Definitions and Basics, Exercise 2

Question

The Gaussian function

$f (x) = \exp (- \frac{x^{2}}{2})$

is well known in statistics and is extensively used in science for error calculations. Its integral is of great importance, but unfortunately it cannot be expressed in terms of the elementary functions. One way to deal with it is to use power series.

Express the Gaussian in power series (Maclaurin)!

Integrate the Gaussian $I (x) = \int f d x$ !

What is necessary for I(x) to be a function with defined parity and what is the parity?

For I(x) write a general expression of the n-th derivative at x=0 !

Reminder

... the N-th approximation, which can be written in a general form as

$\begin{array}{l} y_{N} = f_{0} + \sum_{n = 1}^{N} \frac{1}{n!} {(\frac{d^{n} f}{d x^{n}})}_{0} {(x - x_{0})}^{n} = \\ = \sum_{n = 0}^{N} \frac{1}{n!} {(\frac{d^{n} f}{d x^{n}})}_{0} {(x - x_{0})}^{n} where {(\frac{d^{0} f}{d x^{0}})}_{0} = f_{0} \end{array}}$	(1.3.5.4)

The expression of (1.3.5.4) is also called the Taylor polynomial.

... a simple example:

$f (x) = \exp x and x_{0} = 0$	(1.3.5.8)

By the choice of x₀=0 the powers of (x−x₀) from the Taylor polynomial (1.3.5.4) become simply powers of x and such polynomial is commonly called after the name of Maclaurin.

From (1.3.5.4) and (1.3.5.8) one obtains:

$y_{N} = 1 + \sum_{n = 1}^{N} \frac{x^{n}}{n!}$	(1.3.5.9)

$\int x^{p} d x = \frac{x^{p + 1}}{p + 1} + C$	(2.1.1.10)

Parts 1-2

Solution of question 1

From (1.3.5.9) we know that $\exp x = \sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$ It is also known that this series is converging for any value of x.
From part 1 follows that one can substitute x from the series with $- \frac{x^{2}}{2}$ and obtain: $\exp (- \frac{x^{2}}{2}) = \sum_{n = 0}^{\infty} \frac{{(- \frac{x^{2}}{2})}^{n}}{n!} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n}}{n! 2^{n}}$

Part 3

Solution of question 2

With the help of (2.1.1.10) one obtains

I (x) = \int d x \exp (- \frac{x^{2}}{2}) = C + \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{n! 2^{n} (2 n + 1)}

Part 4

Solution of question 3

Since in the series of part 3 there are only odd powers of x, the function I(x) is odd if C=0.

Parts 5-8

Solution of question 4

From (1.3.5.4) one have for the Maclaurin power series of a function:

$f (x) = f_{0} + \sum_{n = 1}^{\infty} \frac{x^{n}}{n!} {(\frac{d^{n} f}{d x^{n}})}_{0}$

and one has to compare the terms with the same power of x with the series of part 3. The user could make the comparison with the series corresponding to the derivative of I (x), but this just complicates things.
Since the even powers of x do not appear in I (x) we have that ${(\frac{d^{2 n} I}{d x^{2 n}})}_{0} = 0 for n = 1, 2, 3, ...$
The comparison of the terms with odd powers of x yields:

$\frac{x^{2 n + 1}}{(2 n + 1)!} {(\frac{d^{2 n + 1} I}{d x^{2 n + 1}})}_{0} = \frac{{(- 1)}^{n} x^{2 n + 1}}{n! 2^{n} (2 n + 1)}$
and finally one obtains

${(\frac{d^{2 n + 1} I}{d x^{2 n + 1}})}_{0} = \frac{{(- 1)}^{n} (2 n)!}{n! 2^{n}} for n = 0, 1, 2, 3, ...$

Score

Questions 1,2 give credit of 2 points each.

Question 3 gives credit of 1 point.

Question 4 gives credit of 5 points.

By parts:

Parts 1,2,4,5,6,8 are worth 1 point each.

Part 3,7 are worth 2 points each.

Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 2: Integration; Section 1: Indefinite Integrals; page 1