Calculus II

Marie1309

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Hello, could someone help me doing this exercice please? I’m really struggling.

image.jpg
 
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I suspect the issue is that the images are sideways, and not with your handwriting, which is actually very nice. If you're uploading images from a phone, then it's not really your fault...I see a lot of sideways images on forums from mobile users. Is this the question?

a) Find the radius and interval of convergence of:

\[\sum_{n=0}^{\infty}\frac{n(3x-1)^n}{6^n}\]
 
This is how I started. Is it correct?
 

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I suspect the issue is that the images are sideways, and not with your handwriting, which is actually very nice. If you're uploading images from a phone, then it's not really your fault...I see a lot of sideways images on forums from mobile users. Is this the question?

a) Find the radius and interval of convergence of:

\[\sum_{n=0}^{\infty}\frac{n(3x-1)^n}{6^n}\]
Yes that’s correct this is how I started but I’m not sure if it is correct

1455A4BC-839E-4731-9AF8-EC57A0E43D70.jpeg
 
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Then use the "limit ratio test": \(\displaystyle \sum a_n\) converges absolutely if \(\displaystyle \lim \left|\frac{a_{n+1}}{a_n}\right|< 1\).
I did but I’m not sure if it is correct
 

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Yes that’s correct this is how I started but I’m not sure if it is correct

View attachment 11480

What we want is:

[MATH]\frac{|3x-1|}{6}<1[/MATH]
or:

[MATH]\left|x-\frac{1}{3}\right|<2[/MATH]
And so the radius of convergence is 2, and the interval of convergence is:

[MATH]\left(\frac{1}{3}-2,\frac{1}{3}+2\right)[/MATH]
or:

[MATH]\left(-\frac{5}{3},\frac{7}{3}\right)[/MATH]
 
Your work is very sloppy. When you replace n with n+1 you must write (n+1). The powers of (3x-1) are both n+1in your work. But this can't be since in one case n stays n and in the other case n becomes (n+1). You were told to take lim |{an+1/an|<1, yet you did not take the absolute value nor did you write limit. Also how did you compute that the radius of convergence is 6?
 
Your work is very sloppy. When you replace n with n+1 you must write (n+1). The powers of (3x-1) are both n+1in your work. But this can't be since in one case n stays n and in the other case n becomes (n+1). You were told to take lim |{an+1/an|<1, yet you did not take the absolute value nor did you write limit. Also how did you compute that the radius of convergence is 6?
I think I got it now. So IS The radius of convergence 2?
 
What we want is:

[MATH]\frac{|3x-1|}{6}<1[/MATH]
or:

[MATH]\left|x-\frac{1}{3}\right|<2[/MATH]
And so the radius of convergence is 2, and the interval of convergence is:

[MATH]\left(\frac{1}{3}-2,\frac{1}{3}+2\right)[/MATH]
or:

[MATH]\left(-\frac{5}{3},\frac{7}{3}\right)[/MATH]
Hello, Thank you so much for your help and I did not understand how did you find the radius equals to 2?
 
If a power series converges for |x-a|<R and diverges for |x-a|>R, then the power series is centred at point a and has radius r=R. In your case a=1/3 and R=2. The interval of convergence is (a-R, a+R) = (-5/3,7/3) . The series diverges for both x=-5/3 and x=7/3 so the interval is open.
 
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1st compute lim| an+1/an|

Set this lim <1
Then solve for x.

Example: |4x-6|<1
Factor out a 4
4|x-3/2| <1
|x - 3/2| < 1/4

So the interval of convergence is 1/4
 
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Hello, Thank you so much for your help and I did not understand how did you find the radius equals to 2?

We have:

[MATH]\frac{|3x-1|}{6}<1[/MATH]
And we want to get this in the form:

[MATH]|x-a[<r[/MATH]
And so we factor out a 3 from the numerator and divide out common factors to get:

[MATH]\frac{\left|x-\dfrac{1}{3}\right|}{2}<1[/MATH]
Multiply through by 2:

[MATH]\left|x-\frac{1}{3}\right|<2[/MATH]
And so we have it in the form we want, where \(\dfrac{1}{3}\) is the center of the interval of convergence, whose radius is \(2\).

Does that make sense?
 
We have:

[MATH]\frac{|3x-1|}{6}<1[/MATH]
And we want to get this in the form:

[MATH]|x-a[<r[/MATH]
And so we factor out a 3 from the numerator and divide out common factors to get:

[MATH]\frac{\left|x-\dfrac{1}{3}\right|}{2}<1[/MATH]
Multiply through by 2:

[MATH]\left|x-\frac{1}{3}\right|<2[/MATH]
And so we have it in the form we want, where \(\dfrac{1}{3}\) is the center of the interval of convergence, whose radius is \(2\).

Does that make sense?
RIGHTTT. Makes a lot of sense. Thank you so much for your time.
 
If a power series converges for |x-a|<R and diverges for |x-a|>R, then the power series is centred at point a and has radius r=R. In your case a=1/3 and R=2. The interval of convergence is (a-R, a+R) = (-5/3,7/3) . The series diverges for both x=-5/3 and x=7/3 so the interval is open.
Hey. Thanks for helping. But may I know why it diverges for both? . Thanks
 
Hey. Thanks for helping. But may I know why it diverges for both? . Thanks

When we use the ratio test, we find the limit:

[MATH]L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|[/MATH]
The ratio test states that:
  • if \(L<1\) then the series converges absolutely;
  • if \(L>1\) then the series is divergent;
  • if \(L=1\) or the limit fails to exist, then the test is inconclusive, because there exist both convergent and divergent series that satisfy this case.
And so, because we require \(L<1\), the interval of convergence is necessarily open, that is, we do not include the end-points.
 
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Hey. Thanks for helping. But may I know why it diverges for both? . Thanks
When [MATH]x=\frac{7}{3}[/MATH] you have L=1 which is the case when the ratio test does not allow to conclude. In this case you replace x
by [MATH]\frac{7}{3}[/MATH] in [MATH]a_n[/MATH] and use the divergence test. Use the same step for [MATH]x=\frac{-5}{3}[/MATH]. The interval of convergence is the set of points where the series converges.
 
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