Question on sequnces of functions.

[MathematicalPhysicist]

1)let's define a sequence of functions f_n(x) for every x>0 by induction:
f_1(x)=sqrt(x) f_{n+1}=sqrt(x+f_n(x)).
prove that f_n(x)->f(x) as n->$$\infty$$.
2) we have: f_n(x)=(n^2x^2)/(1+n^2x^2), prove that:
$$lim \int_{-1}^{1}f_n(x)dx=\int_{-1}^{1}lim f_n(x)dx$$
as n->$$\infty$$
with the second question i tried this way:
$$\int_{-1}^{1} f_n(x)dx=\int_{-1}^{1}\frac{1}{(1/(x^2n^2))+1}=\left[arctg(1/nx)\right]_{-1}^{1}$$
which equals 0 as n approaches infinity, while the rhs in the former equation equals 2, i reackon there's a problem with my last integral.

with the first question, i tried to it this way:
let e>0, for every x>0 there exists N which depends on e and x, such that for every n>N |f_n(x)-f(x)|<e, but how do i define N as a function of e and x, and how do i employ it in f_N(x)?

thanks.

 

[benorin]

1) I think you put: for x>0, $$f_{1}(x)=\sqrt{x},f_{n+1}(x)=\sqrt{x+f_{n}(x)},$$ correct? Well, if

$$f(x)=\lim_{n\rightarrow\infty}f_{n}(x)=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}},$$

then $$f^2(x)=x+f(x)$$ or $$f^2(x)-f(x)-x=0,$$ which is quadratic in f(x), so $$f(x)=\frac{1\pm\sqrt{1+4x}}{2},$$ but clearly we have f(x)>0, so, of these, choose $$f(x)=\frac{1+\sqrt{1+4x}}{2}$$.

Does that help?

 

[MathematicalPhysicist]

okay, but i need to express N by e and x (perhaps with f(x)) and that way prove that |f_n(x)-f(x)|<e.

i think that your way, also assumes that the limit exists while i need to prove it.

 

[matt grime]

In what sense of convergence of functions are you talking? Pointwise? Uniform?

The real problem with benorin's post is that he puts a final sqrt(x) symbol in the infinitely long expression for f.

To show pointwise convergence, if that is what you need, fix an x and consider the sequence of real numbers f_n(x), it suffices to show that this is increasing and bounded above or decreasing and bounded below.

 

[MathematicalPhysicist]

i don't understand also how did you arrive at f(x) cause:
$$(\sqrt{x+\sqrt{x+\sqrt{x+...}}})^2-\sqrt{x+\sqrt{x+\sqrt{x+...}}}-x=0$$

 

[HallsofIvy]

1)let's define a sequence of functions f_n(x) for every x>0 by induction:
f_1(x)=sqrt(x) f_{n+1}=sqrt(x+f_n(x)).
prove that f_n(x)->f(x) as n->$$\infty$$.

Without knowing what f(x) is? Was the question to show that the sequence converges to some f(x) (my first guess) or determine what f(x) (benorin's). In either case, benorin's suggestion is good.

2) we have: f_n(x)=(n^2x^2)/(1+n^2x^2), prove that:
$$lim \int_{-1}^{1}f_n(x)dx=\int_{-1}^{1}lim f_n(x)dx$$
as n->$$\infty$$
with the second question i tried this way:
$$\int_{-1}^{1} f_n(x)dx=\int_{-1}^{1}\frac{1}{(1/(x^2n^2))+1}=\left[arctg(1/nx)\right]_{-1}^{1}$$
which equals 0 as n approaches infinity, while the rhs in the former equation equals 2, i reackon there's a problem with my last integral.

with the first question, i tried to it this way:
let e>0, for every x>0 there exists N which depends on e and x, such that for every n>N |f_n(x)-f(x)|<e, but how do i define N as a function of e and x, and how do i employ it in f_N(x)?

thanks.

Yex, if $f_n(x)= \frac{n^2x^2}{1+ n^x^2}$, dividing both numerator and denominator by n² gives $\frac{x^2}{\frac{1}{n^2}+ x^2}$ which has limit 1 as n goes to infinity and so
$$\int_{-1}^1 lim_{x\rightarrow \infty}f_n(x)dx= \int_{-1}^1 dx= 2$$.

As far as your:
$$\int_{-1}^{1} f_n(x)dx=\int_{-1}^{1}\frac{1}{(1/(x^2n^2))+1}=\left[arctg(1/nx)\right]_{-1}^{1}$$
yes, dividing both numerator and denominator of $f_n(x)= \frac{n^2x^2}{1+ n^x^2}$ by n²x² gives $\frac{1}{1+ \frac{1}{n^2x^2}$ but the integral of that is not "$arctan(\frac{1}{x})$". In order to get an arctan you have to have the integrand of the form $\frac{1}{1+ u^2}$ which means that, here, you are making a substitution of the form $u= \frac{1}{nx}$. But then $du= -\frac{1}{nx^2}dx$.

Instead make the substitution u= nx in the original integral $\int_{-1}{1}\frac{n^2x^2}{1+ n^2x^2}dx$. Then du= ndx so the integral becomes $\frac{1}{n}\int_{-n}^n \frac{u^2du}{1+ u^2}$.
Use the fact that $\frac{u^2}{u^2+ 1}= 1+ \frac{u-1}{u^2+ 1}$ to integrate that.

 

[MathematicalPhysicist]

In what sense of convergence of functions are you talking? Pointwise? Uniform?

The real problem with benorin's post is that he puts a final sqrt(x) symbol in the infinitely long expression for f.

To show pointwise convergence, if that is what you need, fix an x and consider the sequence of real numbers f_n(x), it suffices to show that this is increasing and bounded above or decreasing and bounded below.

pointwise, when you say fix an x, what do you mean?
do you mean i should take an arbitrarily x0, and to show that f_n(x0)<=M=f(x0) (M the supremum of f_n(x), and then by the defintion of the supremum: f(x)-e<f_n(x)-> e>|f_n(x0)-f(x0)|. correct?

 

[MathematicalPhysicist]

halls you must mean: u^2/(1+u^2)=1-1/(1+u^2), thanks halls.
about the first question, the question was prove that the limit exist, without any given about what it is.

 

[matt grime]

pointwise, when you say fix an x, what do you mean?
do you mean i should take an arbitrarily x0, and to show that f_n(x0)<=M=f(x0) (M the supremum of f_n(x), and then by the defintion of the supremum: f(x)-e<f_n(x)-> e>|f_n(x0)-f(x0)|. correct?

I mean that you shuold just fix x and show that f_n(x) converges to something which we shall label f(x). There are two particularly nice ways that one can hope to show a sequence converges: increasing and bounded above or decreasing and bounded below. If the former case is what happens, yes, f(x) is the sup of the set {f_n(x) : n in N}.

 

[MathematicalPhysicist]

so should i employ here the epsilon definition?

 

[benorin]

i don't understand also how did you arrive at f(x) cause:
$$(\sqrt{x+\sqrt{x+\sqrt{x+...}}})^2-\sqrt{x+\sqrt{x+\sqrt{x+...}}}-x=0$$

Consider that $$f_{n+1}(x)=\sqrt{x+f_{n}(x)}\Rightarrow f_{n+1}^{2}(x)-f_{n}(x)-x=0$$ now let $$n\rightarrow\infty$$ and consider that $$\lim_{n\rightarrow\infty}f_{n}(x)=\lim_{n\rightarrow\infty}f_{n+1}(x)=f(x),$$ to arrive at $$f^2(x)-f(x)-x=0.$$

 

[matt grime]

You are assuming that the pointwise limit exists, though.