Determining the interval of convergence

[lxman]

Homework Statement

f(x)=x^{0.4}

Construct a power series to represent the function and determine the first few coefficients. Then determine the interval of convergence.

The Attempt at a Solution

Determining the first few coefficients is simple enough. Take the first few derivatives, etc. etc.

Where I am having an issue is determining the interval of convergence.

As I take the successive derivatives, they begin to form a pattern of:

\frac{r}{x^{s}}

In order to determine the interval of convergence I need to identify the pattern and phrase it in sigma notation so that I can apply the ratio test.

Determining the s in the denominator is simple enough. It is simply n-1.4. So I now have a series with the form:

\frac{r}{x^{n-1.4}}

Determining the r, though, is where I am having difficulty. I made a list of the first 10 values of r (skipping the first few terms as they don't seem to follow the general pattern). They are:

.24
.384
.9984
3.59424
16.533504
92.5876224
611.07830784
4644.195139584
39940.07820042239
383424.750724055

This series (the successive values of r) represents a recursion, since, as we take the successive derivatives, r is the product of the previous n-1.4 and the previous r. I can't seem to figure out how to write this in sigma notation so that I can use it in a ratio test. The following appears true, but not very helpful:

\sum^{\infty}_{n=0}(n-.4)\ S_{n-1}

How would I express a recursive sequence like this without using previous terms in the summation formula?

 

[eczeno]

hello,

first, where have you chosen to center your power series? the coefficients will be different for different choices.

there will not be a nice way to write the coefficients since they will involve products of many fractions, but you can just sort of keep track of what is being multiplied at each value of n. so if f(x)=x^{\frac{-1}{2}} then (if the center is at x=1) the nth coefficient would be:

a_n=\frac{1}{n!} (\frac{-1}{2})(\frac{-1}{2}-1)\bullet\bullet\bullet(\frac{-1}{2}-(n-1))

which works for n \geq 1.

it will only be the first term in your series that does not fit the pattern, the rest will. if your write it this way, it is not hard to apply the ratio test.

cheers.

 

[lxman]

Okay, that makes good sense. I have to define a center since depending on where I pick the center, the curve will be different and the interval of convergence will be different.

So, my center is 7 in this case. What I understand from what you have written is that each successive numerator will be given by:

(0.4),(0.4)(0.4-1),(0.4)(0.4-1)(0.4-2),(0.4)(0.4-1)(0.4-2)(0.4-3)...

which does indeed generate the correct terms. I still do not understand, though, how to go from this to:

\sum^{+\infty}_{n=0}something_{n}

 

[eczeno]

right. since our series will not converge everywhere, the coefficients will depend on our choice of center.

recall that the coefficients are given by the formula:

a_n=\frac{f^{(n)}(c)}{n!}

so your expression is not complete.

in our case,

f^{(n)}(x) = (.4)(.4-1)\bullet\bullet\bullet(.4-(n-1))x^{(.4-n)}

so,

a_n=\frac{(.4)(.4-1)\bullet\bullet\bullet(.4-(n-1))7^{(.4-n)}}{n!}

and that is probably the best you are going to do as far as a formula for the coefficients. it is not as pretty as usual, but it will work fine. you can apply the ratio test to it quite easily.

hope this helps

 

[lxman]

Okay, so looking at what I have to work with, I can calculate that:

\frac{a_{n+1}\ (x-7)^{.4-n-1}}{a_{n}\ 7^{.4-n}}=\frac{.4-n}{n+1}\ (x-7)

Now, the ratio test is looking at \lim_{n\rightarrow\infty} which would be x-7. Therefore, I need \left|x-7\right|&lt;1\Rightarrow 6<x<8, and I still have to test my endpoints.

Am I correct to this point?

 

[lxman]

Nope, that should be:

\left|\frac{1}{x-7}\right|&lt;1

which tells me x>8 or x<6

 

[eczeno]

oh, no, you were right the first time! 6<x<8. just check the endpoints.
cheers

 

[lxman]

Hmm, thanks much for the help, but I tried giving that to my professor and he says "not correct." I thought it would be (6,8], but just in case I was wrong, I tried [6,8], [6,8) and (6,8). He's not agreeing with any of them. I thought I understood things all the way, but I'm confused again.

 

[eczeno]

well, i know that i might be wrong, but i am quite confident it is one of those intervals. we have found the taylor series for x^.4 centered at 7 and it's radius of convergence is 1. the radius of convergence would be 1 no mater where we centered it as long as c>1, so it is very puzzling what your professor wants.

 

[lxman]

Okay, I'm definitely not questioning you. Let me start from scratch with the problem definition directly from the homework and maybe I missed something or left something out.

Represent the function f(x)=x^0.4 as a power series:

\sum^{\infty}_{n=0}c_{n}(x-7)^{n}

Find the following coefficients c_{0}, c_{1}, c_{2}, c_{3}

Find the interval of convergence.

I've already submitted the coefficients. They are (graded as correct):

7^{0.4}, 0.4(7^{-0.6}), -0.12(7^{-1.6}), 0.064(7^{-2.6})

Does any of this information change anything?

 

[eczeno]

hmm... yes, i see the problem. there will be a factor of 1/7 left in the limit when you apply the ratio test. so the radius of convergence will actually be 7. sorry i missed that earlier.

 

[lxman]

Okay, [0,14] did the trick. Thanks for hanging in there with me.

I have a couple of questions still left. First off, the inclusion of zero in the set. The way that I am seeing it, the coefficients themselves make up a series which is alternating and decreasing (actually beginning with term one, not term zero). I would therefore conclude that the coefficients, as a series, would converge absolutely. At x = 0, however, we multiply these coefficients by -7^n, which, as it happens, also alternates, but opposite to the coefficients. The end result is that after the zero term, all of the other terms are negative. As my objective is to construct a series to give me a reasonably accurate representation of x^0.4, and since my first term is actually x^0.4, if I begin subtracting values off without putting anything back, isn't this a problem?

I actually see this occurring for any x < 7.

Then there is the case of x = 7. If I examine this case, I see that I now have 0^n, which on the zero term evaluates to 0^0. Undefined I believe? Is this an issue? Or do we just change our series numbering or leave out the zero term, or something of that nature to fix the (apparent) issue?