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Math Help - analysis questions

  1. #1
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    analysis questions

    Okay so here are some questions I am having trouble with.

    I did this proof and the professor marked it completely wrong, but I don't think I did anything wrong. If you could point out where I went astray that would be fantastic.

    Question: Let x_1=a \geq 2, and x_{n+1}=1/4(x_n + 5) . Show that x_n is a decreasing sequence.
    My proof:

    We will show \{x_n\} is a decreasing sequence using induction. Assume that x_k \geq x_{k+1} for all k \leq n . x_1>x_2 says 2>\frac{7}/{4} which is true, so it holds for n=1. Assume that x_k \geq x_{k+1} . This says that 4x_{k+1}-5 \geq x_{k+1} . So 4x_{k+1} \geq x_{k+1}+5, which says x_{k+1} \geq \frac{1}{4}(x_{k+1}+5) . Thus x_{k+1} \geq x_{k+2} . Therefore by induction, x_n \geq x_{n+1} , which by definition says that \{x_n\} is a decreasing seq.

    And here are some general questions I am having trouble answering, so any hints are appreciated.

    1)Assuming that the function f(x)= 2^x is continuous for all x, prove that there is an x \in (0,1) with x2^x=1 .

    x2^x=1 says 2^x =\frac{1}{x} , so the plan of attack I have for this proof is to let g(x)=\frac{1}{x} and prove that f(x)=g(x) for some x in the interval somehow. I am guessing it will involve using the definition of continuity somewhere along the line.

    2) Prove that if f:[a,b] \rightarrow \Re is continuous and f(x)>0 for x \in [a,b], then there is a constant k>0 so that f(x) \geq k for all x \in [a,b].
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  2. #2
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    Do you mean x_{n+1}=\frac{1}{4(x_n+5)}?

    Here is the correct LaTeX code.
    [tex]x_{n+1}=\frac{1}{4(x_n+5)}[/tex]
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  3. #3
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    Quote Originally Posted by Plato View Post
    Do you mean x_{n+1}=\frac{1}{4(x_n+5)}?

    Here is the correct LaTeX code.
    [tex]x_{n+1}=\frac{1}{4(x_n+5)}[/tex]
    No, what I meant was x_{n+1}=\frac{1}{4}(x_n+5)
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  4. #4
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    I've been sitting down and working on these, and I think I came up with a soultion to #2, tell me if it's right.

    Okay, so we know that f is continuous over [a,b], which means that is defined for all x in [a,b]. We also know that f(x)>0 for all x in [a,b]. By the density of real numbers, there is some rational number q such that 0<q<f(x) for all x in [a,b], since f(x)>0 for all x in [a,b]. This rational number satisfies what we need for k, and I'm done.

    Does that sound about right?

    EDIT: Another solution I just thought of, perhaps the more orthodox one the professor was looking for, is to use the extreme value theorem. Since [a,b] is a closed & bounded interval, and f is continuous over [a,b], then by the EVT, f is bounded on I, meaning f has a lower bound k. THe only trouble I would have now is to prove somehow that said k is greater than 0.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    1.) Choose  0<\epsilon<1 such that  \epsilon is arbitrarily small.

    It is easy to see  2^\epsilon < \frac{1}{\epsilon} .

    In addition,  2^1 > \frac{1}{1} .

    Let  f(x) = 2^x-\frac{1}{x} . We know  f(x) is continuous on  (0,1) . Also  f(\epsilon) < 0 and  f(1) > 0 .

    By the Intermediate Value Theorem  \exists \; c \in (\epsilon,1) \subset (0,1) such that  f(c) = 0 .
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    For #2, try proving the contrapositive.
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  7. #7
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    Quote Originally Posted by chiph588@ View Post
    1.) Choose  0<\epsilon<1 such that  \epsilon is arbitrarily small.
    Look, just note that f(x)=x2^x is continuous.
    So f(0)=0~\&~f(1)=2 giving te result at once.
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  8. #8
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    Quote Originally Posted by Plato View Post
    Look, just note that f(x)=x2^x is continuous.
    So f(0)=0~\&~f(1)=2 giving te result at once.
    man that was deceptively simple, thank you!
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  9. #9
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    For what it is worth here is my solution to the sequence part:

    Show that x_{n+1}=\frac{1}{4}(x+5), with x_1\geq2, is deacreasing.

    Lets assume that x_n>\frac{5}{3}

    Now it follows that

    x_{n+1}=\frac{1}{4}(x+5)>\frac{1}{4}(\frac{5}{3}+5  )=\frac{1}{4}\frac{20}{3}=\frac{5}{3}.

    So x_{n+1}>\frac{5}{3}. Now since x_1>\frac{5}{3} it follows that x_n>\frac{5}{3} for all n\geq1.

    Since we know that x_n>\frac{5}{3}, then also 3x_n>5. So:

    x_{n+1}=\frac{1}{4}(x_n+5)<\frac{1}{4}(x_n+3x_n)=x  _n

    So x_{n+1}<x_n, and there for the seqence is decreasing (and bounded).


    I really cant tell you what is wrong with your solution, I don't see it.

    Hope that helps.
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