I discovered this notation in the problem outlined below. My confusion lies in the subscript notation used on the I's, i.e. the $n$ and $n-1$ in $I_n$ and $I_{n-1}$

Show that if $I_n$ is defined by the integral

$$I_n = \int_{0}^\infty x^ne^{-x} dx$$

where $n$ is a positive integer, then

$$I_n = nI_{n-1}$$

After integrating $I_n$ by parts, the final line of the proof is as follows:

$$I_n = 0+n\int_{0}^\infty x^{n-1}e^{-x} dx = nI_{n-1}$$

With this result, I would assume that the subscript notation says to replace $n$ with $n-1$. But a more formal definition of what this notation means is what I'm looking for.

  • $\begingroup$ It is usually used in reduction formulas for integrals. $\endgroup$ – Karl Jan 14 '17 at 22:16
  • $\begingroup$ The subscript is there to emphasise the functional dependence of the value of the integral on $n$. You see, for different $n$s, the integral evaluates to different values. You may write it as $I(n)$ if you like, as in the usual notations for functions in $n$. $\endgroup$ – user1551 Jan 14 '17 at 22:18
  • $\begingroup$ It saves you time and effort computing new integrals from scratch if you already have computed a related one. $\endgroup$ – Karl Jan 14 '17 at 22:21
  • $\begingroup$ The author there defined a sequence of real numbers, that's all! This sequence is $(I_n)$, $n$ being a positive integer. The value of the $n$-th term of the sequence is given by the (definte) integral in question. It's easy, but perhaps that integral makes the notation a little more scary... $\endgroup$ – Pythagoricus Jan 14 '17 at 22:22
  • 3
    $\begingroup$ Thinking of $I_n$ as $I(n)$ seems like the most logical thing to me $\endgroup$ – HumptyDumpty Jan 14 '17 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.