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Thread: Integration by Parts

  1. #16
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    Quote Originally Posted by mkeg1 View Post
    I am confused because the original problem gives the integral with n=0,1,2,.... then it asks for a recurrence formula for y sub k to y sub k-1 for k 1,2,3,.... am I still plugging in the same number?
    You are meant to get an expression for y_n in terms of y_{n-1} and n. Please go back and carefully review the thread. Everything is there, all you have to do is put together.
    Last edited by mr fantastic; Sep 27th 2010 at 08:39 PM. Reason: Fixed typo. The OP should know what was meant ....
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  2. #17
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    Quote Originally Posted by mr fantastic View Post
    You are meant to get an expression for y_n in terms of y_n and n. Please go back and carefully review the thread. Everything is there, all you have to do is put together.
    y_n should be in terms of y_{n-1} and n
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  3. #18
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    y_{n-1}=-t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt} for n=1,2,3....

    ?
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  4. #19
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    Remember that y_n = \int{e^{-t}t^{n}\,dt}.

    You have shown \int{e^{-t}t^n\,dt} = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}.


    So y_n = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}.


    Can you see that \int{e^{-t}t^{n-1}\,dt} is almost the same as your original integral - the only difference being that n is now n-1.

    Surely this is y_{n-1}.


    So what do you think y_n is in terms of y_{n-1}?
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  5. #20
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    y_{n-1}=-t^{n-1}e^{-t} + (n-1)\int{e^{-t}t^{n-2}\,dt}
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  6. #21
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    Quote Originally Posted by mkeg1 View Post
    y_{n-1}=-t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt} for n=1,2,3....

    ?
    Did you read post #12 at all? A lot of time is starting to be spent saying the same things. Please .... take some time to think about the probelm and what has been posted in this thread.
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  7. #22
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    Yes, I read post 12, I am not very good at math as you can see. Maybe I just need to spend a little more time evaluating the problem
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  8. #23
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    Quote Originally Posted by mkeg1 View Post
    y_{n-1}=-t^{n-1}e^{-t} + (n-1)\int{e^{-t}t^{n-2}\,dt}
    I suggest you read post number 19...
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  9. #24
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    Quote Originally Posted by Prove It View Post
    I suggest you read post number 19...
    y_n is a definite integral not an indefinite integral.
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  10. #25
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    y_n = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}

    Kind of a shot in the dark because I sill dont understand, but would i replace the integral part with y sub n-1?
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  11. #26
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    Quote Originally Posted by mkeg1 View Post
    since dv=e^{-t}dt i took the integral and got -(1/e^{t}loge)

    I know integral of e^x is e^x but doesn tthe negative sign change it?
    That "log" is the natural logarithm and log(e)= 1!

    I suspect you looked up the integral of a^x. You should know that \int e^x dx= e^x+ C without having to look it up. If you let u= -x, then du= -dx so that \int e^{-x}dx= -\int e^u du= -e^u+ C= -e^{x}+ C.
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  12. #27
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    Just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,



    ... is lazy integration by parts, doing without u and v.



    Now, of course, state y_{k} in terms of y_{k-1} and k.
    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  13. #28
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    Hey guys I really appreciate all the help.

    I understand the integration, I just dont understand what to do when applying y_nin terms of y_{n-1}
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  14. #29
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    Can you at least see the pattern in

    y_{1} = y_{0} - e^{-1}

    y_{2} = 2y_{1} - e^{-1}

    y_{3} = 3y_{2} - e^{-1}

    and write it as

    y_{k} = ... ?

    (or y_{n} = ... ?)

    I don't think any more is required than that. (And the starting value, for which, see above.)
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  15. #30
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    Quote Originally Posted by mkeg1 View Post
    y_n = -t^ne^{-t} + n\int{e^{-t}t^{n-1}\,dt}

    Kind of a shot in the dark because I sill dont understand, but would i replace the integral part with y sub n-1?
    No. For what seems like the one millionth time: y_n is a DEFINITE integral!

    \displaystyle y_n = \left[-t^ne^{-t}\right]_{0}^{1} + n \int_0^1 {e^{-t}t^{n-1}\,dt}.

    Now the integral on the right hand side can be replaced with y_{n-1}. The first term on the right hand side needs to be evaluated in the usual way.
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