Dummy Index

Posted on February 10, 2024 by ateixeira

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 when we aren’t used to thinking in these terms. Lets us prove that an odd function that is being integrated in symmetric limits yields a null 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, there might be some confusion on the variables that we are using. The reason for this possibility, is the change of variable that we made  first t=-x. First it is t=-x, and now he it is t=x. 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.

About ateixeira

Experienced professional with a track record on strategic roles in several industries in Angola and other emerging markets. Right now I am getting back to Physics and this blog is a way to keep track of my progress and struggles
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