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2) The Fourier expansion theorem proves that any periodic function F(x) can be expanded in terms of sines and cosines. If the function happens to be even, only cosines are needed and the expansion has the form F(x) = sum_{n=0}^{infty} A_n cosleft(frac{2pi n x}{lambda} ight), where lambda is the wavelength of the function. We will now find the coefficients A_n. a. Prove that A_0 = frac{1}{lambda} int_0^{lambda} F(x),dx b. Prove that for m > 0, A_m = frac{2}{lambda} int_0^{lambda} F(x)cosleft(frac{2pi m x}{lambda} ight),dx c. If our periodic function F(x) is odd, the Fourier expansion of such a function requires only sine functions: F(x) = sum_{n=1}^{infty} B_n sinleft(frac{2pi n x}{lambda} ight). Following the procedures for part (b), prove that B_m = frac{2}{lambda} int_0^{lambda} F(x)sinleft(frac{2pi m x}{lambda} ight),dx. Why is there no n = 0 term here?

          2) The Fourier expansion theorem proves that any periodic function F(x) can be expanded in terms of sines and cosines. If the function happens to be even, only cosines are needed and the expansion has the form F(x) = sum_{n=0}^{infty} A_n cosleft(frac{2pi n x}{lambda}
ight), where lambda is the wavelength of the function. We will now find the coefficients A_n.

a. Prove that A_0 = frac{1}{lambda} int_0^{lambda} F(x),dx

b. Prove that for m > 0, A_m = frac{2}{lambda} int_0^{lambda} F(x)cosleft(frac{2pi m x}{lambda}
ight),dx

c. If our periodic function F(x) is odd, the Fourier expansion of such a function requires only sine functions: F(x) = sum_{n=1}^{infty} B_n sinleft(frac{2pi n x}{lambda}
ight). Following the procedures for part (b), prove that B_m = frac{2}{lambda} int_0^{lambda} F(x)sinleft(frac{2pi m x}{lambda}
ight),dx. Why is there no n = 0 term here?

$2) The Fourier expansion theorem proves that any periodic function F(x) can be expanded in terms of sines and cosines. If the function happens to be even, only cosines are needed and the expansion has the form F(x) = sumn=0^infty An cosleft(frac2pi n xlambda ight), where lambda is the wavelength of the function. We will now find the coefficients An. a. Prove that A0 = frac1lambda int0^lambda F(x),dx b. Prove that for m > 0, Am = frac2lambda int0^lambda F(x)cosleft(frac2pi m xlambda ight),dx c. If our periodic function F(x) is odd, the Fourier expansion of such a function requires only sine functions: F(x) = sumn=1^infty Bn sinleft(frac2pi n xlambda ight). Following the procedures for part (b), prove that Bm = frac2lambda int0^lambda F(x)sinleft(frac2pi m xlambda ight),dx. Why is there no n = 0 term here?$

Added by Pilar V.

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