Pick any exercise in 8.8 of the calculus textbook. Solve and post your solution. If you have issues doing so, discuss them.
Key Idea 8.8.1 gives the \(n^{th}\) term of the Taylor series of common functions. In Exercises 3–6, verify the formula given in the Key Idea by finding the first few terms of the Taylor series of the given function and identifying a pattern.
\[f(x)=e^x; c=0\]
\[e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}\]
\[1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...\]
\[(-\infty, \infty)\] Note: The derivative of \(e^x\) is also \(e^x\)
\[f(x) = e^x,\ thus\ f'(x) = e^x\ and\ f''(x) = e^x \] So that being confirmed, then
\[f(0)= 1 = f'(0) = f''(0) = f'''(0) = f''''(0)\] As the question defines the center at 0, this will follow a Maclaurin series.
\[p_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^n(0)}{n!}x^n\] So substitute for \(f(0)\) and derivatives, which results in for \(e^x\)
\[p_n(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + ... + \frac{1}{n!}x^n\] Solving for \(p_5(x)\) to approximate the value of \(e\) when \(x = 1\)
\[p_5 = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5\] Thus
\[p_5(1) = 1 + x + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} = \frac{163}{60} = 2.71667\]
Given the pattern observed from \(p_n(x)\), the resulting summation is equal to the key idea series from 8.8.1
\[e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}\]
Using the taylor function from the pracma library
library(pracma)
x <- function(x) exp(x)
result <- taylor(x, 0, 5)
result## [1] 0.008334245 0.041666573 0.166666726 0.499999996 1.000000000 1.000000000sum(result)## [1] 2.716668The summation 2.7166675 equals the calculated value above.
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