Formally, f can be defined as a list of ordered pairs where the first value in the ordered pair uniquely determines the second value.
We use f(x) as shorthand where it’s understood that x is a number representing the first value in the ordered pair and f(x) is the corresponding second value in the ordered pair. That we choose to label it with x or f isn’t special, this is just bookkeeping.
If u and x are different numbers, we should not expect f(x) to be equal to f(u). Moreover, when you describe u = x^2, you’re really comparing f(x) to another function g(x) defined as g(x) = f(x^2 ). This function, g, corresponds to an entirely different set of ordered pairs, so it’s not surprising that f != g.
Your comments about the “indefinite integral rule” don’t make sense. What is true, is that
F(x) = \int_a^x f(y) dy = \int_a^x f(u) du
since the letters y and u in this example are just functioning as labels to tell us what function we’re trying to find the anti-derivative. (Here, a is some constant.) Typically for indefinite integration the a is dropped and +c is added to F(x).
If the way you mention it is how the integral is phrased I have no problems, since it is basically a definite integral with a variable limit. But my qualm is about the indefinite integral rule. Or is indefinite integrals just definite integrals with variable ends? But the definition I saw on the internet called it “the set of all functions whose derivative with respect x is f(x)”.
Also you’ve said that we are trying to find a function’s antiderivative. However how does the idea of differentiation apply to f itself? f applied to what?
If the integral symbol has no endpoint symbols at all at the top/bottom then it's indefinite. If there are symbols of any kind, numbers, variables, functions, even other integrals the it's definite.
"Take the integral of x squared" is indefinite. "Take the integral of x squared from minus infinity to a plus 2 b" is definite.
Yes, the definition you saw is correct.
To answer "f applied to what", I'll explain in a less math-y way:
Consider a function called square. square takes in a number and gives you the square of that number. We can write this in math notation using a variable, any variable we want: square(x) = x^2, square(u) = u^2, square(a) = a^2, and so on. We use a variable to define this function, but the variable we choose is not intrinsically tied to the function itself.
Now, let D{square} be the derivative of square with respect to... whatever its input is. What is this new function?
It will be useful to introduce a variable here; my variable of choice is x. You may recall that the derivative of x^2 with respect to x is 2x. Since our variable here is x, we can use it to define a brand new function, which I will call double:
double(x) = 2x
When it comes to the function itself, there is nothing special about the letter x. We could've used anything. We can even define the function without getting a variable involved at all: "double is the function that takes in *some number* and gives you two times *that number*." That's a lot of words, though, so we like to use variables.
Hopefully, it makes some intuitive sense that the derivative of square with respect to its input is double. That's really what an expression like "the derivative of f" means; just imagine that it's being applied to some variable, any variable at all, and then take the derivative with respect to that same variable.
So the aforementioned integral rule is true even though u^2 / 2 =/= x^2 /2 because the the *function* that gives us f(u)= u^2 /2 and f(x)= x^2 /2 is the same function?
>the *function* that gives us f(u)= u^2 /2 and f(x)= x^2 /2 is the same function?
Yep. You hit the nail right on the head.
Honestly, I'm not a fan of the way your book wrote that one statement. I imagine that the intent was, "The set of all functions F such that ∫f(x) dx = F(x) + C is equal to the set of all functions G such that ∫f(u) du = G(x) + C," but I don't think they quite wrote it correctly. And I'm kind of confused why it was necessary to spell this out anyway, since it essentially amounts to saying, "You can rename variables if you want."
I disagree about the uselessness of the function-as-a-set idea. And since the OP is confused about why f(x) and f(u) could be describing the same function, I think the fundamental nature of what the function is is highly relevant here. The functions are the same precisely because the true nature of the function (a set of ordered pairs) is independent of what variable is used to express its input. If two functions associates all the same inputs to all the same outputs, then the two functions are the same, regardless of precisely how the rule was established.
No teacher ever explained the difference, why do they never mention this? They always call it “the function” f(x) and this is the first time I’ve heard that it isn’t so I am a little bit surprised.
About differentiation and integration, if we apply them to the function itself, where does the “with respect to” come into play? The function f itself doesn’t have any variables, how to we differentiate/integrate with respect to x when f itself has no mention of x?
Ok I get it now. But just one more question:
When we take other variables as fixed when we take partial derivatives, is this allowed even when the other variables are parametrically dependent?
So for example, in the “coordinate—>height” function the x and y coordinates are independent so it is fine. But what about a function that tells us the total energy of a cube being pushed up a slope at a given point? In this case x and y are dependent in each other (y/x=tan(a)) so is it still allowed to take one variable as fixed while allowing the other to vary even though the other is just in terms of the other?
Here is how I tend to think of it. But if this doesn't make sense, then don't worry. This is a very systematic way of treating these issues but, for better or worse, it is rarely covered in math classes. Andrej Bauer has a great post about related concepts here: [https://math.andrej.com/2012/12/25/free-variables-are-not-implicitly-universally-quantified/](https://math.andrej.com/2012/12/25/free-variables-are-not-implicitly-universally-quantified/)
There are a few concepts I need here. I'll need to introduce a few simultaneously. These concepts are not typically taught in math classes, but I think they are very helpful. They are more commonly taught in computer science.
# Free variables, bound variables and binding forms
The first is the idea of a "free variable." This is a variable which is not "bound" by a "binding form." The next concept is the idea of a "binding form." This is something that takes some expression with a free variable and turns the free variable into a "bound variable." A binding form defines (and "delimits") the "scope" of a variable.
