Maths exam in a few hours, need a bit of help:

Daniel S

Basically I can't remember how I did a few questions:

One was interval of convergence, when I work out:
Lim as n goes to infinity of |An+1|/|An|

Taking out the "x's and numbers" the "n"s become 1.

I'm left with the following on different questions and I've no idea how I did this:

Q1. |x|/8 became -8<x<8
Q2. |x+4|/9 became -13<x<5
Q3. |x+2|/4 became -6<x<2
Q4. 2|x-3| became 2.5<x<3.5
Q5 |x-5|/3 became 2<x<8
Q6. |x-10|/10 became 0<x<20
Q7. |x-1|/4 became -3<x<5
Q8 |x+1|/2 became -3<x<1

If anyone knows how to do that step or can tell me how I did it please tell me, I'll be borderline on this test, if I fail it I'm out of the course.

sponsoredwalk

What you're doing is trying to find the radius of convergence of a power series f(x) = ∑ᵢ₌₁°°aᵢxⁱ in order to find it's "interval of convergence".

The reason why all of this works in the first place is due to "Abel's theorem": If the power series f(x) = ∑ᵢ₌₁°°aᵢxⁱ converges at the value x₀ then it converges for all x s.t. |x| < |x₀| (or, equivalently, all x such that |x|/|x₀| < 1)
Proof:
1: Expand the power series - f(x) = ∑ᵢ₌₁°°aᵢxⁱ = a₀ + a₁x + a₂x² + ...
2: Multiply each term by 1 (a convenient 1, a term that you can use to bound the power series, the term in your hypothesis) -
a₀ + a₁x + ... = a₀ + a₁x(|x₀|/|x₀|) + a₂x²(|x₀|/|x₀|)² + ... = a₀ + (a₁|x₀|)(x/|x₀|) + a₂|x₀|²(x/|x₀|)² + ...
3: Notice each term (x/|x₀|)ⁱ is less than 1 since |x| < |x₀| thus we almost have a geometric series except for the (a₁|x₀|) terms, thus we use the fact that these are all constants & find some M bigger than all of them -
f(x) = a₀ + (a₁|x₀|)(x/|x₀|) + a₂|x₀|²(x/|x₀|)² + ... < g(x) = Ma₀ + M(x/|x₀|) + M(x/|x₀|)² + ...
4: Now we have an infinite geometric series with ratio (x/|x₀|) < 1 which converges thus the power series converges as long as (x/|x₀|) < 1. If you want to be more pedantic you can say that since f(x) < g(x) & since g(x) converges when (x/|x₀|) < 1, by the comparison test f(x) must also converge.
So |x|/8 became -8<x<8 because |x|/8 < 1 → |x| < 8. Note that here x₀ = 8. The rest of your examples are the same...
The whole idea is just to bound f using the number x₀ that you know the power series will converge at. Just multiply the power series by your convenient 1 = x/x₀ = (x/x₀)² = ... then bound the constants by the same M to turn it into an infinite geometric series. The |x| < |x₀| part gives you the interval of convergence (-x₀,x₀) = - x₀ < x < x₀.

Now that you know that the power series will converge in an interval once you know that it converges for some value x₀ (e.g. x₀ = 8 in your example, thus whatever the actual power series was you know it converges for all x s.t. |x|/8 < 1), you want to find the largest interval in which the power series will converge. This is called finding the radius of convergence & that is when the idea of " Lim as n goes to infinity of |An+1|/|An|" comes into the picture.

Again the idea is really just to bound the power series, you want to know when you can be sure that:
f(x) = ∑ᵢ₌₁°°aᵢxⁱ = a₀ + a₁x + a₂x² + ... < g(x) = |a₀| + |a₁||x| + |a₂||x|² + ...
What are the conditions on x that force f(x) < g(x)?
Here you use D'Alembert's Ratio test to find the conditions.
Assume that the limit (as n goes to infinity) of the ratio of two succeeding n'th terms of the series exists:
lim [((a_n+1) xⁿ⁺¹)/a_n xⁿ] = lim [(a_n+1xⁿ ·x)/a_n xⁿ] = lim [(a_n+1)/a_n]x = Lx
then the series converges if this limit is less than 1, Lx < 1. This is just the statement of the ratio test. But you can use it to write x < (1/L).
This (1/L) = lim(a_n/a_n+1) is the radius of convergence.
Basically the radius of convergence is found by finding the limit of the ratio of a_n over a_n+1. You can also use other idea's but I sincerely doubt you'd be asked that kind of stuff...

