Change of Variables: Rewrite g(s) as g(t)?

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Homework Help Overview

The discussion revolves around the change of variables in a mathematical context, specifically transitioning from variable s to variable t in the function g(s). The original poster is uncertain whether they can simply rewrite g(s) as g(t) or if they need to modify the function further.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants question the validity of directly substituting g(s) with g(t) without considering the continuity of the function. Others suggest that the relationship between s and t may require a more nuanced approach, particularly when constants are involved.

Discussion Status

The discussion is active, with participants exploring different interpretations of the change of variables. Some have provided examples to illustrate their points, while others have asked for clarification on the original problem context. There is no explicit consensus yet, but guidance has been offered regarding the complexities of the substitution.

Contextual Notes

The original poster mentions having a function g(s+) and g(s-) and a relation linking these variables, indicating that the problem may involve additional complexities such as differentiation.

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Homework Statement


I am performing a change of variables, s --> t and am wondering can I just write g(s) --> g(t) or do I have to alter the function e.g. g(s) --> a*g(t).


Homework Equations


s = (a/b)* t
g(s) is to be found numerically and therefore we do not have it's definition.

So Option one:
just rewrite g(s) as g(t)

Option two:
g(s) = (a/b)*g(t)

Thanks
 
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Um, generally, if s -> t, then g(s) does NOT go to g(t). If you know that g is continuous, then you may assume that. Otherwise, you will have to work off the official definition of limits.
 
I would assume that if s => ta /b, g(s) => g(ta/b) if a,b are arbitrary / real constants. I could be wrong though.
 
That's definitely not true. Consider the function

g(x) = x + 1 for x > 0 and x - 1 for x <= 0. Then as s -> 0 from the right, g(x) => 1, even though g(0) = -1.
 
who_ said:
That's definitely not true. Consider the function

g(x) = x + 1 for x > 0 and x - 1 for x <= 0. Then as s -> 0 from the right, g(x) => 1, even though g(0) = -1.
You are misunderstanding the notation. The question is not about limits. The OP is asking 'If we change variable s to variable t, can we write f(s) as simply f(t)?'

MathMonster, there are times when that is acceptable but I would avoid it. Technically if, for example, f(s)= s^2- 3s+ 2 then f(t)= t^2- 3t+ 2 but I don't believe that is what you are trying to do. If f(s)= s^2- 3s+ 2[/tex] then replacing s with t= s-2 gives, since s= t+2, f(s)= f(t+ 2)= (t+ 2)^2- 3(t+ 2)+ 2= t^2+ 4t+ 4- 3t- 6+ 2= t^2+ t[/tex] but notice that is f(t+2), not f(t). If you like you say explicitely that g(t)= f(t+ 2) and then use g(t).
 
I see - my apologies. Nevertheless, you should not be moving constants out of expressions, g(at/b) =/= (a/b) * g(t). So unless you would like to simply g(t) with some change of variable (again, this as the above advised, this is not suggested - there are many scenarios where this is invalid - an introductory book on Real Analysis can help clarify the conditions under which you may do so), there isn't too much of a point of doing so.
 
Hi!

Thank you for the reply! Yes I am talking about a change in variables not limits - sorry that's my fault I didn't think of the notation I was using.

HallsofIvy - thank you! I thought that was the case but was confused how to so it. Basically I've got a function g(s+) and g(s-) and a relation linking s+ & s-. I then have a formula with the two 'g's' in which I need to write as s+ in order to solve it.

So basically, I will just write g(s-) = g((a/b)*s+) and leave it at that? There's a bit of differentiation that will come in which will take into account the change.

Thanks!
 
Could you possibly post the question you are working on?
 

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