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Cal 1 homework. No one on my course can do it. watch

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    Group chats with 300 folks in and they can't do it.
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    Question 2c....

    You have no idea how many people you could make happy
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    (Original post by GeorgeMath)
    Question 2c....

    You have no idea how many people you could make happy
    Use integration by parts - the integrand is a product of two functions.
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    (Original post by GeorgeMath)
    Question 2c....

    You have no idea how many people you could make happy
    You must be on one weakass course because this took me a minute and I did far harder questions at school.

    Just integrating by parts immediately gives you I(n) = a/(n+1) I(n+1) which gives you the inductive step.
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    (Original post by GeorgeMath)
    Question 2c....

    You have no idea how many people you could make happy
    You derived the reduction formula in 2b

    I_n = \frac n a I_{n-1}

    And you have I_0. How can you use this?

    Also imo induction isn't really needed. If you evaluate I_1, then I_2 etc. you'll see what's happening.
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    Everyone: when someone posts "plz help, no1 on my course can do this" and posts an easy question - consider that you're probably being "click-baited" into doing someone's homework for them.

    I wouldn't rise to it at all until they demonstrate some effort on their side but for God's sake don't end up doing the whole question because your so eager to prove you're brighter than 300 people who probably don't exist.
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    (Original post by DFranklin)
    Everyone: when someone posts "plz help, no1 on my course can do this" and posts an easy question - consider that you're probably being "click-baited" into doing someone's homework for them.

    I wouldn't rise to it at all until they demonstrate some effort on their side but for God's sake don't end up doing the whole question because your so eager to prove you're brighter than 300 people who probably don't exist.
    I gathered, though "You have no idea how many people you could make happy" made me chuckle a bit.
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    (Original post by DFranklin)
    Everyone: when someone posts "plz help, no1 on my course can do this" and posts an easy question - consider that you're probably being "click-baited" into doing someone's homework for them.

    I wouldn't rise to it at all until they demonstrate some effort on their side but for God's sake don't end up doing the whole question because your so eager to prove you're brighter than 300 people who probably don't exist.
    I do realize that and I don't really care whether it's one person who's stuck or 300. Also I deliberately just gave the recurrence relationship at the heart of the proof rather than the full derivation.
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    (Original post by _gcx)
    You derived the reduction formula in 2b

    I_n = \frac n a I_{n-1}

    And you have I_0. How can you use this?

    Also imo induction isn't really needed. If you evaluate I_1, then I_2 etc. you'll see what's happening.
    Well, induction is needed if you actually want a proof, rather than a guess. And unless I'm being silly, all the necessary work took place in the second step, so I don't see why people are suggesting IBP :confused:
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    (Original post by atsruser)
    Well, induction is needed if you actually want a proof, rather than a guess. And unless I'm being silly, all the necessary work took place in the second step, so I don't see why people are suggesting IBP :confused:
    Spoiler:
    Show



    I think the pattern I_n = \frac n a I_{n-1} = \frac {n(n-1)} {a^2} I_{n-2} = \ldots is obvious/clear enough to avoid the use of induction in a proof but if the OP's question is requiring the use of it, rather than just suggesting it, then I'd go with it. (Though induction would only take an extra line or so, since the base case has been completed in a previous part)



    Yup, it's a show that question too, on the previous part, so even if the OP couldn't complete 2b, they could've made an attempt at 2c.
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    A group chat with 300 people all trying to do the same calculus problem sounds fun
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    Is your course English Lit or something? Actually, scratch that, out of 300 English Lit students I'd imagine one of them has done A-level maths to a reasonable standard and could do this..
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    (Original post by math42)
    Is your course English Lit or something? Actually, scratch that, out of 300 English Lit students I'd imagine one of them has done A-level maths to a reasonable standard and could do this..
    tbf it is calc I which is the first time integration would be introduced so I can somewhat see students having difficulty.
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    (Original post by _gcx)
    tbf it is calc I which is the first time integration would be introduced so I can somewhat see students having difficulty.
    Right. This may be maths for biologists or something - we can't assume that everyone will have seen induction before or even have done A level maths.
 
 
 
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