Question

Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\cos x, \quad c=\frac{\pi}{4} $$

Short answer

The Taylor series centered at \(c = \frac{\pi}{4}\) for \(f(x) = \cos x\) is \(\frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2}(x-\frac{\pi}{4}) -\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^2}{2!} + \frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^3}{3!} +\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^4}{4!}-...\).

Step-by-step solution

Understand the pattern of derivatives

Find the first few derivatives of the cosine function and observe the pattern:1st derivative (f'): \(-\sin(x)\),2nd derivative (f''): \(-\cos(x)\),3rd derivative (f'''): \(\sin(x)\),4th derivative (f''''): \(\cos(x)\).The pattern then begins to repeat.

Calculate the derivatives at \(c = \frac{\pi}{4}\)

Now, calculate the values of each derivative at \(c = \frac{\pi}{4}\) using the known values of sine and cosine at \(\frac{\pi}{4}\), which are equivalent to \(\frac{\sqrt{2}}{2}\):Derivative 0: \(f(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\),1st derivative: \(f'(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\),2nd derivative: \(f''(\frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\),3rd derivative: \(f'''(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\),4th derivative: \(f''''(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\). It's clear that the pattern repeats after the 4th derivative.

Incorporate the terms into the Taylor series

Add the terms to the general Taylor series formula, dividing by the factorial of the term's degree: \(f(x) = f(a)+f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} +...\) This would yield the series centered at \(\frac{\pi}{4}\):\(f(x) = \frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2}(x-\frac{\pi}{4}) -\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^2}{2!} + \frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^3}{3!} +\frac{\sqrt{2}}{2}\frac{(x-\frac{\pi}{4})^4}{4!}-...\)