# Changing the argument of a function

• I
• dyn
In summary, the conversation discusses the use of the chain rule in physics and the abuse of notation when defining functions. It is important to note the difference between mathematical and physical notation and to provide clear context when discussing mathematical concepts. The conversation also touches on the concept of change of coordinates and its relevance in physics.
dyn
Hi.
If I have a function f(x) = √(x+1) and I define u=x+1 is it correct to state f(u) = √u ?

dyn said:
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}\,.##

VVS2000
Thanks , what would f(u) be then ?

dyn said:
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.

jim mcnamara
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

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\,##.

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

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.

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

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:
dyn said:
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

dyn said:
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 textbooks 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##.

dyn
dyn said:
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 $f : x \mapsto x^2$ and $g : x \mapsto x + 1$ then $h = f \circ g : x \mapsto (x + 1)^2$. $h$ is not the same function as $f$.

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

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

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

dyn
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

PeroK
The difference is basically the following:

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

Thanks for all your replies. I think the problem is the difference between the way physicists and mathematicians are strict/not strict about notation

dyn said:
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.

fresh_42 said:
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

## 1. What does it mean to change the argument of a function?

Changing the argument of a function refers to modifying the input value that is passed into the function. This can affect the output or result of the function, making it a useful tool for manipulating data or performing different calculations.

## 2. How do you change the argument of a function?

To change the argument of a function, you can simply modify the value that is passed into the function when it is called. This can be done by assigning a new value to the variable or parameter that represents the argument, or by using a different value altogether.

## 3. Can changing the argument of a function affect its return value?

Yes, changing the argument of a function can definitely affect its return value. Since the argument is used as the input for the function, modifying it can change the way the function processes the data and ultimately result in a different return value.

## 4. What are some reasons for changing the argument of a function?

There are many reasons why one might want to change the argument of a function. Some common reasons include performing different calculations or operations on the data, filtering or manipulating the data in a specific way, or customizing the output of the function based on different inputs.

## 5. Are there any limitations to changing the argument of a function?

Yes, there can be limitations to changing the argument of a function. Depending on the function's design, there may be specific data types or formats that are required for the argument, or certain constraints on the range of values that can be used. Additionally, changing the argument may not always result in a meaningful or useful output, so it is important to understand the function's purpose and how it processes data before making any modifications.

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