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freckles

  • one year ago

Hey! I want to think about evaluating: \[\int_0^\infty \lfloor x\rfloor e^x dx \text{ where } \lfloor . \rfloor\text{ is greatest integer function }\]

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  1. freckles
    • one year ago
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    trying to figure out the floor function thingy

  2. freckles
    • one year ago
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    got it that is a long command for the floor function

  3. freckles
    • one year ago
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    anyways... \[n \le x<n+1 \text{ then } \lfloor x \rfloor=n \text{ where } n \text{ is integer }\] now I think I want to do integration by parts...

  4. freckles
    • one year ago
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    trying to think how to integrate the greatest integer function

  5. freckles
    • one year ago
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    \[\int\limits_0^\infty \lfloor x \rfloor e^x dx \\ \int\limits_0^1 0 \cdot e^x dx+\int\limits_1^2 1 \cdot e^x + \int\limits_2^3 2 \cdot e^x + \cdots + \int\limits_n^{n+1} n \cdot e^{x} dx +\cdots \]

  6. freckles
    • one year ago
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    \[\sum_{n=1}^{\infty} \int\limits _n^{n+1} n \cdot e^x dx\]

  7. freckles
    • one year ago
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    \[\sum_{n =1}^\infty n \int\limits _n^{n+1} e^x dx \\ \sum_{n=1}^\infty n e^x|_n^{n+1} \\ \sum_{n=1}^\infty n[e^{n+1}-e^{n}]\]

  8. freckles
    • one year ago
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    actually I guess I don't need integration by parts

  9. ganeshie8
    • one year ago
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    thats looks very neat, but it doesn't converge right

  10. freckles
    • one year ago
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    yeah wolfram says it doesn't diverge I thought this was suppose to converge those for some reason did i make a mistake above

  11. freckles
    • one year ago
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    I mean it does diverge sorry

  12. freckles
    • one year ago
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    you know what maybe if the exponent on the x was negative it would have converged

  13. ganeshie8
    • one year ago
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    yeah exponential overtakes any polynomial, floor(x) is cheaper than a linear polynomial

  14. freckles
    • one year ago
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    did something wrong

  15. freckles
    • one year ago
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    oops didn't mean to delete that without copying it

  16. freckles
    • one year ago
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    \[\sum_{n=1}^{\infty}n(e^{-(n+1)}-e^{-n})\] whatever I know this converges I just need to find the sum now I was thinking this maybe a telescoping series

  17. ganeshie8
    • one year ago
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    \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx \\ =\sum_{n=1}^{\infty} \int\limits_{n}^{n+1} n e^{-x} dx \\ =\sum_{n=1}^{\infty} -n e ^{-x}|_n^{n+1} \\ \\ =\sum_{n=1}^{\infty} -n[e^{-(n+1)}-e^{-n}] \\ \text{ ratio test: } \\ \lim_{n \rightarrow \infty} \frac{-(n+1)[e^{-(n+2)}-e^{-(n+1)]}}{-n [e^{-(n+1)}-e^{-n}]}\] \[\lim_{n \rightarrow \infty}-(1+\frac{1}{n}) \frac{ e^{-(n+2)+(n+1)-e^{-(n+1)+(n+1)}}}{e^{-(n+1)+(n+1)}-e^{-n+(n+1)} } \\ -(1+0) \frac{e^{-1}-e^{0}}{e^{0}-e^{1}} =-1 \frac{e^{-1}-1}{1-e} =-1 \frac{1-e}{e-e^2} =\frac{e-1}{e-e^2} \\ =\frac{e-1}{e(1-e)}=\frac{-1}{e}\]

  18. freckles
    • one year ago
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    \[(e^{-2}-e^{-1})+2(e^{-3}-e^{-2})+3(e^{-4}-e^{-3})+4(e^{-5}-e^{-4}) +\cdots \] not exactly a telescoping series the way it is written at last

  19. ganeshie8
    • one year ago
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    i pulle that from ur deleted replies

  20. freckles
    • one year ago
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    thanks @ganeshie8 now people can know what problem I'm talking about I'm trying to evaluate: \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx\]

  21. ganeshie8
    • one year ago
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    mimicing ur earlier work, \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx = (e-1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \]

  22. xapproachesinfinity
    • one year ago
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    so the first floor (x) e^x diverge or not?

  23. freckles
    • one year ago
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    i showed it diverge so I think it diverge (unless I made some mistake)

  24. freckles
    • one year ago
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    well wolfram showed my sum diverged :p

  25. xapproachesinfinity
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @freckles \[\sum_{n =1}^\infty n \int\limits _n^{n+1} e^x dx \\ \sum_{n=1}^\infty n e^x|_n^{n+1} \\ \sum_{n=1}^\infty n[e^{n+1}-e^{n}]\] \(\color{blue}{\text{End of Quote}}\) from tis result ?

