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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pka

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Please post a readable question.
 

Marie1309

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Is it clearer now?

27553DD4-8ED8-467B-81C9-3F2D8FF6BF8F.jpeg
 
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MarkFL

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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:

\[\sum_{n=0}^{\infty}\frac{n(3x-1)^n}{6^n}\]
 

Marie1309

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Yes that’s correct.
 

HallsofIvy

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Then use the "limit ratio test": \(\displaystyle \sum a_n\) converges absolutely if \(\displaystyle \lim \left|\frac{a_{n+1}}{a_n}\right|< 1\).
 

Marie1309

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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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MarkFL

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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:

\(\displaystyle \frac{|3x-1|}{6}<1\)

or:

\(\displaystyle \left|x-\frac{1}{3}\right|<2\)

And so the radius of convergence is 2, and the interval of convergence is:

\(\displaystyle \left(\frac{1}{3}-2,\frac{1}{3}+2\right)\)

or:

\(\displaystyle \left(-\frac{5}{3},\frac{7}{3}\right)\)
 

Jomo

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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?
 

Marie1309

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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?
 

Marie1309

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What we want is:

\(\displaystyle \frac{|3x-1|}{6}<1\)

or:

\(\displaystyle \left|x-\frac{1}{3}\right|<2\)

And so the radius of convergence is 2, and the interval of convergence is:

\(\displaystyle \left(\frac{1}{3}-2,\frac{1}{3}+2\right)\)

or:

\(\displaystyle \left(-\frac{5}{3},\frac{7}{3}\right)\)
Hello, Thank you so much for your help and I did not understand how did you find the radius equals to 2?
 

MathProfessor

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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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Jomo

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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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MarkFL

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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:

\(\displaystyle \frac{|3x-1|}{6}<1\)

And we want to get this in the form:

\(\displaystyle |x-a[<r\)

And so we factor out a 3 from the numerator and divide out common factors to get:

\(\displaystyle \frac{\left|x-\dfrac{1}{3}\right|}{2}<1\)

Multiply through by 2:

\(\displaystyle \left|x-\frac{1}{3}\right|<2\)

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?
 

Marie1309

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We have:

\(\displaystyle \frac{|3x-1|}{6}<1\)

And we want to get this in the form:

\(\displaystyle |x-a[<r\)

And so we factor out a 3 from the numerator and divide out common factors to get:

\(\displaystyle \frac{\left|x-\dfrac{1}{3}\right|}{2}<1\)

Multiply through by 2:

\(\displaystyle \left|x-\frac{1}{3}\right|<2\)

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.
 

Marie1309

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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
 

MarkFL

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Hey. Thanks for helping. But may I know why it diverges for both? . Thanks
When we use the ratio test, we find the limit:

\(\displaystyle L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\)

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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MathProfessor

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Hey. Thanks for helping. But may I know why it diverges for both? . Thanks
When \(\displaystyle x=\frac{7}{3}\) you have L=1 which is the case when the ratio test does not allow to conclude. In this case you replace x
by \(\displaystyle \frac{7}{3}\) in \(\displaystyle a_n\) and use the divergence test. Use the same step for \(\displaystyle x=\frac{-5}{3}\). The interval of convergence is the set of points where the series converges.
 
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