How to solve and graph the functional equation f(x)=f(y)+cx

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Homework Statement


I was wondering if anyone can tell me about this equation:

f(x)=f(y)+cx

How can I select points for this equation and graph it? What values are constant and what values are changing? Thanks

Homework Equations



f(x)=f(y)+cx

The Attempt at a Solution



Well... maybe choose a fixed value for y and for c?
 
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lonewolf219 said:

Homework Statement


I was wondering if anyone can tell me about this equation:

f(x)=f(y)+cx

How can I select points for this equation and graph it? What values are constant and what values are changing? Thanks

Homework Equations



f(x)=f(y)+cx

The Attempt at a Solution



Well... maybe choose a fixed value for y and for c?

What happens if you put y = 0?

[Edit-Added]Are you sure you have stated the problem completely and correctly? Putting y=0 gives f(x) = f(0) + cx. Putting y = t gives f(x) = f(t) + cx. Subtracting those says f(0) = f(t) so f is constant. But that doesn't work unless c = 0.
 
Last edited:
lonewolf219 said:
:smile:OK, that's a possibility, thanks LCKurtz

Make sure you notice my edited comment.
 
Oh thanks for that message, I did not see your edited response...You are right, there is more! :smile: The equation is [itex]\alpha[/itex]^(-1)(t)=[itex]\alpha[/itex]^(-1)(m)+(1/2[itex]\pi[/itex])(β)(t). This is for the running of the couplings, haven't done any of this stuff until this summer... Hope you can shed some light!
 
lonewolf219 said:
The equation is [itex]\alpha[/itex]^(-1)(t)=[itex]\alpha[/itex]^(-1)(m)+(1/2[itex]\pi[/itex])(β)(t).
I assume you want to arrive at y as a function of x, y=y(x) (or in the revised version m = m(t)).
Please clarify what alpha is here. Judging from the original post it is an arbitrary unknown function. If so, I see no hope of saying much about m(t). If you assume it's invertible then you can show m(0) = 0.