Maclaurin series for 3x^2/(1+x^3) -- radius of convergence

[h2sbf7]

Hi, Free Math Help. I'm trying to do a Maclaurin series problem and I'm having trouble.

The problem (AP calculus FRQ #6 style my teacher wrote) asks us to find the first four non-zero terms for the Maclaurin series for 3x^2/(1+x^3) as well as the series's general term. As far as I can tell, I did this correctly and got 3x^2-3x^5+3x^8-3x^11 as my first four terms and [sum from 1 to infinity] of (-1)^(n+1) * 3x^(3x-1) as my general.

Then the problem asks me to find the radius/interval of convergence. I did the ratio test as I'm wont to do in this situation, and I got something I can't really work with. I got:

lim n>infinity: |[(-1)^(n+2)3x^(3n+2)]/[(-1)^(n+1)3x^(3n-1)]|

This reduced to lim n>infinity: |[3x^3(n+1)]|

Problem is that I can't effectively get the n by itself so I can take its limit. Any ideas? Thanks for any help you can give!

 

[Deleted member 4993]

Hi, Free Math Help. I'm trying to do a Maclaurin series problem and I'm having trouble.

The problem (AP calculus FRQ #6 style my teacher wrote) asks us to find the first four non-zero terms for the Maclaurin series for 3x^2/(1+x^3) as well as the series's general term. As far as I can tell, I did this correctly and got 3x^2-3x^5+3x^8-3x^11 as my first four terms and [sum from 1 to infinity] of (-1)^(n+1) * 3x^(3x-1) as my general.

Then the problem asks me to find the radius/interval of convergence. I did the ratio test as I'm wont to do in this situation, and I got something I can't really work with. I got:

lim n>infinity: |[(-1)^(n+2)3x^(3n+2)]/[(-1)^(n+1)3x^(3n-1)]|

This reduced to lim n>infinity: |[3x^3(n+1)]|

Problem is that I can't effectively get the n by itself so I can take its limit. Any ideas? Thanks for any help you can give!

How did you get:

$\displaystyle \displaystyle{\sum_{n \to \infty}\dfrac{3x^2}{1 + x^3} \ = \ 3\left (x^2 - x^5 + x^8 - x^{11} ....\right )}$

If I were to do this problem, I would first note that:

$\displaystyle \displaystyle{\dfrac{3x^2}{1 + x^3} \ = \ \dfrac{d}{dt}[ln(1+x^3)}]$

Then I'll expand ln(1+x^3) and differentiate it....

 

[stapel]

...find the first four non-zero terms for the Maclaurin series for:

. . . . .$\displaystyle \dfrac{3x^2}{1\, +\, x^3}$

...as well as the series's general term. As far as I can tell, I did this correctly and got:

. . . . .$\displaystyle 3x^2\,-\,3x^5\,+\,3x^8\,-\,3x^{11}$

...as my first four terms and:

. . . . .$\displaystyle \displaystyle \sum_{n\, =\, 1}^{\infty}\, (-1)^{n+1}\,\cdot\, 3\, x^{3x-1}$

...as my general.

I agree with your summation form and first four terms. I've type-set a bit, but thank you for showing your work so completely!

Then the problem asks me to find the radius/interval of convergence. I did the ratio test as I'm wont to do in this situation, and I got something I can't really work with. I got:

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg| \, \dfrac{(-1)^{n+2}\, \cdot\, 3\, x^{3n+2}}{(-1)^{n+1}\, \cdot\, 3\, x^{3n-1}}\,\bigg|$

This reduced to:

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg| \,3\,x^3\,(n\, +\, 1)\, \bigg|$

I'm not sure how you arrived at this...? Let's work through the steps:

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg| \, \dfrac{(-1)^{n+2}\, \cdot\, 3\, x^{3n+2}}{(-1)^{n+1}\, \cdot\, 3\, x^{3n-1}}\,\bigg|$

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg| \, \dfrac{(-1)^{n+1}\,(-1)\, \cdot\, 3\,\cdot\, x^{3n-1}\,x^3}{(-1)^{n+1}\, \cdot\, 3\, \cdot\, x^{3n-1}}\,\bigg|$

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg|\, \left(\dfrac{(-1)^{n+1}}{(-1)^{n+1}}\right)\, (-1)\,\cdot\, \left(\dfrac{3}{3}\right)\, \cdot\, \left(\dfrac{x^{3n-1}}{x^{3n-1}}\right)\, x^3\, \bigg|$

You should end up with something more like:

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg|\, (-1)\, x^3\, \bigg|$

Problem is that I can't effectively get the n by itself so I can take its limit.

In this case, does that matter? Since the convergence must not depend upon the value of "n" (since there is no "n" in the expression), you don't need to worry about that. Besides, you're trying to find the x-values for the interval. (here)

You need to have |x^3| be less than 1. What then must be the convergence interval?

 

[h2sbf7]

How did you get:

$\displaystyle \displaystyle{\sum_{n \to \infty}\dfrac{3x^2}{1 + x^3} \ = \ 3\left (x^2 - x^5 + x^8 - x^{11} ....\right )}$

If I were to do this problem, I would first note that:

$\displaystyle \displaystyle{\dfrac{3x^2}{1 + x^3} \ = \ \dfrac{d}{dt}[ln(1+x^3)}]$

Then I'll expand ln(1+x^3) and differentiate it....

Thanks for your reply. If you were to find the general Maclaurin series for ln(1+x^3), how would you go about doing that?

 

[h2sbf7]

I agree with your summation form and first four terms. I've type-set a bit, but thank you for showing your work so completely!


I'm not sure how you arrived at this...? Let's work through the steps:

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg| \, \dfrac{(-1)^{n+2}\, \cdot\, 3\, x^{3n+2}}{(-1)^{n+1}\, \cdot\, 3\, x^{3n-1}}\,\bigg|$

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg| \, \dfrac{(-1)^{n+1}\,(-1)\, \cdot\, 3\,\cdot\, x^{3n-1}\,x^3}{(-1)^{n+1}\, \cdot\, 3\, \cdot\, x^{3n-1}}\,\bigg|$

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg|\, \left(\dfrac{(-1)^{n+1}}{(-1)^{n+1}}\right)\, (-1)\,\cdot\, \left(\dfrac{3}{3}\right)\, \cdot\, \left(\dfrac{x^{3n-1}}{x^{3n-1}}\right)\, x^3\, \bigg|$

You should end up with something more like:

. . . . .$\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,$$\displaystyle \bigg|\, (-1)\, x^3\, \bigg|$

Problem is that I can't effectively get the n by itself so I can take its limit.

In this case, does that matter? Since the convergence must not depend upon the value of "n" (since there is no "n" in the expression), you don't need to worry about that. Besides, you're trying to find the x-values for the interval. (here)

You need to have |x^3| be less than 1. What then must be the convergence interval? [/QUOTE]

That was really helpful. Thank you! I'm assuming that the IoC would simply be -1<x<1, then, and the RoC would be 1?

 

[Prove It]

Hi, Free Math Help. I'm trying to do a Maclaurin series problem and I'm having trouble.

The problem (AP calculus FRQ #6 style my teacher wrote) asks us to find the first four non-zero terms for the Maclaurin series for 3x^2/(1+x^3) as well as the series's general term. As far as I can tell, I did this correctly and got 3x^2-3x^5+3x^8-3x^11 as my first four terms and [sum from 1 to infinity] of (-1)^(n+1) * 3x^(3x-1) as my general.

Then the problem asks me to find the radius/interval of convergence. I did the ratio test as I'm wont to do in this situation, and I got something I can't really work with. I got:

lim n>infinity: |[(-1)^(n+2)3x^(3n+2)]/[(-1)^(n+1)3x^(3n-1)]|

This reduced to lim n>infinity: |[3x^3(n+1)]|

Problem is that I can't effectively get the n by itself so I can take its limit. Any ideas? Thanks for any help you can give!

Just write it as \(\displaystyle \displaystyle \begin{align*} \frac{3\,x^2}{1 + x^3} &= 3\,x^2\,\left[ \frac{1}{1 - \left( -x^3 \right)} \right] \end{align*}\)

The stuff in the square brackets is a geometric series with \(\displaystyle \displaystyle \begin{align*} r = -x^3 \end{align*}\) and so is convergent for \(\displaystyle \displaystyle \begin{align*} \left| -x^3 \right| < 1 \implies \left| x \right| < 1 \end{align*}\). The series is

\(\displaystyle \displaystyle \begin{align*} 3\,x^2 \, \left[ \frac{1}{1 - \left( -x^3 \right) } \right] &= 3\,x^2 \sum_{n = 0}^{\infty} \left( -x^3 \right) ^n \\ &= 3\,x^2 \sum_{n = 0}^{\infty} \left( -1 \right) ^n \,x^{3\,n} \\ &= 3\,\sum_{n = 0}^{\infty} \left( -1 \right) ^n \, x^{3\,n + 2} \end{align*}\)