Question

Write the following power series in summation (sigma) notation. $$-\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\frac{x^{8}}{4 !}-\cdots$$

Short answer

Question: Rewrite the power series -x²/1! + x⁴/2! - x⁶/3! + x⁸/4! - ... in summation notation. Answer: The power series can be rewritten in summation notation as: $$\sum_{n=0}^{\infty} (-1)^n \cdot \frac{x^{2n}}{n!}$$

Step-by-step solution

Identify patterns in the series

Observe the given power series: $$-\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\frac{x^{8}}{4 !}-\cdots$$ Notice that there are the following patterns: 1. The sign of each term alternates between positive and negative. 2. The exponent of x in each term doubles in value as the series progresses, with only even exponents included. 3. The denominator contains factorials that are consecutively numbered.

Find the general term

We can create a general term for the series based on the patterns identified in Step 1. Since the exponents of x are even, we can represent an arbitrary exponent as \(2n\). Since the denominator contains consecutively numbered factorials, we can represent an arbitrary factorial as \(n!\). To account for the alternating sign, we can use \((-1)^n\). Combining these components, we obtain the general term: $$a_n = (-1)^n \cdot \frac{x^{2n}}{n!}$$

Form the summation notation

Now, using the general term \(a_n\), we can write the power series in summation (sigma) notation by adding all the terms generated by varying n: $$\sum_{n=0}^{\infty} (-1)^n \cdot \frac{x^{2n}}{n!}$$ The given power series in summation notation is: $$-\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\frac{x^{8}}{4 !}-\cdots = \sum_{n=0}^{\infty} (-1)^n \cdot \frac{x^{2n}}{n!}$$

Key concepts

Sigma Notation

Sigma notation is a powerful shorthand used in mathematics to express the summation of a sequence of terms. The Greek letter \( \Sigma \) symbolizes the summation, letting us efficiently write sequences that might otherwise be cumbersome.
For example, the notation \( \sum_{n=0}^{\infty} a_n \) represents the infinite sum of the sequence \( a_n \), starting from \( n = 0 \) and continuing indefinitely.

Here are key elements of sigma notation:

• Lower bound of summation: The value below the sigma, \( n=0 \), is the starting point (or index) of the summation.
• Upper bound of summation: The value above the sigma, often infinity (\( \infty \)), represents where the summation ends. For infinite series, this continues indefinitely.
• General term: This is \( a_n \), which is the rule defining each term in the sequence. This expression changes with each increment of \( n \).

In the solution provided, the power series is transformed into the sigma notation \( \sum_{n=0}^{\infty} (-1)^n \cdot \frac{x^{2n}}{n!} \), which methodically sums the terms based on the rules defined for \( a_n \).
This transformation makes complex series more compact and easier to manipulate mathematically.

Factorials

Factorials, denoted by the symbol \( n! \), represent the product of all positive integers less than or equal to \( n \). They are essential in various areas of mathematics, particularly in series and probability.
For example:

• \( 0! \) is defined as \( 1 \) to maintain consistency across mathematical frameworks.
• \( 1! \) is \( 1 \).
• \( 3! = 3 \times 2 \times 1 = 6 \).
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow very quickly with increasing numbers and are often found in the denominators of terms in a series to relate to combinations or permutations.

In the given power series, they appear as \( n! \) in the denominator of each term, indicating that as the power of \( x \) increases, the factorial also increases. This helps control the growth of each term, preventing the series from diverging rapidly. The general term in sigma notation demonstrates this by having \( \frac{x^{2n}}{n!} \), allowing each term to adjust according to both \( x \)'s power and the factorial's growth.

Alternating Series

An alternating series is a series where the sign of each term alternates between positive and negative. This alternating sign pattern is cleverly captured using powers of \((-1)^n\).
Here's why that's important:

• With \( (-1)^n \), when \( n \) is even, the term \( (-1)^n \) becomes positive.
• When \( n \) is odd, the term \( (-1)^n \) becomes negative.

This regular change from positive to negative is what defines an *alternating series*. Such series often converge more reliably due to the cancellation effects of opposing signs, which helps stabilize the sum.

In our problem, the alternating nature ensures that terms like \(-\frac{x^{2}}{1 !}\) and \(+\frac{x^{4}}{2 !}\) counterbalance one another. The presence of the term \((-1)^n\) in the sigma notation successfully encapsulates this behavior, ensuring that the series reflects the pattern of alternating signs seen in the original problem, \( \sum_{n=0}^{\infty} (-1)^n \cdot \frac{x^{2n}}{n!} \).