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(1 point) The Taylor series for \( f(x)=\cos (x) \) at \( \mathrm{a}=\frac{\pi}{2} \) is \( \sum_{n=0}^{\infty} c_{n}\left(x-\frac{\pi}{2}\right)^{n} \). Find the first few coefficients. c 0


=
c 1

=
c 2

=
c 3

=
c 4

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User Cade Roux
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1 Answer

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To find the coefficients of the Taylor series, we need to follow these steps:

1. The Taylor series expansion of a function around a point can be written as: f(x) ≈ f(a) + f'(a)(x-a)/1! + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ... So, we will need to find the value of the function and its derivatives at the point a = π/2.

2. Let's start with f(x) = cos(x). The value of cos(x) at x = π/2 is 0. So, the first coefficient, c0 = f(a) = f(π/2) = cos(π/2) = 0.

2. Next, we need to find the derivatives of cos(x) and calculate their values at x = π/2:

- The first derivative of cos(x) is -sin(x). So c1 = f'(a) = -sin(π/2) = -1.

- The second derivative of cos(x) is -cos(x). So c2 = f''(a) = -cos(π/2) = 0.

- The third derivative of cos(x) is sin(x). So c3 = f'''(a) = sin(π/2) = 1.

- The fourth derivative of cos(x) is cos(x) again. So c4 = f''''(a) = cos(π/2) = 0.

However, it is important to remember a key aspect of Taylor series - the derivatives in the numerator must be divided by the respective factorial in the denominator (e.g. divided by 1! =1, 2! =2, 3! =6, and 4! =24).

After dividing the calculated values by their respective factorials, we find that all coefficients c1, c2, c3, and c4 for our series turn out to be zero, with the correct scaling.

So, our Taylor series, for cos(x) expanded at π/2 to the fourth order, only has c0 term and all the subsequent terms are zero. Which means, c0 equals the order of the polynomial O(1, (x, 15707963267949/10000000000000)).

Therefore, the coefficients for the Taylor series expansion of cos(x) around the point π/2 are: c0 = O(1, (x, 15707963267949/10000000000000)), c1 = 0, c2 = 0, c3 = 0, c4 = 0.

User Rupok
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