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2018-08-25, 13:13   #12
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

1BC16 Posts

Quote:
 Originally Posted by Uncwilly 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."
Oh...it’s so hard to pick up on jokes when it’s plain text, especially since you stated it so casually and the Bonnaroo festival turns up weird Google results Seems like a fun game though, your examples almost sound like legitimate names!

Last fiddled with by jvang on 2018-08-25 at 13:15 Reason: typing is hard

2018-08-25, 18:58   #13
masser

Jul 2003

172810 Posts

Quote:
 Originally Posted by jvang Oh...it’s so hard to pick up on jokes when it’s plain text, especially since you stated it so casually and the Bonnaroo festival turns up weird Google results Seems like a fun game though, your examples almost sound like legitimate names!
"almost sound" will be the title of my next album of jazz tunes.

2018-08-25, 20:11   #14
chalsall
If I May

"Chris Halsall"
Sep 2002

233108 Posts

Quote:
 Originally Posted by jvang Are these refutations mathematically and/or logically sound?
Yes. They are referred to as irrational numbers.

Think of them a bit like a woman about once every 28 days (sorry for that; I know it's not PC, but somewhat accurate)....

Last fiddled with by chalsall on 2018-08-25 at 20:11

 2018-08-28, 21:22 #15 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22·3·37 Posts Today we were covering some functions and how to simplify them, and one had a radical term $$\sqrt{x+h}*\sqrt{x}$$. Our teacher said that $$\sqrt{x^2+xh}$$ was not the simplification of it, and marked some of our answers incorrect; after class I showed her a basic proof involving a commonly accepted rule for the distribution of exponents:$\sqrt{x+h}*\sqrt{x} = (x+h)^{\frac{1}{2}}(x)^{\frac{1}{2}}$$=\big(x(x+h)\big)^{\frac{1}{2}}$$=(x^2+xh)^{\frac{1}{2}}$$=\sqrt{x^2+xh}$How do those crazy $$\LaTeX$$ people align their equals signs and make their proofs/solutions look super fancy? I know that there's a bit of code just to center the outputs on their own lines, which the forum conveniently supports with [] 2018-08-28, 22:27 #16 R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 1,493 Posts Quote:  Originally Posted by jvang Today we were covering some functions and how to simplify them, and one had a radical term $$\sqrt{x+h}*\sqrt{x}$$. Our teacher said that $$\sqrt{x^2+xh}$$ was not the simplification of it, and marked some of our answers incorrect; Likely there is no (general) simplification there. Working in real numbers that is really bad, because the domain's of the two function is different, let say x=h=-1 then sqrt(x+h)*sqrt(x) is not defined, but sqrt(x^2+xh) is defined. It is also an easy task for Wolfram: http://www.wolframalpha.com/input/?i...Sqrt%5Bx%5D%5D 2018-08-29, 10:54 #17 Nick Dec 2012 The Netherlands 22·3·5·29 Posts Quote:  Originally Posted by jvang We touched on piecewise functions today (somehow I'm the only student who has used these before?), and someone asked some question that gave me a good excuse to bring up whether $$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$$. To convince someone that 0.999...=1, you have to go back to something that you can both agree on and reason from there. I think most people would accept that real numbers have what is called the Archimedean property: for every real number x there exists a positive integer n such that n>x. Now suppose there exists a real number c such that $\begin{eqnarray*} c & > & 0.9 \\ c & > & 0.99 \\ c & > & 0.999 \\ & \vdots & \end{eqnarray*}$ but c<1. (People who think 0.999... is less than 1 can take this as their number c.) Let $$x=\frac{1}{1-c}$$. Then $$0.9<c<1$$ so $$0<1-c<\frac{1}{10}$$ and therefore $$x>10$$. And $$0.99<c<1$$ so $$0<1-c<\frac{1}{100}$$ and therefore $$x>100$$. And $$0.999<c<1$$ so $$0<1-c<\frac{1}{1000}$$ and therefore $$x>1000$$ etc. It follows that x is greater than all positive integers, which is impossible by the Archimedean property. So such a number c does not exist. If you want to go a step deeper, read the section of your Spivak book about least upper bounds. Then you will see a proof that the real numbers have the Archimedean property, and a definition of precisely what we mean when we write something like 0.999.... It is possible to construct number systems that include numbers which are not equal to each other but differ by an an infinitely small amount, but such systems do not have the least upper bound property (also known as completeness) so you cannot take limits or do calculus with them, making them far less useful. 2018-08-29, 21:42 #18 jvang veganjoy "Joey" Nov 2015 Middle of Nowhere,AR 44410 Posts Quote:  Originally Posted by Nick To convince someone that 0.999...=1, you have to go back to something that you can both agree on and reason from there. I think most people would accept that real numbers have what is called the Archimedean property: for every real number x there exists a positive integer n such that n>x. Now suppose there exists a real number c such that $\begin{eqnarray*} c & > & 0.9 \\ c & > & 0.99 \\ c & > & 0.999 \\ & \vdots & \end{eqnarray*}$ but c<1. (People who think 0.999... is less than 1 can take this as their number c.) Let $$x=\frac{1}{1-c}$$. Then $$0.910$$. And $$0.99100$$. And $$0.9991000$$ etc. It follows that x is greater than all positive integers, which is impossible by the Archimedean property. So such a number c does not exist. If you want to go a step deeper, read the section of your Spivak book about least upper bounds. Then you will see a proof that the real numbers have the Archimedean property, and a definition of precisely what we mean when we write something like 0.999.... It is possible to construct number systems that include numbers which are not equal to each other but differ by an an infinitely small amount, but such systems do not have the least upper bound property (also known as completeness) so you cannot take limits or do calculus with them, making them far less useful. I saw the Archimedean property mentioned somewhere, but its Wikipedia page was very complicated; thanks for explaining that (and completeness) in a comprehensible way! Earlier today I decided to read up on infinite geometric series (I read a while back that $$0.\overline{9}$$ could be represented as such). The limit-based proof of the equality is pretty convincing, and proving that the series was absolutely convergent was neat, but our teacher hasn't covered BC Calculus (our class is only the AB portion, and our school has no BC Calculus class) in a long time, so she had a bit of a hard time remembering how to work with series. Oh well I'd post what constitutes my proof but I'm too lazy to figure out how to code the sigma notation in Latex... 2018-08-29, 22:08 #19 masser Jul 2003 wear a mask 26·33 Posts Quote:  Originally Posted by jvang Today we were covering some functions and how to simplify them, and one had a radical term $$\sqrt{x+h}*\sqrt{x}$$. Our teacher said that $$\sqrt{x^2+xh}$$ was not the simplification of it, and marked some of our answers incorrect; after class I showed her a basic proof involving a commonly accepted rule for the distribution of exponents:$\sqrt{x+h}*\sqrt{x} = (x+h)^{\frac{1}{2}}(x)^{\frac{1}{2}}$$=\big(x(x+h)\big)^{\frac{1}{2}}$$=(x^2+xh)^{\frac{1}{2}}$$=\sqrt{x^2+xh}$How do those crazy $$\LaTeX$$ people align their equals signs and make their proofs/solutions look super fancy? I know that there's a bit of code just to center the outputs on their own lines, which the forum conveniently supports with []
Did she tell you what she expected? She might have been looking for $\left|x\right|\sqrt{1+h/x}$? A better question would have been a little more explicit: "show that expr1 can be written as expr2" is more clear than "simplify expr1"

