Find the Taylor polynomialof order 5 appoximation to sin(1.5)
sinx = x - x^3/3! + x^5/5! - ...
sin 1.5 = 1.5 - (1.5)^3/6 + (1.5)^5/120
= 1.5 - .5625 + .06328 -
= 1.00078
my calculator said
tan 1.5 = .997495
so clearly 3 terms are not sufficient for good accuracy, (sine can't have a value > 1)
Taking one more term, -(1.5)^7/7! would bring it to .9974
To find the Taylor polynomial of order 5 approximation for sin(1.5), we can use the Taylor series expansion of the sine function. The general formula for the Taylor series expansion of a function f(x) centered at x = a is given by:
f(x) ≈ f(a) + f'(a)(x - a)/1! + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ... + f^n(a)(x - a)^n/n!,
where f'(x) is the first derivative of f(x), f''(x) is the second derivative, and so on, and f^n(x) is the nth derivative of f(x).
For the sine function, the Taylor series expansion centered at x = 0 is given by:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
To find the Taylor polynomial approximation of order 5, we need to find the values of the derivatives of sin(x) up to the 5th derivative evaluated at x = 0.
Derivatives of sin(x):
f(x) = sin(x)
f'(x) = cos(x)
f''(x) = -sin(x)
f'''(x) = -cos(x)
f''''(x) = sin(x)
f'''''(x) = cos(x)
Now, let's substitute these values into the Taylor series expansion formula:
sin(x) ≈ sin(0) + cos(0)(x - 0)/1! - sin(0)(x - 0)^2/2! - cos(0)(x - 0)^3/3! + sin(0)(x - 0)^4/4! + ...
Simplifying this expression, we have:
sin(x) ≈ x - x^3/3! + x^5/5! - x^7/7! + x^9/9! - ...
Now, to find the Taylor polynomial of order 5 approximation to sin(1.5), we substitute x = 1.5 into the equation:
sin(1.5) ≈ 1.5 - (1.5)^3/3! + (1.5)^5/5! - (1.5)^7/7! + (1.5)^9/9!
Evaluating this expression, we get:
sin(1.5) ≈ 1.5 - (1.5)^3/3! + (1.5)^5/5! - (1.5)^7/7! + (1.5)^9/9!
Note that if you would like a more precise approximation, you can include more terms in the Taylor series expansion.
To find the Taylor polynomial of order 5 approximation to sin(1.5), we can use the Taylor series expansion of the sine function.
The Taylor series expansion of the sine function is given by:
sin(x) = x - (x^3 / 3!) + (x^5 / 5!) - (x^7 / 7!) + ...
To find the Taylor polynomial of order 5, we need to consider the terms up to x^5. Let's calculate the coefficients for these terms:
Coefficient of x: The coefficient of x in the Taylor series for sin(x) is 1.
Coefficient of x^3: The coefficient of x^3 in the Taylor series for sin(x) is (-1^1 / 3!) = -1 / 6.
Coefficient of x^5: The coefficient of x^5 in the Taylor series for sin(x) is (1^2 / 5!) = 1 / 120.
Now, let's substitute these coefficients into the Taylor series expansion:
sin(x) ≈ x - (x^3 / 3!) + (x^5 / 5!)
To find the approximation for sin(1.5), we substitute x = 1.5 into the Taylor polynomial:
sin(1.5) ≈ 1.5 - (1.5^3 / 3!) + (1.5^5 / 5!)
Calculating these values:
sin(1.5) ≈ 1.5 - (1.5^3 / 6) + (1.5^5 / 120)
≈ 1.5 - (3.375 / 6) + (3.375 / 120)
≈ 1.5 - 0.5625 + 0.028125
≈ 0.965625
Therefore, the Taylor polynomial of order 5 approximation to sin(1.5) is approximately 0.965625.