Alternate Series Estimation Theorem
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- This calculus 2 video tutorial provides a basic introduction into the alternate series estimation theorem also known as the alternate series remainder. It explains how to estimate the sum of the infinite series correct to three decimals places by calculating the number of terms needed by using the remainder estimate theorem.
Integral Test For Divergence:
• Calculus 2 - Integral ...
Remainder Estimate - Integral Test:
• Remainder Estimate For...
P-Series:
• P-series
Direct Comparison Test:
• Direct Comparison Test...
Limit Comparison Test:
• Limit Comparison Test
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Alternating Series Test:
• Alternating Series Test
Alternate Series Estimation Theorem:
• Alternate Series Estim...
Absolute & Conditional Convergence:
• Absolute Convergence, ...
The Ratio Test:
• Ratio Test
The Root Test:
• Root Test
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Series Tests - Practice Problems:
• Calculus 2 - Geometric...
Taylor & Maclaurin Polynomials:
• Taylor Polynomials & M...
Taylor's Remainder Theorem:
• Taylor's Remainder The...
Power Series - Interval Convergence:
• Power Series - Finding...
Power Series - Derivatives & Integrals:
• Power Series - Differe...
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OCT ur the best, but you should re-record this video since it's simply causing too much confusion with everyone. More time is spent scrolling to the comments and getting confused there too (since some are arguing against the errors). It's almost easier to just look for another video in this case.
taking the inverse of an inequality at 5:25 should flip the sign since they are both the same sign
yuuup right
yeah his real mistake was writing the inequality wrong to begin with - should be > not
@@zachb9440 He didn't do any mistakes in this video, let's see this with basic numbers 2 < 4 if we take inverse it will be 1/4 < 1/2, which is not the same case in the video. In the video, the case is 1/2 < 1/3 and if we take inverse it will be 2
@@MohcenMaher He didn't do any mistakes in this video, let's see this with basic numbers 2 < 4 if we take inverse it will be 1/4 < 1/2, which is not the same case in the video. In the video, the case is 1/2 < 1/3 and if we take inverse it will be 2
@@slasargol1390 Tam olarak anlamadığım nokta. 1/2
Why wasn't the direction of sign
The signs should have flipped.
Many people have noticed this, and the general question may be.
How did OCT still get the correct answer despite this mistake, and what exactly happened?
The issue appears to be, the value of the remainder was plugged into the inequality, as if Rn = 0.0001
Remember it was not originally, the remainder naturally needs to be LESS than 0.0001 to be accurate to four decimal places.
So the inequality should have been set out with Rn
I spotted that too and was confused at first
This comment section points out so many errors in this video that im scared to watch the video😂
They all lead to the same error, so if you take mind of that, you should be able to look at the video just fine.
ファッカーマスター ikk i was being over dramatic
What do you do if you have an n! as the denominator?
Death Punch gotta plug in numbers. That’s all you can do
n! = n(n-1)!
Did you got the answer? I'm thinking (n+1)!
I dont like this example because non of my problems are over 1 so the multiply by -1 power doesnt do anything.
Your videos are usually really helpful but this one isn't unfortunately, at least in my case :(
So n represents the interger where the series converges?
nope, it's where the partial sum of the series is correct to the infinite sum up to a certain percentage value
Thank you so much, besides the errors which were annoying and which I had to read through the comments to figure out, this video was very helpful. I had no clue how to do these types of problems and I have a midterm tomorrow, but now I think I understand how to do them!
I'm not gonna lie, I probably now have the chance of passing bc of you
Was it always this easy?
Just realized man🙂🙂
Fucking phd profs , make proof look hardass shit
It will be easy thanks to the lecturer of this channel 👌💎😁
is the error not suppose to be 0.005?
You have incorrect calculations in 2 steps. First, a(n+1) must be less than 0.001. Second, when you raise both sides to the negative 1 power, or taking the reciprocal of both sides, you must invert the inequality sign. You missed the inequality sign twice, so your answer is correct, but that doesn't make your argument valid.
