The Chain Rule and Function Composition

[PFuser1232]

This is a problem that has been bugging me for ages. I just can't wrap my head around this weird result. I know I went wrong somewhere [as a matter of fact, that was the answer I was hoping for], but most sources, (including, but not limited to, wikipedia), suggest otherwise.

I will cut to the chase.

A function f is defined by $$f(x)=x^2$$

The derivative, f', is then $$f'(x)=2x$$

So far so good.

Here is where all hell broke loose for me:

The composition of f and f' yields $$f[f'(x)]=(2x)^2=4x^2$$

I will now switch to Leibniz Notation for convinience.
u=f(x), and v=f'(x)

According to the chain rule

$\frac{du}{dx}$=$\frac{du}{dv}$$\frac{dv}{dx}$

$$f'(x)=f'[f'(x)]f''(x)$$

Well, we know what f'(x) is, it is 2x as shown above.
du/dv is (2)(2x)=4x
And dv/dx is simply 2.
Putting all of this in the above equation, for nonzero x: $$2x=8x$$ $$2=8$$

My speculation as to why this bizarre result popped out was the fact that my substitution of f'(x) for x in f(x) to get f[f'(x)] was wrong. But after going through several websites, I came to realize that there is nothing mathematically wrong with this substitution.

Can someone please point out to me where I went wrong with this?
I would be more than grateful.

 

[Fredrik]

I will now switch to Leibniz Notation for convinience.
u=f(x), and v=f'(x)

According to the chain rule

$\frac{du}{dx}$=$\frac{du}{dv}$$\frac{dv}{dx}$

$$f'(x)=f'[f'(x)]f''(x)$$

That's not what the chain rule says. In your notation, you would have to write u=f(f'(x)) instead of u=f(x). Now you have $\frac{du}{dx}=\frac{du}{dv}\frac{dv}{dx}$. I prefer the following notation: The chain rule says that $(g\circ h)'(x)=g'(h(x))h'(x)$. If you apply this with g=f and h=f', you get $(f\circ f')'(x)=f'(f'(x))f''(x)$.

 

[PFuser1232]

Makes sense now, thank you for pointing this out! :)
But one question remains regarding function composition. When it comes to physical systems (the simplest of which is a case in kinematics), I have trouble making sense of what happens while changing the argument of the function I'm looking at.
For instance, the velocity of a body, v m/s, measured at time t s is given by: $$v(t)=3+5t$$
I am given to understand that to find velocity as a function of another variable, say, acceleration (a m/s^2), we do this simple substitution: $$v(a)=3+5a$$
But isn't this a wrong result? I want your thoughts on this please.

 

[Fredrik]

If v(t)=3+5t for all t, then v(a)=3+5a for all a. For example, if you input 2 into v, the output will be 13 regardless of which of the symbols $a$ or $t$ that you used to represent the number 2. So the first formula (interpreted as a "for all" statement) certainly implies the latter.

But this isn't a way to find "velocity as a function of acceleration" in general. It only accomplishes that goal when the relationship between acceleration and time is t=a (which doesn't really make sense unless we somehow use the same units for acceleration and time, but I'll ignore that issue for now). If you have a formula that gives you a way to determine the value of t from the value of a, something like t=f(a), then you can certainly write 3+5t=3+5f(a). This is however not the same as plugging in a into v. The right-hand side is equal to v(t) (because that's what the left-hand side is equal to). It's not equal to v(f(a)) unless f(a)=t.

 

[PFuser1232]

I am glad you brought up the part about t=a being the constraint we assume if we are to find velocity as a function of acceleration in this manner.
Can't we use this same counterargument for any composite function $f(/circ)g$? The mere act of substituting g(x) for x implies that $g(x)=x$, doesn't it?

 

[PFuser1232]

Sorry, I meant $f \circ g$

 

[Fredrik]

Can't we use this same counterargument for any composite function $f(/circ)g$? The mere act of substituting g(x) for x implies that $g(x)=x$, doesn't it?

