Question

Approximating \(f(x)=e^{x}\) In calculus, it can be shown that $$ f(x)=e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !} $$ We can approximate the value of \(f(x)=e^{x}\) for any \(x\) using the following sum $$ f(x)=e^{x} \approx \sum_{k=0}^{n} \frac{x^{k}}{k !} $$ for some \(n\). (a) Approximate \(f(1.3)\) with \(n=4\). (b) Approximate \(f(1.3)\) with \(n=7\). (c) Use a calculator to approximate \(f(1.3)\) (d) Using trial and error, along with a graphing utility's SEOuence mode, determine the value of \(n\) required to approximate \(f(1.3)\) correct to eight decimal places.

Short answer

a) 3.519. b) 3.66968. c) 3.669296. d) n ≈ 13.

Step-by-step solution

Understanding the Series Expansion

The function we are working with is the exponential function, which can be approximated by a series sum. This series is: f(x) = e^x ≈ k = 0 (kxk!)

Approximating f(1.3) with n=4

We need to approximate f(1.3) by summing the first 5 terms (from k=0 to k=4) of the series. Substituting x = 1.3: f(x) ≈ k = 0 (1.3k! = 1+(1.3() +(1.3)2(2+ + 1.34(4) = 1+1.3+ 0.845 +0.283+ 0.091 = 3.519 .

Approximating f(1.3) with n=7

We need to approximate f(1.3) by summing the first 8 terms (from k=0 to k=7) of the series. Substituting x = 1.3: f(x) ≈ k = 0 (1.3k! = 1+1.3+ 0.845 +0.283+ 0.091 + 0.01652(5+ 0.0032+0.00059 + = 3.66968 .

Using a Calculator to Approximate f(1.3)

Use a calculator to find e1.3. The actual value is: f(1.3) ≈ 3.669296 .

Determining Required n for Eight Decimal Places Correctness

Using trial and error, calculate the series sum for different values of n until the approximation matches f(1.3) to eight decimal places. Carry out repetitive calculation using a graphing utility’s sequence mode until f(1.3)'s value stabilizes to 3.66929593. This happens when n is around 13. Hence, n ≈ 13.

Key concepts

series expansion

The idea behind series expansion is to represent a function as a sum of infinitely many terms. Consider the exponential function, \(e^{x}\). It can be represented as a power series: \[e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!}\]. Each term in this series includes a factorial in the denominator which makes the series converge fast.
This means, for small values of \(|x|\), the terms get significantly smaller as k increases.
When doing calculations, we often truncate this series to a finite number of terms, thus approximating the function.

Taylor series

A specific type of series expansion is the Taylor series. It's used to approximate functions around a point. The general form is: \[f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x - a)^{k}\]. For the exponential function, the Taylor series around 0 (also known as the Maclaurin series) looks as follows: \[e^{x} = \sum_{k=0}^{\infty} \frac{x^{k}}{k!}\].
The beauty of the exponential function is that all its derivatives are \(e^x\). Thus, the series becomes straightforward. Depending on the approximation needed, we sum up the first n terms.

convergence

Convergence refers to how closely a series approximation gets to the actual function value. The more terms we include in the Taylor series, the closer we get to the true value of the function.
For the exponential function: \(e^{x}\), it converges quite fast because each term becomes very small quickly. Finding how many terms (n) are required for a desired accuracy involves trial and error.
In our example, to match \(f(1.3)\) to eight decimal places, we needed about 13 terms. This means that the Taylor series sum up to n=13 provides the precision required for the approximation.