Find the Taylor and maclaurin series for the function f(x)=cosx at c=π/6 and c=0 respectively
The Taylor series for a function f(x) centered at c is given by:
f(x) = f(c) + f'(c)(x-c)/1! + f''(c)(x-c)^2/2! + f'''(c)(x-c)^3/3! + ...
First, let's find the Taylor series for f(x) = cos(x) centered at c = π/6:
f(π/6) = cos(π/6) = √3/2
f'(x) = -sin(x)
f'(π/6) = -sin(π/6) = -1/2
f''(x) = -cos(x)
f''(π/6) = -cos(π/6) = -√3/2
f'''(x) = sin(x)
f'''(π/6) = sin(π/6) = 1/2
The terms involving f^(k)(π/6) for k ≥ 4 are all 0.
Using these values in the Taylor series formula:
f(x) = f(π/6) + f'(π/6)(x-π/6) + f''(π/6)(x-π/6)^2/2! + f'''(π/6)(x-π/6)^3/3! + ...
= (√3/2) + (-1/2)(x-π/6) + (-√3/2)(x-π/6)^2/2! + (1/2)(x-π/6)^3/3! + ...
Simplifying further:
f(x) = (√3/2) - (x-π/6)/2 + (-√3/2)(x-π/6)^2/2 + (x-π/6)^3/6 + ...
This is the Taylor series for f(x) = cos(x) centered at c = π/6.
Next, let's find the Maclaurin series for f(x) = cos(x) centered at c = 0:
Using the same derivative values as before:
f(0) = cos(0) = 1
f'(x) = -sin(x)
f'(0) = -sin(0) = 0
f''(x) = -cos(x)
f''(0) = -cos(0) = -1
f'''(x) = sin(x)
f'''(0) = sin(0) = 0
The terms involving f^(k)(0) for k ≥ 4 are all 0.
Using these values in the Taylor series formula:
f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + ...
= 1 + 0x + (-1)x^2/2! + 0x^3/3! + ...
Simplifying further:
f(x) = 1 - x^2/2 + x^4/24 - x^6/720 + ...
This is the Maclaurin series for f(x) = cos(x) centered at c = 0.
To find the Taylor and Maclaurin series for the function f(x) = cosx at c = π/6 and c = 0, we can start by finding the derivatives of the function and evaluating them at the given points.
The general formula for the nth derivative of cosx is:
f^(n)(x) = cos(x + nπ/2)
Let's start by finding the Taylor series at c = π/6:
1. Find the derivatives of f(x) = cosx:
f'(x) = -sinx
f''(x) = -cosx
f'''(x) = sinx
f''''(x) = cosx
2. Evaluate the derivatives at c = π/6:
f(π/6) = cos(π/6) = √3/2
f'(π/6) = -sin(π/6) = -1/2
f''(π/6) = -cos(π/6) = -√3/2
f'''(π/6) = sin(π/6) = 1/2
f''''(π/6) = cos(π/6) = √3/2
3. Write the Taylor series using the derivatives evaluated at c = π/6:
f(x) = f(π/6) + f'(π/6)(x - π/6) + f''(π/6)(x - π/6)^2/2! + f'''(π/6)(x - π/6)^3/3! + f''''(π/6)(x - π/6)^4/4! + ...
Substituting the derivatives evaluated at c = π/6, we have:
f(x) = √3/2 - 1/2(x - π/6) - (√3/2)(x - π/6)^2/2! + 1/2(x - π/6)^3/3! + (√3/2)(x - π/6)^4/4! + ...
This is the Taylor series for f(x) = cosx centered at c = π/6.
Now, let's find the Maclaurin series for f(x) = cosx:
Since Maclaurin series are simply Taylor series centered at c = 0, we can use the same derivatives found earlier and evaluate them at c = 0.
1. Evaluate the derivatives at c = 0:
f(0) = cos(0) = 1
f'(0) = -sin(0) = 0
f''(0) = -cos(0) = -1
f'''(0) = sin(0) = 0
f''''(0) = cos(0) = 1
2. Write the Maclaurin series using the derivatives evaluated at c = 0:
f(x) = f(0) + f'(0)x + f''(0)x^2/2! + f'''(0)x^3/3! + f''''(0)x^4/4! + ...
Substituting the derivatives evaluated at c = 0, we have:
f(x) = 1 + 0x - x^2/2! + 0x^3/3! + x^4/4! + ...
This is the Maclaurin series for f(x) = cosx centered at c = 0.