Question

Show that \[e ^{x} \ge 1+x\] for all real numbers x

Answer 1

I have the idea that I could use the McLauren expansion for e^x
\[e ^{x}=\sum_{n=0}^{\infty}\frac{ x ^{n} }{ n! }\]

Answer 2

So the two first expansions would be \[1+x\] but since I have to add the Lagrange reminder it would be proven bigger than 1+x

Answer 3

Would that be somewhat correct?

Answer 4

\[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(n+1)}(\xi) }{(n+1)! }x ^{n+1}\]
\[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(2)}(\xi) }{(2)! }x ^{2} \]
\[f(x)=e ^{x}; f \prime(x)=e ^{x}; f \prime \prime(x)=e {^x}\] so that gives the remainder \[\frac{ e ^{\xi}x ^{2}}{2! }\] and \[e ^{x}=1+x+\frac{ e ^{\xi}x ^{2}}{2! }\ \ge 1+x\]
Is this a genuine proof that\[e^{x}\ge 1+x\] ?

Answer 5

Oh forgot to write that \[\xi \in (0,x)\]

Answer 6

i think we can use by calculus principle...

Answer 7

differential calculus i meant

Answer 8

What about the Mclaurinexpansion I made, doesn't it prove the fact that it's bigger?

Answer 9

sorry i forgot about Mclaurinexpansion, i f by differential i got it

Answer 10

How did you show it using differentials?

Answer 11

e^x ≥ x + 1
e^x - 1 - x ≥ 0
let given f(x) = e^x - 1 - x, so
f '(x) = e^x - 1
criticals points of f, hapended when saat f '(x) = 0
so, e^x - 1 = 0
get x = 0
to knowing the kind (max or min) of f, use the 2nd derivative
so, f ''(x) = e^x
for x=0, gives
f ''(0) = 1 > 0

because f ''(0) > 0, therefore its kind is minimum
the minimum value of f is f(0) = e^0 - 1 - 0 = 0
because f has the minimum value, is 0
so,, obviously
f(x) ≥ 0

e^x - 1 - x ≥ 0

e^x ≥ x + 1 (proof) :)

Answer 12

Oh that's a good way to show it, pretty simple to, thank you RadEn! :)

Answer 13

very, welcome.... :)