- #1
spaghetti3451
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- 34
Consider the Taylor series expansion of ##e^{-x}## as follows:
##\displaystyle{e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\dots}##
For ##x>0##, the partial sums ##1##, ##1-x##, ##\displaystyle{1-x+\frac{x^{2}}{2}}## bound ##e^{-x}## from above and from below alternately.
How do I prove this?
##\displaystyle{e^{-x}=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+\dots}##
For ##x>0##, the partial sums ##1##, ##1-x##, ##\displaystyle{1-x+\frac{x^{2}}{2}}## bound ##e^{-x}## from above and from below alternately.
How do I prove this?