20180817, 00:40  #1 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·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 precalculus 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: \[(3x2)^{4}(x+3)+(x+3)^2(3x2)^{3}\]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 \((3x2)^3(x+3)\), 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 \((3x2)^4(x+3)\), 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 20180819 at 13:23 Reason: typing is hard 
20180821, 04:10  #2 
"Curtis"
Feb 2005
Riverside, CA
5005_{10} Posts 
It's common when teaching factorbyGCF 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. 
20180821, 04:21  #3 
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.

20180821, 22:27  #4  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Quote:
Quote:
Last fiddled with by jvang on 20180821 at 22:29 Reason: typing is hard 

20180823, 02:03  #5 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·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:\[x^2(x  1)  x(x  1)  6(x  1)\]\[(x^2  x  6) (x1)\]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, timeconsuming, and inaccurate
However, it seems to me that the grouping and the synthetic division solutions are only effective for "neat" higherdegree 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 
20180823, 03:35  #6  
"Dylan"
Mar 2017
586_{10} Posts 
Quote:
Code:
$\mathbb{Z}$ For future reference most of the symbols in mathematics (along with the LaTex commands to produce them) are listed here. Last fiddled with by Xyzzy on 20180824 at 13:31 Reason: We have attached the referenced file. 

20180823, 05:12  #7  
"Curtis"
Feb 2005
Riverside, CA
5·7·11·13 Posts 
Quote:
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. 

20180823, 23:09  #8  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}×3×37 Posts 
Quote:
Quote:


20180824, 13:42  #9 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
43·233 Posts 

20180824, 23:46  #10  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
674_{8} Posts 
Quote:
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 20180824 at 23:52 Reason: typing is hard 

20180825, 01:33  #11  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
43·233 Posts 
Quote:
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 sheshed. 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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