# Analysis Questions-Thanks for helping me!

1. Jul 7, 2009

### BustedBreaks

Analysis Questions--Thanks for helping me!

I have a final in an introductory analysis class that covers mostly sequences and series on Friday and I'm a bit behind in the course work. I have a bunch of questions so instead of making a different thread for each one I figured I would just use this one. All the posts with questions will be in bold to distinguish from responses. Thanks in advance for all your help and here's the first question:

Let S be the sequence $<x_{n}>_{n}$ where:

A.$$x_{n}=1+\frac{(-1)^{n}}{(n+1)}}$$, for n=0,3,6,9,...

B.$$x_{n}=-1+\frac{1}{n}}$$, for n=1,4,7,10,...

C.$$x_{n}=\frac{1}{2}+\frac{3}{n}$$, for n=2,5,8,11,...

Determine
1. The limsup of S.
2. The liminf of S.
3. Whether S has a limit.

The limit as n goes to infinity for A is 1
The limit as n goes to infinity for B is -1
The limit as n goes to infinity for C is 1/2

So there is no limit, to answer 3, because the limits for each subsequence do not equal each other, correct?

I'm not sure what to do about liminf and limsup...

I want to say that the liminf is -1 and the lim sup is 1, but I'm not sure if my reasoning is correct. I just took the least and greatest of the limits as it approaches infinity. Another answer could be that the lim inf is 0 because that is the smallest first number of the sequence, but I don't think that's right because by that logic you can find smaller numbers by using n greater than 0 etc which is why I think taking the infinite limit is the right answer.

Last edited: Jul 7, 2009
2. Jul 7, 2009

### HallsofIvy

Staff Emeritus
Re: Analysis Questions--Thanks for helping me!

Well, what is the definition of "limsup" and "liminf"? The limsup of a sequence is the supremum of the set of all "subsequential limits" and liminf is the infimum of that set. And, of course, for a finite set of numbers, "sup" and "inf" are just the largest and smallest numbers in the set. Can you see that because every number in the sequence is in one of those three subsequences, that converge to -1, 1/2, and 1, those are the only "subsequential limits"?

3. Jul 8, 2009

### BustedBreaks

Re: Analysis Questions--Thanks for helping me!

Yes, that's what I think I was thinking. So then the Limsup would be 1 and the liminf would be -1 because these are the greatest and least limits of the subsequences which compose all of S? What does that make 1/2? Is it relevant to limsup/ liminf?

4. Jul 8, 2009

### Billy Bob

Re: Analysis Questions--Thanks for helping me!

Yes.

Just another subsequential limit.

No.

5. Jul 8, 2009

### BustedBreaks

Re: Analysis Questions--Thanks for helping me!

The next question is: Determine real numbers $$a_{n}, n=0,1,2,...$$ such that
$$sin x + cos x =\sum^{\infty}_{n=0}a_{n}x^{n}$$

Could I generalize the sum of the series for sin x and cos x?
Here's what I'm trying:

$$sin x =\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+...$$

$$cos x =\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n}}{(2n)!}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+...$$

$$1+x-\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}-\frac{x^{6}}{6!}-\frac{x^{7}}{7!}$$

$$(1+x)-(\frac{x^{2}}{2!}+\frac{x^{3}}{3!})+(\frac{x^{4}}{4!}+\frac{x^{5}}{5!})-(\frac{x^{6}}{6!}+\frac{x^{7}}{7!})$$

still working... am I on the right track? THANKS!

6. Jul 8, 2009

### Dick

Re: Analysis Questions--Thanks for helping me!

Sure, that's the series alright. You don't really need to regroup the terms. To determine the radius of convergence, I'd suggest a ratio test.

7. Jul 8, 2009

### BustedBreaks

Re: Analysis Questions--Thanks for helping me!

It seems to work if I just use

$$\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n}}{(2n)!}+\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$

as my power series, but I don' t think that is in the form the question is asking for - Determine real numbers $$a_n, n=0,1,2,...$$ such that: $$sinx + cosx = \sum^{\infty}_{n=0}a_{n}x^{n}$$

8. Jul 8, 2009

### Dick

Re: Analysis Questions--Thanks for helping me!

Well, a0=1, a1=1, a2=(-1/2!), a3=(-1/3!), a4=(1/4!), etc etc. That's what I read from your series. What's wrong with that?

9. Jul 9, 2009

### JG89

Re: Analysis Questions--Thanks for helping me!

Sorry to interrupt -- I don't mean to hijack to the thread -- but in post #5 the OP regroups the terms in the sum $$\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n}}{(2n)!}+\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$$

so that the x^2 term precedes the x^3 term, and the x^3 term precedes the x^4 term and so on. Now, I know that if a series is conditionally convergent, then we can make the series sum to any number A, or even make it diverge, just by rearranging it in a specific order. Can the rearranging that the OP did possibly effect the convergence of the series to the sum sin(x) + cos(x)?

10. Jul 9, 2009

### snipez90

Re: Analysis Questions--Thanks for helping me!

I don't think so, the theorem you are referring to is generally applied to numerical series. The theorems that are used to prove the convergence of sin(x) and cos(x) in elementary calculus involve the concept of uniform convergence.

