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April 12th, 2018, 03:57 AM   #1
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English

What is the name of the thing that mark with ...?
mx + n = y
n is ...
ax^2 + bx + c = y
c is ...
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April 12th, 2018, 05:03 AM   #2
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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.
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April 12th, 2018, 05:15 AM   #3
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The technical term is "ellipsis". It isn't related to integration.
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April 12th, 2018, 05:18 AM   #4
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I think he is refering to the "constant term".
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April 12th, 2018, 05:20 AM   #5
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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.
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April 12th, 2018, 07:58 AM   #6
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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.
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April 13th, 2018, 02:01 AM   #7
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Quote:
Originally Posted by skipjack View Post
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.
What are the circumstances?
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April 13th, 2018, 02:20 PM   #8
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Quote:
Originally Posted by policer View Post
What are the circumstances?
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 "anti-derivative", 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 anti-derivative 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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