How do you find the taylor series series for f (x) = 5 cos πx at a=1/2?

1 Answer
Jul 4, 2015

The general formula for writing out a Taylor series is:

sum_(n = 0)^(oo) (f^((n))(a))/(n!) (x-a)^n

Note that according to this formula, you need to take the derivative of the function n times. Let's say n = 4.

f^((0))(x) = f(x) = 5cospix

f'(x) = -5pisinpix

f''(x) = -5pi^2cospix

f'''(x) = 5pi^3sinpix

f''''(x) = 5pi^4cospix

Thus, the truncated Taylor series can be written as:

= (f^((0))(a))/(0!)(x-1/2)^0 + (f'(a))/(1!)(x-1/2)^1 + (f''(a))/(2!)(x-1/2)^2 + (f'''(a))/(3!)(x-1/2)^3 + (f''''(a))/(4!)(x-1/2)^4 + ...

= 5cospia + (-5pisinpia)(x-1/2) + ((-5pi^2cospia)/(2))(x-1/2)^2 + ((5pi^3sinpia)/(6))(x-1/2)^3 + ((5pi^4cospia)/(24))(x-1/2)^4 + ...

Notice how cos(pi/2) = 0 and sin(pi/2) = 1:

= cancel(5cos(pi/2)) - 5picancel(sin(pi/2))^(1)(x-1/2) -cancel((5pi^2cos(pi/2))/2(x-1/2)^2) + (5pi^3cancel(sin(pi/2))^(1))/(6)(x-1/2)^3 + cancel((5pi^4cos(pi/2))/24(x-1/2)^4) - ...

As a result, this becomes a series with alternating signs and only odd terms (similarly, if you did this with 5sinpix, you'd have alternating signs and only even terms).

= color(blue)(-5pi(x-1/2) + (5pi^3)/(6)(x-1/2)^3 - ...)

Just remember these things:
- Only the x in f(x) approaches a, not the one in (x - a)^n
- Plug in a after taking the derivative
- a does not vary, but n does