# Homework Help: Power series and taylor

1. Dec 22, 2007

### jacobrhcp

[SOLVED] power series and taylor

1. The problem statement, all variables and given/known data

Let f be a function defined by $$f(x)=\frac{1+c x^2}{1+x^2}$$, and let x be an element of R

for $$c\neq1$$, find the taylor series around the point a, and find the radius of convergence of the taylor series

2. Relevant equations

for power series, $$\sum|c_k|(x-a)^k$$, the radius of convergence is given by $$\rho=\frac{1}{limsup |c_k|^{1/k}}$$

the taylor expansion is given by $$f(x)=\sum\frac{f^k(a) (x-a)^k}{k!}$$

3. The attempt at a solution

I tried writing out the taylor series;

$$f(x)=\frac{1+c a^2}{1+a^2}+\frac{2a^2(c-1)}{(1+a^2)^2}(x-a)+\frac{4a(c-1)(1-a^4)}{2(1+a^2)^4}(x-a)^2+etc...$$

I did this for the first 4 terms but there was no clear pattern to simplify to an infinite summation, in which case I can use the formula for radius of convergence of power series.

Last edited: Dec 22, 2007
2. Dec 22, 2007

### HallsofIvy

Another thing you could do is write
$$\frac{1+ cx^2}{1+ x^2}= \frac{1}{1+ x^2}+ \frac{cx^2}{1+ x^2}$$
as two separate series, using the fact that the sum of the geometric series $a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^n\cdot\cdot\cdot$ is
$$\frac{a}{1- r}[/itex] For the first series, take a= 1, $r= -x^2$. For the second take $a= cx^2$, $r= -x^2$. Then combine the two series. 3. Dec 22, 2007 ### Dick Hey, Halls, that gives you a nice way of getting an expansion around x=0, but jacobrhcp wants to expand around x=a. I know the answer to the radius of convergence question, it's the distance to the nearest pole in the complex plane from x=a. Since the poles are at z=+/-i, that's easy. But I can't find a neat way of expressing this without using complex variables. I also can't find a nice way of expressing the nth term in the taylor series and so I definitely can't find a way to use that expression for radius of convergence. Last edited: Dec 22, 2007 4. Dec 22, 2007 ### Gib Z To get the expansion about x=a from the expansion of x=0 (which Halls provided a method for) wouldn't it suffice to replace every "x" with (x-a) ? 5. Dec 22, 2007 ### Dick You should know better than that, Gib. 6. Dec 22, 2007 ### Gib Z Never mind me, terrible, terrible day today :( 7. Dec 23, 2007 ### jacobrhcp Now I can write the function as a summation using the geometric series, yes (thanks by the way). And the geometric series have radius of conversion 1, so x=±i. but I still have no clue how to find the taylor expansion, and to my shame I don't even see how Halls provided a way to find the expansion around zero. (isn't the expansion around zero stopped after the first term, because there is a factor 'a' in all other terms but the first?) Moreover, is in this case the radius of convergence of the taylor series equal to the radius of convergence of the geometric series? I suppose so, but I have no clue how to prove it. Last edited: Dec 23, 2007 8. Dec 23, 2007 ### jacobrhcp me neither, but where did you get that from? 9. Dec 23, 2007 ### morphism I decided to delete my post. I don't really think what I posted was helpful at all, but I got it from playing around with the expansion of 1/(1+x^2) about a, and isolating the terms in x^k. Here it is again in case anyone is curious: [tex]\frac{1}{1+x^2} = \sum_{n=0}^{\infty} \left[ \sum_{k=n}^{\infty} \frac{(-1)^k}{(1-a)^{k+1}} {k \choose n} a^{k-n} \right] x^{2n}$$

10. Dec 23, 2007

### jacobrhcp

I've thought about it some more, and I am stuck on the exact same problems as Dick, and it bugs me.

11. Dec 24, 2007

### Gib Z

morphism: how did you get that expression?

To all: If that expression is valid, it solves the problem, since the original function is merely this one with additive and multiplicative constants.

12. Dec 24, 2007

### jacobrhcp

why? it's not more usefull than the standard geometric summation halls gave, is it? It also doesn't give a taylor expansion, nor does it give the radius of convergence easily (though I'm no expert on radius of convergence yet, so I might be wrong on that last part)

thanks by the way, to all.

13. Dec 24, 2007

### Gib Z

The expression is a Taylor series centered around a, the method Halls gave was not. The only reason one might not like it is because the coefficient of the x term is given by another series, but that is perfectly acceptable. Just like you are probably fine with $$\int^5_3 \ln x dx$$ but might not like $$\int^5_3 \int^x_1 \frac{1}{t} dtdx$$.

EDIT: By the way, Christmas morning, Merry Christmas everyone =]

EDIT: O wait, I'm quite behind :( Oh well =]

Last edited: Dec 24, 2007
14. Dec 24, 2007

### dynamicsolo

I suppose this is just another county heard from, but here are my suggestions.

First, the original function might be dealt with by polynomial division, since the degrees of the polynomials in the rational function are equal, as

$$\frac{1+ cx^2}{1+ x^2}= c + \frac{1-c}{1+ x^2} , c\neq1 ,$$

so the expansion for $$f(x) = \frac{1}{1+ x^2}$$ need only be dealt with once.

As for the derivatives $$f^{(n)}(x)$$, it seems what makes this unpleasant is the differentiation of products that soon emerges. Perhaps we might differentiate $$\frac{1}{1+ u}$$ , with u = x^2 , instead and without putting everything over a common denominator. The next two derivatives are then

f'(x) = (-1) · (1+u)^(-2) · u'
and
f''(x) = (-1)(-2) · (1+u)^(-3) · u' + (-1) · (1+u)^(-2) · u'' .

The possible virtue of dealing with the derivatives this way is that
u' = 2x, u'' = 2, and all the higher derivatives vanish, so the further derivatives of f(x) do not unfold beyond two terms. Moreover, these two terms can be generalized in a reasonably clear fashion.

f'''(x) = (-1)(-2)(-3) · (1+u)^(-4) · u' + (-1)(-2) · (1+u)^(-3) · u''
+ (-1)(-2) · (1+u)^(-3) · u'' + 0
= (-1)(-2)(-3) · (1+u)^(-4) · u' + 2·(-1)(-2) · (1+u)^(-3) · u'' ;

f^(4) (x) = (-1)(-2)(-3)(-4) · (1+u)^(-5) · u' + (-1)(-2)(-3) · (1+u)^(-4) · u''
+ 2·(-1)(-2)(-3) · (1+u)^(-4) · u'' + 0
= (-1)(-2)(-3)(-4) · (1+u)^(-5) · u' + 3·(-1)(-2)(-3) · (1+u)^(-4) · u'' .

So the nth derivative of f(x) at a becomes

$$f^{(n)} (a) = \frac{(-1)^n · (n!) · (2a)}{(1+a^2)^{n+1}} + \frac{(-1)^{(n-1)} · (n-1) · (n-1)! · (2)}{(1+a^2)^{n}}$$

I've checked this against the derivatives of $$\frac{1}{1+ x^2}$$ and it appears to work correctly. You could put this all over a common denominator from here, but it doesn't simplify much.

Last edited: Dec 24, 2007
15. Dec 24, 2007

### Gib Z

We love you dynamicsolo =]

16. Dec 25, 2007

### jacobrhcp

wow you're my hero

merry christmas

I have just one last question, and then this problem will be completely solved: