Question

Use Taylor's Formula to determine the number of terms of the Maclaurin series for e^x that should be used to estimate e^0.1to within 0.00001.

Short answer

The number of terms of the Maclaurin series is 4

Step-by-step solution

Definition of the Maclaurin series

A function's Taylor series is an infinite sum of terms stated in terms of the derivatives of the function at a single point. Near this point, the function and the sum of its Taylor series are equivalent for most typical functions.

Finding the number of terms

Let us take, \[{\rm{f(x) = }}{{\rm{e}}^{\rm{x}}}\]

Then the nth derivative will be e^x

\[{{\rm{f}}^{{\rm{(n)}}}}{\rm{ = }}{{\rm{e}}^{\rm{x}}}\]

For e^0.1 we can look at the Taylor inequality at x=0.1

For Maclaurin series, a=0.

We can start at n=2 and move up by 1 unit the remainder is less than 0.00001

So,

\[\begin{array}{c}\left| {{f^{(n)}}(0.1)} \right| \le M\\{e^{0.1}} \le M\\\left| {{R_n}(x)} \right|{\rm{ }} \le \frac{M}{{(n + 1)!}}|x - a{|^{n + 1}}\\\left| {{R_2}(x)} \right| \le \frac{{{e^{0.1}}}}{{(2 + 1)!}}|0.1 - 0{|^{2 + 1}} \le 0.0002\\\left| {{R_3}(x)} \right| \le \frac{{{e^{0.1}}}}{{(3 + 1)!}}|0.1 - 0{|^{3 + 1}} \le 0.000005\end{array}\]

So, \[|{R_3}(x)|\] is less than 0.00001,

Therefore, we need 4 terms

That is \[\left( {n = 0,1,2,3} \right)\]

Hence, the number of terms of the Maclaurin series is 4.