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 2018-08-17, 00:40 #1 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22×3×37 Posts Introductory Calculus Discussion Thread My hardest class this year is AP Calculus AB. Here I’ll be posting questions and discussing some of the fundamentals that we learn. For the first few weeks/months our class will be focusing on pre-calculus concepts, like trigonometry and algebra fundamentals, and drawing parallels between simple concepts and their calculus counterparts. The first subject that we touched on was factoring polynomials. Most everything the teacher assigned was “apply formula” stuff, like sum/difference of squares/cubes, but the last one was neat, and was our first relation between simpler concepts and those of calculus: $(3x-2)^{-4}(x+3)+(x+3)^2(3x-2)^{-3}$This was confusing to most of the class, but she showed that, as we do with other polynomials (such as $$a^2b+b^2c$$), we should look for common factors between the terms. The way she did it, she factored out $$(3x-2)^-3(x+3)$$, the greatest common factor between the two terms. However, the resulting simplification was a lot of work and required tricky fractions of polynomials I talked with our teacher after class and asked her about factoring out $$(3x-2)^-4(x+3)$$, leaving nothing but a 1 for the first term and (theoretically) multiplying the second term by more than its exponent. This worked much better and avoided much use of fractions; she said that it made more sense, but she chose the otherwise because it might be confusing for the others who think that you are not allowed to factor out more than the greatest common factor. So at least I’m doing well so far! Last fiddled with by jvang on 2018-08-19 at 13:23 Reason: typing is hard
 2018-08-21, 04:10 #2 VBCurtis     "Curtis" Feb 2005 Riverside, CA 506610 Posts It's common when teaching factor-by-GCF to explain the method as: factor out the smallest exponent of each piece. In this case, -4 is the smallest exponent of that piece, rather than -3. I like your way better, and I think it's easier to teach than your teacher's way.
 2018-08-21, 04:21 #3 CRGreathouse     Aug 2006 3×1,993 Posts I would also have done it your way. But your teacher has more experience in teaching students with backgrounds like your fellow students', and she probably had good reason to teach the way she did. But good for you for seeing the other (standard) way of doing this.
2018-08-21, 22:27   #4
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

44410 Posts

Quote:
 Originally Posted by VBCurtis It's common when teaching factor-by-GCF to explain the method as: factor out the smallest exponent of each piece. In this case, -4 is the smallest exponent of that piece, rather than -3.
Quote:
 Originally Posted by CRGreathouse I would also have done it your way. But your teacher has more experience in teaching students with backgrounds like your fellow students', and she probably had good reason to teach the way she did. But good for you for seeing the other (standard) way of doing this.
Ohhhhhhh, I see the problem now. Around here, factoring-by-GCF is taught as factoring out the smallest exponent (how else would you do it?) but the kids automatically see “smaller” as “closer to zero” without consciously realizing it. If one were to ask them if $$-3 \leq -4$$ is true, most would not agree, but in the context of the problem where they pay less attention to that they forget.

Last fiddled with by jvang on 2018-08-21 at 22:29 Reason: typing is hard

 2018-08-23, 02:03 #5 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22×3×37 Posts Something neat we (re)covered in class regarding factoring. We were factoring $$x^3 - 2x^2 - 5x + 6$$, for which some basic methods do not work (and we haven't learned the cubic formula). Our teacher showed us a neat way to group the terms, by breaking them up into smaller, but still equivalent, terms:$x^3 - x^2 - x^2 + 1x - 6x + 6$From which we can factor out:$x^2(x - 1) - x(x - 1) - 6(x - 1)$$(x^2 - x - 6) (x-1)$Then we can factor the quadratic expression by normal means. Back in Algebra II, our teacher had us use synthetic division randomly until we found a term ($$x \pm n$$, $$n \in$$ Ints*. We would start from $$\pm 1$$ and proceed in both directions) that we could factor out. Very tedious, time-consuming, and inaccurate However, it seems to me that the grouping and the synthetic division solutions are only effective for "neat" higher-degree functions that factor with whole number solutions. Is the cubic formula the way to factor such functions regardless of their "neat"-ness, as with quadratics, or are there other ways? And for "neat" cubics, are the aforementioned methods effective or too situational/slower than another way? *: I was trying to figure out the fancy symbol for the set of integers but for some reason Google really doesn't like my computer and restricts access to searches after a very small amount of queries. It makes no sense; I wouldn't mind it if the recaptcha that it gives you worked, but it doesn't
2018-08-23, 03:35   #6
Dylan14

"Dylan"
Mar 2017

10010010102 Posts

Quote:
 Originally Posted by jvang *: I was trying to figure out the fancy symbol for the set of integers but for some reason Google really doesn't like my computer and restricts access to searches after a very small amount of queries. It makes no sense; I wouldn't mind it if the recaptcha that it gives you worked, but it doesn't
The symbol that you are referring to is the Zahlen symbol, which denotes the set of integers. If you use LaTex and you type the following:

Code:
$\mathbb{Z}$
it will produce this: $\mathbb{Z}$. That should be the correct symbol.
For future reference most of the symbols in mathematics (along with the LaTex commands to produce them) are listed here.
Attached Files
 Symbols.pdf (255.3 KB, 192 views)

Last fiddled with by Xyzzy on 2018-08-24 at 13:31 Reason: We have attached the referenced file.

