The MacLaurin formula for #e^x# is:
#100e^x = 100sum_(n=0)^oo x^n/(n!)#
Based on Lagrange remainder theorem, if we use the sum of order #N#, the error is:
#100e^(1/2) = 100(sum_(n=0)^N (1/2)^n/(n!)+ xi^(N+1)/((N+1)!))#
where #0 <= xi <= 1/2#.
As #xi^(N+1)# is a monotone increasing function its maximum in the interval occurs for #x=1/2#, so:
#xi^(N+1)/((N+1)!) <= (1/2)^(N+1)/((N+1)!)#
and we have an approximation within #0.01# if:
#100(1/2)^(N+1)/((N+1)!) < 1/100#
#1/(2^(N+1)) < ((N+1)!)/10^4#
#2^(N+1)((N+1)!) > 10^4#
By attempts, we can see that for #N=4#:
#2^5 xx 5! = 3840#
and for #N=5#:
#2^6 xx 6! = 46080#
Then:
#100e^(1/2) ~= 100sum_(n=0)^5 1/(2^n(n!)) = 100(1 + 1/2+1/8+1/48+1/348+1/3840) = 100/3840(3840+1920+480+80+10+1) = 31655/192#
In fact using a calculator:
#31655/192 = 164.869791667#
#100e^(1/2) = 164.87212707#