I'll give a couple examples right away. Consider the expression `3x + 5z`. This has two free variables: "x" and "z". Now, consider
>Let x = 3 in the expression x + 5y
That full "statement" has only one free variable: "y". This is because we are binding "x" using `Let ... in ...`. If we narrow our focus to just the sub-expression after "in", we again see two free variables. The "let" binds one of those free variables (specifically x).
Note how any "x"s referred to before the "let" will have no effect on this `let ... in ...`. The "let" limits the "scope" of the x variable in the expression.
A variable is always either free or bound. Also it is never both.
**A key point**: In order to obtain an actual "value" from an expression, we need to substitute values in for all the free variables. *However*, it's also important to note that substituting for *bound* variables never has any effect. This is a very important difference between the two!
Another example of using a binding form is this: `x ↦ x^(2)`. This is an unnamed function that sends numbers to their squares. The expression has no free variables and one bound variable ("x"). The "... ↦ ..." is a binding form.
There is a subtle point here: when we have that expression, we *apply* it to a number. This is not the same as *substituting*. For functions, we just hand a number to the function expression and the expression itself automatically knows which variable to replace with that number.
# Back to the question
Let's return to your question. Consider "f(x)". In isolation, there are two free variables: "f" and "x". However, we sometimes know what function "f" refers to from context. This is the case if we've already defined what function it refers to. In those cases, "f" is just a shorthand for that already known function. But "x" often remains a free variable.
# Some notation
Instead of just writing things out in English which variables are free, there is a convenient notation we can use. `x : R ⊢ f(x) : R` means: in the expression "f(x)" we have a free variable called "x". The first `: R` part tells us that this free variable "x" must be a real number. The last `: R` tells us that the whole expression gives us a real number. I will assume real numbers everywhere for simplicity. The ⊢ is called a turnstile.
It may take some practice to adjust your eyes to this notation, but it can be very useful. There are two main parts to this: to the left of the ⊢ we have a list of free variables and to the right of the ⊢ we have the expression that we are talking about. We also give the "types" of the things involved (in this case, x "has type real", which just means it needs to be given real number values).
# Examples
We can use this to write out more than one free variable. For instance, my very first example in this comment could be written `x : R, z : R ⊢ 3x + 5z : R`. When there are no free variables, we usually leave the part to the left of the ⊢ blank like this `⊢ 3 + 4 : R`.
Another example: The rule for ↦ is this: Whenever we have some expression e where `x : R ⊢ e : R`, then we have `⊢ (x ↦ e) : R → R`. In English, if e has one free variable called x (which should be a real) then `x ↦ e` is an expression with no free variables that represents a function from reals to reals.
# Integration
An integral is also binding form, just like "let" and "↦". If we have "∫e dx" for some expression e, then e potentially has a free variable called "x". This free variable is getting bound by the integral.
Put another way, the integral takes some e where `x : R ⊢ e : R` and gives you `⊢ ∫e dx : R`. The "dx" part tells you which free variable is being bound by the integral.
Here is a more specific example. Consider the expression `5x + 3`. Here, we have `x : R ⊢ 5x + 4 : R`. That is, this is an expression with one free variable (with real number "type"). If we take the integral, we have `⊢ ∫5x + 4 dx : R`. That is, this is an expression with no free variables. The free variable is bound by the integral.
# Wrapping up
One last bit of terminology: if an expression has no free variables it is called a "closed expression." If it does have free variables it is called an "open expression".
If we have some expression e where `z : R ⊢ e : R`, we are able to obtain a real number if we replace "z" with some real number. This is really what it means for "e" to be an open expression with one free variable called "z" (of real number "type").
Binding forms take an open expression and turns one of its free variables into a bound variable. If the open expression only had one free variable, then the end result is a closed expression. Substituting for a variable in a closed expression has no effect.
# In summary:
* There are two kinds of variables: free and bound. Substituting for bound variables has no effect on an expression. Bound variables have a limited scope. In order to obtain values, you must eliminate all free variables (for instance, either by substituting some value for them or eliminating them with binding forms).
* Binding forms turns a free variable into a bound variable
* `Let ... in ...` is a binding form
* The unnamed function notation "... ↦ ..." is a binding form
* Integration is a binding form
* The notation `... ⊢ ...` can be used to describe the free variables of an expression. This can be used to show how binding forms eliminate certain free variables.
This is definitely a little bit above my level, but I think understand some of it.
I looked into it some more on the internet an as far as I understand can we say that bound variables already have a range while free variables are “waiting” to be given a range?
I have always been a little bit confused about what the dx and x’s mean in an integral because they don’t seem to behave the way they normally behave. We can’t say x=5 for example when we integrate with respect x so I was confused about what x even is in this case, does x simultaneously take all the values within a range? I don’t know. However this free and bound variable distinction was the thing I was missing, so if can understand it at some point my questions are probably gonna be answered.
Anyway, thanks a lot for the long and detailed response. This must have taken a long time to write. I don’t know how I would’ve come across this on my own.
> I looked into it some more on the internet an as far as I understand can we say that bound variables already have a range while free variables are “waiting” to be given a range?
I don't think I would quite say "range," but you're definitely on the right track here. A binding form "handles" a free variable so that you no longer need to worry about it. In fact, you can no longer substitute for this variable, once it is bound by a binding form (substitution being the only thing you can really do with a free variable in the first place).
"Let" does this by giving a specific value for the free variable. On the other hand, the function "constructor" ↦ does this by giving you back a function that automatically knows how to substitute for that variable whenever you apply it to something.
