Absolute and conditional convergence (KristaKingMath)

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Krista King

Krista King

Күн бұрын

Пікірлер: 30
@rostyloco1
@rostyloco1 6 жыл бұрын
everyone that goes online to look for conditional convergences gets stuck looking through shit loads of absolute convergence and hardly any conditional convergence. still, good vid. ty.
@brendam9374
@brendam9374 8 жыл бұрын
you are so good at explaining! such a talent!
@kristakingmath
@kristakingmath 8 жыл бұрын
+Brenda M Thank you so much!
@hg2.
@hg2. 8 жыл бұрын
Go Krista!
@olle6343
@olle6343 4 жыл бұрын
Can you take away the absolute value bars on a function, that involves x, and see if it is only conditionally convergent on intervals of x values? Or do you recommend another test for the purpose of finding conditional convergence? Thank you!
@teshamartin3353
@teshamartin3353 8 жыл бұрын
Your videos are very helpful...Thanks for the great explanations
@kristakingmath
@kristakingmath 8 жыл бұрын
Thanks! I'm glad they're helping. :)
@draft378
@draft378 10 жыл бұрын
This is exactly what I'm doing in my BC calc class thanks :)
@MrYusufgovani
@MrYusufgovani 10 жыл бұрын
Very good explanation! Are you planning to do a course for coursera or edex ?
@kristakingmath
@kristakingmath 10 жыл бұрын
I have my own course at my website :) integralcalc.com/
@muzamilanvar4564
@muzamilanvar4564 3 жыл бұрын
wow, what a useful content well explained clear voice thanks
@kristakingmath
@kristakingmath 3 жыл бұрын
You're welcome, Muzamil! I'm so glad you liked it! :)
@bibhuprasad1970
@bibhuprasad1970 8 жыл бұрын
Thank you ,for good explanation.
@kristakingmath
@kristakingmath 8 жыл бұрын
Glad you liked it!
@mattbartlett3049
@mattbartlett3049 7 жыл бұрын
So is L the value it converges to? If i get a value of L=0.5 is that where my series converges to?
@kristakingmath
@kristakingmath 7 жыл бұрын
No, L isn't the value it converges to. You can only tell whether or not it converges based on the value of L.
@HariKrishnan-bi6zt
@HariKrishnan-bi6zt 8 жыл бұрын
great explanation thank you miss
@kristakingmath
@kristakingmath 8 жыл бұрын
Thanks!
@raajduby993
@raajduby993 6 жыл бұрын
Thank u fo well explanation of series problem
@kristakingmath
@kristakingmath 6 жыл бұрын
You're welcome, Raaj! :)
@sabithasasikumar239
@sabithasasikumar239 11 жыл бұрын
I still don't get the part about 4^(2n+3), shouldn't it just be 4^(2n+2)??
@kristakingmath
@kristakingmath 11 жыл бұрын
no, because you start with 4^(2n+1), and then you plug n+1 in for n, and you get 4^[2(n+1)+1]. distributing the 2, you get 4^[2n+2+1], which simplifies to 4^(2n+3). :)
@ramansb1008
@ramansb1008 10 жыл бұрын
could you please make proof videos opposed to examples.
@silbridge1155
@silbridge1155 10 жыл бұрын
You are beautifully amazing !!
@anthonym4366
@anthonym4366 10 жыл бұрын
Those limits were not simple to do but I understand now convergent must be less than 1 .
@muzamilanvar4564
@muzamilanvar4564 3 жыл бұрын
i study in portuguese but i understood all the steps
@MrYaseen100
@MrYaseen100 8 жыл бұрын
i feel like a series will never converge conditionally, because of the absolute value, is the like the ultimate weapon lol
@bebarshossny5766
@bebarshossny5766 5 жыл бұрын
If i was a rich millionare i would devote every penny i have to funding you as well as khan academy trevtutor patrickjmt and every other math/science channel who literally just make videos to help students without getting anything in return Fuck my shitty professor man You people are saving my ass every damn day
@kristakingmath
@kristakingmath 5 жыл бұрын
I'm just happy that I have the opportunity to help! :D
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