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frx

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

  • one year ago
  • one year ago

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  1. frx
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    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
  2. frx
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    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
  3. frx
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    Would that be somewhat correct?

    • one year ago
  4. frx
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    \[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(n+1)}(\xi) }{(n+1)! }x ^{n+1}\] \[e ^{x}=\sum_{n=0}^{1} \frac{ x ^{n} }{ n! }=1+x +\frac{ f ^{(2)}(\xi) }{(2)! }x ^{2} \] \[f(x)=e ^{x}; f \prime(x)=e ^{x}; f \prime \prime(x)=e {^x}\] 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
  5. frx
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    Oh forgot to write that \[\xi \in (0,x)\]

    • one year ago
  6. RadEn
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    i think we can use by calculus principle...

    • one year ago
  7. RadEn
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    differential calculus i meant

    • one year ago
  8. frx
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    What about the Mclaurinexpansion I made, doesn't it prove the fact that it's bigger?

    • one year ago
  9. RadEn
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    sorry i forgot about Mclaurinexpansion, i f by differential i got it

    • one year ago
  10. frx
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    How did you show it using differentials?

    • one year ago
  11. RadEn
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    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
  12. frx
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    Oh that's a good way to show it, pretty simple to, thank you RadEn! :)

    • one year ago
  13. RadEn
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    very, welcome.... :)

    • one year ago
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