For those of you who are curious about how he proved R4 < 0.04. He grouped the terms in pairs again (-1/36 + 1/49) + (-1/64 + 1/81) ... and realized that this term must be negative because each pairs sum is negative. Then he realized R4 = 1/25 + that negative number. This meant R4 < 1/25. Hence R4 < 0.04. He also realized that R4 couldn't be negative because he proved earlier that R4 > 0. Very clever Sal! Great work!
Matthew, That's all very well, but it rests on a crucial assumption, that the infinite series that has been bracketed has an even number of terms, and that the last terms in the series is 1/[some odd square]. Suppose this isn't true? Suppose, instead, that R4 is -1/36 +(1/49 - 1/64) + (1/81 - /100).... Then we are faced with the question of whether all that bracketed stuff, call it r5, which is clearly positive, is more or less than 1/36. Wey-yull, it's the series ~1/209 + ~1/426 +~1/757 ... All positive numbers, and sadly, not in any fixed proportion to one another... :-) Any thoughts? I think David Hilbert said something like "The craziest damn things happen when you mess around with infinities." Cheers, -dlj.
This is a crazy way to think about this. Just, take your truncated derivative's value and that's your answer? Why doesn't my teacher explain it like this? Is there a point i could miss on the free response for doing this? (I'm in AP Calc BC, exams are coming, and I've been studying over Spring break :P )
I've got a silly question here.. what if I put parentheses around -1/36 and 1/ 49 and so on so that every brackets after 1/25 would be negative? How to prove then that the absolute value of the sum of those brackets is smaller than 1/25 ?
can't put parenthesis around -1/36 and -1/49 because it changes the values; it's - (1/36 - 1/49). You have to distribute the negative and it'll become - 1/36 + 1/49 in brackets (a negative value) which u add onto 1/25. Since the value inside the brackets is negative, you're still subtracting from 1/25 and the bound is still < 0.04.
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For those of you who are curious about how he proved R4 < 0.04.
He grouped the terms in pairs again (-1/36 + 1/49) + (-1/64 + 1/81) ... and realized that this term must be negative because each pairs sum is negative.
Then he realized R4 = 1/25 + that negative number. This meant R4 < 1/25. Hence R4 < 0.04.
He also realized that R4 couldn't be negative because he proved earlier that R4 > 0.
Very clever Sal! Great work!
Matthew,
That's all very well, but it rests on a crucial assumption, that the infinite series that has been bracketed has an even number of terms, and that the last terms in the series is 1/[some odd square].
Suppose this isn't true?
Suppose, instead, that R4 is -1/36 +(1/49 - 1/64) + (1/81 - /100).... Then we are faced with the question of whether all that bracketed stuff, call it r5, which is clearly positive, is more or less than 1/36. Wey-yull, it's the series ~1/209 + ~1/426 +~1/757 ... All positive numbers, and sadly, not in any fixed proportion to one another...
:-)
Any thoughts? I think David Hilbert said something like "The craziest damn things happen when you mess around with infinities."
Cheers,
-dlj.
2:30, quick maths
Thanks, great video.
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What a nice explanation! Thank you!
This is a crazy way to think about this. Just, take your truncated derivative's value and that's your answer? Why doesn't my teacher explain it like this? Is there a point i could miss on the free response for doing this? (I'm in AP Calc BC, exams are coming, and I've been studying over Spring break :P )
I've got a silly question here.. what if I put parentheses around -1/36 and 1/ 49 and so on so that every brackets after 1/25 would be negative? How to prove then that the absolute value of the sum of those brackets is smaller than 1/25 ?
can't put parenthesis around -1/36 and -1/49 because it changes the values; it's - (1/36 - 1/49). You have to distribute the negative and it'll become - 1/36 + 1/49 in brackets (a negative value) which u add onto 1/25. Since the value inside the brackets is negative, you're still subtracting from 1/25 and the bound is still < 0.04.
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You went quite a distance for a request like that, seeing as how you're in a Calc 2 subject comment thread.
@@ghlegend195 some high schools in NYC give linear algebra after AP Calc BC