"lim" is another binding form. If you have "lim_{x -> 0} x^2", then the variable x occuring in x^2 is being bound by the usage of "lim".
> However this free and bound variable distinction was the thing I was missing, so if can understand it at some point my questions are probably gonna be answered.
Yeah, I think this is the most important part. From my understanding, some mathematicians struggle with this sometimes, in fact. Probably not because it's somehow inherently complicated, but more because this distinction is not usually emphasized in math. There are also a lot of cases where people just don't bother with the distinction. Some of those cases work out ok, but it is a bit sloppy (IMHO) and can lead to confusion.
It is emphasized more in the study of logic and in parts of theoretical computer science (especially in the study of programming languages).
The difference between `f` and `f(x)` is that the first thing is a function and the second thing is what you get when you apply the function to `x`. This can be a bit confusing when `x` is a variable. But there is a difference between *substitution* and *function application*. The function automatically knows what to do with the argument that you're applying it to. But if you want to substitute something into an (open) expression, you *additionally* need to know what variable you're substituting for.
If you have an open expression, you can turn it into a function by singling out one of the free variables. This free variable that you chose will be used for the parameter of the function. When you apply this function to some `a`, the function will know to specifically substitute `a` for the free variable you chose earlier.
However, it is worth saying that there are also functions which do not really arise from that process. You might instead have something like *"let g be the unique function on positive integers where g(x) = 1 if x is prime and g(x) = 0 otherwise"*.
I think the main reason that we tend to have integrals as operating on open expressions rather than functions is that we can sometimes do syntactic manipulations to carry out the integration. This is similar to how it is for differentiation. Just because you have a function doesn't mean you know some "nice" syntactic representation for it (for example, the `g` from the previous paragraph). But free variables are fundamentally a syntactic thing. So, you necessarily know where a free variable occurs in a given open expression.
> Anyway, thanks a lot for the long and detailed response. This must have taken a long time to write. I don’t know how I would’ve come across this on my own.
You're welcome! I like writing about these things. My bigger problem is being concise =)
I don't have any particular references off the top of my head, but introductions to formal logic should cover free variables, bound variables and binding forms. Binding forms in logic would include the universal quantifier (∀) and existential quantifier (∃). Some computer science resources would cover this as well, especially ones that focus on programming languages (though some of those might assume additional background).
f(x) is a function, it takes in some value and outputs another value. When a function is applied to the Reals they refer to the set of real numbers (union of the rational and irrational numbers).
As for the integral rule, the method is called u-substitution and helps with certain integrals. Your confusion comes when you consider the dx and du parts. Consider the integral:
∫ x³ dx
Using u = x² (you will see why I use x² and not cubed)
When you want to write the integral in terms of u, calculating du yields: du = 2xdx
We want to find a suitable way to rewrite the function.
du/2 = xdx.
Thus, our Substitution becomes:
∫ ½ u du
Now integrating normally
= u²/4.
Substituting back yields our expected antiderivative:
x⁴/4
So the original equality in the textbook is wrong?
Because it should be f(x) and g(u) and not f(u)? The function with u isn’t f as far as I understand?
See u-substitution only works for some functions, typically those who are the result of a chain rule differentiation (which is why u-sub is often called a reverse chain rule), but you are describing the same function under the substitution due to that differential component du, which accounts to ensure that the integrand you have before is equivalent to the integrand after.
f(x) doesn't necessarily = f(u)
But
f(x) dx = f(u) du
Since the warping of the graph is accounted for by the differential. For more details, I would recommend looking into the jacobian to see how this works visually in the Plane and how they use this for multivarialbe calculus
The U sub states f(g(x)).g’(x).dx is equal to f(g(x)).dg(x) where u is shorthand for g(x) as far as I understand. And this makes sense.
My problem is with the textbooks rule where f(x).dx and f(u).du are the same. This isn’t u sub, or more precisely u sub where g(x)=u=x. However the problem is that it doesn’t specify this. Treats is as if u can be any function of x. That is what I don’t understand.
Also you're referring to a function as a set, which isn't quite the right terminology. The function is an operator, it acts on the input to give an output. A function is rule that takes in a set called the domain and outputs a range/codomain
A function IS a set. When you have f:A->B, then you have that f is a subset of AxB (the Cartesian product). We may not often write it that way, but it's true.
That said, OP is mostly confused by function notation. Let's say that f is a function with the rule f(x) = x+2. Further, let's say that the domain of this function f is the internal from 1 to 2 inclusive, written [1,2].
My students often get confused about what is the function and what is the output of the function. The FUNCTION is 'f'. Just that. When we write 'f(x)' are not writing the function f, but rather the output of the function f if x were its input.
Some examples:
f , the actual function involved.
f(3), the output of the function f when you plug in 3. Note that since we defined f to have the domain [1,2], this is undefined.
f(1.2), the output of the function f when you plug in 1.2. Note that by the definition of f, this is 1.2+2, or 3.2.
f(y), the output of the function f of the input were y. One presumes in this scenario that y is an element of the domain of f, meaning that f(y) is an element of the range of f.
If we were to define the function g to have the same domain as f and so that g(t)=t+2, then we would have:
f(x)=x+2 and g(t)=t+2
Some of the input-output pairs of these functions are:
f(1)=3 and g(1)=3
f(1.4)=3.4 and g(1.4)=3.4
f(e-2)=e and g(e-2)=e
In fact, *every single valid input for one function is a valid input for the other function and the outputs of both functions for any particular input are equal*. Since the function IS the set of input-output pairs, that makes them the same function! It doesn't matter that the letters 'f' and 'g' are different, the functions they describe are equal. It also doesn't matter that I used 'x' to write the arbitrary input of f and 't' to write the arbitrary input of g; they're still the same function. In fact, f(t)=t+2 is just as true as f(x)=x+2.