I agree with you. +1
Could you please explain why a(n+1) must be less than 0.001? Since the a(n+1) is the first term we leave off the summation (which, therefore, has the greatest absolute value of remaining left-off terms), doesn't make sense that our error term (0.001) is less than or equal to a(n+1)? However, I do agree with your second point, so the provided answer should have the opposite sign. 31 >= n
So, why aren't we flipping the inequality sign whenever we take the inverse? we know that both values have to be positive
Exactly my question also
You are supposed to flip the inequality only when you either multiply or divide by -1. Anything else doesn't change the inequality.
Pratham Patil Taking the reciprocal also flips the sign, if you do the math and cross multiply then youll see why
@@Marcosss-7-years-ago How would this make sense in the context of the problem? N
@@anasasim3856 you guys must be kidding. Just apply to (1/2)>(1/3). Take the power -1 for both sides and boom! the answer should have been 2
I have no idea how am I supposed to do this while I'm not allowed to use calculator
In my experience, they will cook the numbers. On his last example he does .0001 but realistically you will have to do .001 so 1/.001 = 1000 and the cube root of 1000 is 10= n+1 so then n=9 and then you can do nine sums because 1/1= 1 1/2 squared = 1/4 and so on nine times. Add them all up in your head and you wont have to simplify! Best of luck!
SchmidtyBangBang thank u
In Maclaurin expansion of Sinx, find the remainder term after 3 term and find the value of θ ?
🔊🔊🔊
I am confused, when trying to prove 1/n^2 is convergent, you cannot perform divergence test.. the divergence test only confirms if it is divergent if it is not equal to 0 and is inconclusive otherwise. I'm not sure how that can be applied in this problem unless someone wants to inform me on something I'm missing?
Additionally, why were the comparison signs not flipped in the calculations?
JanAcademy you can use p series to prove that it is convergent
he did the "alternating series test", not the "divergence test"
@@charlesdonofrio9044 he clearly stated "so it passes the divergence test" ?
@@zaidalsalihi9301 the alternating series test states: the FIRST rule that the series An must pass is "the divergence test of An must equal 0" even tho when the divergence test is 0 it is inconclusive the alternating series test dosent care that its inconclusive it just wants the divergence test equal to 0 :) to be able to proceed. Furthermore even tho it is inconclusive that just means it CAN diverge OR it CAN converge either or but if we get any other answer other then 0 we automatically know it diverges with no chance of convergence. hope this made sense
enequality must be flipped.
How did you know to do it correct to 2 decimal places? Why not, 10, 15, or 11?
what if our Sn is Ssub567? is there a way to get that partial sum easily?
Do you also have a video on solving rubics cube? ur voice resemble to that kid
Tf? R u high
4:32 I don't think you can raise -1. The inequality flips
You should have flipped the less than or equal to sign for a greater than or equal to sign when adding "to the power of minus one" on both sides
nice video 🤽♂️
At 5:27 if I get n
Izco Mdz you would round up to 5 because 4 would still not provide an adequately accurate answer
make it 5, always choose the closest, greater integer!
Professor Organic Chemistry Tutor, thank you for another exceptional video/lecture on the Alternate Series Estimation Theorem in Calculus Two. I will review this topic for a clear and deep understanding of the Alternate Series Estimation Theorem and its error estimation. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
winton overwat
@@ngamer7202 winton
There is no way i could do this in my calc class because they want and error of .00005 and we cant use a calculator
To do the math without a calculator, convert the decimal to a fraction. For example, rewrite .00005 as 5/100,000 which simplifies to 1/20,000. Much easier to deal with than decimals.
mcmaster squad?
@@riteshpatel4767 yea boi
how did you get that for an+1?? can you please start showing your algebraic steps
oh its cause you solved for bn+1.....
At 5:23 if n can only be an integer, then wouldn't truncation i.e. rounding down also work as well? If not then why?
and at 0:00, you can see that if you scroll down in the comments, there is no need for yours
I know that the time in which you needed this answer is long passed but its because if you need at LEAST 7.23 terms then 7 terms would not satisfy that need, so you round up to 8
how do we choose the error? why 0.001 instead 0.00(and any other digit)
its the smallest number. ik its a late reply its for other people who are also wondering