This is an incredibly helpful video! I sent this video to one of my students for extra practice, and he noted that, in question 19, you need n = 20 for the error to be less than the given quantity, but the question asks for the value of k. Since n = 20 is the first term left off, would that mean that the k value is 19?
Could you explain the intuition for question 11 - was it just seeing that there is a 2x in the numerator and understanding that it is a result of the chain rule being applied, so you know to integrate it, find that series, then take the derivative of that series to get the series for the original term provided?
When I'm trying to write a series for a given function (especially one that has a constant minus a variable in the denominator) I always start with the thought that the series might just be geometric. If the function isn't immediately giving me a geometric series I always consider that the derivative or antiderivative might give a function that I could write a geometric series for. At that point it's a question of should I integrate or find a derivative. (One thing that can complicate this is the 2x could just be a geometric times 2x or it could be that the 2x does come from the chain rule.) --I kind of think everything I've said here is basically what you said but more words...hope something in here is a little more insight though!
you doing limit comparison? if you do that you get lim 1/(2n+3)/(1/n) to infinity is 1/2 which is positive and finite, so both do the same thing. even if you flip it, you'd get a limit of 2 which is also positive and finite. (unless someone forces me to use direct comparison, i'm using limit comparison 100% of the time) good luck!!!
Good question! In that problem we're not given the nth term of the series, we're given the the formula for the nth partial sum. So, for example, if you use 1 term, you'll get an approximation of 3/9. If you use 2 terms, you'll get an approximation of 6/14, if you use 3 terms, you get 9/19, etc. So if you let the number of terms go to infinity you'll get the sum of the series: 3/5 in this case. If you were given the nth term of the series then the limit not being 0 would make the series diverge. Hope this helps!
the general term of a taylor series is (nth deriv at a)(x-a)^n / n!, so in this case we just look for (x-a)^2 and know the coefficient is f''(a)/2! and solve from there. This is the way to do this particular type of problem. If you imagine (or do) find f''(x)'s approximation by taking the derivative twice, you'll see that everything that comes with (x-a)^3 still has an (x-a) next to it, so subbing in a will cancel it out at the level of the second derivative. Hope this helps, I don't think I explained it well...
It's a compilation of questions from or based on exams from many years. All the questions are still very relevant to the exam. Good luck with your studies!
This is an incredibly helpful video!
I sent this video to one of my students for extra practice, and he noted that, in question 19, you need n = 20 for the error to be less than the given quantity, but the question asks for the value of k. Since n = 20 is the first term left off, would that mean that the k value is 19?
thanks! yeah, honestly i mess that kind of question up way more often than i should...
@@turksvids Yep, I make that kind of mistake all the time. But seriously, really helpful material and video. Thanks again!
For number nineteen wouldn’t the answer also be nineteen because the error is the n plus one term?
That’s what I was thinking
Yeah, same here!!
exactly what i was saying.. i believe you're correct, right?
I also was thinking this @turkvids could you please answer
Yes, that's correct. There's another comment where I thought it had been addressed. Sorry!
THIS IS AMAZING THANKS!
For 20, d can be instantly ruled out because for it to equal k, it would need to converge which is what part c is
Could you explain the intuition for question 11 - was it just seeing that there is a 2x in the numerator and understanding that it is a result of the chain rule being applied, so you know to integrate it, find that series, then take the derivative of that series to get the series for the original term provided?
When I'm trying to write a series for a given function (especially one that has a constant minus a variable in the denominator) I always start with the thought that the series might just be geometric. If the function isn't immediately giving me a geometric series I always consider that the derivative or antiderivative might give a function that I could write a geometric series for. At that point it's a question of should I integrate or find a derivative. (One thing that can complicate this is the 2x could just be a geometric times 2x or it could be that the 2x does come from the chain rule.) --I kind of think everything I've said here is basically what you said but more words...hope something in here is a little more insight though!
This was super helpful!
i didn't get the question 4. my benchmark series = 1/n and it diverges. but the comparison rule says if 0
you doing limit comparison? if you do that you get lim 1/(2n+3)/(1/n) to infinity is 1/2 which is positive and finite, so both do the same thing. even if you flip it, you'd get a limit of 2 which is also positive and finite. (unless someone forces me to use direct comparison, i'm using limit comparison 100% of the time) good luck!!!
For question 11 when you’re writing out the terms what happened to the factorial denominators? (0!, 1!, 2!, 3! Etc)
For question 19, wouldnt the answer be b? The error is the value of the next term, which is the twentieth. And so k would be the one before 20, or 19?
check the other comments for a discussion about this. (i agree)
Sorry, didn’t see that comment when i looked through them last time.
no problem, i don't expect people to read the comments, really, but it's easier to point to them than answer again! good luck with your studies!
for question 8, wouldn't the series diverge because the limit of the sequence does not equal 0?
Good question! In that problem we're not given the nth term of the series, we're given the the formula for the nth partial sum. So, for example, if you use 1 term, you'll get an approximation of 3/9. If you use 2 terms, you'll get an approximation of 6/14, if you use 3 terms, you get 9/19, etc. So if you let the number of terms go to infinity you'll get the sum of the series: 3/5 in this case. If you were given the nth term of the series then the limit not being 0 would make the series diverge. Hope this helps!
@@turksvids Thank you, that clears the problem up. Great video by the way!!
at 3, why didn't you count the ^2 parts at the (x+2)^3/252? Shouldn't the x^2 part be (6/252-3/25)x^2?
the general term of a taylor series is (nth deriv at a)(x-a)^n / n!, so in this case we just look for (x-a)^2 and know the coefficient is f''(a)/2! and solve from there. This is the way to do this particular type of problem. If you imagine (or do) find f''(x)'s approximation by taking the derivative twice, you'll see that everything that comes with (x-a)^3 still has an (x-a) next to it, so subbing in a will cancel it out at the level of the second derivative. Hope this helps, I don't think I explained it well...
For 5, couldn’t you say that this is a geometric series with ratio (2x+1)/4 and the absolute value of the ratio must be less than 1?
Yup. That works!
great video
For 4 isn't it 4
Four doesn't actually alternate because you end up with (-1)^(2n) = 1, so there's no alternating.
@@turksvids ohhhh thank you very much
What year is this from?
It's a compilation of questions from or based on exams from many years. All the questions are still very relevant to the exam. Good luck with your studies!
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