Long story short, your examples are just exemplifying the "then it converges for all x s.t. |x| < |x₀| (or, equivalently, all x such that |x|/|x₀| < 1)" part of Abel's theorem.

Daniel S

sponsoredwalk wrote: »

What you're doing is trying to find the radius of convergence of a power series f(x) = ∑ᵢ₌₁°°aᵢxⁱ in order to find it's "interval of convergence".

The reason why all of this works in the first place is due to "Abel's theorem": If the power series f(x) = ∑ᵢ₌₁°°aᵢxⁱ converges at the value x₀ then it converges for all x s.t. |x| < |x₀| (or, equivalently, all x such that |x|/|x₀| < 1)
Proof:
1: Expand the power series - f(x) = ∑ᵢ₌₁°°aᵢxⁱ = a₀ + a₁x + a₂x² + ...
2: Multiply each term by 1 (a convenient 1, a term that you can use to bound the power series, the term in your hypothesis) -
a₀ + a₁x + ... = a₀ + a₁x(|x₀|/|x₀|) + a₂x²(|x₀|/|x₀|)² + ... = a₀ + (a₁|x₀|)(x/|x₀|) + a₂|x₀|²(x/|x₀|)² + ...
3: Notice each term (x/|x₀|)ⁱ is less than 1 since |x| < |x₀| thus we almost have a geometric series except for the (a₁|x₀|) terms, thus we use the fact that these are all constants & find some M bigger than all of them -
f(x) = a₀ + (a₁|x₀|)(x/|x₀|) + a₂|x₀|²(x/|x₀|)² + ... < g(x) = Ma₀ + M(x/|x₀|) + M(x/|x₀|)² + ...
4: Now we have an infinite geometric series with ratio (x/|x₀|) < 1 which converges thus the power series converges as long as (x/|x₀|) < 1. If you want to be more pedantic you can say that since f(x) < g(x) & since g(x) converges when (x/|x₀|) < 1, by the comparison test f(x) must also converge.
So |x|/8 became -8<x<8 because |x|/8 < 1 → |x| < 8. Note that here x₀ = 8. The rest of your examples are the same...
The whole idea is just to bound f using the number x₀ that you know the power series will converge at. Just multiply the power series by your convenient 1 = x/x₀ = (x/x₀)² = ... then bound the constants by the same M to turn it into an infinite geometric series. The |x| < |x₀| part gives you the interval of convergence (-x₀,x₀) = - x₀ < x < x₀.

Now that you know that the power series will converge in an interval once you know that it converges for some value x₀ (e.g. x₀ = 8 in your example, thus whatever the actual power series was you know it converges for all x s.t. |x|/8 < 1), you want to find the largest interval in which the power series will converge. This is called finding the radius of convergence & that is when the idea of " Lim as n goes to infinity of |An+1|/|An|" comes into the picture.

Again the idea is really just to bound the power series, you want to know when you can be sure that:
f(x) = ∑ᵢ₌₁°°aᵢxⁱ = a₀ + a₁x + a₂x² + ... < g(x) = |a₀| + |a₁||x| + |a₂||x|² + ...
What are the conditions on x that force f(x) < g(x)?
Here you use D'Alembert's Ratio test to find the conditions.
Assume that the limit (as n goes to infinity) of the ratio of two succeeding n'th terms of the series exists:
lim [((a_n+1) xⁿ⁺¹)/a_n xⁿ] = lim [(a_n+1xⁿ ·x)/a_n xⁿ] = lim [(a_n+1)/a_n]x = Lx
then the series converges if this limit is less than 1, Lx < 1. This is just the statement of the ratio test. But you can use it to write x < (1/L).
This (1/L) = lim(a_n/a_n+1) is the radius of convergence.
Basically the radius of convergence is found by finding the limit of the ratio of a_n over a_n+1. You can also use other idea's but I sincerely doubt you'd be asked that kind of stuff...

Long story short, your examples are just exemplifying the "then it converges for all x s.t. |x| < |x₀| (or, equivalently, all x such that |x|/|x₀| < 1)" part of Abel's theorem.

Wow, that's a lot of effort to go to. Thanks for that!