$\Large \int\limits \frac{ (1+e^x)^2 }{ e^x } dx$

Prove by induction that for $ x \ge 0 $ and any natural number $n$, $\large e^x \ge 1+x+\frac{ x^2 }{ 2! } + ... + \frac{ x^n }{ n! }$

Question

$\Large \int\limits \frac{ (1+e^x)^2 }{ e^x } dx$

Prove by induction that for $ x \ge  0 $ and any natural number $n$, $\large e^x \ge 1+x+\frac{ x^2 }{ 2! } + ... + \frac{ x^n }{ n! }$

anonymous · Answer

two problems.

loser66 · Answer

for the first one, to me, I break it down to (1+2e^x +e^x^2)/e^x= e^(-x) +2 +e^x, then take integral term by term.

anonymous · Answer

yes,that idea does word :/ ...

and for second one?

zarkon · Answer

for the 2nd one 

1) do the basis step
2) assume $\large e^x \ge 1+x+\frac{ x^2 }{ 2! } + ... + \frac{ x^n }{ n! }$ is true for some $n$

let $\displaystyle f(x)= e^x -\left(1+x+\frac{ x^2 }{ 2! } + ... + \frac{ x^n }{ n! }+\frac{ x^{n+1} }{ (n+1)! }\right)$

show that $f(x)$ is an increasing function (by taking the derivative) that is only zero when $x=0$