A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

frx
 2 years ago
Best ResponseYou've already chosen the best response.0So 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

frx
 2 years ago
Best ResponseYou've already chosen the best response.0Would that be somewhat correct?

frx
 2 years ago
Best 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\[e^{x}\ge 1+x\] ?

frx
 2 years ago
Best ResponseYou've already chosen the best response.0Oh forgot to write that \[\xi \in (0,x)\]

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.1i think we can use by calculus principle...

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.1differential calculus i meant

frx
 2 years ago
Best ResponseYou've already chosen the best response.0What about the Mclaurinexpansion I made, doesn't it prove the fact that it's bigger?

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.1sorry i forgot about Mclaurinexpansion, i f by differential i got it

frx
 2 years ago
Best ResponseYou've already chosen the best response.0How did you show it using differentials?

RadEn
 2 years ago
Best ResponseYou've already chosen the best response.1e^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) :)

frx
 2 years ago
Best ResponseYou've already chosen the best response.0Oh that's a good way to show it, pretty simple to, thank you RadEn! :)
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.