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Continuity in an Interval

State and exp...

Question

State and explain fouriers theorem determine the coefficient of fouriers series derive the formula

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Solution

$A series of sines and cosines of an angle and its multiples of the form of \frac{a_{0}}{2} + a_{1} \cos x + a_{2} \cos 2 x + a_{3} \cos 3 x + . . . . . + a_{n} \cos nx + . . . . b_{1} \sin x + b_{2} \sin 2 x + . . . . . + b_{n} \sin nx + . . . . . . = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} a_{n} \cos nx + \sum_{n = 1}^{\infty} b_{n} \sin nx is called Fourier Series where a_{0}, a_{1}, a_{2}, . . . . . a_{n}, . . . . b_{1}, b_{2}, . . . . b_{n}, . . . . are constants . DETERMINATION OF FOURIER COEFFICIENTS : f (x) = \frac{a_{0}}{2} + a_{1} \cos x + a_{2} \cos 2 x + a_{3} \cos 3 x + . . . . . + a_{n} \cos nx + . . . . b_{1} \sin x + b_{2} \sin 2 x + . . . . . + b_{n} \sin nx + . . . . . . - - - - (1) To find a_{0} : Integrating both sides of (1) from x = 0 to x = 2 π, we get \int_{0}^{2 π} f (x) dx = \frac{a_{0}}{2} \int_{0}^{2 π} dx + a_{1} \int_{0}^{2 π} \cos x dx + a_{2} \int_{0}^{2 π} \cos 2 x dx + . . . + a_{n} \int_{0}^{2 π} \cos nxdx + . . . + b_{1} \int_{0}^{2 π} \sin x dx + b_{2} \int_{0}^{2 π} \sin 2 x dx + . . . + b_{n} \int_{0}^{2 π} \sin nxdx + . . . \Rightarrow \int_{0}^{2 π} f (x) dx = \frac{a_{0}}{2} \int_{0}^{2 π} dx + 0 + 0 + . . . 0 + . . . + 0 + 0 + . . . . . . + 0 + . . . . . [as, \int_{0}^{2 π} \cos nx dx = \int_{0}^{2 π} \sin nx dx = 0] \Rightarrow \int_{0}^{2 π} f (x) dx = \frac{a_{0}}{2} {[x]}_{0}^{2 π} \Rightarrow \int_{0}^{2 π} f (x) dx = \frac{a_{0}}{2} (2 π) \Rightarrow a_{0} = \frac{1}{π} \int_{0}^{2 π} f (x) dx To find a_{n} : Multiply each side of (1) by \cos nx and Integrating both sides of (1) from x = 0 to x = 2 π, we get \int_{0}^{2 π} f (x) \cos nx dx = \frac{a_{0}}{2} \int_{0}^{2 π} \cos nx dx + a_{1} \int_{0}^{2 π} \cos nx . cosx dx + . . . + a_{n} \int_{0}^{2 π} \cos^{2} nx dx + . . . + b_{1} \int_{0}^{2 π} \cos nx \sin x dx + b_{2} \int_{0}^{2 π} \cos nx \sin 2 x dx + . . . + b_{n} \int_{0}^{2 π} cosnx \sin nx dx + . . . = \frac{a_{0}}{2} \times 0 + a_{1} \times 0 + . . . . + a_{n} \int_{0}^{2 π} \cos^{2} nx dx + . . . . + b_{1} \times 0 + b_{2} \times 0 + . . . . . . . + b_{n} \times 0 + . . . . . = a_{n} \int_{0}^{2 π} \cos^{2} nx dx [as, \int_{0}^{2 π} \cos mx . \cos nx dx = \int_{0}^{2 π} \sin mx . \cos nx dx = \int_{0}^{2 π} \cos nx \sin nx dx = 0] = a_{n} \int_{0}^{2 π} [\frac{1 + \cos 2 nx}{2}] dx = \frac{a_{n}}{2} {[x + \frac{\sin 2 nx}{2 n}]}_{0}^{2 π} = \frac{a_{n}}{2} [(2 π + \frac{\sin 4 nπ}{2 n}) - 0] = \frac{a_{n}}{2} [(2 π + 0) - 0] = a_{n} π So, \int_{0}^{2 π} f (x) \cos nx dx = a_{n} π \Rightarrow a_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \cos nx dx To find b_{n} : Multiply each side of (1) by \sin nx and Integrating both sides of (1) from x = 0 to x = 2 π, we get \int_{0}^{2 π} f (x) \sin nx dx = \frac{a_{0}}{2} \int_{0}^{2 π} \sin nx dx + a_{1} \int_{0}^{2 π} \sin nx . cosx dx + . . . + a_{n} \int_{0}^{2 π} \sin nx cosnx dx + . . . + b_{1} \int_{0}^{2 π} \sin nx \sin x dx + b_{2} \int_{0}^{2 π} \sin nx \sin 2 x dx + . . . + b_{n} \int_{0}^{2 π} \sin^{2} nx dx + . . . = \frac{a_{0}}{2} \times 0 + a_{1} \times 0 + . . . . + a_{n} \times 0 + . . . . + b_{1} \times 0 + b_{2} \times 0 + . . . . . . . + b_{n} \int_{0}^{2 π} \sin^{2} nx dx + . . . . . = b_{n} \int_{0}^{2 π} \sin^{2} nx dx [as, \int_{0}^{2 π} \cos mx . \cos nx dx = \int_{0}^{2 π} \sin mx . \cos nx dx = \int_{0}^{2 π} \cos nx \sin nx dx = 0] = b_{n} \int_{0}^{2 π} [\frac{1 - \cos 2 nx}{2}] dx = \frac{b_{n}}{2} {[x - \frac{\sin 2 nx}{2 n}]}_{0}^{2 π} = \frac{b_{n}}{2} [(2 π - \frac{\sin 4 nπ}{2 n}) - 0] = \frac{b_{n}}{2} [(2 π - 0) - 0] = b_{n} π So, \int_{0}^{2 π} f (x) \sin nx dx = b_{n} π \Rightarrow b_{n} = \frac{1}{π} \int_{0}^{2 π} f (x) \sin nx dx$

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