Question

Diberikan fungsi f dengan f(x) = cos x, x anggota [0,2π]. Ekspansikan fungsi f di sekitar x = π/4
Mohon bantuannya ya, caranya juga heheh

Answer 1

Ekspansi deret taylor :

f(x) = (x - a)^0.f(a)/0! + (x - a)^1.f'(a)/1! + (x - a)^2.f''(a)/2! + ... + (x - a)^n.f_n(a)/n!

f(x) = cosx
f(π/4) = √2 / 2
f'(x) = -sinx
f'(π/4) = -√2 / 2
f''(x) = -cosx
f'' (π/4) = -√2 / 2
f'''(x) = sinx
f'''(π/4) = √2 / 2
.
.
.

maka,

f(x) = (x - a)^0.f(a)/0! + (x - a)^1.f'(a)/1! + (x - a)^2.f''(a)/2! + ... + (x - a)^n.f_n(a)/n!

f(x) = ((x - π/4)^0.√2/2)/0! + ((x - π/4)^1.-(√2/2))/1! + ((x - π/4)^2.-(√2/2))/2! + ...

f(x) = √2/2 - √2/2.(x - π/4) - √2/2(x - π/4)^2/2! + √2/2(x - π/4)^3/3! + √2/2(x - π/4)^4/4! + ..., x∈[0, 2π]