  26. freckles
    • one year ago
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    I could have use the ratio test to come up with that know

  27. freckles
    • one year ago
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    ues from that last line

  28. xapproachesinfinity
    • one year ago
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    ah ok i guess ration test can testify divergence

  29. ganeshie8
    • one year ago
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    we can simply use the limit test

  30. xapproachesinfinity
    • one year ago
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    oh yes! that limit does not go to 0

  31. ganeshie8
    • one year ago
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    \[\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{-nx}=\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{-nx}= \frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\cdots\]

  32. freckles
    • one year ago
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    and we know \[\frac{d}{dx} \frac{e^{-x}}{1-e^{-x}} \\ \frac{d}{dx} \frac{1}{e^x-1} \\ \frac{d}{dx}(e^x-1)^{-1} \\ -1(e^x-1)^{-2}(e^x) \\ - \frac{e^x}{(e^x-1)^2}\] but we should evaluate this at x=1 since we had that we really wanted to evaluate \[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx = (e-1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \] \[(e-1) \frac{-e}{(e-1)^2}=\frac{-e}{e-1}\] but I think we are to get 1/(e-1)

  33. ganeshie8
    • one year ago
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    Oh right \[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{-x} dx =-\sum_{n=1}^\infty n[e^{-(n+1)}-e^{-n}] = \frac{e-1}{e}\sum\limits_{n=1}^{\infty} \frac{n}{e^n}\]

  34. ganeshie8
    • one year ago
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    algebra mistakes

  35. freckles
    • one year ago
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    thanks @ganeshie8 ok I was sorta failing at the algebra a little too but I see it now

  36. freckles
    • one year ago
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    I can't remember who gave me this problem but @Loser66 this is also on your gre practice exam if you want to look at this.

  37. ganeshie8
    • one year ago
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    i remember working this problem differently sometime back here in openstudy but google isn't helping at the moment..

  38. freckles
    • one year ago
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    And I bet you have answered so many questions since this that going back through your old questions is horrifying and it also doesn't work to well sometimes. Sometimes it automatically throws me out of the looking through my old question thing. Not sure if you know what I mean.

  39. Loser66
    • one year ago
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    Thanks so mmmmmmmmuch

  40. xapproachesinfinity
    • one year ago
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    truly i could not see how you threw differentiation in there @gabylovesu how is that sum equal to that differentiation ?

  41. xapproachesinfinity
    • one year ago
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    @ganeshie8

  42. xapproachesinfinity
    • one year ago
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    wrong tagging lol

  43. anonymous
    • one year ago
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    Have you tried parameterizing and applying the Laplace transform? I'm thinking something along the lines of \[I(s)=\int_0^\infty \lfloor x\rfloor e^{-sx}\,dx=\mathcal{L}\{\lfloor x\rfloor\}\] The transform definitely exists because \(\lfloor x\rfloor\) is piecewise continuous and of exponential order.

  44. ganeshie8
    • one year ago
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    \[\color{red}{\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=-\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{-nx}}=-\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{-nx}= -\frac{d}{dx}\frac{e^{-x}}{1-e^{-x}}=\cdots\] I think it would be easier to make sense of it by differentiating \(\large \color{red}{e^{-nx}}\) and seeing that you get back the starting expression...

  45. freckles
    • one year ago
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    hey @SithsAndGiggles I haven't thought of that but I would be interested in seeing that way if you wanted to show that way. I honestly I haven't seen any laplace transform action in like 10 years so I kind of forgot all of that. :p

  46. anonymous
    • one year ago
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    The idea would be to express \(\lfloor x\rfloor e^{-x}\) in terms of the step function \[\theta(x-c)=\begin{cases}1&\text{for }x\ge c\\0&\text{for }x<c\end{cases}\]

  47. xapproachesinfinity
    • one year ago
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    ooh i see it now

  48. anonymous
    • one year ago
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    \[\lfloor x\rfloor e^{-x}=\sum_{c=1}^\infty ce^{-x}\Bigg(\theta(x-c)-\theta(x-c-1)\Bigg)\] So \[\begin{align*}\mathcal{L}\left\{\lfloor x\rfloor\right\}&=\int_0^\infty \lfloor x\rfloor e^{-sx}\,dx\\ &=\int_0^\infty \sum_{c=1}^\infty ce^{-sx}\Bigg(\theta(x-c)-\theta(x-c-1)\Bigg)\,dx\\ &=\sum_{c=1}^\infty c\int_0^\infty e^{-sx}\Bigg(\theta(x-c)-\theta(x-c-1)\Bigg)\,dx\\ &=\sum_{c=1}^\infty c\mathcal{L}\{\theta(x-c)-\theta(x-c-1)\}\\ &=\sum_{c=1}^\infty c\frac{e^{-cs}-e^{-(c+1)s}}{s} \end{align*}\] Setting \(s=1\) gives the series @ganeshie8 was working with. Nothing new I suppose :P

  49. freckles
    • one year ago
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    does seem similar but still pretty @SithsAndGiggles anyways this is making me hungry so peace and thanks for the fun

  50. freckles
    • one year ago
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    I will come back later though to analyze more deeply what you said

  51. anonymous
    • one year ago
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    As for the actual result: \[\sum_{c=1}^\infty c(e^{-c}-e^{-(c+1)})=\sum_{c=1}^\infty c\left(\frac{1}{e^c}-\frac{1}{e^{c+1}}\right)=\frac{1}{e}+\frac{1}{e^2}+\frac{1}{e^3}+\cdots=\frac{1}{1-\frac{1}{e}}-1\]

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