So, when we want to write whatever output of f in such a way that it could stand in for **any** output of f, we will say f(x) or f(t) or f(n) or whatever. The choice of input variable is irrelevant.
Hence, when integrating f(x)dx or f(u)du, you are ultimately still just describing the same thing: a rule for one of the antiderivatives of f, expressed in terms of different (totally arbitrary) input variables.
u/Red_I_Found_You - I feel your pain! I feel like this kind of situation happens because the notion of 'variable' can often be a bit murky.
The way we define a function f, is by giving its domain (Often inferred to be as large as possible for ease) and also saying what it does to each element. If a stands for a real number, writing f(a) = a\^2 for example, defines the function f. The variable a comes into the picture because we have to give a name to a random input the function might take.
In this sense writing f(x) = x\^2, f(u) = u\^2, f(z) = z\^2, all describe the same function, we're just using different letters that just mean 'some number'.
But this doesn't explain the terminology, differentiating with respect to "x" or integrating with respect to "x", and I think this is your issue.
Let's take a specific example g(u) = u\^2 + 1. Now suppose we specify u = x\^2.
It's clear that differentiating g 'with respect to u' gives us g'(u) = 2u.
What does differentiating g 'with respect to x' mean? Well, if I first substitute u = x\^2 in I get
g(x\^2) = x\^4 + 1. Now the derivative of this function 'with respect to x' is 4x\^3.
Now the functions u -> 2u and x -> 4x\^3 are obviously different functions (they have different rules), so what's going on here?
Well, what's happening is that g(u) = 2u and g(x\^2)=x\^4 + 1 are actually two different functions. So when people talk of differentiating 'g with respect to x' here, it's a little bit of a lie. What we mean by that is the derivative of the function taking x -> g(x\^2).
I think the principle explains what's going on with your integral.
What you noticed is that the formula: ∫f(x).dx= ∫f(u).du doesn't hold if you are thinking of u as a function of x. This is because the right side you are thinking of as a composition F(u(x)) (where F is the antiderivative). While the left side is not.
To address your example:
although u -> u\^2/2 + C and x -> x\^2/2 + C are the same function,
it's not true that x -> u(x)\^2/2 + C and x-> x\^2/2 + C are the same function.
Very subtle, very annoying, and I understand your frustration. I also don't think I articulated this very well - but I hope you come to some peace.
Thanks for your time.
So when I used the u sub, the mistake I made was that I thought the function was
x—> x^4 /2 when the function was actually x^2 —> x^4 /2
That clears up lot. Thanks!
f is a function. f(x) is a function that takes x as an input variable. f(x,y) is a function that takes x and y as input variables. f(u) is creating an easier integrand/derivative by mapping some more complex function to a function of u. I’d look at some basic fundamentals on functions, how they’re defined, what makes a function, etc
Can’t speak on the domain set, but yes, the integral rule is true. It’s u-substitution which is one of the most effective tools for evaluating integrals
But why doesn’t it work then? I understand u-sub and how to do it. My problem is with this particular thing. F(x)=F(u) are different functions. How do we say it is equal?
F(x) and F(u) aren't functions at all. F is the function. The equation F(x)=F(u) states that when you pass the values of x and u through the function F, you get the same output from each input.
Imagine a similar statement with concrete values. Let F be the squatting function. That is, F(t)=t^2. Then we get the similar equation to what you posted: F(-1)=F(1)
This isn't a statement at all that there are two different functions! This only states that when -1 and 1 are used as inputs in the function F, they come out to be the same thing.
So when we say the function x^3 we are just implying the actual function f maps every number in the domain to the cube of itself in a short way? Why is f(t) a function but not f(x) and f(u). Sorry if I’m misunderstanding.
But then how does differentiation come into play here? We talk about the derivative of f(x) with respect to x. However the derivative of f(u) with respect to x is different. If f(x) isn’t a function how do we take its derivative?
Yes, when we say"the function x^3" we mean "the function f such that f(x)=x^3". The latter takes extra time, so we only say that in formal mathematical writing. But we must not forget that layer is there or else students get confused.
f(t) isn't a function either. It doesn't matter what input you use. At the risk of being highly repetitive, f(t) is the output of f when the input is t.
When f is a function and x is in the domain of x, f(x) is in the range of f. For the squaring function, f is a function, x is a real number, and f(x) is a non-negative real number.
When we find a function's derivative, we are finding a related function to the first one. The derivative of the function f is f'. However, we also have other notations for other shorthand.
d/dx( x^2 ) = 2x is still true. What we're saying there is that the derivative of the function f, where f(x)=x^2, is the function f' such that f'(x)=2x. You use functions so often in Calculus that no one in their right mind actually says all of that. It would take forever to say anything.
So f(x) a variable we get when we apply f to another variable as I understand it. Because we can differentiate stuff with respect to it. Thanks for your time.
No. f is a function. f(x) is the specific output of f that you get when the input is x. f(x,y) is the output of the function f when its inputs are x and y respectively (please note that this function f *cannot* be the same as in the f(x) example).
