f(x) = e^(3x), f(0)=1
df/dx = 3e^(3x), f'(0)=3
d²f/dx² = 9e^(3x), f''(0)=9
d³f/dx³ = 27e^(3x), f'''(0)=27
f(x)=f(0)+f'(0)x+(1/2)f''(0)x²+(1/6)f'''(0)x³+.
=1+3x+(3/2)x²+(9/2)x³+.
MacLaurin Series for e^x:
e^x = 1/0! + x/1! + x²/2! + x³/3! + x⁴/4! + .+ xⁿ/n!
e^9 = 1/0! + 9/1! + 9²/2! + 9³/3! + 9⁴/4! + .+ 9ⁿ/n!
= 1+9+40.5+121.5+273.375+492.075+738.113+949.002+1067.627+.
这个级数非常有欺骗性,按照上式计算,它看上去是一个发散级数(Divergent Series),事实上,麦克劳林级数和泰勒级数(Taylor Series)都有这样的现象,
因为楼主并未给出精确到几位数的要求,为了准确,本人在此假设精确到0.01,
因为9^25/25!=0.046,所以,e^9估计如下：
e^9 = 1/0! + 9/1! + 9²/2! + 9³/3! + 9⁴/4! + .+ 9ⁿ/n!
≈ 1+9+40.5+121.5+273.375+492.075+738.113+949.002+1067.627+
+1067.627+96.843+786.162+589.621+408.199+262.414+157.448
+88.564+46.887+23.444+11.105+4.997+2.142+0.867+0.34+0.046
≈8103