2018-08-23, 05:12   #7
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

2×17×149 Posts

Quote:
 Originally Posted by jvang Back in Algebra II, our teacher had us use synthetic division randomly until we found a term ($$x \pm n$$, $$n \in$$ Ints*. We would start from $$\pm 1$$ and proceed in both directions) that we could factor out. Very tedious, time-consuming, and inaccurate However, it seems to me that the grouping and the synthetic division solutions are only effective for "neat" higher-degree functions that factor with whole number solutions. Is the cubic formula the way to factor such functions regardless of their "neat"-ness, as with quadratics, or are there other ways? And for "neat" cubics, are the aforementioned methods effective or too situational/slower than another way?
If a polynomial has a rational root, synthetic division and guess-and-check is the usual plan.
We can do better than randomly guessing, though: note that in your example, each root is a factor of the constant term of the original cubic. This happens every time the leading term is 1: every possible rational root must be a factor of the constant term. In your example, you would need check only 1,2,3,6 (positive and negative).

If the leading term is not 1, any factor of the leading term could be a denominator of a possible rational root. See "rational root theorem" on e.g. wiki for a better explanation.

2018-08-23, 23:09   #8
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22·3·37 Posts

Quote:
 Originally Posted by Dylan14 The symbol that you are referring to is the Zahlen symbol, which denotes the set of integers. If you use LaTex and you type the following: Code: $\mathbb{Z}$ it will produce this: $\mathbb{Z}$. That should be the correct symbol. For future reference most of the symbols in mathematics (along with the LaTex commands to produce them) are listed here.
Oh, cool. Thanks for the link!

Quote:
 Originally Posted by VBCurtis If a polynomial has a rational root, synthetic division and guess-and-check is the usual plan. We can do better than randomly guessing, though: note that in your example, each root is a factor of the constant term of the original cubic. This happens every time the leading term is 1: every possible rational root must be a factor of the constant term. In your example, you would need check only 1,2,3,6 (positive and negative). If the leading term is not 1, any factor of the leading term could be a denominator of a possible rational root. See "rational root theorem" on e.g. wiki for a better explanation.
Ohhhhhhh, I forgot about the rational root theorem We did use that back in Algebra II, rather than random guessing. But it sure was a lot of writing and seemed random! Perhaps I should mention that to our teacher, since (now that I remember correctly) it was pretty useful for cubics and whatnot.

2018-08-24, 13:42   #9
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

10,103 Posts

Quote:
 Originally Posted by jvang our teacher had us use synthetic division randomly until we found a term
Random Synthetic Division is playing on the Who Stage at Bonaroo in 2019.

2018-08-24, 23:46   #10
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22·3·37 Posts

Quote:
 Originally Posted by Uncwilly Random Synthetic Division is playing on the Who Stage at Bonaroo in 2019.
From what Google tells me, this seems very eccentric. Either that or I'm bad at reading

We touched on piecewise functions today (somehow I'm the only student who has used these before?), and someone asked some question that gave me a good excuse to bring up whether $$0.999... = 1$$ (alternatively $$0.\overline{9} = 1$$). Proofs that I've seen that support the equality include:$\frac{1}{3} = 0.\overline{3}, \frac{1}{3} * 3 = 1, 0.\overline{3} * 3 = 1$ and$x = 0.\overline{9}, 10x = 9.\overline{9}$Subtract $$x$$ from both sides to get$9x = 9, x = 1$.

My teacher had a refutation for both; for the first, she said that we are unable to express $$\frac{1}{3}$$ as a decimal quantity, so $$\frac{1}{3} \neq 0.\overline{3}$$ and the rest of that proof is unfounded.

I can go along with that, but I wasn't sure about the second one. She said that, since $$0.\overline{9}$$ consists of infinitely many decimal places, we are unable to accurately use it in operations, such as multiplication and subtraction. Are these refutations mathematically and/or logically sound?

Edit: I briefly mentioned one of the few other arguments against the equality back in the Learning to Learn thread here. However, that argument is based upon the hyperreal number set, which includes infinitely large and infinitely small quantities with the real numbers. It asserts that there is an infinitesimal difference $$h$$ between $$0.\overline{9}$$ and $$1$$.

Last fiddled with by jvang on 2018-08-24 at 23:52 Reason: typing is hard

2018-08-25, 01:33   #11
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

10,103 Posts

Quote:
 Originally Posted by jvang From what Google tells me, this seems very eccentric. Either that or I'm bad at reading
Many music bands have odd made up names. The trend is to have an adjective (or adjectival phrase) and a noun. Strawberry Alarm clock, Led Zeppelin, are 2 old examples. The game is: when someone says something that sounds like it could be a trendy band name, the hearer's response should be, "They are playing at [music festival]" or "I saw them at [trendy small club] last spring" or the like.

I have been collecting things that I have heard (often on a show or in a podcast) that sound like band names. Here are some examples:
Dolphin Rumspringa
Therapy Racoon
Sunglass Monocle
Jane Austin Truthers
Horses with Putin
Pareidolia and the Moon
Mexican Prison Pizza Party (yes those actual words in that order, in a sentence, with nothing between them)
Swedish Death Cleaning

I have many others. It is a fun and infectious game to play. Frequently when someone makes an allusion to something fun results can ensue. Completely new and made up example.
"So, Janet had a big rug hanging on the back wall of her she-shed. For a while I thought it was her Shawshank poster, she always seems to vanish when the baby needs changing."
"Shawshank Poster is opening up on Saturday at Bumbershoot this year."

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