Dummy Index

This is a concept that appears in Mathematics and it is a pretty useful one. Basically it means that the symbol we are using doesn’t really matter. What matters is the context it’s being used in. For instance in
the following summations \displaystyle\sum_{i=0}^{5}1, \displaystyle\sum_{j=0}^{5}1 the summing index is said to be a dummy one. The result of the sum is the same regardless of whether one uses an i, a k, or whatever other symbol.

Now, are all indexes dummy indexes? No, they are not. For instance in \displaystyle\sum_{j=0}^{l}1 the l index isn’t a dummy one, since the value of the sum changes as one changes the value of  l.

That being said let’s look at a simple proof where some confusion may arise in some people,not really used to think in these terms. Lets us prove that an odd function that is being integrated in symmetric limits yields a 0 integral.

A function,  f(x), is said to be odd if f(-x)=-f(x), and we are trying to calculate \displaystyle\int_{-a}^{a}f\left(x \right)dx.

We know that

\displaystyle\int_{-a}^{a}f\left(x \right)dx=\displaystyle\int_{-a}^{0}f\left(x \right)dx+\displaystyle\int_{0}^{a}f(x)dx
Let’s take a look at the first integral in the right hand side of the equality:
\displaystyle\int_{-a}^{0}f\left(x \right)dx

Let t=-x, hence dt=-dx. Moreover

\displaystyle\int_{-a}^{0}f\left(x \right)dx=\displaystyle\int_{a}^{0}f\left(-t \right)\left(-dt \right)=-\displaystyle\int_{a}^{0}f\left(-t \right)dt=\displaystyle\int_{0}^{a}f\left(-t \right)dt
Remembering the definition of an odd function and the fact that we assumed that f was such a function we have: f(-t)=-f(t) and so
\displaystyle\int_{0}^{a}f\left(-t \right)dt=-\displaystyle\int_{0}^{a}f\left(t \right)dt

Now lets us think for a while. The t variable in that integral is a dummy one or not? If we, in a given function, exchange the t for any given symbol, will that change the value of the integral? No ,it won’t! So t is in fact a dummy variable. Thus we can, for example, change it for x.

At this point I want you to have some furious outcries in your head. That is, of course, if you aren’t used to this type of reasoning. If you are , everything is perfectly normal. The reason for this possible outcry is the change of variable that we made  first t=-x. First this guy tells us that t=-x, and now he tells us that t=x. So what’s it gonna be Mr. Smarty Pants?! Both choices are mutually exclusive (except in the trivial case were t=x=0) and we need a resolution.

The resolution is the fact that the second equality is just us noticing that for effects of computing the integral what matters is the functional form of the function and the limits of integration. So we just changed the symbol in the functional form (not the function itself) and the limits of integration stayed the same. Taking this into account, the result is
-\displaystyle\int_{0}^{a}f\left(t \right)=-\displaystyle\int_{0}^{a}f\left(x \right)dx

and this leads us to:

\displaystyle\int_{-a}^{a}f\left(x \right)dx=-\displaystyle\int_{0}^{a}f\left(x \right)dx+\displaystyle\int_{0}^{a}f\left(x \right)dx=0
Which is the result we expected all along.

Ps: I’ve assumed the reader is familiar with some calculus and if you don’t get all of this don’t despair. I’ll make this a physics course from the bottom up (I’m just trying some things for now and see if I like what I get. ) and in due time I’ll address calculus and it’s foundations. As a matter of fact calculus and some more basic math will be the first things treated on this blog.

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10 Responses to “Dummy Index”

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  2. I like your style of taking the position of the confused reader and asking the questions he/she might ask.

  3. Thanks a lot for the comment Mike. Also sorry for not updating this blog in a very long time (actually the blog should be finished by now), but this time I’m really getting back to it.

    Take care

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