet, is that a big problem? It seems to me that those doing that are wasting their education. Either figure it out yourself, or ask a friend to explain it in a way that you understand. I don't see why anyone would do different. Perhaps I am supremely naive.

[OutOfThyme]

With regards to people copying answers of the internet, is that a big problem? It seems to me that those doing that are wasting their education. Either figure it out yourself, or ask a friend to explain it in a way that you understand. I don't see why anyone would do different. Perhaps I am supremely naive.

Yeah, it is. I'm still in college, but the amount of copying people do on engineering work is astounding. Fortunately, we're at the point where the online sources have about a 50/50 shot of being right (even the official solutions are occasionally wrong). It's not ethically sound, but people who copy their homework tend to generally get screwed on tests. Some people have figured out that the best way to study for the test is to actually do the homework without taking any corners.

It's extremely wasteful, but people are lazy.

[snowblizz]

Perhaps I am supremely naive.
Unfortunately that is the correct answer.




It's extremely wasteful, but people are lazy.
The amount of time people do not spend studying their chosen degree is astounding at times.

I remember one university class where it really struck me. It was a short course with a visiting professor who was trying to engage the class and literally noone was responding. I was struck by the insight that we are all sitting here voluntarily purposefully avoiding to interact with someone who came here specifically to teach us something "special". Mind blowing stuff.

A symptom of chasing credits not knowledge in a world where having a degree is more important than actually knowing.

You see the same thing in grade school in the "when am I ever going to use XXX in real life?" question.

[Chen]

You see the same thing in grade school in the "when am I ever going to use XXX in real life?" question.

To be fair, a huge amount of university education that you get tested on, closed book, during exams will also never be used in your job. Either there will be a shortcut or you'll have programs/reference sheets to do almost all the difficult math work.

[Kato]

So, as has been pointed out, the problem with people who copy answers from online is that (usually) those are the people who think they can just cheat their way through the class. (of course there are other reasons, like terrible educators who give tasks that require you to use other sources because they suck at teaching you what you need for your homework)
And these people waste resources. They take away seats in classes, they take away time from their teachers (even more so because they tend to disrupt class) they have them check homework they didn't do. And all that effort is wasted and cannot go towards students who actually want to learn.

Also, this tangent is going pretty far away from the initial question.