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York

Answered question

2021-01-28

Use the definition of Taylor series to find the Taylor series, centered at c, for the function.

f (x) = \cos x, c = \frac{- π}{4}

grbavit

Skilled2021-01-29Added 109 answers

f (x) = \cos x, c = \frac{- π}{4}

The Taylor polynomial for a function f(x) is

f (x) = f (c) + f^{'} (c) (x - c) + \frac{f^{″} (c)}{2!} (x - c)^{2} + \frac{f^{‴} (c)}{3!} (x - c)^{3} + . . . + f^{(n)} \frac{c}{n!} (x - c)^{n}

Now is

f (c) = \cos (\frac{π}{4})

= \frac{1}{\sqrt{2}}

f^{'} (c) = - \sin (\frac{π}{4})

= - \frac{1}{\sqrt{2}}

f^{″} (c) = - \cos (\frac{π}{4})

= \frac{- 1}{\sqrt{2}}

f^{‴} (c) = \sin (\frac{π}{4})

= \frac{- 1}{\sqrt{2}}

Then is

\cos x = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2} (x - \frac{π}{4})} - \frac{\sqrt{2}}{(2!) (x - \frac{π}{4})^{2}} + \frac{\sqrt{2}}{(3!) (x - \frac{π}{4})^{3}} + . . . + \frac{f^{n} (c)}{(n!) (x - \frac{π}{4})^{n}}

= \frac{1}{\sqrt{2}} + \frac{- 1}{\sqrt{2} (x - \frac{π}{4})} + \frac{- 1 (2)}{(2!) (x - \frac{π}{4})^{2}} + \frac{\sqrt{2}}{(3!) (x - \frac{π}{4})^{3}} + . . . + \frac{f^{n} (c)}{(n!) (x - p i / 4)^{n}}

= \sum_{n = 0}^{\infty} \frac{1}{\sqrt{2}} \frac{(- 1)^{\frac{n (n + 1)}{2}}}{n!} (x - \frac{π}{4})^{n}

Result

\cos x = \sum_{n = 0}^{\infty} \frac{1}{\sqrt{2}} \frac{(- 1)^{\frac{n (n + 1)}{2}}}{n!} (x - \frac{π}{4})^{n}

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