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I do not understand how dummy variables work in math.

Suppose we have:

$$I_1 = \int_{0}^{\infty} e^{-x^2} dx$$

How is this equivalent to:

$$I_2 = \int_{0}^{\infty} e^{-y^2} dy$$

How does this dummy variable system work?

Since $y$ is the dependent variable for $I_1$ How can $y$ itself be and independent variable for $I_2$

??

Thanks!

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    $\begingroup$ You are mistaken, $y$ is not a dependent variable for $I_1$. Why do you think that? $\endgroup$ – Mark Fantini Jan 5 '15 at 15:36
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    $\begingroup$ The $y$ in $I_2$ is not necessary this one in $y = e^{-x^2}$. $\endgroup$ – Alex Silva Jan 5 '15 at 15:36
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    $\begingroup$ See @AlexSilva's comment. $I_1$ is just a number, it has no functional relation. If you had $$I_1 = \int_0^{\infty} e^{-y x^2} \, dx,$$ then now you'd have $I_1$ as a function of $y$. But this would be the same as $$I_1 = \int_0^{\infty} e^{-y t^2} \, dt.$$ The dummy variable changed but all is the same. $\endgroup$ – Mark Fantini Jan 5 '15 at 15:40
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    $\begingroup$ If you are not in a setting where you've stated explicitly that $y = e^{-x^2}$, then there is no connection between $x$ and $y$ at all. And even if you were, the integral $\int e^{-x^2}dx$ is equal to the integral $$ \int e^{-y^2}dy = \int e^{-e^{-2x^2}}d(e^{-x^2}) $$ This is what variable substitution is all about. (I cannot use limits here, since $e^{-x^2}$ never becomes $0$ or $\infty$.) $\endgroup$ – Arthur Jan 5 '15 at 15:41
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    $\begingroup$ As I always tell my students: you need to break free from the chains of labels, and see them as a vehicle to speak of something, not the ontological manifestation of something. Whether $y$ is a dependent variable, independent variable, coordinate function or a scream of agony during finals depends entirely on context, not on the particular shape of the symbol or its common use in one course or another. $\endgroup$ – guest Jan 5 '15 at 21:39
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$$ \sum_{j=1}^3 j^4 = 1^4 + 2^4 + 3^4 = \sum_{k=1}^3 k^4. $$ In the first term, $1^4$, we can say that $j=1$; in the second term $2^4$ we have $j=2$, and in the third term, $3^4$ we have $j=3$. But when we call the index $k$ rather than $j$, then in the first term $k=1$, in the second $k=2$, and in the third $k=3$.

$j$ or $k$ is a bound variable (also sometimes called a dummy variable). The value of the whole expression $1^4+2^4+3^4$, which is $98$, does not depend on the value of a bound variable.

If we write $\displaystyle\sum_{j=1}^3 (j \cos (j+m))^4$, then $j$ is bound and $m$ is free. The sum is $$(1\cos (1+m))^4 + (2\cos (2+m))^4 + (3\cos(3+m))^4.\tag 1$$ Its value depends on the value of the free variable $m$, but not on the value of anything called $j$. Accordingly we do not see $j$ in $(1)$. We could rename $j$ and call it $k$, and the whole thing would still be equal to the expression $(1)$ in which we also do not see $k$.

See this Wikipedia article.

Another example is expressions like $\displaystyle\lim_{h\to0}\frac{(x+h)^3-x^3}h=3x^3$. The value of this expression depends on the value of the free variable $x$, but not of the bound (or "dummy") variable $h$. We could have said $\displaystyle\lim_{k\to0}\frac{(x+k)^3-x^3}k=3x^3$ and it would still be $3x^2$.

