Question

How would you use the Maclaurin series for \[{{e}^{-x}}\] to calculate \[{{e}^{0.1}}\]?

Answer 1

Hint: Maclaurin series, a type of series in which all terms are nonnegative integer powers of the variable. It can also be termed as a Taylor series that is expanded about the reference point zero and that takes the form\[f(x)=f(0)+\dfrac{f\prime (0)}{1!}x+\dfrac{f\prime \prime (0)}{2!}{{x}^{2}}+\cdots +\dfrac{f[n](0)}{n!}{{x}^{n}}\]. It should also hold the conditions of a Taylor series. The value of \[f(0)=1\]

Complete step by step answer:
As per the given question, we have to find the value of an expression using the Maclaurin series. Here, we have to find \[{{e}^{0.1}}\] using the Maclaurin series of \[{{e}^{-x}}\].
The first derivative of \[{{e}^{-x}}\] is \[-{{e}^{-x}}\], on substituting \[x=0\]we get \[{{f}^{1}}(0)=-1\]. Similarly, we calculate other derivatives.
Let \[f(x)={{e}^{-x}}\] then Maclaurin series would be \[{{e}^{-x}}=1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+\cdots \]
It is an alternating series and for \[|x|<1\], the series is absolutely convergent. So error handled by cutting off after the \[{{n}^{th}}\] term is negligible than the higher order term left.
We need to calculate \[{{e}^{0.1}}\], so x will be \[0.1\].
Let us assume the precision \[\delta =0.00001\] then n=4 would suffice. After that, \[{{e}^{0.1}}=\dfrac{1}{{{e}^{-0.1}}}\]
\[\Rightarrow {{e}^{0.1}}={{e}^{-(-0.1)}}\]
\[\Rightarrow {{e}^{0.1}}=1-(-0.1)+\dfrac{{{(-0.1)}^{2}}}{2!}-\dfrac{{{(-0.1)}^{3}}}{3!}+\dfrac{{{(-0.1)}^{4}}}{4!}+\left( higher\text{ }terms \right)\]
\[\Rightarrow {{e}^{0.1}}\approx 1+0.1+\dfrac{0.01}{2}+\dfrac{0.001}{6}+\dfrac{0.0001}{24}\]
\[\Rightarrow {{e}^{0.1}}\approx 1.1+0.005+0.00016667+0.000041667\]
\[\Rightarrow {{e}^{0.1}}\approx 1.10517\]
If we use the calculator and find the value \[{{e}^{0.1}}\] we will get the value as \[{{e}^{0.1}}=1.10517091....\]
The answers we got from the Maclaurin series and from the calculator are nearly the same.
\[\therefore \] The value of \[{{e}^{0.1}}\] using Maclaurin series of \[{{e}^{-x}}\] is \[1.10517\].

Note:
In the above way we can find expansions of many functions and also find the values of exponential functions easily. While calculating derivatives make sure we calculate the inner derivative of functions. Check whether the function holds the conditions of Taylor series. Maclaurin series helps us to approximate the functions with polynomials. We must avoid calculation mistakes to get the correct result or solution.