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Professor Organic Chemistry Tutor, thank you for explaining the Remainder Estimate for the Integral Test in Calculus Two. I am somewhat confused on this estimate for the Integral Test; however, doing problems is the best way to increase my understanding of the material. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
what kind of calculator are you using? I entered the expression sum(1 / x^2, x, 1, 1e6) // sum from one to a million in wolfram alpha and it took about three seconds.
@@cornetapluspluseleven9296 Or if you know a bit of code and are on a computer, you can write an easy for loop and accumulator and it'll go far faster than any calculator (provided you aren't on a dinosaur of a computer, my time for the exact summation that he was doing was .756 seconds just on a basic ass loop).
In part d) you find that n>200. But in part c) the sum is bounded to within 0.005 (the maximum of 1.649768-1.64522 and 1.64522-1.640677) so in fact only 10 terms are needed. To ensure an error less than 0.005 the remainder estimate says that you need more than 200 terms, but part c) says you need only 10. I think it would be good to explain that n>200 only ensures that the error will be less than 0.005 but it doesn't say how much less. If you were to compute the error in part c) with n=200 it will be much less than 0.005.
you're right! the sign should have been switched because he took the inverse of both sides of the inequality. so n should be greater than or equal to 200 :)
It can go either way, 1/n < .005 , deviding n to the right, then deviding .005 from .005n to the left is the same as .005 < 1/n being powered by a negative 1
for part d the video is wrong, you have (integral from n to infinity) > Rn, then you know that if 0.005 > (the integral from n to infinity) then the error must also be less than 0.005. so work out the integral, assuming it's 1/n, you would work with 0.005>1/n>Rn and this is true for n>1/0.005 = 200, so n > 200.
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Professor Organic Chemistry Tutor, thank you for explaining the Remainder Estimate for the Integral Test in Calculus Two. I am somewhat confused on this estimate for the Integral Test; however, doing problems is the best way to increase my understanding of the material. This is an error free video/lecture on UA-cam TV with the Organic Chemistry Tutor.
"this might take a hot minute"
thanks man! we appreciate your efforts a lot and God bless you!
"This might take a hot minute" 12:45 lmao you right. thanks!
what kind of calculator are you using? I entered the expression
sum(1 / x^2, x, 1, 1e6) // sum from one to a million
in wolfram alpha and it took about three seconds.
😭😭
@@cornetapluspluseleven9296 Or if you know a bit of code and are on a computer, you can write an easy for loop and accumulator and it'll go far faster than any calculator (provided you aren't on a dinosaur of a computer, my time for the exact summation that he was doing was .756 seconds just on a basic ass loop).
In part d) you find that n>200. But in part c) the sum is bounded to within 0.005 (the maximum of 1.649768-1.64522 and 1.64522-1.640677) so in fact only 10 terms are needed. To ensure an error less than 0.005 the remainder estimate says that you need more than 200 terms, but part c) says you need only 10. I think it would be good to explain that n>200 only ensures that the error will be less than 0.005 but it doesn't say how much less. If you were to compute the error in part c) with n=200 it will be much less than 0.005.
Why when you raise both sides to the power of negative 1 you don't switch the inequality sign also?
Saving us again god bless❤️
Shouldn't we flip the inequality when we raise both sides by -1? 15:34
no?
@@erikdahlen9140 yeah we should it's basic math
15:43 you raised the negative power without changing inequality ? Are you kidding???
Thank you for making this video. It is great!
The last inequality is giving me some trouble. 0.005
you're right! the sign should have been switched because he took the inverse of both sides of the inequality. so n should be greater than or equal to 200 :)
I also noticed that one
Recall 1/(n+1)=200 and n>=199....so to satisfy both it should be n>=200
@@albertlegaspi9614 actually appreciate this explanation - it actually showed the error he made (which was not just forgetting to flip the sign)
Why does the inequality sign change around 15:00 ?
For part D, shouldn't you solve 1/n < 0.005 instead of 0.005
It can go either way, 1/n < .005 , deviding n to the right, then deviding .005 from .005n to the left is the same as .005 < 1/n being powered by a negative 1
@@kamves8822 If you devidde the left side by n you get 1/n^2 not 1
Would the inequality switch as you take the reciprocal of both sides?
Yes
for part d the video is wrong, you have (integral from n to infinity) > Rn, then you know that if 0.005 > (the integral from n to infinity) then the error must also be less than 0.005. so work out the integral, assuming it's 1/n, you would work with 0.005>1/n>Rn and this is true for n>1/0.005 = 200, so n > 200.
Not my favorite video feels like there could be a better structured explanation. Love your videos btw.
Nice :> Now I can approximate the Zeta function that way :)
this guy is amazing!
The inequality doesnt change in part d, if you multiply both sides by n and divide both sides by 0.005 you should get n
Cool fact: that infinite series
(1/x^2) actually converges to pi squared over 6
Never knew
Allah razı olsun, God bless you
Very understandable
thank you so much!
actual answer is exactly pi^2/6
thank you
Isn't it confusing that n should be greater than 200 cause 1/n should be greater than 1/200?
thank you :]
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Why are yall quoting “hot minute” what’s so crazy about that?
I don't get it
i love you
you're a fucking gem man
목소리 존나 좋다