Changing the argument of a function

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

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  • #2
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Hi.
If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?
No. If ##u=x+1## then ##x=u-1## and you get ##f(x)=f(u-1)=\sqrt{u}=\sqrt{x+1}\,.##
 
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  • #3
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Thanks , what would f(u) be then ?
 
  • #4
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Thanks , what would f(u) be then ?
##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.
 
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  • #5
dyn
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I am a bit rusty but I seem to remember seeing examples like the one I originally quoted many times.
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
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Could it be that you confuse it with an integral? Here we have ##\int_a^b \sqrt{x+1}\,dx = \int_{a+1}^{b+1} \sqrt{u}\,du## and if the integral limits aren't noted, it looks as if the function's argument would just have been shifted: ##\int \sqrt{x+1}\,dx =\int \sqrt{u}\,du\,##.
 
  • #7
dyn
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No , there was no integral. Another similar example is differentiation using the chain rule , for example if
##y(x) = (x+1)^2## then using u=x+1 , ##y(u) = u^2## and then this is differentiated using the chain rule
 
  • #8
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These are two different functions. If ##y_1(x)=(x+1)^2## and ##y_2(u)=u^2## then ##y_1(0)=1## and ##y_2(0)=0## so they cannot be the same function.
 
  • #9
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But that is how the chain rule is used as far as I know and I have been using it for many years and it always worked
 
  • #10
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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.
 
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  • #11
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Consider f(z) = √ (z+1) , now define u = z+1 then f(u) = √u ,
Again, no. ##f(u) = \sqrt{u + 1}##, as already explained by @fresh_42.

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
This part isn't relevant to what you asked about.
 
  • #12
PeroK
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Hi.
If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?
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:

##\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##.
 
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  • #13
pasmith
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No , there was no integral. Another similar example is differentiation using the chain rule , for example if
##y(x) = (x+1)^2## then using u=x+1 , ##y(u) = u^2## and then this is differentiated using the chain rule
This is an abuse of notation.

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.
 
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  • #14
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Yes , I have a physics background and all the notes and books I use are mainly from the physics side of things which might explain the abuse of notation but it always seem to turn out right for physicists
 
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  • #15
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The difference is basically the following:
245330

which means it is a matter of coordinates: a shift by +1.
 
  • #16
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Thanks for all your replies. I think the problem is the difference between the way physicists and mathematicians are strict/not strict about notation
 
  • #17
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Thanks for all your replies. I think the problem is the difference between the way physicists and mathematicians are strict/not strict about notation
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
dyn
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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.
I was quite happy with the answers before this post but the above post does not make any sense
 

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