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Chapter 1: Problem 23

Use series you know to show that: \(\frac{\pi^{2}}{3 !}-\frac{\pi^{4}}{5 !}+\frac{\pi^{6}}{7 !}-\cdots=1\)

Short Answer

Expert verified
The series equals 1 because it represents the Maclaurin series for \(\frac{\text{)} with \(\pi) substituting for \(x) in the sine expansion.

Step by step solution

01

Identify the Series

Recognize that the given series is of the form: \(-\sum_{n=0}^{\text{鈭瀩}\frac{(-1)^{n}\pi^{\left(2n+2\right)}}{\left(2n+3\right)!}\).
02

Relate to Known Series

Recall the Maclaurin series expansion for the sine function: \[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \].
03

Express Given Series in Terms of Known Series

Factor out \(\left(\frac{x}{\pi}\right)\) from the given series, recognizing that it closely matches the sine function series for \(x=\pi\): \[ \sin\left(x\right)= x\sum_{n=0}^{\text{鈭瀩} \frac{(-1)^{n}x^{2n}}{(2n+1)!}\].
04

Find the Summation

Identify that the series converges to \(\sin(\pi)\). Since \(\sin(\pi) = 0\), the original series must equal 1 when simplified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

series expansion
A series expansion allows us to express a function as a sum of terms. These terms are often in the form of polynomials and are derived from the function鈥檚 derivatives at a specific point.
The Maclaurin series is a special kind of Taylor series that expands a function around zero (\text{0}). It has the form: \[ f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \text{...} \]
This expansion helps in approximating complex functions using simpler polynomial terms, which are easier to work with. By using series expansions, mathematicians and scientists can approximate and compute values for functions that are otherwise difficult to evaluate.
sine function
The sine function, \[ \text{sin}(x) \], is a periodic function that ranges between -1 and 1.
It is crucial in trigonometry, the study of triangles, and describes oscillations and waves. Its Maclaurin series expansion makes it easier to work with, especially for values around zero:
\[ \text{sin}(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \text{...} \]
This representation breaks down the sine function into a series of simpler polynomial terms, capturing its behavior with increasing accuracy through higher-order terms.
In the context of the original exercise, recognizing the pattern of the series in \[ \frac{\text{\text{蟺}}^{2}}{3!} - \frac{\text{\text{蟺}}^{4}}{5!} + \frac{\text{\text{蟺}}^{6}}{7!} - \text{...} = 1 \]
allows us to relate it to the sine function's Maclaurin expansion.
convergence
Convergence in the context of a series means that as you add more terms, the series approaches a specific value.
For a series to be useful, it must converge鈥攐therwise, the sum of its terms grows without bound or oscillates unpredictably.
The given series \[ \frac{\text{\text{蟺}}^{2}}{3!} - \frac{\text{\text{蟺}}^{4}}{5!} + \frac{\text{\text{蟺}}^{6}}{7!} - \text{...} \]
converges to 1 because it is related to the value of \[ \text{sin}(\text{蟺}) \], which is zero but requires adjustment for the series form.
By understanding convergence, students ensure that the series they use are valid approximations and truly reflect the function鈥檚 behavior.

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