20180825, 13:13  #12  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Quote:
Last fiddled with by jvang on 20180825 at 13:15 Reason: typing is hard 

20180825, 18:58  #13 
Jul 2003
Behind BB
1761_{10} Posts 
"almost sound" will be the title of my next album of jazz tunes.

20180825, 20:11  #14 
If I May
"Chris Halsall"
Sep 2002
Barbados
10045_{10} Posts 
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 20180825 at 20:11 
20180828, 21:22  #15 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·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 [$$]

20180828, 22:27  #16  
"Robert Gerbicz"
Oct 2005
Hungary
2·3·11·23 Posts 
Quote:
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 

20180829, 10:54  #17  
Dec 2012
The Netherlands
2×3×293 Posts 
Quote:
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 \(x=\frac{1}{1c}\). Then \(0.9<c<1\) so \(0<1c<\frac{1}{10}\) and therefore \(x>10\). And \(0.99<c<1\) so \(0<1c<\frac{1}{100}\) and therefore \(x>100\). And \(0.999<c<1\) so \(0<1c<\frac{1}{1000}\) 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. 

20180829, 21:42  #18  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
444_{10} Posts 
Quote:
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 limitbased 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... 

20180829, 22:08  #19  
Jul 2003
Behind BB
3×587 Posts 
Quote:
When I want the equal signs to line up, I usually use an equation array. 

20180831, 00:51  #20  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Quote:
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&errored 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 nonexistent 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 \(g(x) = \dfrac{x1}{x1}\). 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? 

20180831, 02:51  #21 
"Curtis"
Feb 2005
Riverside, CA
11715_{8} Posts 
We can use separate values, when a onesided 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 onesidedness, 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 onesided limit, but DNE for the regular (twosided) limit. 
20180831, 06:26  #22  
Romulan Interpreter
"name field"
Jun 2011
Thailand
10011001010000_{2} Posts 
Quote:
\[f(x)=\lbrace{{x1}\atop {x^3}}{:\ \atop:\ }{{\ x>1}\atop{\ x<1}}\] or use a matrix to align them \[f(x)=\lbrace\begin{array}{cc} x1 & : & x<1 \\ x^3 & : & x>1 \end{array}\] Last fiddled with by LaurV on 20180831 at 06:39 

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