Page 2 of 2 FirstFirst 12
Results 16 to 26 of 26
Like Tree10Thanks

Math Help - Prove x > ln(1+x) for all x > 0

  1. #16
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    If a>b, lna>lnb
    Follow Math Help Forum on Facebook and Google+

  2. #17
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,817
    Thanks
    700

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by Hartlw View Post
    If a>b, lna>lnb
    That statement is false over the Reals. It is only true over the positive reals. So, if a>b>0, then ln a > ln b. Even so, if the original poster has not proven that the natural logarithm is a strictly increasing function over the positive reals, it must be proven.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by SlipEternal View Post
    That statement is false over the Reals. It is only true over the positive reals. So, if a>b>0, then ln a > ln b. Even so, if the original poster has not proven that the natural logarithm is a strictly increasing function over the positive reals, it must be proven.
    The statement is true: see OP, x>0. And anyhow, lna only defined for a positive. It is a strictly increasing function because either by def of lnx as integral of 1/t from 1 to x, x>0, or most easily by d(logx)/dx = 1/x and x>0. see OP.


    ie, lna>lnb if a>b because derivative is pos.

    Why make everything so complicated?

    I give a 3 line answer to the OP and you turn a mole-hill into a mountain.
    Follow Math Help Forum on Facebook and Google+

  4. #19
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,817
    Thanks
    700

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by Hartlw View Post
    Why make everything so complicated?

    I give a 3 line answer to the OP and you turn a mole-hill into a mountain.
    I did not make everything so complicated. I answered the OP's second question (which you ignored to start a rant that proofs don't need supporting arguments). You can read the OP's second question here. It is rude to hijack threads and troll those attempting to actually help.
    Thanks from Plato and topsquark
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by SlipEternal View Post
    I did not make everything so complicated. I answered the OP's second question (which you ignored to start a rant that proofs don't need supporting arguments). You can read the OP's second question here. It is rude to hijack threads and troll those attempting to actually help.
    I proved OP with

    x>in(1+x)
    e^x>1+x
    e^x = 1+x+.... >1+x

    Every step is trivial and would have been taken for granted in an introductory calculus course.

    But I was asked to show: e^x>e^y -> x > y which I did by showing that
    If a>b, lna>lnb which I considered trivial at this level of discussion (obviously derivative of ln>0 without a lot of blah-blah, or look at the graph). Actually, this step was hi-jacked by someone else with an exagerated discussion in abstract terms to prove this trivial statement.

    What did I hi-jack? One of the other posts? you’re kidding.

    And yet a statement like substitute into Napier's inequity and the answer falls right out goes unquestioned.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,817
    Thanks
    700

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by Hartlw View Post
    I proved OP with

    x>in(1+x)
    e^x>1+x
    e^x = 1+x+.... >1+x
    I was not going to respond to you the first time that you posted this. Then, as I continued replying to the OP, you made the exact same post a second time and asked, "what's wrong with this?". I mistook it for a genuine question. Obviously my mistake. I apologize for whatever I said that put you on the defensive.

    Quote Originally Posted by Hartlw View Post
    I was asked to show: e^x>e^y -> x > y which I did by showing that
    If a>b, lna>lnb
    Thank you. This was constructive to the thread.

    Quote Originally Posted by Hartlw View Post
    And yet a statement like substitute into Napier's inequity and the answer falls right out goes unquestioned.
    Napier invented logarithms. Not only is Napier's inequality pertinent, it is also very interesting that the answer does fall right out. I provided one method of proof, Plato provided a second, and you provided a third (although I still think the order of your three statements should be reversed). Again, the problem started because I mistook what you said as a question. It is difficult to read inflection in a post. I apologize once more for putting you on the defensive.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1

    Re: Prove x > ln(1+x) for all x > 0

    Unless the OP has more to post/ask here, I think the thread is pretty much finished.

    Many of the posts were good ones, but please keep the arguments to PMs.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #23
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    You are correct. Three sols were given. The first in the first post.

    1) f(x) = x-ln(1+)x and f’(x)>0. Good answer.
    2) Napier’s inequality. Answer
    My only objection to this answer is it proves one unproved inequality with another unproved inequality. Couldn’t find any reference to it in a calculus book. I vaquely recall seeing it in statistics, so possibly inappropriate to subject. So the answer depends on happening to remember a formula from somewhere.
    3) x>lnx, e^x>1+x, e^x=1+x+…>1+x. Answer

    I missed 1) and 2) because I was thrown off by the endless pages of math which I had no inclination to follow and which led me to believe that the respondents were still fishing for an answer.

    3) was posted after a considerable lull and then followed soon after by another huge chunk of math out of the clear blue sky in response to an unreferenced “hint.” I could only conclude without wasting a lot of time that the answer was still being sought, which puzzled me since an answer had just been given.

    The suggestion that I hi-jacked an answer is outrageous. All three answers are clearly independent.
    Follow Math Help Forum on Facebook and Google+

  9. #24
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by Hartlw View Post
    You are correct. Three sols were given. The first in the first post.

    1) f(x) = x-ln(1+)x and f’(x)>0. Good answer.
    2) Napier’s inequality. Answer
    My only objection to this answer is it proves one unproved inequality with another unproved inequality. Couldn’t find any reference to it in a calculus book. I vaquely recall seeing it in statistics, so possibly inappropriate to subject. So the answer depends on happening to remember a formula from somewhere.
    3) x>lnx, e^x>1+x, e^x=1+x+…>1+x. Answer

    .
    Whoops. 1) is not the answer. f'(x)>0 means f(x) increasing but not necessarily positive. So you have to show it's positive, ie, x>ln(1+x), for x -> 0. Or did I miss something? Was that the point of all those pages of math? Or was there another answer buried somewhere else?

    That really only leaves 3) in the spirit of the question which can be shortened to:

    e^x=1+x+…>1+x

    EDIT: OK, and the inverse (ln) of an increasing function is increasing. I like the essence of a solution without distractions. That's why I liked 1) before I realized it wasn't the essence of the solution. But at least it was clear.
    Last edited by Hartlw; October 5th 2013 at 04:28 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #25
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    Let f(x) = x-ln(1+x), x>0 as in post 1.

    OK: f(0)=0 and f'(x)>0, x>0 -> f(x)>0, x>0 is obvious.

    Sorry, I got hung up on f'(0)=0.
    Last edited by Hartlw; October 5th 2013 at 05:45 AM.
    Follow Math Help Forum on Facebook and Google+

  11. #26
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Prove x > ln(1+x) for all x > 0

    Quote Originally Posted by Hartlw View Post
    Let f(x) = x-ln(1+x), x>0 as in post 1.

    OK: f(0)=0 and f'(x)>0, x>0 -> f(x)>0, x>0 is obvious.

    Sorry, I got hung up on f'(0)=0.
    Intuitiveley it seems obvious, but can't get over f'(0)=0

    I find the question interesting in its own right so I started a new topic in calculus forum where you can fire away.

    given: f'(x)>0, x>0; f(0)=0, f'(0)=0. Show f(x)>0, x>0
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Prove this ...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 10th 2009, 09:47 AM
  2. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  3. prove
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 24th 2008, 07:15 PM
  4. Prove
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 24th 2008, 07:12 AM
  5. Prove
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 22nd 2008, 09:32 AM

Search Tags


/mathhelpforum @mathhelpforum