Jbreezy said:Homework Statement
Question. Did I do this OK?
Homework Equations
The Attempt at a Solution
A_n = Ʃ e^(n^2) x^n from n = 1 to ∞
So I tried the root test. After you take the nth root you have x e^n so then I took the limit of this as n-->∞ and I got infinity. I then said OK |x| < 1 for this to converge so the interval of convergence is (-1,1) ?? This OK?
LCKurtz said:Why is there an ##n## on the left side and not the right side?
I don't see it sorry. Where?
No, it isn't OK. It would help if you would write equations instead of prose. Show us your work instead of just declaring ##|x|<1##.
{A_n} = Ʃ e^(n^2) x^n from n = 1 to ∞
LCKurtz said:Why is there an ##n## on the left side and not the right side?
Jbreezy said:I don't know what prose is.
I just did
(e^(n^2) x^n)^(1/n) = e^n (x)
Took the limit as n --> infinity I got infinity.
This is when I said |x| < 1
LCKurtz said:The ##n## is right there. And please don't put your question (where's the ##n##) as part of my quote.
So look it up in a dictionary.
I know that's what you said. It's wrong. Explain how you got |x| < 1.
Who is "it"?Jbreezy said:Didn't know it was in the quote. I got |x| <1 because by the root test it converges for those values.
Jbreezy said:0 < 1. It is what they did in my book when they got infinity for the limit. They took whatever was in abs like |x-3| or in this case |x| and did |x| <1 and found the interval of convergence from it.
Mark44 said:Who is "it"?
Show us your work for the root test. That's what LCKurtz is asking for.
Regarding his comment about prose, what he's saying is more math, and less words.
Jbreezy said:This is what I did
(A_n)^(1/n) = (e^(n^2) x^n)^(1/n) = e^n (x)
|(A_n)^(1/n)| = |x| lim n---> ∞ e^n = ∞ = L
OK the series diverges but if I want it to converge I need|x| < 1
So that's where it came from
Mark44 said:So you to find the values of x for which
$$|x|\lim_{n \to \infty}e^n < 1$$
|x| < 1 isn't going to cut it.
Jbreezy said:I;m lost.
This is $$\lim_{n \to \infty}e^n = ∞ $$
This is $$|x| < 1 $$ for the series to converge? Yes. I swear my book has this. I don't understand why this is wrong. Is it not true?
You've misunderstood what the book is saying, so the conclusion you're drawing based on your misconception isn't making sense to us. Try going over the example in the book again more carefully. If you're still confused, post the example from the book and explain to us what you think it's saying.Jbreezy said:Didn't know it was in the quote. I got |x| <1 because by the root test it converges for those values. 0 < 1. It is what they did in my book when they got infinity for the limit. They took whatever was in abs like |x-3| or in this case |x| and did |x| <1 and found the interval of convergence from it.
A power series is an infinite series of the form ∑n=0∞ an(x-c)n, where an is a coefficient and c is a constant. It is a type of mathematical series that is useful for representing functions as an infinite sum of polynomials.
The convergence of a power series refers to the values of x for which the series converges, or approaches a finite value. This can be determined by using the ratio test or the root test, which evaluate the limit of the ratio or root of the terms in the series as n approaches infinity.
Power series can be used to approximate functions by truncating the series at a certain point and using the remaining terms to approximate the function. This is particularly useful for functions that are difficult to evaluate or integrate analytically.
The radius of convergence of a power series is the distance from the center point c at which the series will converge. It is determined by the convergence tests and can be a finite or infinite value.
Power series have many practical applications, such as in physics, engineering, and economics. They can be used to model various phenomena, such as motion, electromagnetic fields, and financial data. They are also used in numerical methods for solving differential equations and other mathematical problems.