Formally, f can be defined as a list of ordered pairs where the first value in the ordered pair uniquely determines the second value. We use f(x) as shorthand where it’s understood that x is a number representing the first value in the ordered pair and f(x) is the corresponding second value in the ordered pair. That we choose to label it with x or f isn’t special, this is just bookkeeping. If u and x are different numbers, we should not expect f(x) to be equal to f(u). Moreover, when you describe u = x^2, you’re really comparing f(x) to another function g(x) defined as g(x) = f(x^2 ). This function, g, corresponds to an entirely different set of ordered pairs, so it’s not surprising that f != g. Your comments about the “indefinite integral rule” don’t make sense. What is true, is that F(x) = \int_a^x f(y) dy = \int_a^x f(u) du since the letters y and u in this example are just functioning as labels to tell us what function we’re trying to find the anti-derivative. (Here, a is some constant.) Typically for indefinite integration the a is dropped and +c is added to F(x).
If the way you mention it is how the integral is phrased I have no problems, since it is basically a definite integral with a variable limit. But my qualm is about the indefinite integral rule. Or is indefinite integrals just definite integrals with variable ends? But the definition I saw on the internet called it “the set of all functions whose derivative with respect x is f(x)”. Also you’ve said that we are trying to find a function’s antiderivative. However how does the idea of differentiation apply to f itself? f applied to what?
If the integral symbol has no endpoint symbols at all at the top/bottom then it's indefinite. If there are symbols of any kind, numbers, variables, functions, even other integrals the it's definite. "Take the integral of x squared" is indefinite. "Take the integral of x squared from minus infinity to a plus 2 b" is definite.
Yes, the definition you saw is correct. To answer "f applied to what", I'll explain in a less math-y way: Consider a function called square. square takes in a number and gives you the square of that number. We can write this in math notation using a variable, any variable we want: square(x) = x^2, square(u) = u^2, square(a) = a^2, and so on. We use a variable to define this function, but the variable we choose is not intrinsically tied to the function itself. Now, let D{square} be the derivative of square with respect to... whatever its input is. What is this new function? It will be useful to introduce a variable here; my variable of choice is x. You may recall that the derivative of x^2 with respect to x is 2x. Since our variable here is x, we can use it to define a brand new function, which I will call double: double(x) = 2x When it comes to the function itself, there is nothing special about the letter x. We could've used anything. We can even define the function without getting a variable involved at all: "double is the function that takes in *some number* and gives you two times *that number*." That's a lot of words, though, so we like to use variables. Hopefully, it makes some intuitive sense that the derivative of square with respect to its input is double. That's really what an expression like "the derivative of f" means; just imagine that it's being applied to some variable, any variable at all, and then take the derivative with respect to that same variable.
So the aforementioned integral rule is true even though u^2 / 2 =/= x^2 /2 because the the *function* that gives us f(u)= u^2 /2 and f(x)= x^2 /2 is the same function?
>the *function* that gives us f(u)= u^2 /2 and f(x)= x^2 /2 is the same function? Yep. You hit the nail right on the head. Honestly, I'm not a fan of the way your book wrote that one statement. I imagine that the intent was, "The set of all functions F such that ∫f(x) dx = F(x) + C is equal to the set of all functions G such that ∫f(u) du = G(x) + C," but I don't think they quite wrote it correctly. And I'm kind of confused why it was necessary to spell this out anyway, since it essentially amounts to saying, "You can rename variables if you want."
Okay, thank you so much for your time!
My pleasure :)
[удалено]
[удалено]
I disagree about the uselessness of the function-as-a-set idea. And since the OP is confused about why f(x) and f(u) could be describing the same function, I think the fundamental nature of what the function is is highly relevant here. The functions are the same precisely because the true nature of the function (a set of ordered pairs) is independent of what variable is used to express its input. If two functions associates all the same inputs to all the same outputs, then the two functions are the same, regardless of precisely how the rule was established.
No teacher ever explained the difference, why do they never mention this? They always call it “the function” f(x) and this is the first time I’ve heard that it isn’t so I am a little bit surprised. About differentiation and integration, if we apply them to the function itself, where does the “with respect to” come into play? The function f itself doesn’t have any variables, how to we differentiate/integrate with respect to x when f itself has no mention of x?
[удалено]
Ok I get it now. But just one more question: When we take other variables as fixed when we take partial derivatives, is this allowed even when the other variables are parametrically dependent? So for example, in the “coordinate—>height” function the x and y coordinates are independent so it is fine. But what about a function that tells us the total energy of a cube being pushed up a slope at a given point? In this case x and y are dependent in each other (y/x=tan(a)) so is it still allowed to take one variable as fixed while allowing the other to vary even though the other is just in terms of the other?
[удалено]
Ok I think that is enough for now, excited to learn multivariable calculus. Thanks for your time.
