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how to verify if an equation already exists (and whom to credit for it)

 
 
Ray C
 
Reply Fri 13 Apr, 2018 11:54 am
Say I’m writing an article on a mathematical topic…and I wish to crisply articulate how the distance between two consecutive squares is the sum of the two square roots (( ex: 20^2 is 400, which is 39 more than 19^2 (361), and 39 = 20 + 19 )).

So I include the equation: (x+1)^2 – (x)^2 = 2x + 1

Now, because this equation involves rather elementary subject matter, I assume someone else has most likely authored such an equation.

But how does one verify for sure if someone else has already authored a particular equation ?

And, moreover, how does one ascertain whom to credit when one wishes to include this equation in an article ?

I entered this equation (surrounded by quotation marks) into the Google and did not get a direct hit.

Might there be a database of existing equations or some similar resource ?
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Type: Question • Score: 1 • Views: 390 • Replies: 5
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engineer
 
  2  
Reply Fri 13 Apr, 2018 12:35 pm
@Ray C,
You don't need to attribute basic math.
0 Replies
 
ekename
 
  0  
Reply Fri 13 Apr, 2018 09:29 pm
@Ray C,
You could try searching using verbal description.

Difference of two squares.

https://en.wikipedia.org/wiki/Difference_of_two_squares

https://wikimedia.org/api/rest_v1/media/math/render/svg/5d52e4cff7a7157a34586d9a29df412a7d37f574
0 Replies
 
maxdancona
 
  1  
Reply Fri 13 Apr, 2018 10:49 pm
@Ray C,
I agree with Engineer. This math is so basic that it needs no attribution; same with 512 + 42 = 654
0 Replies
 
maxdancona
 
  1  
Reply Sun 15 Apr, 2018 06:35 am
I did a little research, searching for the history of polynomials and their multiplication. It is clear that Renee Descarte knew about this equation in 1637; it was in his book "La Géométrie". Other sites suggest that this knowledge was developed much earlier.

https://en.wikipedia.org/wiki/La_G%C3%A9om%C3%A9trie
Ray C
 
  1  
Reply Fri 20 Apr, 2018 11:55 am
@maxdancona,
Oh, okay, fair enough. Thanks for letting me know, guys.
0 Replies
 
 

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