11. Jul 9, 2009

### JG89

Re: Analysis Questions--Thanks for helping me!

I haven`t learned about uniform convergence yet nor about series converging to functions (though I have learned about the Taylor series of a function, but not in much depth as of yet), so I will ask this question anyway -- if I need to learn more before this question can be answered properly, please let me know:

Say x = 1, then $$\sum^{\infty}_{n=0}\frac{(-1)^{n}x^{2n}}{(2n)!}+\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} = \sum^{\infty}_{n=0}\frac{(-1)^{n}}{(2n)!}+\frac{(-1)^{n}}{(2n+1)!}$$.

This should converge to sin1 + cos1 and this is a numerical calculation. Is it possible that rearranging the series now would affect the sum of it?

Last edited: Jul 9, 2009
12. Jul 9, 2009

### snipez90

Re: Analysis Questions--Thanks for helping me!

Sorry, I should have been less vague. No, it will not, because there is a theorem that deals with the convergence of power series which states that a few conditions, series such as sin(x) will converge uniformly and absolutely at any point.

13. Jul 9, 2009

### JG89

Re: Analysis Questions--Thanks for helping me!

Thanks snipez. Sorry for derailing the thread!

14. Jul 9, 2009

### BustedBreaks

Re: Analysis Questions--Thanks for helping me!

Here's another question:

Determine the power series expansion about $$x=1/2$$ of the function $$f(x)$$ where:

$$f(x)=\frac{1}{(1-x)}$$

Okay, I know that the power series expansion of

$$f(x)=\frac{1}{(1-x)}$$

is

$$1+x+x^{2}+...$$

I saw my teacher do something where he wanted to get the power series expansion about
a so he made $$x=a+(x-a)$$ and plugged that in for x and then did some other things I don't remember because I didn't write it down...

15. Jul 9, 2009

### Dick

Re: Analysis Questions--Thanks for helping me!

A series expansion around x=1/2 is an expansion in powers of (x-1/2). So maybe you should rewrite 1/(1-x) so it contains an (x-1/2). Also wouldn't be a bad idea to write something down sometimes.

16. Jul 9, 2009

### BustedBreaks

Re: Analysis Questions--Thanks for helping me!

The funny thing about that is used to try to copy everything all of my teachers ever wrote down and I found that I would never use my notes to learn material nor would I remember what it was I was writing, so for this class I decided to try hard not to take too many notes and pay more attention...

This is what I get:

$$f(x)=\frac{1}{1-(\frac{1}{2}+(x-\frac{1}{2}))}=1+[\frac{1}{1-(\frac{1}{2}+(x-\frac{1}{2}))}]+[\frac{1}{1-(\frac{1}{2}+(x-\frac{1}{2}))}]^2 ...$$

but still not sure what to do now...

17. Jul 9, 2009

### Dick

Re: Analysis Questions--Thanks for helping me!

Write it as 1/(1/2-(x-1/2)). Now what? You want to get it in the form A/(1-B*(x-1/2)). Then the expansion goes A*(1+(B*(x-1/2))+(B*(x-1/2))^2+...).

18. Jul 9, 2009

### BustedBreaks

Re: Analysis Questions--Thanks for helping me!

So I have

$$\frac{1}{(\frac{1}{2})-(x-\frac{1}{2})}$$

then I turn it into the form $$\frac{a_n}{1-(x)}$$ by multiplying by 2/2 and get

$$\frac{2}{(1-(2x-1)}$$

then sub u for $$(2x-1)$$

$$\frac{2}{1-u}=\sum^{\infty}_{n=0}2u^{n}, u=2x-1$$

$$=2(2x-1)^{0}+2(2x-1)^{1}+2(2x-1)^{2}...$$

$$=2+2\cdot2(x-\frac{1}{2})+2\cdot2\cdot2(x-\frac{1}{2})^{2}...$$

$$=\sum^{\infty}_{n=0}2^{n+1}(x-\frac{1}{2})^{n}$$

19. Jul 9, 2009

### Dick

Re: Analysis Questions--Thanks for helping me!

That's it.

20. Jul 9, 2009

### BustedBreaks

Re: Analysis Questions--Thanks for helping me!

Here's another question:

Suppose that $$<x_{n}>_{n}$$ is a sequence of integers such that for all $$n: 0\leq t_{n}\leq9.$$ Show that the series:

$$\sum^{\infty}_{n=0}t_{n}10^{-1}$$

converges and determine lower and upper bounds.

Here is what my TA wrote for me:

$$0\leq a_{n}\leq b_{n}$$

$$\sum b_{n}$$ converges $$\Rightarrow \sum a_{n}$$ exists.

$$a_{n}=t_{n}10^{-1}$$

$$a_{n}\leq \frac{9}{10^{n}}=b_{n}$$ (this is off to the side: $$\sum_{n}a_{a}x^{n}$$)

and $$\sum_{n}b_{n}=9\frac{1}{1-\frac{1}{10}}$$

So $$\sum a_{n}$$ converges.

I'm having a little trouble understanding why he chose $$\frac{9}{10^{n}}$$ for $$b_{n}$$ and where he went from there.

Thanks