Respuesta :
Answer:
[tex]f(x) = x +\frac{1}{3}x^{3}+....[/tex]
Step-by-step explanation:
As per the question,
let us consider f(x) = tan(x).
We know that The Maclaurin series is given by:
[tex]f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......[/tex]
So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).
Therefore,
f'(x) = sec²x
f''(x) = 2 × sec(x) × sec(x)tan(x)
= 2 × sec²(x) × tan(x)
f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)
= 4sec²(x) tan²(x) + 2sec⁴(x)
= 6 sec⁴x - 4 sec² x
We then substitute x with 0, and find the values
f(0) = tan 0 = 0
f'(0) = sec²0 = 1
f''(0) = 2 × sec²(0) × tan(0) = 0
f'''(0) = 6 sec⁴0- 4 sec² 0 = 2
By putting all the values in the Maclaurin series, we get
[tex]f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......[/tex]
[tex]f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......[/tex]
[tex]f(x) = x +\frac{1}{3}x^{3}+....[/tex]
Therefore, the expansion of tan x at x = 0 is
[tex]f(x) = x +\frac{1}{3}x^{3}+....[/tex].
