91Ó°ÊÓ

The Taylor series expansion for \(e^x\) is given by: \[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots\] Since we want to find the value of \(e\), we need to calculate the Taylor series for \(x = 1\): \[e \approx 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots\]

To approximate \(e\) to four decimal places, we need to find the number of terms of the Taylor series such that the difference between the approximation and the actual value of \(e\) is less than \(0.0001\). Let's start with \(n = 1\) and keep increasing the value of \(n\) until the difference is less than \(0.0001\). When calculating the differences, consider the remainder term (the term after the one we stop with) as a rough estimate of the error. For \(n = 1\), we have: \[e \approx 1 + 1 = 2\] The error is \(\frac{1}{2!} = 0.5\), which is larger than \(0.0001\). Let's proceed to the next value of \(n\). For \(n = 2\), we have: \[e \approx 1 + 1 + \frac{1}{2!} = 2 + 0.5 = 2.5\] The error is \(\frac{1}{3!} = \frac{1}{6} \approx 0.167\), which is still larger than \(0.0001\). We will keep increasing \(n\) until the error is less than \(0.0001\). For \(n = 3\), we have: \[e \approx 1 + 1 + \frac{1}{2!} + \frac{1}{3!} = 2.5 + \frac{1}{6} = 2.6667\] The error is \(\frac{1}{4!} = \frac{1}{24} \approx 0.0417\), which is still larger than \(0.0001\). For \(n = 4\), we have: \[e \approx 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} = 2.6667 + \frac{1}{24} = 2.7083\] The error is \(\frac{1}{5!} = \frac{1}{120} \approx 0.0083\), which is still larger than \(0.0001\). For \(n = 5\), we have: \[e \approx 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} = 2.7083 + \frac{1}{120} = 2.7167\] The error is \(\frac{1}{6!} = \frac{1}{720} \approx 0.0014\), which is still larger than \(0.0001\). Finally, for \(n = 6\), we have: \[e \approx 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} = 2.7167 + \frac{1}{720} = 2.7181\] The error is \(\frac{1}{7!} = \frac{1}{5040} \approx 0.0002\), which is less than \(0.0001\). Now that we have found the number of terms necessary for the required accuracy, we will use this approximation for the value of \(e\).

The approximate value of \(e\) to four decimal places is \(2.7181\).