Nonlinear function to linear function conversion (as part of linear regression)

[Agent Smith]

How do I change a nonlinear functiom to a linear function?

I'm given $y = x^2$ (parabola/quadratic) and asked to convert it into a linear function.
Something to do with $x^2$, but when I try to graph this (on desmos), I still get a parabola.
$f(x) = x^2$ (parabola)
$g(x) = 3f(x) = 3x^2$ (which I expected to be linear, because it looks like $g(z) = 3z$), is not)

Gracias

 

[Agent Smith]

A problem in the practice section:
$y = x^2$ and $y = 3x^2$. The former is a quadratic and the latter is, they say, linear. However, when I plot these two, I get both as parabolas.

 

[Agent Smith]

 

[Dr.Peterson]

Please show exactly what "they" say, in context, including the entire example. You can't change a non-linear function to a linear function, since it isn't. But maybe some particular sort of transformation is allowed, such as a log? We can't know without context.

 

[Agent Smith]

@Dr.Peterson


And yes they do with exponential functions as well (with logarithms).

 

[Dr.Peterson]

I don't see anything said there, just two ambivalent pictures. Please show the whole context, including whatever is actually said.

 

[Agent Smith]

@Dr.Peterson , it's difficult to keep taking screenshots and uploading them.

The gist of the idea seems to be that

1. If the graph for function f vs. x² is a straightline, f is a quadratic: f = mx² + b

2. If the graph for log f and x is a straight line, f is an exponential function: f = $n^x$

3. If the graph for f and x is a straight line, f is a linear function: y = mx + b

So possibly, if I have f = mx² + b (a quadratic) I can turn it into a linear function. I tried this with f(x) = 3x² + 0 = 3x², but desmos plots it as a parabola/quadratic, not a straight line.


From my notes

 

[BeansNRice]

if I'm understanding this, which it's a fair bet I'm not, OP intends something like this

$f(x) \text{ is non-linear, however } \exists g(x) \ni f(g(x)) \text{ is linear.}$

My reaction is ... ok so?

 

[Agent Smith]

@BeansNRice I believe you nailed it. Can we do that (convert nonlinear to linear) in desmos. I tried with y = mx² + b: y = 3x² + 2 is not linear, it's still a parabola.

 

[Agent Smith]

I tried many things on desmos but to no avail.

 

[Dr.Peterson]

The gist of the idea seems to be that

1. If the graph for function f vs. x² is a straightline, f is a quadratic: f = mx² + b

2. If the graph for log f and x is a straight line, f is an exponential function: f = nxn^xnx

3. If the graph for f and x is a straight line, f is a linear function: y = mx + b

So there are no actual WORDS you can copy so people can see what someone is supposedly SAYING? And no link you can give for your source?

How do I change a nonlinear functiom to a linear function?

I'm given $y = x^2$ (parabola/quadratic) and asked to convert it into a linear function.
Something to do with $x^2$, but when I try to graph this (on desmos), I still get a parabola.

Show us the actual problem as you see it.

A problem in the practice section:
$y = x^2$ and $y = 3x^2$. The former is a quadratic and the latter is, they say, linear. However, when I plot these two, I get both as parabolas.

Show us the actual wording of the problem, in context.

If you can't give specific context, we can't be sure we're suggesting the right answers.

 

[Agent Smith]

@Dr.Peterson

 

[Dr.Peterson]

Yes, AI can find a variety of different contexts in which one can do something that can be called "converting a nonlinear relationship to a linear relationship". I think it's just trying to guess what you are asking for, by suggesting possibilities.

What's missing here is any information about what the original problem you were asking about wants you to do. Why do you refuse to provide that problem and its context? If you would do that, then someone could answer. All I'm trying to do is to elicit this necessary information. I don't feel like guessing, though I have several ideas about what they might mean.

 

[Agent Smith]

I would like to know how to plot the linear avatar of a nonlinear function with desmos. I can't seem to do that.
From the AI overview, we see that if $y = 3x^2 + 2$, we can take $x^2 = a$ and get a linear equation $y = 3a + 2$. I tried this with desmos, doesn't work. With google sheets, it works.

 

[Agent Smith]

Isn't this $y = 3x^2$ and then $x^2 =a$ and then $y = 3a$, function composition? So desmos should've accepted this $f(x) = x^2, g(f(x)) = 3f(x)$, but


@Dr.Peterson

 

[BeansNRice]

Is this the sort of thing you're after?

 

[Dr.Peterson]

Isn't this $y = 3x^2$ and then $x^2 =a$ and then $y = 3a$, function composition? So desmos should've accepted this $f(x) = x^2, g(f(x)) = 3f(x)$, but
View attachment 38889

@Dr.Peterson

It isn't clear what you expect Desmos to graph here; what you wrote doesn't tell it. You have to speak a particular program's language.

I would like to know how to plot the linear avatar of a nonlinear function with desmos. I can't seem to do that.
From the AI overview, we see that if $y = 3x^2 + 2$, we can take $x^2 = a$ and get a linear equation $y = 3a + 2$. I tried this with desmos, doesn't work. With google sheets, it works.

If you have a function $f(x)=3x^2+2$, and want to express it as a function of $u=g(x)=x^2$, then you can use $h(u)=f(g^{-1}(u))=f(\sqrt{u})=3(\sqrt{u})^2+2=3u+2$ (for $u\ge0$):

​

Unfortunately, Desmos doesn't understand inverses, so I had to make $g^{-1}(u)=\sqrt{u}$ manually.

But what we're doing here is decomposing function f as a composition $f=h\circ g$, and h can be obtained as $h=f\circ g^{-1}$.

 

[Agent Smith]

It isn't clear what you expect Desmos to graph here; what you wrote doesn't tell it. You have to speak a particular program's language.


If you have a function $f(x)=3x^2+2$, and want to express it as a function of $u=g(x)=x^2$, then you can use $h(u)=f(g^{-1}(u))=f(\sqrt{u})=3(\sqrt{u})^2+2=3u+2$ (for $u\ge0$):

View attachment 38892​

Unfortunately, Desmos doesn't understand inverses, so I had to make $g^{-1}(u)=\sqrt{u}$ manually.

But what we're doing here is decomposing function f as a composition $f=h\circ g$, and h can be obtained as $h=f\circ g^{-1}$.

Can you link me to your desmos graph? Thanks

 

[Dr.Peterson]

Can you link me to your desmos graph? Thanks

No, I deleted it because it's easy enough to replicate. I typed in just what you see. The ideas are more important.

But I suppose I could do more if you would decide to provide the original problem that I've asked for repeatedly, so I could be sure whether this is what you need.

 

[Agent Smith]

@Dr.Peterson I'm just trying to clarify my doubts. At first Google sheets graphs squared perfectly with what was taught in the lessons but then I couldn't get Desmos to do the correct thing. I don't know why $g(f(x)) = 3f(x)$ doesn't return an error and also fails to show a graph.

It's not that I haven't defined the functions