How would you use the Maclaurin series for #e^-x# to calculate #e^0.1#?

1 Answer
Nov 29, 2016

# e^(0.1) ~~ 1.10517 (5dp) #

Explanation:

We know that

#e^(-x) = 1 - x + x^2/(2!)-x^3/(3!)+cdots#

is an alternating series and for #absx < 1# this series is absolutely convergent, so the error handled by cutting off after the #n# term is smaller than the higher order term left.

If we need to calculate #e^(-0.1)# within a precision #delta# then we solve first

#delta le x^n/(n!)# or #delta le (0.1)^n/(n!)# for instance, if #delta = 0.00001# then #n=4# would suffice. After that, #e^(0.1) = 1/e^(-0.1)#

# :. e^(0.1) = e^(-(-0.1)) #
# :. e^(0.1) = 1-(-0.1)+(-0.1)^2/(2!)-(-0.1)^3/(3!)+(-0.1)^4/(4!) +#(higher terms)
# :. e^(0.1) ~~ 1+0.1+0.01/2+0.001/6+0.0001/24 #
# :. e^(0.1) ~~ 1.1+0.005+0.00016667+0.000041667 #
# :. e^(0.1) ~~ 1.10517083 ... #
# :. e^(0.1) ~~ 1.10517 (5dp) #

Compare with the calculator answer #e^0.1=1.10517091 ...#