# Radius of convergence (Power series)

1. Oct 29, 2011

### ibysaiyan

1. The problem statement, all variables and given/known data
Hi there,
I have just started taylor series for my course.. most seems arlgiht so far, however when it comes to validating a given series( tayor or maclaruin), I have an idea on how to find out the x value.. but I don't know what I am doing wrong.

Take for example: The following function (1+x)^-1 . Expand the maclaurin series of it and give values of x for which the above series is valid:

2. Relevant equations
F(x) = f(0) + f'(0)x /1! + f'' (0)x^2/ 2! + f'''(0)x^3 /3! + f^n (0)x^n/n!

3. The attempt at a solution

Here's my working:

Let f(x) = (1+x)^-1 so f(0) = (1+0)^-1 = 1
f'(x) = -1. (1+x)^-2 f' (0) = -1 . (1)^-2

f''(x) = 2.1 (1+x)^-3 = > f''(0) = 2.1 (1)^-3

f'''(x) = -3.2.1. (1+x)^-4 => f'''(0) = -3.2.1 (1)^-4

Now initially i was puzzled on how to predict (next term or the general pattern) but I think next term will be f'4 = 4.3.2.1 (1x)^-5 so f'4 (0) = 4.3.2.1 (1)^-5

Now I know test ratio can be done to find the radius of convergence i.e for a given function say anx^n ratio would be => (an+1)x^(n+1) / anx^n....

I tried doing the same for the above series.. My general formula is : f(n) : (-1)^n . n! (1+x)^ -(n+1) .... by manipulation i got : 1.(n+1) x [ 1^-n-2] / 1^-n-1]

= > 1. (n+1) but since n => infinity , this function diverges ?
Any constructive input will be appreciated. Thanks!

Last edited: Oct 29, 2011
2. Oct 29, 2011

### spamiam

Hint: think of geometric series.

3. Oct 29, 2011

### ibysaiyan

I know geometric series goes like : 1+ q+ q^2 +q^3 ... and is in the form of a/1-r but how do I approach this problem ? I am confused .. can't I use ratio method to get value/s of x which converge.

4. Oct 29, 2011

### ibysaiyan

Ok.. so the general form of my series is :

$F^{n}$ = $$(-1)^n (n!) (1+x)^{-(n+1)}$$

So we can ignore $$(-1)^n$$ as n -> infinity.

I am left with : $$(n!) (1+x)^{-(n+1)}$$

$F^{n+1}$ / $F^{n}$= $$(n+1!) (1+x)^{-(n+2)} / n! (1+x)^{-(n+1)}$$

I end up with the expression 1/ (1+x)....

EDIT: I don't mean to spam or brea any forum rules, I can find series and predict nth term.. it's only the convergence bit confusing me :(

Last edited: Oct 29, 2011
5. Oct 29, 2011

### spamiam

How can you express $\frac{1}{1+x}$ as a geometric series? If what you've given above is what you think the nth term of the series is, then it's incorrect. It looks more like $f^{(n)}(x)$, but you can avoid calculating this by using my hint. If you're having trouble, look up geometric series in Wikipedia.

For what values of x does this series converge? I don't think you need to use the ratio test--facts about geometric series should be sufficient.

6. Oct 29, 2011

### ibysaiyan

Well... a /1-r = 1/ x+1 ... I am totally lost...
hm..

7. Oct 29, 2011

### spamiam

If my hint isn't helping you, maybe you should go back to your original approach. Afterward I can show you the faster method I was hinting at if you're interested.

What do you get for $\displaystyle \frac{f^{(n)}(0)}{n!}$?

8. Oct 29, 2011

### ibysaiyan

Coming back to my original approach.. Is the general form of the series wrong ? hm.. I am determined to finish ( and understand it) this problem , tonight.
Thanks for your patience with me.

EDIT: Oh i think i see where i may have gone wrong.
Previous

$F^{n}$ = $$(-1)^n (n!) (1+x)^{-(n+1)}$$

Shouldn't the expression $$(1+x)^{-(n+1)}$$ actually be $$(1+x)^{-n-1}$$ ?
The form which you have posted looks like a general form of maclaurin or taylor series hm...

EDIT: Actually i will give a little read(google) over the 'geometric approach' hm..