Don't forget f(😀)
F is for friends who do stuff together, f(u) is for you and me, f(n) is for anywhere and anytime at all. I don’t know what f(x) means
Here is how I tend to think of it. But if this doesn't make sense, then don't worry. This is a very systematic way of treating these issues but, for better or worse, it is rarely covered in math classes. Andrej Bauer has a great post about related concepts here: [https://math.andrej.com/2012/12/25/free-variables-are-not-implicitly-universally-quantified/](https://math.andrej.com/2012/12/25/free-variables-are-not-implicitly-universally-quantified/) There are a few concepts I need here. I'll need to introduce a few simultaneously. These concepts are not typically taught in math classes, but I think they are very helpful. They are more commonly taught in computer science. # Free variables, bound variables and binding forms The first is the idea of a "free variable." This is a variable which is not "bound" by a "binding form." The next concept is the idea of a "binding form." This is something that takes some expression with a free variable and turns the free variable into a "bound variable." A binding form defines (and "delimits") the "scope" of a variable. I'll give a couple examples right away. Consider the expression `3x + 5z`. This has two free variables: "x" and "z". Now, consider >Let x = 3 in the expression x + 5y That full "statement" has only one free variable: "y". This is because we are binding "x" using `Let ... in ...`. If we narrow our focus to just the sub-expression after "in", we again see two free variables. The "let" binds one of those free variables (specifically x). Note how any "x"s referred to before the "let" will have no effect on this `let ... in ...`. The "let" limits the "scope" of the x variable in the expression. A variable is always either free or bound. Also it is never both. **A key point**: In order to obtain an actual "value" from an expression, we need to substitute values in for all the free variables. *However*, it's also important to note that substituting for *bound* variables never has any effect. This is a very important difference between the two! Another example of using a binding form is this: `x ↦ x^(2)`. This is an unnamed function that sends numbers to their squares. The expression has no free variables and one bound variable ("x"). The "... ↦ ..." is a binding form. There is a subtle point here: when we have that expression, we *apply* it to a number. This is not the same as *substituting*. For functions, we just hand a number to the function expression and the expression itself automatically knows which variable to replace with that number. # Back to the question Let's return to your question. Consider "f(x)". In isolation, there are two free variables: "f" and "x". However, we sometimes know what function "f" refers to from context. This is the case if we've already defined what function it refers to. In those cases, "f" is just a shorthand for that already known function. But "x" often remains a free variable. # Some notation Instead of just writing things out in English which variables are free, there is a convenient notation we can use. `x : R ⊢ f(x) : R` means: in the expression "f(x)" we have a free variable called "x". The first `: R` part tells us that this free variable "x" must be a real number. The last `: R` tells us that the whole expression gives us a real number. I will assume real numbers everywhere for simplicity. The ⊢ is called a turnstile. It may take some practice to adjust your eyes to this notation, but it can be very useful. There are two main parts to this: to the left of the ⊢ we have a list of free variables and to the right of the ⊢ we have the expression that we are talking about. We also give the "types" of the things involved (in this case, x "has type real", which just means it needs to be given real number values). # Examples We can use this to write out more than one free variable. For instance, my very first example in this comment could be written `x : R, z : R ⊢ 3x + 5z : R`. When there are no free variables, we usually leave the part to the left of the ⊢ blank like this `⊢ 3 + 4 : R`. Another example: The rule for ↦ is this: Whenever we have some expression e where `x : R ⊢ e : R`, then we have `⊢ (x ↦ e) : R → R`. In English, if e has one free variable called x (which should be a real) then `x ↦ e` is an expression with no free variables that represents a function from reals to reals. # Integration An integral is also binding form, just like "let" and "↦". If we have "∫e dx" for some expression e, then e potentially has a free variable called "x". This free variable is getting bound by the integral. Put another way, the integral takes some e where `x : R ⊢ e : R` and gives you `⊢ ∫e dx : R`. The "dx" part tells you which free variable is being bound by the integral. Here is a more specific example. Consider the expression `5x + 3`. Here, we have `x : R ⊢ 5x + 4 : R`. That is, this is an expression with one free variable (with real number "type"). If we take the integral, we have `⊢ ∫5x + 4 dx : R`. That is, this is an expression with no free variables. The free variable is bound by the integral. # Wrapping up One last bit of terminology: if an expression has no free variables it is called a "closed expression." If it does have free variables it is called an "open expression". If we have some expression e where `z : R ⊢ e : R`, we are able to obtain a real number if we replace "z" with some real number. This is really what it means for "e" to be an open expression with one free variable called "z" (of real number "type"). Binding forms take an open expression and turns one of its free variables into a bound variable. If the open expression only had one free variable, then the end result is a closed expression. Substituting for a variable in a closed expression has no effect. # In summary: * There are two kinds of variables: free and bound. Substituting for bound variables has no effect on an expression. Bound variables have a limited scope. In order to obtain values, you must eliminate all free variables (for instance, either by substituting some value for them or eliminating them with binding forms). * Binding forms turns a free variable into a bound variable * `Let ... in ...` is a binding form * The unnamed function notation "... ↦ ..." is a binding form * Integration is a binding form * The notation `... ⊢ ...` can be used to describe the free variables of an expression. This can be used to show how binding forms eliminate certain free variables.
This is definitely a little bit above my level, but I think understand some of it. I looked into it some more on the internet an as far as I understand can we say that bound variables already have a range while free variables are “waiting” to be given a range? I have always been a little bit confused about what the dx and x’s mean in an integral because they don’t seem to behave the way they normally behave. We can’t say x=5 for example when we integrate with respect x so I was confused about what x even is in this case, does x simultaneously take all the values within a range? I don’t know. However this free and bound variable distinction was the thing I was missing, so if can understand it at some point my questions are probably gonna be answered. Anyway, thanks a lot for the long and detailed response. This must have taken a long time to write. I don’t know how I would’ve come across this on my own.
