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RolyPoly Group Title

How to get started? Consider \(\ln x \) in [1, 1+t] (t>0), show that \( \frac{x}{x+1} < \ln (1+x) < x\) .

  • 2 years ago
  • 2 years ago

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  1. mukushla Group Title
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    start with defining a function for each part for example\[f(x)=\ln(1+x)-\frac{x}{x+1}\]and show that f(x) is strictly increasing (for the other part decreasing)

    • 2 years ago
  2. mukushla Group Title
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    make sense?

    • 2 years ago
  3. RolyPoly Group Title
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    I'm sorry. I don't quite understand. Why are you defining that function? Or, I should ask what you are doing. (Sorry again!)

    • 2 years ago
  4. mukushla Group Title
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    see u have 2 works to do here one is showing\[\ln(1+x)>\frac{x}{1+x}\]or\[\ln(1+x)-\frac{x}{1+x}>0\]so defining\[f(x)=\ln(1+x)-\frac{x}{x+1}\]and showing that f(x) is increasing is what u need here

    • 2 years ago
  5. mukushla Group Title
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    other part is showing\[\ln(1+x)<x\]or\[x-\ln(1+x)>0\]for this one let\[g(x)=x-\ln(1+x)\]and show that g(x) is strictly increasing

    • 2 years ago
  6. RolyPoly Group Title
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    (Shhh~~~) Would you mind letting me try the second half without giving me the hint? Thanks for your help. I start to understand it!

    • 2 years ago
  7. mukushla Group Title
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    yeah im finished here :)

    • 2 years ago
  8. RolyPoly Group Title
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    But wait. If I'm doing this way, why do I have to consider lnx in the interval [1, 1+t] for t>0?

    • 2 years ago
  9. RolyPoly Group Title
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    @mukushla Sorry! I'm calling you back :P

    • 2 years ago
  10. mukushla Group Title
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    no problem rolypoly :) yeah thats the domain for which we want prove that inequality

    • 2 years ago
  11. RolyPoly Group Title
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    Two questions: 1. To show f(x) is increasing, we do differentiation, right? (First derivative) 2. If we just take derivative, we're not really consider the domain, and also lnx.

    • 2 years ago
  12. mukushla Group Title
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    1. right and showing f'(x)>0 2. we should just consider the given interval [1,1+t]

    • 2 years ago
  13. RolyPoly Group Title
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    How can we ''just consider the given interval''?

    • 2 years ago
  14. mukushla Group Title
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    emm...sorry what do u mean?

    • 2 years ago
  15. RolyPoly Group Title
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    Sorry! You said that we should just consider the given interval. But in the calculation, how can we (show that) we just consider the given interval? Since we're doing it in a ''general'' way.

    • 2 years ago
  16. mukushla Group Title
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    ahh yeah .. when u want to show that f'(x)>0 u must use that

    • 2 years ago
  17. RolyPoly Group Title
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    \[f'(x) = \frac{1}{1+x} - \frac{(x+1) -(x)}{(x+1)^2}=\frac{1}{1+x}-\frac{1}{(x+1)^2} = \frac{x}{(x+1)^2}\] Hmm, so far so good?

    • 2 years ago
  18. mukushla Group Title
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    yes :)

    • 2 years ago
  19. RolyPoly Group Title
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    Then, it's bad. Once I put f'(x) = 0, I get x=0

    • 2 years ago
  20. mukushla Group Title
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    u see f'(x)>0 for given interval

    • 2 years ago
  21. mukushla Group Title
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    why letting it equal to zero?

    • 2 years ago
  22. RolyPoly Group Title
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    Errr, usual practice.

    • 2 years ago
  23. RolyPoly Group Title
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    Oh wait. For the given interval, f'(x)>0 since 1 is positive and 1+t is also positive for t>0. So, f'(x) >0

    • 2 years ago
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