Last edited: Oct 29, 2011
9. Oct 30, 2011

### spamiam

I think what you had for $f^{(n)}(x)$ was right, but then you tried to do the ratio test using it. The ratio test examines $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n} \right|$ for a power series $\sum_{n=0}^\infty a_n x^n$. So, now that you have $f^{(n)}(x)$, compute $\frac{f^{(n)}(0)}{n!}$, which will give you the coefficient $a_n$.

These are the same thing, so both are correct.

You can try looking at http://en.wikipedia.org/wiki/Geometric_series" [Broken] link for geometric series. The idea is to find a way to write $\frac{1}{1-x}$ as a geometric series, and then use this to do the same for $\frac{1}{1+x}$.

Last edited by a moderator: May 5, 2017
10. Oct 31, 2011

### spamiam

Did you get the answer? If you did and still want to see the shortcut using geometric series, I'll post it.

11. Oct 31, 2011

### ibysaiyan

Sorry other assignments have kept me busy.. i will type down my approach in a bit.If everything fails you can post geometric method.Thanks for your concern =]

12. Oct 31, 2011

### ibysaiyan

Can you show me how to get the co-efficient $a_{n}$ ? (as above).. I am confused.. do I plug in n = 1 ?

Oh do I basically find out my power series and then find out it's general form ? the maclaurin series in this case is : 4

$$1-x+x^2 -x^3 +x^4 -x^5$$

Last edited: Oct 31, 2011
13. Oct 31, 2011

### spamiam

It's just as you had it above, $a_n = \frac{f^{(n)}(0)}{n!}$.
That looks like a good start, but you need an infinite series. What is the general pattern? You said before that $f^{(n)}(x) = (-1)^n (n!) (1+x)^{-(n+1)}$. So what is $f^{(n)}(0)$?

14. Nov 2, 2011

### ibysaiyan

I believe I have finally got the solution.. it took me less than a minute to figure it out... the thing which I was over looking at was the fact that I wasn't trying to find out radius of the derivates rather than the series itself. ( As you have pointed out as well).
My series is : $$1-x+x^2 -x^3 +x^4$$
In the general form : $$(-1)^n * x^n$$
This is An.
So the ratio which's in the form of : $a_{n+1}/a{n}$ as n -> infinity becomes:
$$((-1)^n+1) *(x^n+1)/-1^n * x^n$$

after I manipulate them I end up with -x only... which's weird since the limit is given.. how am I suppose to plug in n -> infinity , if n's cancel... hmm

15. Nov 2, 2011

### HallsofIvy

Staff Emeritus
In general, the power series for a function will converge "until there is a problem". Here, the function is 1/(1+ x) which "has a problem" at x= -1. So there is no reason for the series not to converge between 0 and -1 but it obviously cannot converge to a value at x= -1.

Or, as others have pointed out, 1/(1+ x)= 1/(1- (-x)) is the sum of a goemetric series with "common ratio", r= -x. For what "r" does a geometric series converge?

16. Nov 2, 2011

### spamiam

Looks good. (You can add the LaTeX markup \cdots to make it $1-x+x^2 -x^3 +x^4 - \cdots$.)
That's okay. If $\left| \frac{a_{n+1}}{a_n} \right|$ is the same thing for every n (i.e., doesn't depend on n), then taking the limit shouldn't change anything, right? For example, $\lim_{n \to \infty} 2 = 2$.

Read what the ratio test has to say a bit more carefully. First of all, it examines $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$, so don't forget the absolute values. Now, for which values of L does the series converge? For which does it diverge? And for which is the test inconclusive?

17. Nov 2, 2011

### ibysaiyan

Would the r value be r>1 ? since if that's plugged into the equation... i get negative number. I still however am confused since typical ratio test problems have some value of x turning out to be either negative or positive value which tells of divergence,convergence of the function.

Last edited: Nov 2, 2011
18. Nov 2, 2011

### spamiam

How do you get a negative number since you're taking the absolute value?
Like I said, read carefully what the ratio test tells you: http://en.wikipedia.org/wiki/Ratio_test

19. Nov 2, 2011

### ibysaiyan

I see. So in my case L = -x, right ?

20. Nov 2, 2011

### spamiam

Almost, but don't forget to take the absolute value!