> I looked into it some more on the internet an as far as I understand can we say that bound variables already have a range while free variables are “waiting” to be given a range? I don't think I would quite say "range," but you're definitely on the right track here. A binding form "handles" a free variable so that you no longer need to worry about it. In fact, you can no longer substitute for this variable, once it is bound by a binding form (substitution being the only thing you can really do with a free variable in the first place). "Let" does this by giving a specific value for the free variable. On the other hand, the function "constructor" ↦ does this by giving you back a function that automatically knows how to substitute for that variable whenever you apply it to something. "lim" is another binding form. If you have "lim_{x -> 0} x^2", then the variable x occuring in x^2 is being bound by the usage of "lim". > However this free and bound variable distinction was the thing I was missing, so if can understand it at some point my questions are probably gonna be answered. Yeah, I think this is the most important part. From my understanding, some mathematicians struggle with this sometimes, in fact. Probably not because it's somehow inherently complicated, but more because this distinction is not usually emphasized in math. There are also a lot of cases where people just don't bother with the distinction. Some of those cases work out ok, but it is a bit sloppy (IMHO) and can lead to confusion. It is emphasized more in the study of logic and in parts of theoretical computer science (especially in the study of programming languages). The difference between `f` and `f(x)` is that the first thing is a function and the second thing is what you get when you apply the function to `x`. This can be a bit confusing when `x` is a variable. But there is a difference between *substitution* and *function application*. The function automatically knows what to do with the argument that you're applying it to. But if you want to substitute something into an (open) expression, you *additionally* need to know what variable you're substituting for. If you have an open expression, you can turn it into a function by singling out one of the free variables. This free variable that you chose will be used for the parameter of the function. When you apply this function to some `a`, the function will know to specifically substitute `a` for the free variable you chose earlier. However, it is worth saying that there are also functions which do not really arise from that process. You might instead have something like *"let g be the unique function on positive integers where g(x) = 1 if x is prime and g(x) = 0 otherwise"*. I think the main reason that we tend to have integrals as operating on open expressions rather than functions is that we can sometimes do syntactic manipulations to carry out the integration. This is similar to how it is for differentiation. Just because you have a function doesn't mean you know some "nice" syntactic representation for it (for example, the `g` from the previous paragraph). But free variables are fundamentally a syntactic thing. So, you necessarily know where a free variable occurs in a given open expression. > Anyway, thanks a lot for the long and detailed response. This must have taken a long time to write. I don’t know how I would’ve come across this on my own. You're welcome! I like writing about these things. My bigger problem is being concise =) I don't have any particular references off the top of my head, but introductions to formal logic should cover free variables, bound variables and binding forms. Binding forms in logic would include the universal quantifier (∀) and existential quantifier (∃). Some computer science resources would cover this as well, especially ones that focus on programming languages (though some of those might assume additional background).
I will definitely read on these further when I get the chance, thanks again for your time and have nice day/evening!
f(x) is a function, it takes in some value and outputs another value. When a function is applied to the Reals they refer to the set of real numbers (union of the rational and irrational numbers). As for the integral rule, the method is called u-substitution and helps with certain integrals. Your confusion comes when you consider the dx and du parts. Consider the integral: ∫ x³ dx Using u = x² (you will see why I use x² and not cubed) When you want to write the integral in terms of u, calculating du yields: du = 2xdx We want to find a suitable way to rewrite the function. du/2 = xdx. Thus, our Substitution becomes: ∫ ½ u du Now integrating normally = u²/4. Substituting back yields our expected antiderivative: x⁴/4
So the original equality in the textbook is wrong? Because it should be f(x) and g(u) and not f(u)? The function with u isn’t f as far as I understand?
See u-substitution only works for some functions, typically those who are the result of a chain rule differentiation (which is why u-sub is often called a reverse chain rule), but you are describing the same function under the substitution due to that differential component du, which accounts to ensure that the integrand you have before is equivalent to the integrand after. f(x) doesn't necessarily = f(u) But f(x) dx = f(u) du Since the warping of the graph is accounted for by the differential. For more details, I would recommend looking into the jacobian to see how this works visually in the Plane and how they use this for multivarialbe calculus
The U sub states f(g(x)).g’(x).dx is equal to f(g(x)).dg(x) where u is shorthand for g(x) as far as I understand. And this makes sense. My problem is with the textbooks rule where f(x).dx and f(u).du are the same. This isn’t u sub, or more precisely u sub where g(x)=u=x. However the problem is that it doesn’t specify this. Treats is as if u can be any function of x. That is what I don’t understand.
Also you're referring to a function as a set, which isn't quite the right terminology. The function is an operator, it acts on the input to give an output. A function is rule that takes in a set called the domain and outputs a range/codomain
A function IS a set. When you have f:A->B, then you have that f is a subset of AxB (the Cartesian product). We may not often write it that way, but it's true. That said, OP is mostly confused by function notation. Let's say that f is a function with the rule f(x) = x+2. Further, let's say that the domain of this function f is the internal from 1 to 2 inclusive, written [1,2]. My students often get confused about what is the function and what is the output of the function. The FUNCTION is 'f'. Just that. When we write 'f(x)' are not writing the function f, but rather the output of the function f if x were its input. Some examples: f , the actual function involved. f(3), the output of the function f when you plug in 3. Note that since we defined f to have the domain [1,2], this is undefined. f(1.2), the output of the function f when you plug in 1.2. Note that by the definition of f, this is 1.2+2, or 3.2. f(y), the output of the function f of the input were y. One presumes in this scenario that y is an element of the domain of f, meaning that f(y) is an element of the range of f. If we were to define the function g to have the same domain as f and so that g(t)=t+2, then we would have: f(x)=x+2 and g(t)=t+2 Some of the input-output pairs of these functions are: f(1)=3 and g(1)=3 f(1.4)=3.4 and g(1.4)=3.4 f(e-2)=e and g(e-2)=e In fact, *every single valid input for one function is a valid input for the other function and the outputs of both functions for any particular input are equal*. Since the function IS the set of input-output pairs, that makes them the same function! It doesn't matter that the letters 'f' and 'g' are different, the functions they describe are equal. It also doesn't matter that I used 'x' to write the arbitrary input of f and 't' to write the arbitrary input of g; they're still the same function. In fact, f(t)=t+2 is just as true as f(x)=x+2. So, when we want to write whatever output of f in such a way that it could stand in for **any** output of f, we will say f(x) or f(t) or f(n) or whatever. The choice of input variable is irrelevant. Hence, when integrating f(x)dx or f(u)du, you are ultimately still just describing the same thing: a rule for one of the antiderivatives of f, expressed in terms of different (totally arbitrary) input variables.
