Taylor series of&nbsp;x22-1+cosx&nbsp;using power series operations

Question

Taylor series of&amp;nbsp;x22-1+cosx&amp;nbsp;using power series operations

alenahelenash · Accepted Answer

To find the Taylor series of the function f(x) = x^2/2 - 1 + cos x, we can use the formula for the Taylor series:

f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (a)}{n!} (x - a)^{n}

where f^(n)(a) denotes the nth derivative of f evaluated at x=a.
First, let's find the derivatives of f(x):

f^{'} (x) = x - \sin x

f^{″} (x) = 1 - \cos x

f^{(3)} (x) = \sin x

f^{(4)} (x) = \cos x

f^{(n)} (x) = {\begin{matrix} \sin x & n = 3 + 4 k \\ \cos x & n = 4 k \\ 0 & otherwise \end{matrix}

where k is a non-negative integer.
Next, let's evaluate these derivatives at x=0:

f (0) = \frac{0^{2}}{2} - 1 + \cos 0 = - \frac{1}{2} + 1 = \frac{1}{2}

f^{'} (0) = 0 - \sin 0 = 0

f^{″} (0) = 1 - \cos 0 = 0

f^{(3)} (0) = \sin 0 = 0

f^{(4)} (0) = \cos 0 = 1

f^{(n)} (0) = {\begin{matrix} 0 & n = 0, 1, 2, 3 \\ (- 1)^{k} & n = 4 k \\ 0 & n = 4 k + 1, 4 k + 2, 4 k + 3 \end{matrix}

where k is a non-negative integer.
Now we can substitute these values into the Taylor series formula to get:

f (x) = \frac{1}{2} + \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{(4 k)!} x^{4 k}

Therefore, the Taylor series of f(x) is

f (x) = \frac{1}{2} + \sum_{k = 1}^{\infty} \frac{(- 1)^{k}}{(4 k)!} x^{4 k}

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Taylor series of x22-1+cosx using power series operations