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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.

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    It is usually used in reduction formulas for integrals.2017-01-14
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    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 $n$.2017-01-14
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    It saves you time and effort computing new integrals from scratch if you already have computed a related one.2017-01-14
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    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...2017-01-14
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    If you insist on a formal definition you can think of $(I_n)$ as an abbreviation for $I(n)$, where $I$ is a function from $\mathbb{Z}^{+}$ to $\mathbb{R}$, that's the technical way of explaining it.2017-01-14
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    Looks a lot like a gamma function. The notations hints towards its connection to factorials.2017-01-14
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    Thinking of $I_n$ as $I(n)$ seems like the most logical thing to me2017-01-14
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    Once you know that $I_n=nI_{n-1}$ for each $n>0,$ and $I_0=1 ,$ you have $I_1=1\cdot I_0=1, I_2=2I_1=(2)(1), I_3=3I_2=(3)(2)(1),$ et cetera.2017-01-14
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    @user254665 yes, this analysis is very much inline with further questioning by the author of this problem.2017-01-15

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