We can substitute one variable (or a more complicated expression) for another if we know that they're equal (i.e. if they represent the same number). For example, if we know that the variables x and y satisfy the constraint g(y)=x, we can substitute g(y) for x in any formula that involves x. However, if a function maps several numbers to the same number, then you may be able to substitute a variable for one with a different value without causing any problems. For example, in the formula $x^2=1$, you can substitute y for x if x=y OR if x=-y.

 

[PFuser1232]

But we are talking about $g(x)=x$, and not $g(y)=x$
Also, in the above example, is it technically wrong to say that $v(a)$ is velocity as a function of acceleration? Must it be $(v \circ f)(a)$? I am getting confused here.
If inputting g(x) for x is allowed to compute $(f \circ g)(x)$, why is plugging in a for t not allowed?

 

[Fredrik]

Regarding the g(x)=x thing, I don't understand what you want to know. Doesn't the $x^2=1$ example show that sometimes you can substitute a number for a different number?

You can certainly say that v(a) is a function of a. You don't have to know anything about any other variable, or what physical quantity v or a represents, to say that v(a) is a function of a. The notation represents a number. What number it represents depends on the value of a, i.e. what number a represents. That's all you need to know to say that v(a) is a function of a.

There's a subtle difference between the phrases "is a function" and "is a function of" that I should probably explain. When x and y satisfy a constraint like x+y=1, which enables us to determine y from x, we say that y is a function of x. However, y is not a function. It's a real number. The symbol "y" is a variable that represents a real number. We can define a function f by f(x)=1-x for all x. Now the constraint can be written y=f(x). Now f is a function and f(x) is a number in its range...which depends on the value of x, so we say that f(x) is a function of x, but that doesn't mean that f(x) is a function. The function is f. The symbol "f" is a variable that represents a function. The expression "f(x)" represents a number.

It will be easier to discuss your concerns if you come up with a realistic example, something from your physics book.

 

[PFuser1232]

Does that mean that when I say $v=v(t)$, then that second v is a function which maps t to the first v? So basically I can make the argument of v whatever i want, but it would ONLY equal the variable v, velocity, if $v=v(t)$? In other words, v equals v(t), but it does not equal v(a).
Am I getting this right?

 

[PFuser1232]

I think this confusion arises because in Physics it is a common practice to choose the same letter for both the function and the variable.

 

[Fredrik]

Does that mean that when I say $v=v(t)$, then that second v is a function which maps t to the first v? So basically I can make the argument of v whatever i want, but it would ONLY equal the variable v, velocity, if $v=v(t)$? In other words, v equals v(t), but it does not equal v(a).
Am I getting this right?

The teacher I had for the introduction to classical mechanics used notations like "v=v(t)" to say that velocity is a function of time. I think it's a pretty common notation in physics, but mathematicians don't use it. I once used it on a math exam, and the teacher looked like he wanted to slap me when I tried to explain it to him. I stopped using it after that. If you use it, you have to keep in mind that it's an abbreviation for the following much longer statement:

v and t are variables that represent the values of two physical quantities. There's a function f such that v=f(t). We will denote this function by v.​

If you write v(a), where v is that function, then v(a) is the velocity at time a. This is equal to the velocity at acceleration a, if t=a. (This doesn't really make sense because of the units). But if the relationship between t and a is more complicated, then v(a) may not be the velocity of the object when its acceleration is a.

You really need to find a realistic example, where we don't have to ignore that the units are all wrong. Preferably it should be one in which the relationship between the variables isn't just an equality.

 

[PFuser1232]

Actually when I was talking about v, a and t i was referring to dimensionless quantities such that velocity is $v ms^-1$, acceleration is $a ms^-2$ and time is $t s$.

 

[Stephen Tashi]

If we consider the case of the ideal "falling body", the acceleration is constant (with respect to time), so making velocity a function of acceleration is hopeless. Both position and velocity change, so making velocity a function of position is possible. Try that example.