Finding the Radius of Convergence through Ratio Test

[JustinDaniels]

Homework Statement

Let f(x)= (1+x)^4/3 - In this question we are studying the Taylor series for f(x) about x=2.

This assignment begins by having us find the first 6 terms in this Taylor series. For time, I will omit them; however, let's note that as we continuously take the derivative of this function, we are applying the chain rule over and over. The derivative of the 'inside function' is 1, so in essence, we are really just applying the power rule again and again.

(b) Let c_k = f^(k)(2)/k! be the coefficient of (x-2)^k in the Taylor series. It is not so easy to write a closed-form expression for c_k. However, find a recursive formula to express c_k+1 in terms of c_k.

(c) Use your recursive formula, together with the ratio test, to determine which values for x would make the Taylor series Σ_k=0^∞c_k(x-2)^k converge or diverge absolutely.

Homework Equations

The Attempt at a Solution

(b) I note that the coefficient of each function is being changed by the power in which the previous coefficient is raised to--which always contains a (3^-1. Furthermore, we must handle the factorial. As a result, I observe the following formula for the general term:

(4-3k)(1/(k+1))(3^-1)c_k = c_k+1

(c) Applying the ratio test: lim k→∞ |c_k+1|/|c_k|

I note that c_k+1 contains a c_k in it. As a result they cancel out. Leaving us with:

lim k→∞ (4-3k)(1/(k+1))(3^-1)

Here is where I get stuck.. I believe we can pull the (1/3) out from the limit, as it is a constant. The other two parts of the limit approach -∞ and 0 respectively. Thus, we would apply L'Hopital's rule? This gives us

1/3 lim k→∞ [-3(-(k+1)^-2)]

From here, do we pull out the -3, and evaluate the limit? If so, our limit approaches 0 again?

Any help would be appreciated. Here is the next question, to give you an idea of what I need.

(d) You should have observed a range of x values that extends for the same distance on either side of x=2. This distance is called the Radius of Convergence. Try to find some property of the function that explains why the radius of convergence is what it is.

Note: I believe the for 0<x<2 it converges.
At 2 it is inconclusive.
2<x<inf it diverges.

I heard this from a classmate. Any help is appreciated!

Thanks,
Calculus 2 Student.

 

[vela]

I didn't get the same relationship between $c_k$ and $c_{k+1}$ you did. It looks like you're missing a factor of 1/3. You also need a factor of $x-2$ because the terms of the series also contain powers of $x-2$. Finally, you misapplied L'Hopital's rule, but you don't really need to use it.

The limit of the expression you got is more straightforwardly evaluated as follows:
$$ \lim_{k\to\infty} \frac{-3k+4}{3k+3} = \lim_{k\to\infty} \frac{-3+\frac 4k}{3+\frac 3k} = \frac{-3}{3} = -1.$$ Just divide the top and bottom by $k$. Of course, as I noted, your expression is incorrect, so you need to fix that first.

 

[JustinDaniels]

I didn't get the same relationship between $c_k$ and $c_{k+1}$ you did. It looks like you're missing a factor of 1/3. You also need a factor of $x-2$ because the terms of the series also contain powers of $x-2$. Finally, you misapplied L'Hopital's rule, but you don't really need to use it.

The limit of the expression you got is more straightforwardly evaluated as follows:
$$ \lim_{k\to\infty} \frac{-3k+4}{3k+3} = \lim_{k\to\infty} \frac{-3+\frac 4k}{3+\frac 3k} = \frac{-3}{3} = -1.$$ Just divide the top and bottom by $k$. Of course, as I noted, your expression is incorrect, so you need to fix that first.

So I need an additional factor of (1/3)? I have the formula containing a 3^-1, which is 1/3. Also, $c_k$ is the coefficient of the kth term in the series (i.e. the number before the $x-2$), therefore I do not think it needs a factor of $x-2$.

Note, the coefficients of the series are:
f(x) = 1
f'(x) = 4/3 --as a result of the power rule being applied the first time.
f''(x) = 4/9
f'''(x) = -8/27.

Notice, the denominator changes by a factor of (1/3) each time, and the numerator by a factor of the numerator of the previous terms power (4-3k). Finally, we have to handle the factorial each time, and as a result I observe the recursive function to be

$(1/3)(1/(k+1))(4-3k)c_k=c_{k+1}$

I hope that makes sense...

Thank you again. I appreciate your time,
jd

 

[vela]

Your series can't be right. Since $f(x) = (1+x)^{4/3}$, you have $f(2) = 3^{4/3}$. For a series expanded about $x=2$, that value is the constant term. Remember that f and its derivatives have to be evaluated at $x=2$ since that's the point you're expanding about.

It might help you to make the substitution $u=x-2$, which gives $f(x) = f(u+2) = (3+u)^{4/3}$ and expand about $u=0$.

 

[JustinDaniels]

I'm quite confused.. Here is what I get for the first six terms in the Taylor series:

Term 0: $(1+(2))^{4/3}$

Term 1: $4/3 (1+2)^{1/3}(x-2)$

Term 2: $\frac{(4/9) (1+2)^{-2/3}}{2!} (x-2)^2$

Term 3: $\frac{-(8/27) (1+2)^{-5/3}}{3!} (x-2)^3$

Term 4: $\frac{(40/81) (1+2)^{-8/3}}{4!} (x-2)^4$

Term 5: $\frac{-(320/243) (1+2)^{-11/3}}{5!} (x-2)^5$

Term 6: $\frac{(3520/72)9 (1+2)^{-14/3}}{6!} (x-2)^6$

I need to find a formula that gives me the coefficient of some term, in respect to its previous term.

