# Determine a half range cosine series to represent the function given byf(t) = 2t + 3 0 < t < 2 f(t) = f(t + 4)

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The Fourier's half range expansion in cosine series of a function f(t) defined over (0 ,L) is given by:

f(t) = a0/2 + summation {an cos (npt/L) }, where p stands for pi and n =1,2,3....

Where a0 = 2/L* Integral {f(t)} dt , t from0to L,

where an = 2/L* integral { f(t) cos(npt/L) }dt , t from 0 to L.

In this case we have :

g(t) = f(t) = 2t+3 , fort in (0, L), where L = 2.

g(t) = 3-2t , for t in (-L,0), where L = 2

g(t+2L) = g(t+4) = g(t). So g(t) is an even function in (-L L) and is equal to f(t) in (0,L) and so we can expand itas:

g(x) = a0/2 + summation an*(cos (npt/L)).

a0 = (2/L) Integral { 2t+3)dt, t from 0 to L, but L=2.

= 2/2{[t^2+3t] t=2 - {t^2+2t] t= 0} = (4+6)= 10.

an = (2/L) *Integral (3t+2)cos(npt/L) dt , t=0 to L

We use Leibnitz' rule that : Integral (UV)dt = UV1 -U'V2+U''V3-u'''V4+...., where ' stands for differentiation and suffix's denote Integration. So

an = (2/2) *(2t+3)(cos(npt/L)1 - (2t+3)'(cos(npt/L)2+... t from 0 to 2

=(2/(np){[ (2t+3)sin(npt/2) , t=2] - [(2t+3)sin(npt/2), t=0]] +

-{(2/np) (sin(npt/2)2}

=0 - {[(2/np) (-cos (npt)/2)/ (np/2) ] at t =2 ]- [(2/np)(-cos(np/2)/(np/2) at t =0}

=(2/np)^2 { (-1)^n -1}

an = 4/p^2n^2 {(-1)^n - 1}

Therefore the required half range Fourier cosine series is :

g(t) = 10/2 + (4/p^2)* summation {[((-1)^n -1)/n^2] cos (npt/2)} for n =1, 2,3,.....

= 10/2 + (4/p^2) {( -2/1^2) cos (pt/2) - 2/3^2 cos(3pt/2) -(2/5^2)cos(5pt/2)-.................}

= 10/2 -(8/p^2){ (1/1^2)cos(pt/2)+(1/3^3)cos(3pt/2)+(1/5^2)cos(1/^2)(5pt/2)+cos(7pt/2)+.............}