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TheMidnightJudge
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22 Sep 2009, 10:54 pm

What does it mean for an equation to be explicit in X? I can't seem to find anything that explains this.
Anyone help?


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leejosepho
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22 Sep 2009, 11:04 pm

Maybe this will tell you if you have a way to open a .ps file:
http://www2.tcs.ifi.lmu.de/~schoepp/Doc ... 02_talk.ps



CTBill
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23 Sep 2009, 4:41 am

From Merriam-Webster:

explicit (of a mathematical function): defined by an expression containing only independent variables

Makes more sense if you also look at implicit:

implicit (of a mathematical function): defined by an expression in which the dependent variable and the one or more independent variables are not separated on opposite sides of an equation

Hope this helps.



ruveyn
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23 Sep 2009, 7:26 am

TheMidnightJudge wrote:
What does it mean for an equation to be explicit in X? I can't seem to find anything that explains this.
Anyone help?


An expression like A = B(X) where the only free variable in the B is X and the variable X does not occur in A.

In short: X's on one side, everything else on the other.

ruveyn



aleclair
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23 Sep 2009, 11:24 pm

ruveyn wrote:
TheMidnightJudge wrote:
What does it mean for an equation to be explicit in X? I can't seem to find anything that explains this.
Anyone help?


An expression like A = B(X) where the only free variable in the B is X and the variable X does not occur in A.

In short: X's on one side, everything else on the other.

ruveyn


Exactly that.

Are you in a calculus class where you are covering the idea of 'implicit differentiation'? I'm going to base my explanation of explicit equations on that assumption, because typically that's where you see the term first.

So, let's create a couple of equations with y and x:

y = x^2 + 2x + 1
2xy + 3x^2 = 2y + y^2

The first one is going to be "explicit" because you can write everything in terms of one variable (the idea applies to multiple variables as well, but if this is Calc 1, talking about that will confuse you more). So we could call y f(x) (i.e. it is a one-variable function) and then use the standard differentiation rules to differentiate f(x).

The second one is going to be "implicit" because as much as we try to get y onto one side, we essentially fail. I tried and got y = (3x^2)/(2x-2+y) - so if we want to find the value of y for a given x, we still need to do algebra after substituting in the numbers, because there is still y on both sides. When it comes to differentiating these functions, it gets kind of convoluted and you have to use the chain rule. Ugh.

Hope this helps without being too technical.



TheMidnightJudge
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25 Sep 2009, 12:49 pm

Thanks.


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