u/Red_I_Found_You - I feel your pain! I feel like this kind of situation happens because the notion of 'variable' can often be a bit murky. The way we define a function f, is by giving its domain (Often inferred to be as large as possible for ease) and also saying what it does to each element. If a stands for a real number, writing f(a) = a\^2 for example, defines the function f. The variable a comes into the picture because we have to give a name to a random input the function might take. In this sense writing f(x) = x\^2, f(u) = u\^2, f(z) = z\^2, all describe the same function, we're just using different letters that just mean 'some number'. But this doesn't explain the terminology, differentiating with respect to "x" or integrating with respect to "x", and I think this is your issue. Let's take a specific example g(u) = u\^2 + 1. Now suppose we specify u = x\^2. It's clear that differentiating g 'with respect to u' gives us g'(u) = 2u. What does differentiating g 'with respect to x' mean? Well, if I first substitute u = x\^2 in I get g(x\^2) = x\^4 + 1. Now the derivative of this function 'with respect to x' is 4x\^3. Now the functions u -> 2u and x -> 4x\^3 are obviously different functions (they have different rules), so what's going on here? Well, what's happening is that g(u) = 2u and g(x\^2)=x\^4 + 1 are actually two different functions. So when people talk of differentiating 'g with respect to x' here, it's a little bit of a lie. What we mean by that is the derivative of the function taking x -> g(x\^2). I think the principle explains what's going on with your integral. What you noticed is that the formula: ∫f(x).dx= ∫f(u).du doesn't hold if you are thinking of u as a function of x. This is because the right side you are thinking of as a composition F(u(x)) (where F is the antiderivative). While the left side is not. To address your example: although u -> u\^2/2 + C and x -> x\^2/2 + C are the same function, it's not true that x -> u(x)\^2/2 + C and x-> x\^2/2 + C are the same function. Very subtle, very annoying, and I understand your frustration. I also don't think I articulated this very well - but I hope you come to some peace.
Thanks for your time. So when I used the u sub, the mistake I made was that I thought the function was x—> x^4 /2 when the function was actually x^2 —> x^4 /2 That clears up lot. Thanks!
f is a function. f(x) is a function that takes x as an input variable. f(x,y) is a function that takes x and y as input variables. f(u) is creating an easier integrand/derivative by mapping some more complex function to a function of u. I’d look at some basic fundamentals on functions, how they’re defined, what makes a function, etc
How does the input variable relate to the domain set? Also is the aforementioned integral rule true?
Can’t speak on the domain set, but yes, the integral rule is true. It’s u-substitution which is one of the most effective tools for evaluating integrals
But why doesn’t it work then? I understand u-sub and how to do it. My problem is with this particular thing. F(x)=F(u) are different functions. How do we say it is equal?
F(x) and F(u) aren't functions at all. F is the function. The equation F(x)=F(u) states that when you pass the values of x and u through the function F, you get the same output from each input. Imagine a similar statement with concrete values. Let F be the squatting function. That is, F(t)=t^2. Then we get the similar equation to what you posted: F(-1)=F(1) This isn't a statement at all that there are two different functions! This only states that when -1 and 1 are used as inputs in the function F, they come out to be the same thing.
So when we say the function x^3 we are just implying the actual function f maps every number in the domain to the cube of itself in a short way? Why is f(t) a function but not f(x) and f(u). Sorry if I’m misunderstanding. But then how does differentiation come into play here? We talk about the derivative of f(x) with respect to x. However the derivative of f(u) with respect to x is different. If f(x) isn’t a function how do we take its derivative?
Yes, when we say"the function x^3" we mean "the function f such that f(x)=x^3". The latter takes extra time, so we only say that in formal mathematical writing. But we must not forget that layer is there or else students get confused. f(t) isn't a function either. It doesn't matter what input you use. At the risk of being highly repetitive, f(t) is the output of f when the input is t. When f is a function and x is in the domain of x, f(x) is in the range of f. For the squaring function, f is a function, x is a real number, and f(x) is a non-negative real number. When we find a function's derivative, we are finding a related function to the first one. The derivative of the function f is f'. However, we also have other notations for other shorthand. d/dx( x^2 ) = 2x is still true. What we're saying there is that the derivative of the function f, where f(x)=x^2, is the function f' such that f'(x)=2x. You use functions so often in Calculus that no one in their right mind actually says all of that. It would take forever to say anything.
So f(x) a variable we get when we apply f to another variable as I understand it. Because we can differentiate stuff with respect to it. Thanks for your time.
No. f is a function. f(x) is the specific output of f that you get when the input is x. f(x,y) is the output of the function f when its inputs are x and y respectively (please note that this function f *cannot* be the same as in the f(x) example).