Sequence

**dwsmith** · May 22nd 2010, 04:42 PM

$\forall n\in\mathbb{Z}^+$ , let $f_n$ be the function on the interval from [0,1] by $f_n(x)=\frac{x^n}{1+x^n}$ . Which of the following statements are true?

1. The sequence { $f_n$ } converges pointwise on [0,1] to a limit function $f$ .
$\lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\inft y}\frac{nx^{n-1}}{nx^{n-1}}=0=f(x)$

$\lim_{n\to\infty}\frac{1^n}{1+1^n}=\frac{1}{2}=f(x )$
True

2. The sequence { $f_n$ } converges uniformly on [0,1] to a limit function $f$ .
False

3. $\lim_{n\to\infty}\int_0^1f_n(x)dx=\int_0^1(\lim_{n \to\infty}f_n(x))dx$
Don't know

**roninpro** · May 22nd 2010, 05:21 PM

Originally Posted by dwsmith

$\forall n\in\mathbb{Z}^+$ , let $f_n$ be the function on the interval from [0,1] by $f_n(x)=\frac{x^n}{1+x^n}$ . Which of the following statements are true?

1. The sequence { $f_n$ } converges pointwise on [0,1] to a limit function $f$ .
$\lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\inft y}\frac{nx^{n-1}}{nx^{n-1}}=1$
True

This isn't quite correct. You need to consider two cases: either $x\geq 1$ or $x<1$ . Try recalculating your limit.

Also, did you have any ideas on the other two?

**dwsmith** · May 22nd 2010, 05:32 PM

Originally Posted by roninpro

This isn't quite correct. You need to consider two cases: either $x\geq 1$ or $x<1$ . Try recalculating your limit.

Also, did you have any ideas on the other two?

If x is less than 1, there would be $\frac{-x}{-x}$ which is still 1. Same situation if x is positive.

**roninpro** · May 22nd 2010, 05:34 PM

Sorry - I meant to write $|x|<1$ and $|x|\geq 1$ .

The important bit is this: if $x$ is fractional, then $x^n\to 0$ . Therefore, $\frac{x^n}{1+x^n}\to 0$ .

**dwsmith** · May 22nd 2010, 05:37 PM

Originally Posted by roninpro

Sorry - I meant to write $|x|<1$ and $|x|\geq 1$ .

The important bit is this: if $x$ is fractional, then $x^n\to 0$ . Therefore, $\frac{x^n}{1+x^n}\to 0$ .

If it converges to 0 for all $0<x<1$ , 1 is still true though.

**roninpro** · May 22nd 2010, 05:39 PM

Yes, that's right. But your limit was incorrect and will affect the way you do part two and three.

For part two, you might try considering that there is an asymptote at $x=-1$ when $n$ is odd.

**dwsmith** · May 22nd 2010, 05:41 PM

Originally Posted by roninpro

Yes, that's right. But your limit was incorrect and will affect the way you do part two and three.

For part two, you might try considering that there is an asymptote at $x=-1$ when $n$ is odd.

What would be an appropriate Mn for two to use?

**roninpro** · May 22nd 2010, 05:43 PM

What do you mean by Mn?

**dwsmith** · May 22nd 2010, 05:45 PM

Originally Posted by roninpro

What do you mean by Mn?

$|f_n(x)|\leq M_n \ \forall x$

**roninpro** · May 22nd 2010, 05:50 PM

Are you trying to use the Weierstrass M-Test? It isn't applicable here since we are not looking at a function defined by an infinite series.

You need to use the definition of uniform convergence: Given any $\varepsilon>0$ , there is $N$ so that whenever $n>N$ , $|f_n(x)-f(x)|<\varepsilon$ for all $x\in [0,1]$ . (Now I realise that you are restricted to $[0,1]$ - sorry for the confusion earlier.) Can you find such an $N$ ? You should take the point $x=1$ into consideration.

**dwsmith** · May 22nd 2010, 05:55 PM

Originally Posted by roninpro

Are you trying to use the Weierstrass M-Test? It isn't applicable here since we are not looking at a function defined by an infinite series.

You need to use the definition of uniform convergence: Given any $\varepsilon>0$ , there is $N$ so that whenever $n>N$ , $|f_n(x)-f(x)|<\varepsilon$ for all $x\in [0,1]$ . (Now I realise that you are restricted to $[0,1]$ - sorry for the confusion earlier.) Can you find such an $N$ ? You should take the point $x=1$ into consideration.

If $x=1$ , then $f_n(1)=\frac{1}{2}$ .

$|\frac{1}{2}-0|<\varepsilon$
False.

**roninpro** · May 22nd 2010, 06:00 PM

This is the right idea, but still not exactly the case. The function converges pointwise to $ f(x) = \begin{cases} 0 & \text{for $x<1$} \\ \frac{1}{2} & \text{for $x=1$} \end{cases}$ . So, it turns out for all $n$ , $|f_n(1)-f(1)|=0<\varepsilon$ .

Have a look at a plot of the sequence below and give it another shot.

**dwsmith** · May 22nd 2010, 06:07 PM

Originally Posted by roninpro

This is the right idea, but still not exactly the case. The function converges pointwise to $ f(x) = \begin{cases} 0 & \text{for $x<1$} \\ 1 & \text{for $x=1$} \end{cases}$ . So, it turns out for all $n$ , $|f_n(1)-f(1)|=0<\varepsilon$ .

Have a look at a plot of the sequence below and give it another shot.

If x<1, then the limit is 0 and if x=1, then the limit is .5.

x=1
$|\frac{1}{2}-\frac{1}{2}|=0<\varepsilon$

[0,1)
$|[0,1)-0|=[0,1)<\varepsilon$
which isn't always true.

Hence, 2 is false?

**roninpro** · May 22nd 2010, 06:14 PM

Yes, part two is false.

Your argument needs to be refined a little bit: you need to find a point that "floats" away from our function for any $n$ . Since all of the $f_n$ are continuous, there exists a point $y_n\in [0,1]$ so that $f_n(y_n)=\frac{1}{4}$ . Since this is the case, we have $|f_n(y_n)-f(y_n)|=|\frac{1}{4}-0|=\frac{1}{4}$ . Therefore, we can never find $n$ large enough so that $|f_n(x)-f(x)|<\varepsilon=\frac{1}{4}$ for all $x\in [0,1]$ . (i.e. Uniform continuity fails if we try to choose $\varepsilon=\frac{1}{4}$ .)

In part three, try actually evaluating the integral $\lim_{n\to \infty} \int_0^1 \frac{x^n}{1+x^n} \text{ d}x$ by putting a convenient bound on it.

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