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Chapter 7: Q15P (page 358)

Use Problem 5.7to show that $\frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \frac{1}{6^{2} - 1} + 鈰� = \frac{1}{2}$

Short Answer

Expert verified

The resultant expansion is ${鈭�}_{n = 2 k}^{鈭�} \frac{1}{n^{2} - 1} = \frac{1}{2} k = 1, 2, 3, 4$ .

Step by step solution

01

Given data

The given function is $\frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \frac{1}{6^{2} - 1} + 鈰� = \frac{1}{2}$ .

02

Concept of Dirichlet’s theorem

Dirichlet's theorem states that the collection of all numbers of the form $n a + b$ contains an unlimited number of prime numbers, where the constants a and b are integers with no common divisors other than 1 and the variable nis any natural number.

03

Simplify the expression

Use the Problem (5.11) with $x = 0$ .

$f (0) = 0 f (0) = \frac{1}{蟺} + 0 - \frac{2}{蟺} {鈭�}_{n = 2 k}^{鈭�} \frac{1}{n^{2} - 1} f (0) = 0 {鈭�}_{n = 2 k}^{鈭�} \frac{1}{n^{2} - 1} f (0) = \frac{1}{2}$

Therefore, expansion is ${鈭�}_{n = 2 k}^{鈭�} \frac{1}{n^{2} - 1} = \frac{1}{2} k = 1, 2, 3, 4$ .

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Most popular questions from this chapter

Starting with the symmetrized integrals as in Problem 34, make the substitutions $伪 = \frac{2 蟺 p}{h}$ (where pis the new variable, his a constant), $f (x) = 蠄 (x)$ , localid="1664270725133" $g (伪) = \sqrt{\frac{h}{2 蟺}} 蠒 (p)$ ; show that then

$\begin{matrix} 蠄 (x) = \frac{1}{\sqrt{h}} {鈭�}_{鈭� 鈭�}^{鈭�} 蠒 (p) e^{\frac{2 蟺 i p x}{h}} d p \\ 蠒 (p) = \frac{1}{\sqrt{h}} {鈭�}_{鈭� 鈭�}^{鈭�} 蠄 (x) e^{鈭� \frac{2 蟺 i p x}{h}} d x \\ {鈭�}_{鈭� 鈭�}^{鈭�} {| 蠄 (x) |}^{2} d x = {鈭�}_{鈭� 鈭�}^{鈭�} {| 蠒 (p) |}^{2} d p \end{matrix}$

This notation is often used in quantum mechanics.

In each case, show that a particle whose coordinate is (a) $x = R e (z)$ , (b) $y = lm z$ is undergoing simple harmonic motion, and find the amplitude, period, frequency, and velocity amplitude of the motion.

role="math" localid="1659242473978" $- 4 e^{i (2 t + 3 蟺)}$

The functions in Problems 1 to 3 are neither even nor odd. Write each of them as the sum of an even function and an odd function.

(a) $I n | 1 - x |$ (b) $(1 + x) (s i n x + c o s x)$

Use the results ${鈭�}_{a}^{b} s i n^{2} k x d x = {鈭�}_{a}^{b} c o s^{2} k x d x = \frac{1}{2} (b - a)$ to evaluate the following integrals without calculation.

(a) ${鈭�}_{0}^{4 蟺 / 3} s i n^{2} (\frac{3 x}{2}) d x$

(b) ${鈭�}_{- 蟺 / 2}^{3 蟺 / 2} c o s^{2} (\frac{x}{2}) d x$

Show that if (12.2) is written with the factor $\frac{1}{\sqrt{2 蟺}}$ multiplying each integral, then the corresponding form of Parseval鈥檚 (12.24) theorem is ${鈭�}_{- 鈭�}^{鈭�} {| f (x) |}^{2} d x = {鈭�}_{- 鈭�}^{鈭�} {| g (伪) |}^{2} d 伪$ .

See all solutions

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