1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rudin 5.2

  1. Apr 29, 2008 #1
    [SOLVED] Rudin 5.2

    1. The problem statement, all variables and given/known data
    Suppose f'(x)>0 in (a,b). Prove that f is strictly increasing in (a,b), and let g be its inverse function. Prove that g is differentiable, and that

    g'(f(x))=1/f'(x)

    when a<x<b.


    2. Relevant equations



    3. The attempt at a solution
    I can prove that f is strictly increasing. I cannot prove that the inverse exists. Do I have to go back to the epsilon-delta definition of the derivative for this or is there some clevel way I can manipulate the limits...
     
  2. jcsd
  3. Apr 29, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You've proved f is strictly increasing. f is continuous since f' exists, yes? And you can't prove the inverse exists???
     
  4. Apr 30, 2008 #3
    Sorry. I meant I cannot prove that the derivative of the inverse exists.
     
  5. Apr 30, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Define 'cannot'. Did you try? If f(x)=y and f is invertible then f^(-1)(y)=x. The difference quotient is (f(x)-f(y))/(x-y). Try using the notion of an inverse. Note f' is non-zero.
     
  6. Apr 30, 2008 #5
    OK. So, let y be in the range of f (domain of g). Let f(x) = y. Then I want to show that

    [tex]\lim_{t\to y}\frac{f(t) - x}{t-f(x)} = \lim_{s\to x} \frac{s-x}{f(s)-f(x)}[/tex]

    As you pointed out the right limit exists and equals 1/f'(x). I am just stuck trying to prove that equality.

    And cannot is defined as a lack of success after about 20 minutes of trying.
     
    Last edited: Apr 30, 2008
  7. Apr 30, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you've been trying to prove that equality for 20 minutes and failing, maybe it's not true. The limit on the left side is (f(y)-x)/(y-f(x)). That's not even a difference quotient. Try and set that up again.
     
  8. Apr 30, 2008 #7
    [tex]
    \lim_{t\to y}\frac{g(t) - x}{t-f(x)} = \lim_{s\to x} \frac{s-x}{f(s)-f(x)}
    [/tex]
     
  9. Apr 30, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Brilliant. This is far from the most difficult exercise in Rudin. If that's the closest you can come to actually expressing the problem in symbols, I don't think I can help you.
     
  10. Apr 30, 2008 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Apologies for the tone of my previous post. I had forgotten you defined g=f^(-1) and COULD NOT figure out why you changed it like that. That's actually exactly what you want. Since f(x)=y, g(y)=x. So rewrite it as (g(t)-g(y))/(t-y). Sorry again.
     
  11. Apr 30, 2008 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would think using the chain rule on f(g(x))= x would be simplest.
     
  12. Apr 30, 2008 #11
    We need to prove that the inverse of f is differentiable first.
     
  13. Apr 30, 2008 #12
    Well, I wrote it as (g(t)-g(y))/(t-y) at first but then I rewrote it as (g(t)-x)/(t-f(x)). I still have to prove the two limits above are equal.
     
  14. Apr 30, 2008 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You still don't see it? Why not pick t=f(s), s=g(t)? Then the two sides ARE equal, even before the limits are taken.
     
  15. Apr 30, 2008 #14
    Sorry, I still do not see it. What do you mean pick t=f(s)? t and s are totally independent symbols. How does that even prove that

    [tex] \lim_{t\to y}\frac{g(t) - g(y)}{t-y}[/tex]

    even exists? For each \epsilon > 0, we need to find [itex]\delta>0[/itex] so that [itex]|t-y|<\delta[/itex] implies

    [tex]\left|\frac{g(t) - g(y)}{t-y} - \lim_{s\to x} \frac{s-x}{f(s)-f(x)}\right|<\epsilon[/tex]


    What is really hard is that limit inside the limit thing.

    I don't see why the "two sides being equal before taking the limit" is a meaningful statement. Are they supposed to "stay" equal as you take the limit? What in the world does that mean?
     
    Last edited: Apr 30, 2008
  16. Apr 30, 2008 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are complicating this to death. DON'T consider t and s totally independent symbols. DO take t=f(s). NOW the two sides are equal. NOW take the limit s->x. The right side is a difference quotient and converges to 1/f'(x). The left side is a difference quotient for g'(y). What else is there to say?
     
  17. Apr 30, 2008 #16
    OK. I agree that

    [tex]\lim_{s\to x} \frac{g(f(s))-g(y)}{f(s)-y} = \frac{1}{f'(x)}[/tex]

    But the heart of the problem is proving the left hand side is equal to

    [tex] \lim_{t \to y} \frac{g(t)-g(y)}{t-y}[/tex]

    Hmm. Maybe that is not so hard actually. By Theorem 4.2 in Rudin we just need to prove for every sequence converging to y, the difference quotient above converges to 1/f'(x). But if you give me a sequence {t_n} converging to y, I can find a sequence {s_n} converging to x such that f(s_n) = t_n so...yes that works. The reason {s_n=g(t_n)} converges to x is that by g is continuous (Rudin Thm 4.17).
     
    Last edited: Apr 30, 2008
  18. Apr 30, 2008 #17

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes! All of the functions here are continuous.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Rudin 5.2
  1. Rudin 8.6 (Replies: 8)

  2. Rudin 9.14 (Replies: 1)

  3. Rudin √2 (Replies: 1)

Loading...