B Why does f = g in implicit equations?

[etotheipi]

TL;DR Summary

N/A

I saw something in my notes that I didn't understand... we have $y=f(x)$, and consider an implicit equation of the form $g(y) = f(x)$. They then say that $f=g$. Why is that true? I would have thought$$f = \{ (a,f(a)) : a\in \mathbb{R} \} \subseteq \mathbb{R}^2$$whilst $g$ is just$$g = \{ (a,a) : a\in \mathbb{R} \} \subseteq \mathbb{R}^2$$What am I missing? Thanks!

 

[PeroK]

I can't make much sense of that. Explicitly $g(y) = y$.

 

[etotheipi]

I can't make much sense of that. Explicitly $g(y) = y$.

Thanks, I presumed it was a mistake of some variety but wanted to check!

 

[FactChecker]

g and f are not identical. They do not necessarily even have the same domain. If f(x) == 1, then all we know is that g(1)=1. g is undefined for all other input values.

 

[member 587159]

Can you post an image of this section in your notes or type out exactly what it says? There might be hidden assumptions.

 

[etotheipi]

It's essentially:

"It is also possible to differentiate an expression of the form $g(y) = f(x)$ with respect to $x$,$$\frac{dg(y)}{dy} \frac{dy}{dx} = \frac{df}{dx}$$Rearranging, we find that the implicit function $g(y) = f(x)$ gives$$\frac{dy}{dx} = \frac{df}{dx}/\frac{dg}{dy}$$Since the function is defined with $f=g$, this looks like we are simply computing the quotient of two quotients."

The implicit differentiation is fine, but that final remark caught me a little off guard...

 

[member 587159]

It's essentially:The implicit differentiation is fine, but that final remark caught me a little off guard...

Yeah, that final remark looks weird. I would just ignore it.

 

[PeroK]

It's essentially:The implicit differentiation is fine, but that final remark caught me a little off guard...

The last remark doesn't make a lot of sense. If we invert the latter derivative we have:
$$\frac{dy}{dx} = \frac{df}{dx}\frac{dy}{dg}$$ which looks like the chain rule, if we allow the association $f \leftrightarrow g$.

I.e. if we imagine parametrising a curve $x(t), y(t)$, then $f$ and $g$ have the same value on the curve:
$$\hat g(t) \equiv g(y(t)) = f(x(t)) \equiv \hat f(t)$$ and $\hat f = \hat g$.

 

[Office_Shredder]

I agree you can ignore it the rest of the statement is true anyway even if you just delete the f=g part?