Question

Please solve for part (b) and (c) thank you!

1. Consider the function f(x) = e-x defined on the interval 0 < x < 1. (a) Give an odd and an even extension of this function onto the interval -1 < x < 1. Your answer can be in the form of an expression, or as a clearly labelled graph. [2 marks] (b) Obtain the Fourier sine and cosine representation for the functions found above. Hint: use integration by parts. [10 marks] (c) Hence show that 1 1 1+ n2-2 2 1 n=1 Hint: you can evaluate your series at more than one point. [3 marks] (d) In this question, use graphing software of your choice. For both the odd and even extension, plot the extension, and its approximation using the first five non-zero terms in the Fourier series. Include the commands or script that you use. [3 marks] (e) Discuss these representations, and any salient features you notice. [1 mark]

Answer 1

b) f(x) = e-*, 0<x<1 As the half range is 1, so L = 1 Fourier sine representation is given as F(x) 5 bn sin(nax) where n=1 bn - il Sf(x) sin nax dx -25 1 e sin nitx dx 0 lit rv) 0 e 1 = 2 = 2 (-sin nix – NA COS NAX) n272 e (-sin na 1 + n2112 ni cos NT) - 2 1 + n272 (-sin 0 - nn cos 0) 2e-1 2 (-n(-1)") 1+ n2712 2nd (1-(-1)"e-1) 1 + n2712 - 1+ n272 (-17) = *-* = Fy(x) = {1+1272 2nn (1-(-1)"e-1) sin(ntx) ... (1) n=1

The Fourier cosine series is given as ao TX f(x) + Enean cos (bn 0) ao = 2 an COS nx n=1 Where ao = {l-f(x)dx = 2 se-* dx = -2e-*]) = 2(1 - e-). 2 an =- [ f(x) cos naxc dx = 2 21 e cos nax dx 1 = 2 2 [1 * 72 73 (- cos ntx + nn sin nax) x)] 10 e 1 = 2 (- cos nn + nn sinnt) - 2 1+ n272 (- cos 0 + nn sin o) 1 + n2772 2e-1 2 (-(-1)") 1 + n2112 (-1) 1 + n2T12 2 1 + n2712 (1-(-1)"e-) Thus, we have the Fourier series is given by

e-* = F. (x) ) = 4ο + + Σ=1 α, cos nx 2 + (1 – (-1)"e-1) coς ηπχ. (2) 1+12π2 c) Put x = O in (2), then 2 (1 – (-1)*2-1) n=1 e-0 = F. (0) = (1 – e-l) +Σ+ πί η=1 1 = (1 – e-1)+ 2 1+n2π2 (1) + (-(-1)"e-1) 1=1 -Σ+ ηπ? ΕΕ e η=1 η=1 η=1 2 1 2(-1)* (1) (3) Put x = 1 in (2), then 2 e-1 = F. (0) = (1 – e-1)+ Σ 1+ ηλπ2 (1 – (-1)*e-1)(-1) or, 2e-1 = 1+ 2 Στε (e-1) 2 2 -Σ 1+12π2 (e-1) + (2e-1 - 1) ... (4) Using (4) in (3), we get 2 2 (1) (e-1)+ (2e-1 - 1) 1+12π2 (-1)" - Σ1+1 πέ 1=1 1=1 ΟΥ, 1 + ηλπ2 (-1), n=1 η=1 e

= Σε κοπό n=1 2 (1 – e-2) - 2e-2 +e-1 1+ η2π2 Cancelling e-7, from both sides and taking 2e-2 to the left side, we get 2 2e-2 = - (1 – e-2) Στο 1=1 1 1 ΟΥ, Σ 1 +12π2 ρ2 (proved) 1 η=1

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