) Find the third degree Taylor Polynomial, T3(), centered at x = 0 for f(x) = sin x.
b) Evaluate lim Sum by replacing sin x with T3(x) found in part a). 3 +0 X
c) If you used a higher degree Taylor Polynomial would the result change.
The Correct Answer and Explanation is:
a) Find the third degree Taylor Polynomial T3(x)T_3(x) of f(x)=sinxf(x) = \sin x centered at x=0x=0.
The Taylor polynomial of degree 3 for a function ff centered at 0 is: T3(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3T_3(x) = f(0) + f'(0) x + \frac{f”(0)}{2!} x^2 + \frac{f”'(0)}{3!} x^3
For f(x)=sinxf(x) = \sin x:
- f(0)=sin0=0f(0) = \sin 0 = 0
- f′(x)=cosx⇒f′(0)=cos0=1f'(x) = \cos x \Rightarrow f'(0) = \cos 0 = 1
- f′′(x)=−sinx⇒f′′(0)=−sin0=0f”(x) = -\sin x \Rightarrow f”(0) = -\sin 0 = 0
- f′′′(x)=−cosx⇒f′′′(0)=−cos0=−1f”'(x) = -\cos x \Rightarrow f”'(0) = -\cos 0 = -1
Substitute these into the polynomial: T3(x)=0+1⋅x+02!x2+−13!x3=x−x36T_3(x) = 0 + 1 \cdot x + \frac{0}{2!} x^2 + \frac{-1}{3!} x^3 = x – \frac{x^3}{6}
Answer: T3(x)=x−x36\boxed{T_3(x) = x – \frac{x^3}{6}}
b) Evaluate the limit sum by replacing sinx\sin x with T3(x)T_3(x):
You wrote:
Evaluate lim∑\lim \sum by replacing sinx\sin x with T3(x)T_3(x) found in part a). 3 +0 X
I assume you want to evaluate a limit of a sum involving sinx\sin x, say something like limn→∞∑k=1nsinkn\lim_{n \to \infty} \sum_{k=1}^n \sin \frac{k}{n}
or limx→0sinxx\lim_{x \to 0} \frac{\sin x}{x}
But the question is a bit unclear. Could you clarify the exact expression for the sum or limit you want to evaluate?
c) If you used a higher degree Taylor polynomial, would the result change?
Short answer:
If you use a higher degree Taylor polynomial, the approximation becomes more accurate, especially near the center x=0x=0. For many limits or sums involving small xx, increasing the degree can improve the accuracy, but if the limit or sum is finite or the function is well-behaved, the final limit value may remain the same. However, for precise numerical approximations or when xx is not very small, higher degree terms can change the result.
Detailed Explanation
The Taylor polynomial provides a polynomial approximation to a function around a specific point, usually x=0x=0 in this case (Maclaurin series). For f(x)=sinxf(x) = \sin x, the Taylor series expansion is: sinx=x−x33!+x55!−x77!+⋯\sin x = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \cdots
The third-degree polynomial T3(x)=x−x36T_3(x) = x – \frac{x^3}{6} approximates the sine function closely near x=0x=0.
When evaluating limits or sums that involve sinx\sin x, substituting T3(x)T_3(x) instead of sinx\sin x can simplify calculations while maintaining accuracy for sufficiently small values of xx. For example, if evaluating a limit as x→0x \to 0, the error introduced by ignoring terms of degree higher than 3 tends to zero, so the limit value remains unchanged.
However, using a higher degree Taylor polynomial (like T5(x)=x−x36+x5120T_5(x) = x – \frac{x^3}{6} + \frac{x^5}{120}) captures more behavior of the sine function farther from zero. This improved approximation may affect the numerical value of sums or integrals where xx is not infinitesimally small, thereby providing more accurate results.
In the context of limits, the final answer often does not change because higher degree terms vanish faster as x→0x \to 0. But for finite sums or integrals over intervals that are not vanishingly small, higher degree polynomials reduce approximation errors and improve the accuracy of the computed values.
Thus, whether the result changes depends on the context:
