What are the rules for using the limit comparison theorem in power series?

In summary: So it's not a definite conclusion, but rather we have to do further analysis. In summary, the conversation is about preparing for a calculus II final and discussing power series and the limit comparison theorem. The summary includes a link to a practice exam and solutions, as well as a discussion on finding the fifth derivative and using the limit comparison theorem. Further questions are also mentioned and answered.
  • #1
Richter915
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Hi, I'm preparing for my calculus II final and I had a question about Power Series. I'm posting the link for the answer solution to a practice exam (it's in pdf form) and I will ask questions based on that.

http://www.math.sunysb.edu/~daryl/prcsol3.pdf

On problem 12 it asks for the fifth derivative, I'm not sure why they did the last part. I can do the first three steps and solve for the power series of f(x) but I have no clue why they do that last part so if you can clear it up it would be appreciated.

On problem 7 part F...can you please explain the rules for limit comparison theorem. Also, when facing a problem like that, what series are good to use in comparison to the series given to you?

Thanks for any help, more questions will be asked in the future...heh.
 
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  • #2
Richter915 said:
Hi, I'm preparing for my calculus II final and I had a question about Power Series. I'm posting the link for the answer solution to a practice exam (it's in pdf form) and I will ask questions based on that.

http://www.math.sunysb.edu/~daryl/prcsol3.pdf

On problem 12 it asks for the fifth derivative, I'm not sure why they did the last part. I can do the first three steps and solve for the power series of f(x) but I have no clue why they do that last part so if you can clear it up it would be appreciated..
They use the fact that if a function f can be written as a power serie [itex]\sum a_kx^k[/itex], then [itex]a_k = f^{(k)}(0)/k![/itex]. So they only look at the coefficient of x^5 and they're assured that it is [itex]f^{5}(0)/5![/itex].
 
  • #3
Richter915 said:
On problem 7 part F...can you please explain the rules for limit comparison theorem. Also, when facing a problem like that, what series are good to use in comparison to the series given to you?

How it works: You chose a serie of positive n-th term b_n for which you know wheter it converges or not. Then you compute the limit

[tex]\lim_{n\rightarrow \infty}\frac{|a_n|}{b_n}[/tex]

i) If the result is any number that is not 0 or infinity, the two series have the same convergence value. So if b_n diverges, a_n too and if b_n converges, a_n too.

ii) If the limit is 0, and if b_n converges, |a_n| converges.

iii) If the limit is [itex]\infty[/itex] and if b_n diverges, then |a_n| diverges.

Otherwise, you cannot conclude.

Good series to use as b_n are the "Riemman p-series":

[tex]\sum \frac{1}{n^p}[/tex]

it diverges for [itex]p \leq 1[/itex] and converges for [itex]p>1[/itex].
 
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  • #4
http://www.math.sunysb.edu/~leontak/prfinal.pdf

http://www.math.sunysb.edu/~leontak/solpf.pdf

Hi I was unsure of the explanation of the answer for question 13 part a. I understand everything up to where he finds the pattern shown by the recursive relation but after that, I'm not sure how he solves for y(x).

Also, can you please explain the solution to 14b. I don't understand the thing with the long division, everything before that I recognize..please help...thanks.
 
  • #5
quasar987 said:
How it works: You chose a serie of positive n-th term b_n for which you know wheter it converges or not. Then you compute the limit

[tex]\lim_{n\rightarrow \infty}\frac{|a_n|}{b_n}[/tex]

i) If the result is any number that is not 0 or infinity, the two series have the same convergence value. So if b_n diverges, a_n too and if b_n converges, a_n too.

ii) If the limit is 0, and if b_n converges, |a_n| converges.

iii) If the limit is [itex]\infty[/itex] and if b_n diverges, then |a_n| diverges.

Otherwise, you cannot conclude.

Good series to use as b_n are the "Riemman p-series":

[tex]\sum \frac{1}{n^p}[/tex]

it diverges for [itex]p \leq 1[/itex] and converges for [itex]p>1[/itex].
so with points ii and iii...what if b_n was divergent and the lim was 0? What would that mean? And with iii, what if the lim goes to infinity and b_n converges? what happens then?

Thanks.
 
  • #6
Richter915 said:
so with points ii and iii...what if b_n was divergent and the lim was 0? What would that mean? And with iii, what if the lim goes to infinity and b_n converges? what happens then?

These cases are covered in "Otherwise, you cannot conclude."

That is to say, if we get that b_n was divergent and the lim was 0 or that the lim goes to infinity and b_n converges, the test doesn't tell us anything about the convergence of |a_n|, and we have to try comparing it with another b_n or try another test.
 

FAQ: What are the rules for using the limit comparison theorem in power series?

What is a power series?

A power series is a mathematical representation of a function as an infinite sum of polynomial terms. It has the form ∑n=0∞ cn(x-a)n, where cn are the coefficients, x is the variable, and a is the center of the series.

What is the purpose of using a power series to represent a function?

The purpose of using a power series to represent a function is to be able to approximate the function with a polynomial of infinite degree. This allows for a more accurate representation of the function, especially when dealing with complicated or non-analytic functions.

How is a power series evaluated?

A power series can be evaluated by substituting the value of the variable x into the series and then simplifying the resulting expression. The series will converge if the value of x is within the radius of convergence, which can be determined using tests such as the ratio test or the root test.

What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a type of power series that represents a function at a specific center, while a Maclaurin series is a Taylor series at the center a=0. In other words, a Taylor series can be centered at any point a, while a Maclaurin series is centered at 0.

Can a power series represent any function?

No, a power series can only represent a function if it has a convergent radius of convergence. This means that the series will only approximate the function within a certain interval, and outside of that interval, the series will diverge and no longer represent the function accurately.

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