April 12th, 2018, 03:57 AM  #1 
Banned Camp Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3  English
What is the name of the thing that mark with ...? mx + n = y n is ... ax^2 + bx + c = y c is ... 
April 12th, 2018, 05:03 AM  #2 
Banned Camp Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3 
And my next question is: What is the difference between this Thing and the Constant Of Integration? Last edited by skipjack; April 12th, 2018 at 05:06 AM. 
April 12th, 2018, 05:15 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 19,974 Thanks: 1850 
The technical term is "ellipsis". It isn't related to integration.

April 12th, 2018, 05:18 AM  #4 
Senior Member Joined: Oct 2009 Posts: 629 Thanks: 192 
I think he is refering to the "constant term".

April 12th, 2018, 05:20 AM  #5 
Banned Camp Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3  I will it again
What is the name of The Thing? mx + n = y n = The Thing ax^2 + bx + c = y c = The Thing. this is a "free number" in my bad English. If integration of f function is (Integral of f) + C what is the difference between c and the "free number"? Is there a connection between them? Last edited by skipjack; April 12th, 2018 at 07:50 AM. 
April 12th, 2018, 07:58 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,974 Thanks: 1850 
In your equations, n and c appear to be constant terms, as there is nothing to suggest they vary as x varies. When you integrate with respect to x, the C that is added as a constant of integration doesn't vary as x varies, but (in certain circumstances) might vary as something else varies. I haven't come across the term "free number" in this type of context.

April 13th, 2018, 02:01 AM  #7  
Banned Camp Joined: Dec 2017 From: Tel Aviv Posts: 87 Thanks: 3  Quote:
 
April 13th, 2018, 02:20 PM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  If you have a "partial differential equation" that has more than one independent variable is one such circumstance. For example, to solve the differential equation $\displaystyle \frac{\partial^2 z}{\partial x\partial y}= 0$, write it as $\displaystyle \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)= 0$. The "antiderivative", with respect to x, would give $\displaystyle \frac{\partial z}{\partial y}$ plus a "constant". But since this is a partial derivative with respect to x, where we hold y constant, that "constant might in fact be some function of y! That is $\displaystyle \frac{\partial z}{\partial y}+ f(y)$. Now integrating with respect to y, we have $\displaystyle z= F(y)+ "C"$ where F is an antiderivative of f. And, again, since we differentiated with respect to y, that "C" might be some function of x. If we call that function "G(x)", that becomes $\displaystyle z= F(y)+ G(x)$. Last edited by skipjack; April 14th, 2018 at 04:48 AM. 

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