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    $\begingroup$ Replace the last $h$ by $k$. $\endgroup$ – Alex Silva Jan 5 '15 at 15:54
  • $\begingroup$ Thank you (+1), but what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \implies \int dF(x) = \int x dx$$ The LHS then does make sense after $\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? $\endgroup$ – anonymous Jan 5 '15 at 17:44
  • $\begingroup$ In the notation $\int_a^b g(x)\,df(x)$, one is dealing with a Riemann–Stieltjes integral, defined a a limit of $\sum\limits_{x\in\text{partition}} g(x)\,\Delta f(x)$ $=\sum_x g(x_i^{*})(f(x_{i+1}-f(x_i))$ as the mesh of the partition approaches $0$. This is the same as $\int_a^b g(x)f'(x)\,dx$ in cases where $f$ is everywhere differentiable, but it is also defined in cases where $f$ may do a lot of its changing of values in places where it is not differentiable, including jump discontinuties and places where $f$ is continuous but not absolutely continuous. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 5 '15 at 18:47
  • $\begingroup$ But $\int_a^b g(x)\,df(x)$ is the same as $\int_a^b g(w)\,df(w)$, i.e. one can re-name a bound variable. But the functions $g$ and $f$ are particular functions; the value of the integral depends on which function $f$ is, so $f$ is not a bound variable. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jan 5 '15 at 18:49
  • $\begingroup$ @MichaelHardy thanks for replying. But I mean, in the above, we have $f(x) = e^{-x^2}$ right? The vertical axis is $f(x)$ then, we have $f(y) = e^{-y^2}$ right? Then we iterate the integral meaning now we have $dxdy$ then if both are independent variables, how will we look at this model in 3D? Since we cannot have the axes related to each other? I mean $dxdy$ how do we look at this in 3D? sicne $y$ and $x$ are both independent variables, we cannot have the traditional width, length $dxdy$ $\endgroup$ – anonymous Jan 6 '15 at 12:45
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In both cases they are the independent variable. Note in your integrals that $x$ appears nowhere in $I_2$ and $y$ occurs nowhere in $I_1$.

It's common to write $y$ as the dependent variable and $x$ the independent variable. But there's no rule that says you have to. You can write $x = f(y) = y^2 + 2\sin y$ if you wanted to, or put the $y$-axis horizontal and $x$ vertical.

edit: note in any event that $I$ is constant $\left({I = \dfrac {\sqrt{\pi}} 2}\right)$ so it isn't "dependent" in the way you may be used to. Mathematicians will say things like "$I$ is a trivial function of $x$" which means that even though it's technically true that $I = f(x)$, the relationship isn't interesting.

Here's an interesting non-trivial function:

$\displaystyle I(z) = \int_0^z e^{-t^2} \, \mathrm dt$

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  • $\begingroup$ (wrote in Michael Hardy's answer too) what I dont understand is that in integration, so you integrate with respect to an independent variable? But then differential equations such as: $$dF(x)/dx = x \implies \int dF(x) = \int x dx$$ The LHS then does make sense after $\implies$ Because then are you consider $F(x)$ to be an independent variable that you are integrating with respect to? $\endgroup$ – anonymous Jan 5 '15 at 17:45
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IMO this can be understood much better by writing it out in a proper context-free way, rather than standard mathematical notation. The integration symbol $\int_0^\infty$ is basically a higher-order-function, taking a real function1 and returning a single number. $$ \int\limits_0^\infty : (\mathbb{R}\to \mathbb{R}) \to \mathbb{R}. $$ Now, people keep saying stuff like "$\sin(x)$ is a function..." but really this is incorrect. $\sin(x)$ is, for any value of $x$, just a single real number. What's a function is $\sin$ itself, i.e. not applied to anything. For instance, $$ \int\limits_0^\pi \sin = 2 $$ would be quite a reasonable statement. OTOH, it's nonsense to write $$ \int\limits_{-\infty}^0 e^x = 1 $$ because $e^x$ is not a function.

Trouble is, most functions you need to integrate won't have a predefined name. You can't thus write "the function itself" directly, but have to specify those results of the function for general inputs, in form of some algebraic expression. You know, the equation-definitions $$ f(x) = e^{-x^2}. $$ Here it should be quite clear that the $x$ symbol is just an arbitrary choice – it's something you use to signify "anything you can put into the function", and $f(y) = e^{-y^2}$ says obviously exactly the same thing.

Always having to give functions a name before integrating would be tedious, so we have "shortcut syntax" to define an "anonymous function" and use it right away. You can thus think of the $\mathrm{d}x$ symbol much as a lambda function.


1In fact, more accurately and generally, a differential form.

2Again, that's not all there is to it since $\mathrm{d}x$ really is a differential form.

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When you are told to compute some mathematical quantity depending on given data $a$, $\ldots\>$, $q$ the slave (person or computer system) doing the computation for you will introduce certain auxiliary variables whose names, temporary values, etc., will be invisible to you, and will not affect the end result. The latter depends only on the "outer" quantities $a$, $\ldots\>$, $q$. When these auxiliary variables are really varying during the computational process you can call them dummy variables – they will be "burnt" at the end; and when they are uniquely determined by the given data you can call them accessory parameters to the question.

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