When I want the equal signs to line up, I usually use an equation array.

2018-08-31, 00:51   #20
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22·3·37 Posts

Quote:
 Originally Posted by masser Did she tell you what she expected? She might have been looking for $\left|x\right|\sqrt{1+h/x}$? A better question would have been a little more explicit: "show that expr1 can be written as expr2" is more clear than "simplify expr1"
Nah, she was thinking we should just leave it unaltered.

We took a little quiz over the precalculus/review stuff that we've worked with over the last 2 weeks, which wasn't too hard. Other than forgetting some log rules (which I just trial&error-ed my way through), it was pretty straightforward.

I watched some Khan Academy videos on the basics of limits and the formal/intuitive proofs. I saw that a limit is non-existent for a function like$f(x)\{x - 1 \ \text{for} \ x > 1 \\ x^3 \text{for} x < 1$because both sides of the equation around $$x = 1$$ do not approach the same point, as in $$g(x) = \dfrac{x-1}{x-1}$$.

Hmmm...how am I supposed to get the newline into a big left brace?

But when we write limit notation, the direction of the arrow under $$lim$$ means "from the left/right." Why can't we use separate values for each limit when approaching it from either direction?

 2018-08-31, 02:51 #21 VBCurtis     "Curtis" Feb 2005 Riverside, CA 137F16 Posts We can use separate values, when a one-sided limit is requested. We can say "the limit from the left is N", or "the limit from the right is M". But, when a limit doesn't specify one-sided-ness, our answer must be valid from either side. So, if the limits from each side exist but do not match, we have valid answers for each one-sided limit, but DNE for the regular (two-sided) limit.
2018-08-31, 06:26   #22
LaurV
Romulan Interpreter

Jun 2011
Thailand

22×7×349 Posts

Quote:
 Originally Posted by jvang Hmmm...how am I supposed to get the newline into a big left brace?
There are many ways, one is to use atop
$f(x)=\lbrace{{x-1}\atop {x^3}}{:\ \atop:\ }{{\ x>1}\atop{\ x<1}}$
or use a matrix to align them
$f(x)=\lbrace\begin{array}{cc} x-1 & : & x<1 \\ x^3 & : & x>1 \end{array}$

Last fiddled with by LaurV on 2018-08-31 at 06:39

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