Calculus BC - 10.12 Lagrange Error Bound

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[@Thyzt]

Exam tomorrow. Our teacher didn't even bother teaching this too much cos it was a bit tricky LOL so thank you so much!

[@sportmixes9899]

This was by far the hardest section for me to understand

[@SS-pl5ds]

There's no nice way to say this, but I'm finally happy to get rid of you!! Kidding. Your amazing calculus videos are the reason I'm feeling confident about tomorrow's exam. Cannot thank you enough for what you do.

[@stress4087]

how did u do if you dont mind me asking? my AB teacher told me he'd help me with the BC stuff and he didnt so im stuck self studying these videos rn LOL

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new mr. bean video just dropped

[@rishivaddiraju9803]

yeahhhhh

[@schqrr]

i love mr bean

[@christopherhui4109]

Why we need to put x=3 when calculate the error bound?

[@teesuu953]

when do you know to use the equal to sign or to not use the equal to sign?

[@user-.001]

Thankyou for clear explanation of error bound. next week monday is my AP Cal BC D-Day.. Thank you for making this video!

[@arjunyeleshwarapu]

fr actually gonna be the worst day of the year

[@abhiramkidambi6666]

Day of the exam lmao gl

[@arjunyeleshwarapu]

@@abhiramkidambi6666 gl

[@arjunyeleshwarapu]

@@abhiramkidambi6666 how was it

[@mycommentisdumbbut]

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[@JahreadyKnow]

I don't get why you didn't find the max value of sin x WITHIN the interval of 0 and 0.2. You basically just explained it as "collegeboard says use 1 for max".

[@rockk3y413]

Yea same. Plus, it was a negative sign. Would that be the max value of zero or is it maximum magnitude? I didn't think it was the best explanation.

[@ragcrafter8127]

Not true a sin function fluctuates between -1 and 1 on the y axis or output in this case since it’s negative sin when normal sin is negative 1 -sin is at its max here if you wanted the max on the interval it would just be 0 since negative sin is strictly decreasing from 0 to 0.2 sin of 0 is just 0 therefore the max on the interval is 0 since all other values are negative. But sometimes it can be hard to tell which end point or if some value between the end point is the max and there you would need to use the first derivative test. Or you can just know the max value of any normal sinx or cosx is just 1

[@catmom1708]

The absolute value of sine from 0 to 0.2 is increasing. Sin(0.2) is the max within the interval 0 to 0.2. I used the sin(0.2) as the max and get the error bound 5.29785x10^7. The AP test is testing the understanding of Lagrange bound. Either using 1 or sin(0.2) should be considered correct. Using sin(0.2) is more accurate. But it is for a max value if sine, 1 is max without thinking the given interval.

[@shuofengpang1328]

One thing that I don't understand is that why would you use 1 for the max of sin, not 1/2. given that sin(0.2)

[@TheAlgebros]

You don't have to stress about the actual value of sine. For the error bound, you get to simplify things for yourself and just consider the maximum value of sine will never be larger than 1. When the tests are graded, you just need to show that you understand that. You could make it even more restrictive by using sin(0.2), but that's not necessary to use the Lagrange error bound.

[@justinkenneally4988]

@@TheAlgebros So can you use 0.2 and still not lose points?

[@marcocruz6210]

@@justinkenneally4988 yes, hes saying you can use a smaller number to get a closer bound however it is only making it harder for yourself since using 1 as bound is acceptable since an sin or cos will never exceed 1

[@muralidharansomasundaram1509]

Since sin(x) < x, can't we use sin(0.2) < 0.2?

[@shuofengpang1328]

@@muralidharansomasundaram1509 but we don't know what is sin(0.2) with out a calculator.

[@kwamp684]

the best video on this so far. Thank you so much. Im studying myself for the exam and ive been mostly using a book but the authors got lazy near the end. Thanks alot!

[@terryzhao3385]

Clear explanation. Thank you so much!

[@TheAlgebros]

You're very welcome!

[@brianlipiano7113]

i think that since you didnt put the absolute value symbol around some of these they could be marked wrong by someone from college board?? im not sure, can anyone confirm?

[@nara5344]

taking 1as the maximum value for -sin(z) knowing that we have an intervale between 0 to0,2 it's going to width R why take 1??yes is never be larger that 1 but in the intervalle 0 to 0,2 is never going to be bigger than sin0,2

[@TimothyTomtheThird]

It’s taking about the y cord instead of the x cord

[@Acoustic_Mutale]

Best Explanation on the Tube so far. Kudos from Zambia 🇿🇲 what would you like me to send for you from Zambia?

[@ramongonzalezfernandez8904]

If i am to plug in the values 0 and 0.2 for -sin(x), I get -sin(0) and -sin(0.2), which equal 0 and ~ -0.2 respectively. Would i take the value with the largest magnitude, (i.e ~0.19), or the max (0)? When i take the max, i get an error of 0 which is nonsense

[@TheAlgebros]

You don't have to stress about the actual value of sine. For the error bound, you get to simplify things for yourself and just consider the maximum value of sine will never be larger than 1. When the tests are graded, you just need to show that you understand that. You could make it even more restrictive by using the values between -sin(0) and -sin(0.2), but that's not necessary to use the Lagrange error bound.

[@ramongonzalezfernandez8904]

@@TheAlgebrosBut if i were to go for the more restrictive error bound, would i use -0.2 as the maximum, or 0?

[@sadcat7031]

You would use 0.2 since everything is in an absolute value, negative does not matter, meaning 0.2 is the largest possible value. It seems that he is implying that on the AP exam, you can just say 1 and be kinda close, but idk

[@thedxrito]

is there a reason why the second problem isn't written as maclaurin? its a taylor polynomial centered at c=0, right?

[@TheAlgebros]

It is definitely a Maclaurin polynomial. Sometimes the AP test will just give you the information like what we did in in that problem so you don't have to know that it is the same thing as a Maclaurin. So we are just trying to give you a variety of ways they might word it. But yes, we could have also just stated that it is a Maclaurin polynomial.

[@thedxrito]

@@TheAlgebros thank you for answering, makes sense! love your videos :)

[@장재혁-c5y]

my class just finished the last unit and the ap test is next week😂

[@keyz6380]

One thing I don't understand is that if we can find the exact error, why do we need an error bound?

[@zappist751]

Calculus was dumb in the 1700s

[@samueldarenskiy6893]

Don't forget to write your absolute value signs next time

[@TheAlgebros]

No problem! And you don't forget your punctuation at the end of the sentence next time. We'll both be on fleek.

[@samueldarenskiy6893]

@@TheAlgebros I didn't mean this as an insult, English isn't my first language.

[@TheAlgebros]

@@samueldarenskiy6893 It's all good! We wish you an amazing day ;)

[@1lemn]

Thanks......can you solve for us accuracy of cubicRoot(x) in the interval[7,9] while centered at x=8 for a degree 2 expansion??

[@kibme5189]

Isn't the answer of the last question the 4th order?

[@preetamnalla8072]

should be

[@catmom1708]

N=5. It used n+1 in the expression. It is not n+1=5. The table shows when n=5, y=0.6568.

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