mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Blogorrhea > jvang

Reply
 
Thread Tools
Old 2018-08-25, 13:13   #12
jvang
veganjoy
 
jvang's Avatar
 
"Joey"
Nov 2015
Middle of Nowhere,AR

1BC16 Posts
Default

Quote:
Originally Posted by Uncwilly View Post
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
jvang is offline   Reply With Quote
Old 2018-08-25, 18:58   #13
masser
 
masser's Avatar
 
Jul 2003
wear a mask

172810 Posts
Default

Quote:
Originally Posted by jvang View Post
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.
masser is online now   Reply With Quote
Old 2018-08-25, 20:11   #14
chalsall
If I May
 
chalsall's Avatar
 
"Chris Halsall"
Sep 2002
Barbados

233108 Posts
Default

Quote:
Originally Posted by jvang View Post
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
chalsall is offline   Reply With Quote
Old 2018-08-28, 21:22   #15
jvang
veganjoy
 
jvang's Avatar
 
"Joey"
Nov 2015
Middle of Nowhere,AR

22·3·37 Posts
Default

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 [$$]
jvang is offline   Reply With Quote
Old 2018-08-28, 22:27   #16
R. Gerbicz
 
R. Gerbicz's Avatar
 
"Robert Gerbicz"
Oct 2005
Hungary

1,493 Posts
Default

Quote:
Originally Posted by jvang View Post
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
R. Gerbicz is offline   Reply With Quote
Old 2018-08-29, 10:54   #17
Nick
 
Nick's Avatar
 
Dec 2012
The Netherlands

22·3·5·29 Posts
Default

Quote:
Originally Posted by jvang View Post
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.
Nick is offline   Reply With Quote
Old 2018-08-29, 21:42   #18
jvang
veganjoy
 
jvang's Avatar
 
"Joey"
Nov 2015
Middle of Nowhere,AR

44410 Posts
Default

Quote:
Originally Posted by Nick View Post
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.
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...
jvang is offline   Reply With Quote
Old 2018-08-29, 22:08   #19
masser
 
masser's Avatar
 
Jul 2003
wear a mask

26·33 Posts
Default

Quote:
Originally Posted by jvang View Post
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.
masser is online now   Reply With Quote
Old 2018-08-31, 00:51   #20
jvang
veganjoy
 
jvang's Avatar
 
"Joey"
Nov 2015
Middle of Nowhere,AR

22·3·37 Posts
Default

Quote:
Originally Posted by masser View Post
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?
jvang is offline   Reply With Quote
Old 2018-08-31, 02:51   #21
VBCurtis
 
VBCurtis's Avatar
 
"Curtis"
Feb 2005
Riverside, CA

137F16 Posts
Default

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.
VBCurtis is offline   Reply With Quote
Old 2018-08-31, 06:26   #22
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
Jun 2011
Thailand

22×7×349 Posts
Default

Quote:
Originally Posted by jvang View Post
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
LaurV is online now   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Different Career Discussion Thread fivemack Lounge 12 2018-08-11 07:51
Career Discussion Thread jvang Lounge 27 2018-08-10 13:32
P-1 discussion thread Rincewind Five or Bust - The Dual Sierpinski Problem 57 2011-02-06 21:53
PRP discussion thread philmoore Five or Bust - The Dual Sierpinski Problem 83 2010-09-25 10:20
Sieving discussion thread philmoore Five or Bust - The Dual Sierpinski Problem 66 2010-02-10 14:34

All times are UTC. The time now is 05:34.


Sat Oct 16 05:34:07 UTC 2021 up 85 days, 3 mins, 0 users, load averages: 1.79, 1.39, 1.21

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.