Cannot understand dummy variables.

[Amad27]

In integration such as,

$$\int_{a}^{b} f(x) dx$$

It is said that that is the exact same as:

$$\int_{a}^{b} f(y) dy = \int_{a}^{b} f(\theta) d\theta = \int_{a}^{b} f(\alpha) d\alpha = \int_{a}^{b} f(\psi)d\psi$$

An the list goes on forever.

I cannot understand,

How is it justified to change the so-called 'dummy-variables'??

Thank you!

 

[Ishuda]

In integration such as,

$$\int_{a}^{b} f(x) dx$$

It is said that that is the exact same as:

$$\int_{a}^{b} f(y) dy = \int_{a}^{b} f(\theta) d\theta = \int_{a}^{b} f(\alpha) d\alpha = \int_{a}^{b} f(\psi)d\psi$$

An the list goes on forever.

I cannot understand,

How is it justified to change the so-called 'dummy-variables'??

Thank you!

Take a function, say f(x). This just says 'pick something (in the domain) and substitute it for x where ever you see x', So , suppose the domain is the real numbers and
f(x) = 2 * x
To evaluate f(x), follow the rule above. For example pick 2 for x. Then f(2) = 2 * 2 = 4.

Now x in the above is called a dummy variable because we could just as well have said Take a function, say f(something-or-another). Again that would just say ''pick something (in the domain) and substitute it for something-or-another where ever you see something-or-another'. Using the same example as above we have
f(something-or-another) = 2 * something-or-another
If we pick 2 for something-or-another, then we get the same answer; f(2) = 2 * 2 = 4.

Note that we have also followed that first function example. It is just that we picked something-or-another for x and substituted it for x everywhere there was an x.

I hope that helps.

 

[Amad27]

Take a function, say f(x). This just says 'pick something (in the domain) and substitute it for x where ever you see x', So , suppose the domain is the real numbers and
f(x) = 2 * x
To evaluate f(x), follow the rule above. For example pick 2 for x. Then f(2) = 2 * 2 = 4.

Now x in the above is called a dummy variable because we could just as well have said Take a function, say f(something-or-another). Again that would just say ''pick something (in the domain) and substitute it for something-or-another where ever you see something-or-another'. Using the same example as above we have
f(something-or-another) = 2 * something-or-another
If we pick 2 for something-or-another, then we get the same answer; f(2) = 2 * 2 = 4.

Note that we have also followed that first function example. It is just that we picked something-or-another for x and substituted it for x everywhere there was an x.

I hope that helps.

Hey there, old friend!

Your post did indeed help very much, mind I ask a few clear-ups?

We are changing the whole axes though right?

But are all variables dummy variables? So

$f(x) = 2x$

Then we can change:

$f(y) = 2y$

??

Why doesnt this change the value though? Thanks!

 

[Ishuda]

Hey there, old friend!

Your post did indeed help very much, mind I ask a few clear-ups?

We are changing the whole axes though right?

But are all variables dummy variables? So

$f(x) = 2x$

Then we can change:

$f(y) = 2y$

??

Why doesnt this change the value though? Thanks!

All variables are dummy variables in the sense that you could choose to call them x or x$ or a-pot. However, there are conventions which make it easier to communicate with others. It might be something specific such as always using s for distance in certain disciplines, or don't change the symbol for the dummy variable unless it is clear you have done so (and generally with an explanation), i.e. let u = sin(x) so that du = cos(x) dx, or to demonstrate something like in your original post.

Why doesn't it change the value? That's the point of a dummy variable. It really has nothing to do with the value. It is just a place holder. It is a mark/symbol to try to make the situation clear. What is important is that the choice of the value of the dummy variable, no matter what we call it, doesn't change. if we have the function f(x) where f describes a relationship and x comes from a particular set, i.e. the reals, if you decide to call it $f(x$) then $f still describes the same relationship and x$ still comes from the reals.

 

[Amad27]

All variables are dummy variables in the sense that you could choose to call them x or x$ or a-pot. However, there are conventions which make it easier to communicate with others. It might be something specific such as always using s for distance in certain disciplines, or don't change the symbol for the dummy variable unless it is clear you have done so (and generally with an explanation), i.e. let u = sin(x) so that du = cos(x) dx, or to demonstrate something like in your original post.

Why doesn't it change the value? That's the point of a dummy variable. It really has nothing to do with the value. It is just a place holder. It is a mark/symbol to try to make the situation clear. What is important is that the choice of the value of the dummy variable, no matter what we call it, doesn't change. if we have the function f(x) where f describes a relationship and x comes from a particular set, i.e. the reals, if you decide to call it $f(x$) then $f still describes the same relationship and x$ still comes from the reals.

Does that follow from the definition of a variable?

 

[Ishuda]

Does that follow from the definition of a variable?

Yes, but I think you have to be careful about where you draw the line between calling that 'thing' a variable or a dummy variable. From a philosophical point, it seems to me that it can almost be both at the same time and the transition from 'dummy' to 'actual' occurs in that nebulous moment when one links the 'thing' to some purpose/definition - let x be the horizontal distance from the origin, or consider the function f with independent variable x, i.e. consider f(x), or let x be the age of Tommy or ... But don't forget that thing called convention. By convention that x in $\displaystyle \int f(x) dx$ is called a dummy variable no matter how it is defined.

 

[pka]

Does that follow from the definition of a variable?

First, you are using the wrong LaTex delimiters on this forum.

[tex]\int_0^1 {f(t)dt} [/tex]

gives $\displaystyle \int_0^1 {f(t)dt}$.

Second: the answer to your question is both yes and no.
There is a school of analyst dating back to the 19th century that use only $\displaystyle \int_0^1 {f(t)}$ when the dt is not necessary. But it is needed here in $\displaystyle \int_0^1 {(at^2+b)dt}$ to specify the variable.
On the other hand, because the antiderivate is not unique we get sloppy:
both $\displaystyle \int {(4x^3-2)dt}=x^4-2x+10$ and $\displaystyle \int {(4x^3-2)dt}=x^4-2x+1$ could be correct.
So we use a dummy constant: $\displaystyle \int {(4x^3-2)dt}=x^4-2x+\color{blue}\bf{C}$

BUT note that:
$\displaystyle \int_0^1 {(4t^3-2)dt}=\int_0^1 {(4x^3-2)dx}=\int_0^1 {(4\theta^3-2)d\theta}$.

 

[Amad27]

But don't forget that thing called convention. By convention that x in $\displaystyle \int f(x) dx$ is called a dummy variable no matter how it is defined.

I am confused, I have never seen a convention that says:

x in $\displaystyle \int f(x) dx$ is called a dummy variable?

Nevermind the above,

it follows from the definition of a bound variable.

Is this what you would conclude?

 

[Ishuda]

I am confused, I have never seen a convention that says:

x in $\displaystyle \int f(x) dx$ is called a dummy variable?

Nevermind the above,

it follows from the definition of a bound variable.

Is this what you would conclude?

In context, 'dummy variable' would generally be the same as the more formal 'free variable' and when, as mentioned before, "one links the 'thing' to some purpose/definition" it becomes a (bound) variable. The example given in the wikipedia article
http://en.wikipedia.org/wiki/Free_variables_and_bound_variables
$\displaystyle \Sigma_{k=0}^{k=10} f(k,n)$
saying k is bound and n is free just means that k has been bound to some purpose (a counter) and must take values from a specific set whereas n is free and has not been bound to a specific purpose/definition.