Question

Froblem 9.7.8. (i) Obtain the series in terms of sines and cosines for f(x) = el in the interval -1 < < 1. (ii) Repeat for the case f(x) = cosh 2. Show that f(z) = sinh 7 f() = 1+ (cos nr-n sin nr) T + 22 (COS nr - nsi represents the function er in the interval - <I< (and its periodicized version outside.) Show how you can get the series for sinh x and cosh x from the above.

Answer 1

Solution:

To find Fourier series of the function

Part(i):

in the int

f(x) = elal in the interval - 15xs1

Fourier series representation of f(x)

f

f(x) =+ancos (nnx) + b sin (nax

Where

20 = {'redx = f ek dx = Lie="dx+[*e* dx = +1+e=1+e=2(e – 1) e* dx = -1 +e-1+e=2(e - 1)

an= f x) cos(nax) dx le* cos(nax)dx + e* cos(ntx) dx -1 + eCos[np] + en Sin[np] -1 + eCos[17] + enn Sin[n 1] 1+n22 1+nanta 2-1 + eCos[nt] + entrSin[nn] 1+n-12 + I II

Since

sin(nn) = 0, cos(nn) = (-1)" for all n = 1,2,3,4,...

1+

And

bm= Selal sin (ntx) dx = -" n-enCos[17]+Sin[n], na-enCo[n]+eSin[ne] 1+ 2 123

substituting in to the fourier series

Is the required Fourier series of f(x).

f(x) = (e – 1) + Σ-12tecos (ηπα)

Part(ii):

in the interval

f(x) = cosh(x) in the interval - 1<x< 1

Since

x ) usos) = (هرچه -

ao = [ (x)dx = -te

(-1+e) Cos[17] + (1 + entSin[nt] f(x) cos(nnx) dx = e + en 12

a, = 1+ )(-1)

And

bn = 110*** sin(nmx) dx = 0

Hence

(xuu) 50"-1"12+1-, ***3+(2+-) = (x)ųso

Now to find fourier series of ex in the interval

et in the interval - SXST

17 2Sinh [1] - e*dx=-

an= 2n Cosh[1]Sin[17] + 2Cos[nr]Sinh [1] ex cos(ntx) dx = T + n21 TJ-

Substituting into Fourier representation we get

e* = Sinh[7]. (-1) 21+ n(cos(nx) - n sin (nx)

Replacing we get Fourier series representation of e-x

x by - x

Then using we get these representations

cosh(x) = **** , sinh(x) = chat