November 14th, 2017, 01:10 AM  #11 
Senior Member Joined: Nov 2011 Posts: 230 Thanks: 2  Rock on
What is "rock on"?

November 14th, 2017, 01:16 AM  #12 
Senior Member Joined: Nov 2011 Posts: 230 Thanks: 2 
There are 4 definitions that connect one to another: 1. To solve the equation you need to seprate the y. 2. It a function in a range of value that strange from the "normal" function by the range that it defines. It a function with a limit of range of value. 3. It is connected to the Implicit functions Theorem. 4. It has 2 inputs (x & y) and one output [I think it a y but it can be also z]. All the definition [4 in particular] in two "outputs": x and y. The noition of it is F(x, y) = 0. Understoond? 
November 14th, 2017, 01:37 PM  #13 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,888 Thanks: 767 Math Focus: Wibbly wobbly timeywimey stuff.  
November 15th, 2017, 04:06 AM  #14  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
If the "strange font" was $\displaystyle \partial$, as in $\displaystyle \frac{\partial f}{\partial x}$, that is, in fact, the partial derivative. Here, f is a function of more than a single variable (say, x and y) and $\displaystyle \frac{\partial f}{\partial x}$ is the "rate of change of f as x changes but y does not change". That is, we treat y as a constant rather than a variable. Similarly we find $\displaystyle \frac{\partial f}{\partial y}$ by treating x as a variable and thinking of f as a function of y only. For example, if $\displaystyle f(x,y)= x^3y^2$, since the derivative of $\displaystyle Cx^3$ is $\displaystyle 3Cx^2$ for any constant C, the "partial derivative with respect to x" is $\displaystyle \frac{\partial f}{\partial x}= 3x^2y^2$. Similarly, since the derivative of $\displaystyle Cy^2$ is $\displaystyle 2Cy$ for any constant C, $\displaystyle \frac{\partial f}{\partial y}= 2x^3y$. Quote:
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If you have a function of two variables (two inputs), F(x, y), set equal to a constant, F(x, y)= C (possibly 0 but not necessarily) then the "implicit function theorem" says that "if $\displaystyle \frac{\partial f}{\partial y}$ is not 0 in some region around $\displaystyle (x_0, y_0)$ then we can find some function of a single variable, f, such that y= f(x) in that region. That is, F(x, f(x))= C".  
November 16th, 2017, 01:38 AM  #15 
Senior Member Joined: Nov 2011 Posts: 230 Thanks: 2 
Thanks for the maintance. You are welcome!!!! In 2 I still think that the functions is limited. Show me Implicit Function that doesn't limited. In the end of the post it is a writing error that "inputs" become "outputs". Last edited by shaharhada; November 16th, 2017 at 01:41 AM. 
November 16th, 2017, 03:59 AM  #16 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
"Limited"? Do you mean "bounded"? The functions defined implicitly by x+ y= 0 are unbounded.

November 16th, 2017, 04:07 AM  #17 
Senior Member Joined: Nov 2011 Posts: 230 Thanks: 2  
November 16th, 2017, 04:21 AM  #18 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Then what do you mean by "implicit function"? Personally, I would not use the phrase "implicit function" but rather "implicitly defined" function. A function is "implicitly defined" if it is given as f(x, y)= C which can be solved for y as a function of x. Perhaps you want an example where there exist more than one way to define y as a function of x? For example, $f(x,y)= x^2+ y^2= 1$ where y can be either $y= \sqrt{1 x^2}$ or $y= \sqrt{1 x^2}$.

November 16th, 2017, 04:49 AM  #19 
Senior Member Joined: Nov 2011 Posts: 230 Thanks: 2 
Yes!!!


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