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Math Help - limit function

  1. #1
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    limit function

    examine the sequence of functions defined by

    f_n(x) = 2nx/n^p + x^2. for different values of p > 0.

    a) show the function above converges to its limit function uniformly on the interval I = [0, infinity) iff p>2.

    b)Show that when p > 4, M-Test be applied to show that g= SUM(f_n) converges uniformly on the interval I, but fails other p's.

    c)Show that SUM(f_n) converges uniformly on any interval I =[0,a] whenever p > 2.
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  2. #2
    Super Member Showcase_22's Avatar
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    I would just like to point out that i've never done a question like this before so here's my attempt:

    f_n(x)=\frac{2nx}{n^p}+x^2=2n^{1-p}x+x^2

    Suppose p>2.

    Need to find \delta>0 such that |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon.

    Let x=\frac{\delta}{2} and y=\frac{\delta}{3} and fix \delta=\min \{ 1, \epsilon \}

    |2n^{1-p}x+x^2-2n^{1-p}y-y^2|=|2n^{1-p}(x-y)+(x^2-y^2)| \leq |2n^{1-p}(x-y)|+|x^2-y^2|

    But:

    =|2n^{1-p}(x-y)|+|(x+y)(x-y)|< |2n^{1-p}|\delta+|x+y|\delta=|2n^{1-p}|\delta+\left| \frac{5\delta}{6} \right|\delta

    But since 0<\delta \leq 1 \Rightarrow \frac{5 \delta^2}{6} \leq \frac{5 \delta}{6}.

    Hence we have |f(x)-f(y)| < |2n^{1-p}| \delta+\frac{5 \delta}{6}=\delta \left( |2n^{1-p}|+\frac{5}{6} \right)

    Hence define \epsilon=\delta \left( |2n^{1-p}|+\frac{5}{6} \right)

    EDIT: I would really like to point out that i'm not sure if this is right. I haven't really encountered uniform continuity before!!

    (I also can't see where i'm used the property that p>2....)
    Last edited by Showcase_22; July 1st 2009 at 01:41 PM.
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    I would just like to point out that i've never done a question like this before so here's my attempt:

    f_n(x)=\frac{2nx}{n^p}+x^2=2n^{1-p}x+x^2

    Suppose p>2.

    Need to find \delta>0 such that |x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon.

    Let x=\frac{\delta}{2} and y=\frac{\delta}{3} and fix \delta=\min \{ 1, \epsilon \}

    |2n^{1-p}x+x^2-2n^{1-p}y-y^2|=|2n^{1-p}(x-y)+(x^2-y^2)| \leq |2n^{1-p}(x-y)|+|x^2-y^2|

    But:

    =|2n^{1-p}(x-y)|+|(x+y)(x-y)|< |2n^{1-p}|\delta+|x+y|\delta=|2n^{1-p}|\delta+\left| \frac{5\delta}{6} \right|\delta

    But since 0<\delta<1 \Rightarrow \frac{5 \delta^2}{6} \leq \frac{5 \delta}{6}.

    Hence we have |f(x)-f(y)| < |2n^{1-p}| \delta+\frac{5 \delta}{6}=\delta \left( |2n^{1-p}|+\frac{5}{6} \right)

    Hence define \epsilon=\delta \left( |2n^{1-p}|+\frac{5}{6} \right)

    EDIT: I would really like to point out that i'm not sure if this is right. I haven't really encountered uniform continuity before!!

    (I also can't see where i'm used the property that p>2....)

    Showcase ,we have two kinds of convergence:

    1) Point wise convergence and,

    2)Uniform convergence

    For pointwise convergence the definition is:

    Given an ε>0 and an x belonging to the interval of definition of the sequence of functions { f_{n}(x)}, there exists a natural No k such that:

    for all n: if n\geq k,then |f_{n}(x) -f(x)|<\epsilon


    For uniform convergence the definition is:

    Given an ε>0 ,there exists a k such that:

    for all n and for all x belonging to the interval of definition of { f_{n}(x)}: if n\geq k and xεI ( I =interval of definition),then
    |f_{n}(x) -f(x)|<\epsilon.

    Note the difference.

    Both definitions can be found in any analysis book under the title:

    Function spaces.

    And in our case i think by applying the above definitions:

    1) The sequence of functions converges point wise in the interval [ 0,to infinity) to the function f(x) = x^2 ,and



    2) converges uniformly in the interval (0,λ) ,where λ is any No greater than 0
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post


    (I also can't see where i'm used the property that p>2....)

    Let: p>2 ====> p-1>1 ====> ln(n)(p-1)> ln(n) ( for  n\geq 2)<====>  n^{p-1} > n ( ln is a strictly increasing function) <====> \frac{1}{n^{p-1}}< \frac{1}{n} or

    \frac{n}{n^p}< \frac{1}{n}

    You going to need that important inequality somewhere along the proof of pointwise or uniform convergence
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  5. #5
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    You have to interpret the question thus: f_n(x)=\frac{2nx}{n^p+x^2} for x\geq0. Then there are three cases to consider:

    1. If 0<p<1 then there is no pointwise convergence as n\to\infty.

    2. If p=1 then f_n(x)\to 2x as n\to\infty.

    3. If p>1 then f_n(x)\to 0 as n\to\infty.

    In the case p=1 we have |f_n(x)-2x|=\frac{2x^3}{n+x^2} and this can be made arbitrarily large for any given n, so no uniform convergence here.

    When p>1 we have |f_n(x)-0|=\frac{2nx}{n^p+x^2}. A bit of calculus shows that this is maximised when x=n^{p/2}, and the supremum is n^{1-(p/2)}. Therefore uniform convergence occurs iff 1-(p/2)<0, i.e. iff p>2.

    We now know that |f_n(x)|\leq n^{1-(p/2)} for p>2, and we know that \sum_{n=1}^\infty n^{1-(p/2)} converges iff 1-(p/2)<-1, i.e. iff p>4. This gives uniform convergence by the M-test when p>4.

    To show non-uniform convergence for other values of p could be tricky, but I'll try it for the limiting case p=4, which probably implies all others.

    So consider the function f_n(x)=\frac{2nx}{n^4+x^2}. Considered as a function of n this is maximised when n=\surd x and is a decreasing function of n for n>\surd x.

    Choosing the integer m so that m\leq\surd x<m+1, the integral test implies \sum_{n=m+1}^\infty f_n(x)>\int_{m+1}^\infty f_t(x) dt<br />
\geq\int_{1+\surd x}^\infty \frac{2tx}{t^4+x^2}dt

    =\int_{1+\frac2{\surd x}+\frac1x}^\infty \frac1{u^2+1}du=\frac\pi2-\tan^{-1}\left(1+\frac2{\surd x}+\frac1x\right) using the substitution t^2=xu.

    Thus as x\to\infty, the sum exceeds \frac\pi2-\tan^{-1}1=\frac\pi4. However, the functions f_n(x) all tend to 0 as x\to\infty, and if the sum were uniformly convergent then its limit would be 0 as well.

    As for the last part, we see the maximum of f_n(x) occurring outside the interval 0\leq x\leq a when n^{p/2}>a (assuming p>2) and f_n(x) increasing on the interval. Thus for an appropriate integer m we have

    \sum_{n=m}^\infty f_n(x)\leq\sum_{n=m}^\infty f_n(a)=\sum_{n=m}^\infty\frac{2na}{n^p+a^2}<br />
<\sum_{n=m}^\infty\frac{2na}{n^p}<\infty if p>2, so the M-test guarantees uniform convergence.
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