James Tantono By convention, when talking about Power series, we sort of assume 0^0 = 1, simply because it works best in this case. If you'd like, you could think of it as starting the summation at n=1, and adding 1 to the whole sum separately.
@blackpenredpen I have a question. I'm doing some self study using the book Fundamentals of Differential Equations 6th edition by Nagle. In section 8.2 (the section for this video) Nagle says that the Ratio Test is the reciprocal of what you just showed. That is: Limit as n--->(infinity) of I a(n)/a(n+1) I = L where (n) and and (n+1) are the subscripts. He says that if the limit (L) exists then the radius of convergence is (L). There's no mention of any convergence or divergence of the series if L1, or inconclusive if L=1: it simply says that the radius of convergence is (L). It is a bit tough to write this here but it's on page 428, section 8.2. I learned the ratio test as you just showed in this video when I did Calculus 2 but is Nagle referencing the same ratio test as you did here or is it a different version used for a different conclusion? EDIT: I found my answer, it's the same statement put another way when the limit in the usual Ratio test as demonstrated in this video is L < 1 (whenever the series converges).
Very good video! Been struggling with intervals of convergence in particular for a while, this finally made me able to solve some problems!
I am from India 🇮🇳 your style of teaching is awesome. Love from India.
I have an exam in an 1 hour and you literally helped me sooo much!! THANK YOU!!!!!!
How did it go😂
You are such a brilliant young man. Love from Uganda
This is great..the way you deliver your lecture is impressive 👍👍 thank you
Thankyou for this! I have been struggling with series.
Great video! Thank you for going slow and explaining every step!
Glad to help. 😃
i'm watching this in the end of the year 2024 you're explaining was a great thanks a bunch
Thank you very much! Once again you are doing maths with fun :) greetings from Greece!!
Thank you for posting your nice videos, this is brilliant especially because there is not too much theory here
Very, very nice explanation🙏🙏🙏🙏 and very conceptual problem
this is so clear! appreciated!
Thank you…..Weeks of learning this in class, but blackpenredpen gets it in less than 14 minutes
Helped me A LOT!!
Thanks, dude! this makes a lot more sense now.
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@@CharanSingh-jb9sy new Hampshire
i ment that what is your country name...?
@@CharanSingh-jb9sy usa
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I love your lectures
A nice video on this topic :)
Helped me. Thanks.
Excellent explanation!!
Thank you good sir
How about when x=1? The first term will become 0^0 which is undefined
James Tantono By convention, when talking about Power series, we sort of assume 0^0 = 1, simply because it works best in this case. If you'd like, you could think of it as starting the summation at n=1, and adding 1 to the whole sum separately.
0^0 is undefined, but in the general sense the lim as x->0 of x^x=1
@blackpenredpen I have a question. I'm doing some self study using the book Fundamentals of Differential Equations 6th edition by Nagle. In section 8.2 (the section for this video) Nagle says that the Ratio Test is the reciprocal of what you just showed. That is:
Limit as n--->(infinity) of I a(n)/a(n+1) I = L
where (n) and and (n+1) are the subscripts. He says that if the limit (L) exists then the radius of convergence is (L). There's no mention of any convergence or divergence of the series if L1, or inconclusive if L=1: it simply says that the radius of convergence is (L).
It is a bit tough to write this here but it's on page 428, section 8.2. I learned the ratio test as you just showed in this video when I did Calculus 2 but is Nagle referencing the same ratio test as you did here or is it a different version used for a different conclusion?
EDIT: I found my answer, it's the same statement put another way when the limit in the usual Ratio test as demonstrated in this video is L < 1 (whenever the series converges).
Great video!
Not all Heroes wear capes, some just hold microphones for many hours.
So good
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Can you actually include - 1 in the interval? Since it converges there, but doesn't converge absolutely?
yes, at the end the interval results [-1,3)
easiest question of my life
Thanks sir
I do not see how at the endpoint 3 it diverges?
You can do it by direct comparison test or ,limit comparison test,.......
its good but u must be precise
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Hey, thanks.