The value of the constant term is approximately: 4.326748
The value of the coefficient of the (x-1) term is approximately: 1.922999
The value of the coefficient of the $(x+1)^2$ term is approximately: .1068333

Applying the formula I suggested, we observe:
$k=1$
$c_k = 1.922999$
$c_k(4-3k)(1/3)(1/2)=.3204998333$, which does not equal my suggested $c_{k+1}$. As you noted, I am off by a factor of (1/3). This suggests our formula should be:

$c_{k+1}=(4-3k)(1/9)(\frac{1}{k+1}){c_k}$

Let us apply this to the $(x+1)^2$ term and $(x+1)^3$ term's coefficients.

The value of the coefficient of the $(x+1)^2$ term is approximately: .1068333
The value of the coefficient of the $(x+1)^3$ term is approximately: -.00791357788

Applying the new formula, we observe:
$k=2$
$c_k = .1068333$
$c_k(4-(3*2))(1/9)(\frac{1}{2+1})=-.00791357788$, which was the suggested $c_{k+1}$.

Given this, we use the recursive formula, together with the ratio test, to determine which values for x would make the Taylor series $\sum_{k=0}^\infty c_k (x-2)^k$ converge or diverge absolutely. Applying the ratio test, we will note the $c_k$'s cancel out.

$\lim_{k\rightarrow +\infty}$ ${(4-3k)(1/9)}$ $\frac{1}{k+1}$ which can be simplified as:

$\lim_{k\rightarrow +\infty}$ $\frac{-3k+4}{9k+9}$ I am assuming we would again divide by $k$ and then evaluate the limit. In doing so:

$\lim_{k\rightarrow +\infty}$ $\frac{-3+\frac{4}{k}}{9+\frac{9}{k}} = \frac{-3}{9} = \frac{-1}{3}$

I am now confused as to how this helps me know which values this series converges or diverges absolutely for.. It appears as though the series' convergence is not dependent on $x$ and since $\frac{-1}{3}$ is smaller than one, this series converges for all values of x. However, the next question on this assignment suggests that the series only converges for some values on x on either side of 2.

edit: I guess in applying the Ratio Test, we would be taking the absolute value of the limit. As a result, we would observe (1/3) and not a negative (1/3)... this doesn't change the result much however..
Thanks again!
jd

 

[vela]

Reread post #2.

 

[JustinDaniels]

Okay, so I incorrectly defined $a_n$ and $a_{n+1}$; as a result, I misapplied the Ratio Test.

$a_{n+1}$ should be $\frac{1}{9} \frac{1}{k+1} (4-3k) c_{k} (x-2)^{k+1}$

In reapplying the Ratio Test, I observe the following:

$lim_{k \rightarrow \infty} |\frac {\frac{1}{9} \frac{1}{k+1} (4-3k)c_k(x-2)^{k+1}}{c_k(x-2)^k}|$

We can pull the constant multiple of (1/9) out of the limit and cancel out the $c_k$'s. Furthermore, $\frac{(x-2)^{k+1}}{(x-2)^k}$ simplifies to $(x-2)$.

$\frac{1}{9} lim_{k \rightarrow \infty} |\frac{-3k+4}{k+1} (x-2)|$

I will omit much of the algebra; however, we will again divide by k and the resulting $4/k$ and $1/k$ will go to zero. I observe the following:

$\frac{1}{9} lim_{k \rightarrow \infty} |\frac{-3}{1} (x-2)|$

Here is where I am stuck again.. Is the following correct?:

Do we pull the $-3$ out of our limit as a negative or positive 3, due to the absolute value? Given we do, I observe the following:

$\frac{3}{9} lim_{k \rightarrow \infty} |(x-2)|$

The 3/9 simplifies to 1/3. Applying the Ratio Test, the series should converge for x values that make the limit less than one. As a result of the absolute value the following bounds satisfy this requirement.

$(1,3)$--a distance of 2. Do we call this the Radius of Convergence, and have I done everything correct up to this point?

I really appreciate your help. I would be completely lost otherwise.

Thanks again,
jd

 

[JustinDaniels]

Okay, spoke to my professor. The limit of the ratio test evaluates out to $\frac{1}{3} |x-2|$, which I essentially had in the previous post. We note that this value is not dependent on $k$, so that when k goes to infinity the limit stays as $\frac{1}{3} |x-2|$. The range of x values that satisfies the ratio being less than 1 is:

$-1<x<5$

We note that when the x values becomes either -1 or 5 the ratio evaluates to 1--telling us nothing about the convergence on the sequence.

I believe we have solved the problem.

Thank you again for your help.
jd

 

[vela]

Okay, spoke to my professor. The limit of the ratio test evaluates out to $\frac{1}{3} |x-2|$, which I essentially had in the previous post. We note that this value is not dependent on $k$, so that when k goes to infinity the limit stays as $\frac{1}{3} |x-2|$.

Note that the limit can't depend on $k$. You found the limit as $k\to\infty.$

We note that when the x values becomes either -1 or 5 the ratio evaluates to 1--telling us nothing about the convergence on the sequence.

The ratio test tells you nothing, so you have to go back to the original series, plug in the each value and see if the resulting series converge.

 

[JustinDaniels]

I observed the range of values to be $-1 < x < 5$. Thus, the radius of convergence is 3. The next question in this assignment asks us to find a property of the function that explains why the radius is what it is. I see no relationship between the two however.

 

[vela]

Consider what I suggested in post #4.