Principles of Fourier Analysis, 2nd Edition

Chapter 1

1. geometric definition: $v \cdot w = | v | | w | \cos (angle between v and w)$

component formula: $v \cdot w = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3}$

Chapter 6

7a. 2

7b. 3

7c. $\sqrt{13}$

7d. $Arctan (\frac{3}{2})$

7e. $- 5 + 12 i$

7f. $\frac{2}{13}$

7g. $\frac{- 3}{13}$

10a. $Re [f (t)] = {(1 + t^{2})}^{- 1}, Im [f (t)] = t {(1 + t^{2})}^{- 1}$

13a. $\frac{1}{2}$

13b. $[- a + a e^{a x} \cos (b x) + b e^{a x} \sin (b x)] {(a^{2} + b^{2})}^{- 1}$

13c. 0

13d. $\frac{p}{2}$

16a. $1, i, - 1 and - i$

16b. $1, \exp (i \frac{2 π}{3}) = \frac{1}{2} [- 1 + i \sqrt{3}] and \exp (i \frac{4 π}{3}) = \frac{1}{2} [- 1 - i \sqrt{3}]$

16c. $\exp (i \frac{π}{3}) = \frac{1}{2} [1 + i \sqrt{3}], - 1 and \exp (i \frac{5 π}{3}) = \frac{1}{2} [1 - i \sqrt{3}]$

16d. $2 \exp (i \frac{π}{3}) = 1 + i \sqrt{3}, - 2 and \exp (i \frac{5 π}{3}) = 1 - i \sqrt{3}$

16e. $\exp (i \frac{π}{4}) = \frac{1}{\sqrt{2}} [1 + i] and \exp (i \frac{5 π}{4}) = - \frac{1}{\sqrt{2}} [1 + i]$

16f. $\exp (i \frac{3 π}{4}) = \frac{1}{\sqrt{2}} [- 1 + i] and \exp (i \frac{7 π}{4}) = \frac{1}{\sqrt{2}} [1 - i]$

16g. $\exp (i \frac{π}{6}) = \frac{1}{2} [\sqrt{3} + i], \exp (i \frac{5 π}{6}) = \frac{1}{2} [- \sqrt{3} + i] and = - i$

Chapter 7

9b. 2

9b. $\frac{2}{3}$

Chapter 9

5a. $fund . period = 2 , fund ...$

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Principles of Fourier Analysis, 2nd Edition by Kenneth B. Howell