- #1

- 622

- 32

Hi.

If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?

If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?

- I
- Thread starter dyn
- Start date

- #1

- 622

- 32

Hi.

If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?

If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?

- #2

fresh_42

Mentor

- 13,842

- 11,037

No. If ##u=x+1## then ##x=u-1## and you get ##f(x)=f(u-1)=\sqrt{u}=\sqrt{x+1}\,.##Hi.

If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?

- #3

- 622

- 32

Thanks , what would f(u) be then ?

- #4

fresh_42

Mentor

- 13,842

- 11,037

##f(u)=\sqrt{u+1}##. The name of the variable doesn't matter. You could as well write ##f(tree)=\sqrt{tree+1}##, but this would be a bit confusing and too long to use.Thanks , what would f(u) be then ?

- #5

- 622

- 32

I will give you the example I have just encountered ; it concerns complex numbers.

Consider f(z) = √ (z+1) , now define u = z+1 then f(u) = √u ,

this is then used to show that f(u) is multi-valued around u=0 and so f(z) is multi-valued around z= -1

- #6

fresh_42

Mentor

- 13,842

- 11,037

- #7

- 622

- 32

##y(x) = (x+1)^2## then using u=x+1 , ##y(u) = u^2## and then this is differentiated using the chain rule

- #8

fresh_42

Mentor

- 13,842

- 11,037

- #9

- 622

- 32

- #10

fresh_42

Mentor

- 13,842

- 11,037

Yes, the derivatives are ##\dfrac{dy_1}{dx}=2(x+1)=2y_1## and ##\dfrac{dy_2}{du}=2u=2y_2##, however, ##\left. \dfrac{d}{dx}\right|_{a}\,y_1=2a+2## whereas ##\left. \dfrac{d}{du}\right|_{a}y_2=2a## which is not the same.

Last edited:

- #11

Mark44

Mentor

- 34,323

- 5,971

Again, no. ##f(u) = \sqrt{u + 1}##, as already explained by @fresh_42.Consider f(z) = √ (z+1) , now define u = z+1 then f(u) = √u ,

This part isn't relevant to what you asked about.dyn said:this is then used to show that f(u) is multi-valued around u=0 and so f(z) is multi-valued around z= -1

- #12

- 15,063

- 7,246

A number of physics text books will use this shorthand but it is mathematically imprecise. They also skate over the chain rule somewhat. What you are really trying to do is to find a new function ##g## such that:Hi.

If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?

##\forall x: \ g(x + 1) = f(x)##

In this case we have ##g(x) = \sqrt{x}##. Then, with ##u \equiv x + 1##, we have:

##g(u) = g(x+1) = f(x)##

Some physics texts then use ##f## for both ##f## and ##g##.

- #13

pasmith

Homework Helper

- 1,882

- 528

This is an abuse of notation.

##y(x) = (x+1)^2## then using u=x+1 , ##y(u) = u^2## and then this is differentiated using the chain rule

If [itex]f : x \mapsto x^2[/itex] and [itex]g : x \mapsto x + 1[/itex] then [itex]h = f \circ g : x \mapsto (x + 1)^2[/itex]. [itex]h[/itex] is not the same function as [itex]f[/itex].

Now if you write [itex]y = h(x)[/itex] then that is not a function definition; it's a constraint imposed on the otherwise independent variables [itex]x[/itex] and [itex]y[/itex]. It is equivalent to the two constraints [itex]y = f(u)[/itex] and [itex]u = g(x)[/itex].

In Newton's notation the chain rule read [itex](f \circ g)' = (f' \circ g)g'[/itex], and the easiest way to calculate it is indeed to calculate [itex]f'(u)[/itex] and [itex]g'(x)[/itex] and then set [itex]u = g(x)[/itex] to obtain [itex]f'(g(x))g'(x)[/itex].

You'll note that to do this I had to introduce three additional names. That's why people abuse the notation.

- #14

- 622

- 32

- #15

fresh_42

Mentor

- 13,842

- 11,037

The difference is basically the following:

which means it is a matter of coordinates: a shift by +1.

which means it is a matter of coordinates: a shift by +1.

- #16

- 622

- 32

- #17

fresh_42

Mentor

- 13,842

- 11,037

I do not think so. As you see in the picture, there are two different curves. Now if you only consider their slopes, then they behave the same (at different points). If you integrate them, then the results will be the same (within different integration limits). In neither case it is a matter of notation. They were and will be two different functions. This is especially important for physicists, as the frames are important here! A mathematician could say "I don't care, I'm only interested in the geometric object", but a physicist cannot. Change of coordinates is completely uninteresting for mathematicians, physicists do nothing else! So it is not an abuse of notation, it is a lack of description from your part. Your initial question is a strict NO, and then you came up with the chain rule, which didn't make any sense without further context. ##f(u)=\sqrt{u+1}## if ##u=x## and ##f(u-1)=\sqrt{u}## if ##u=x+1##. There is literally nothing which can be abused! ##g(u)=\sqrt{u}## is a different function, in physics as in mathematics. Fullstop.

All what came after post #2 (or #4) is pure guesswork (including mine, with the exception of this one) based on lacking context, information and clarity.

- #18

- 622

- 32

I was quite happy with the answers before this post but the above post does not make any senseI do not think so. As you see in the picture, there are two different curves. Now if you only consider their slopes, then they behave the same (at different points). If you integrate them, then the results will be the same (within different integration limits). In neither case it is a matter of notation. They were and will be two different functions. This is especially important for physicists, as the frames are important here! A mathematician could say "I don't care, I'm only interested in the geometric object", but a physicist cannot. Change of coordinates is completely uninteresting for mathematicians, physicists do nothing else! So it is not an abuse of notation, it is a lack of description from your part. Your initial question is a strict NO, and then you came up with the chain rule, which didn't make any sense without further context. ##f(u)=\sqrt{u+1}## if ##u=x## and ##f(u-1)=\sqrt{u}## if ##u=x+1##. There is literally nothing which can be abused! ##g(u)=\sqrt{u}## is a different function, in physics as in mathematics. Fullstop.

All what came after post #2 (or #4) is pure guesswork (including mine, with the exception of this one) based on lacking context, information and clarity.

- Last Post

- Replies
- 2

- Views
- 801

- Last Post

- Replies
- 10

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 2K

- Replies
- 1

- Views
- 878

- Last Post

- Replies
- 2

- Views
- 7K

- Last Post

- Replies
- 1

- Views
- 426

- Replies
- 3

- Views
- 6K

- Replies
- 2

- Views
- 10K

- Last Post

- Replies
- 12

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 1K