Find the Taylor Series expansion of tan(x) about π/4 up to the term in terms of (x-π/4)^3.
Firstly, set f(x) = tan(x). We can then differentiate this 3 times in order to find f'(x), f''(x) and f'''(x). This will require use of the chain rule and the product rule. We can find:f'(x) = sec2xf''(x) = 2 * sec(x) * sec(x)tan(x) = 2sec2(x)tan(x)f'''(x) = 2 * 2sec2(x)tan(x)tan(x) + 2sec2(x)*sec2(x) = 4sec2(x)tan2(x) + 2sec4(x)We then substitute x with π/4, and find the values of f at these values, and then the coefficient an:f(π/4) = 1 a0 = 1/0! = 1 f'(π/4) = 2 a1 = 2/1! = 2f''(π/4) = 4 a2 = 4/2! = 2f'''(π/4) = 16 a3 = 16/3! = 8/3From this we can conclude tan(x) = 1 + 2(x-π/4) + 2(x-π/4)2 + (8/3)(x-π/4)^3.
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