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frx Group TitleBest ResponseYou've already chosen the best response.0
I have the idea that I could use the McLauren expansion for e^x \[e ^{x}=\sum_{n=0}^{\infty}\frac{ x ^{n} }{ n! }\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Would that be somewhat correct?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(n+1)}(\xi) }{(n+1)! }x ^{n+1}\] \[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(2)}(\xi) }{(2)! }x ^{2} \] \[f(x)=e ^{x}; f \prime(x)=e ^{x}; f \prime \prime(x)=e {^x}\] 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\[e^{x}\ge 1+x\] ?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Oh forgot to write that \[\xi \in (0,x)\]
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
i think we can use by calculus principle...
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
differential calculus i meant
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
What about the Mclaurinexpansion I made, doesn't it prove the fact that it's bigger?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
sorry i forgot about Mclaurinexpansion, i f by differential i got it
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
How did you show it using differentials?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
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) :)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Oh that's a good way to show it, pretty simple to, thank you RadEn! :)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
very, welcome.... :)
 one year ago
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