Probelm 3 Page 496 Verify the formula given in the Key Idea by finding the first few terms of the Taylor serie of the given function and identifying a pattern.
\(f(x) = e^x\); c = 0
So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.
We will calculate upto n=5
From the Key Idea the formula is :- \(e^x = \sum_{n=1}^{\infty} \cfrac {x^n}{n!} \Rrightarrow 1 + x + \cfrac{x^2}{2!} + \cfrac{x^3}{3!} + \cfrac{x^4}{4!} + \cfrac{x^5}{5!}\)
The Taylor series of a function is in the form: \(\sum_{n=0}^{\infty } \frac{f^{(n)}(a)(x-a)^{n}}{n!}\)
For n=0 \(\Rrightarrow \cfrac{f(0)(x-0)^0}{0!} \Rrightarrow \cfrac{e^{(0)}(1)}{0!} \Rrightarrow \cfrac {1}{0!}\)
For n=1 \(\Rrightarrow \cfrac{f^{'}(0)(x-0)^1}{1!} \Rrightarrow \cfrac{e^{(0)}(x)}{1!} \Rrightarrow \cfrac {x}{1!}\)
For n=2 \(\Rrightarrow \cfrac{f^{''}(0)(x-0)^2}{2!} \Rrightarrow \cfrac{e^{(0)}(x^2)}{2!} \Rrightarrow \cfrac {x^2}{2!}\)
For n=3 \(\Rrightarrow \cfrac{f^{'''}(0)(x-0)^3}{3!} \Rrightarrow \cfrac{e^{(0)}(x^3)}{3!} \Rrightarrow \cfrac {x^3}{3!}\)
For n=4 \(\Rrightarrow \cfrac{f^{''''}(0)(x-0)^4}{4!} \Rrightarrow \cfrac{e^{(0)}(x^4)}{4!} \Rrightarrow \cfrac {x^4}{4!}\)
For n=5 \(\Rrightarrow \cfrac{f^{'''''}(0)(x-0)^5}{5!} \Rrightarrow \cfrac{e^{(0)}(x^5)}{5!} \Rrightarrow \cfrac {x^5}{5!}\)
Hence, Both the Formula in the key value and by our calculations are equal.