20190101, 22:17  #100 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{3}·53 Posts 
It's 2019!
Went out to practice driving for the first time today (with my mom in the car). First thoughts:

20190122, 03:09  #101 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
650_{8} Posts 
I was going to post these in the cell phone astrophotography thread, but the cell phone ones were very bad. These were taken with a more legitimate camera:

20190122, 03:20  #102 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
8042_{10} Posts 
Very nice.

20190122, 06:20  #103  
Aug 2006
2^{2}·5·293 Posts 
Quote:
Quote:


20190405, 00:39  #104 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{3}·53 Posts 
My dad wanted me to post about my math observation; a long time ago I noticed that each successive square number is a successive odd number less than the next. So starting from 0, the first odd number is 1, which is the first square. Adding the next odd number, 3, to 1 gives 4, the second square. Add 5, get 9, add 7, get 16, etc. I’m not sure what kind of equation would describe this; it’d be trivial to code and easy to make a series describing it though. Is this a coincidental correlation or does it have some meaning?

20190405, 01:02  #105 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2·4,021 Posts 
How long does this trend last?
Have you looked fro this in oeis? 
20190405, 01:15  #106  
Sep 2002
Database er0rr
2^{2}·11·73 Posts 
Quote:
Last fiddled with by paulunderwood on 20190405 at 01:18 

20190405, 01:41  #107 
"William"
May 2003
New Haven
2·3^{2}·131 Posts 
Have you learned Sigma notation yet? If so, you should be able to use it to write an expression for the sum of the first n odd numbers.
Have you learned inductive proofs yet? If so, you should be able to prove the relationship holds. 
20190405, 08:21  #108 
Dec 2012
The Netherlands
3×457 Posts 
Good observation!
Here's a picture to see what's going on: If we add this to the observation you made earlier we have 2 sequences: \[ \begin{eqnarray*} 1+2+3+\ldots +n & = & \frac{1}{2}n(n+1) \\ \underbrace{1+3+5+\ldots +(2n1)}_n & = & n^2 \end{eqnarray*} \] In both sequences, the difference between terms next to each other remains constant (the numbers go up by 1 each time in the first sequence and by 2 each time in the second one). Sequences with this property are known as arithmetic progressions. There is a formula for the sum of all the terms in any arithmetic progession, which is useful to know: add the first and last terms together, multiply by the number of terms and divide by 2 
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