what is the taylor series of f(x) = sin(x) at a=pi/3
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what is the Taylor series of f(x) = sin(x) at a=pi/3
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How do you use Taylor series for sin(x) at a = pi/3? | Socratic
You will see solution with explanation.
sinx = x - x^3/3! + x^5/5! - x^7/7! + ..... for x = π/3
= π/3 - (π/3)^3/6 + (π/3)^5/120 - (π/3)^7/5040 + ...
= 1.047197... - .191396... + .010494... - .000274012 + .... (the next < 3 decimals)
= appr .86602127
real answer : sin π/3 = .866025403 correct to 9 decimals
error : .000004132 , not bad
Reiny
Your answer is Maclaurin series, but here it is need Taylor series.
Of course Maclaurin series is a Taylor series expansion of a function about 0.
Here it is need Taylor series of a function about π / 3 ≈ 1
Your answer is also correct, but I think something else is answer for this question.
You are right, I skimmed over the "Taylor" part.
I should read more carefully.
To find the Taylor series of the function f(x) = sin(x) at a specific value, we can use the formula for the Taylor series:
f(x) = f(a) + f'(a)(x - a)/1! + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...
In this case, we want to find the Taylor series of f(x) = sin(x) at a = π/3. Let's start by finding the first few derivatives of sin(x):
f(x) = sin(x)
f'(x) = cos(x)
f''(x) = -sin(x)
f'''(x) = -cos(x)
f''''(x) = sin(x)
...
Now, we can substitute a = π/3 and these derivatives into the Taylor series formula:
f(π/3) = sin(π/3)
f'(π/3) = cos(π/3)
f''(π/3) = -sin(π/3)
f'''(π/3) = -cos(π/3)
...
Substituting these values into the formula, we get:
f(x) = sin(π/3) + cos(π/3)(x - π/3)/1! - sin(π/3)(x - π/3)^2/2! - cos(π/3)(x - π/3)^3/3! + ...
Now we can simplify this expression to get the Taylor series of f(x) = sin(x) at a = π/3:
f(x) = √3/2 + (1/2)(x - π/3) - (√3/2)(x - π/3)^2/2! - (1/2)(x - π/3)^3/3! + ...
Therefore, the Taylor series of f(x) = sin(x) at a = π/3 is:
sin(x) ≈ √3/2 + (1/2)(x - π/3) - (√3/2)(x - π/3)^2/2! - (1/2)(x - π/3)^3/3! + ...