Taylor Series & Estimating Error

Question

Compute T2(x) at x=0.8 for y=e^x and use a calculator to compute the error |e^x−T2(x)| at x=−0.4.

T2(x) = ______

|e^x−T2(x)| = _______

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Answer 1

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To find \( T_2(x) \) and the error \( |e^x - T_2(x)| \), we'll compute the Taylor series centered at \( a = 0 \) up to the second-degree term.

1. Compute \( T_2(x) \):

The general formula for the Taylor series of a function \( f(x) \) centered at \( a \) is:

\[ T_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n \]

In this case, \( a = 0 \), and we want \( T_2(x) \), which means we'll compute up to the second-degree term.

The function \( y = e^x \) has a simple Taylor series expansion:

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]

We only need the first three terms for \( T_2(x) \):

\[ T_2(x) = 1 + x + \frac{x^2}{2!} = 1 + x + \frac{x^2}{2} \]

2. Compute the error \( |e^x - T_2(x)| \):

To find the error at \( x = -0.4 \), we'll evaluate both \( e^x \) and \( T_2(x) \) at that point and find the absolute difference:

\[ |e^x - T_2(x)| = |e^{-0.4} - T_2(-0.4)| \]

Calculate \( e^{-0.4} \) and \( T_2(-0.4) \):

\[ e^{-0.4} \approx 0.67032 \]

\[ T_2(-0.4) = 1 - 0.4 + \frac{(-0.4)^2}{2} = 1 - 0.4 + 0.08 = 0.68 \]

Now, find the absolute difference:

\[ |e^{-0.4} - T_2(-0.4)| = |0.67032 - 0.68| = 0.00968 \]

So, the values are:

\[ T_2(x) = 1 + x + \frac{x^2}{2} \]

\[ |e^x - T_2(x)| = 0.00968 \]