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freckles
 one year ago
Hey! I want to think about evaluating: \[\int_0^\infty \lfloor x\rfloor e^x dx \text{ where } \lfloor . \rfloor\text{ is greatest integer function }\]
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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freckles
 one year ago
Best ResponseYou've already chosen the best response.3trying to figure out the floor function thingy

freckles
 one year ago
Best ResponseYou've already chosen the best response.3got it that is a long command for the floor function

freckles
 one year ago
Best ResponseYou've already chosen the best response.3anyways... \[n \le x<n+1 \text{ then } \lfloor x \rfloor=n \text{ where } n \text{ is integer }\] now I think I want to do integration by parts...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3trying to think how to integrate the greatest integer function

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\int\limits_0^\infty \lfloor x \rfloor e^x dx \\ \int\limits_0^1 0 \cdot e^x dx+\int\limits_1^2 1 \cdot e^x + \int\limits_2^3 2 \cdot e^x + \cdots + \int\limits_n^{n+1} n \cdot e^{x} dx +\cdots \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\sum_{n=1}^{\infty} \int\limits _n^{n+1} n \cdot e^x dx\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\sum_{n =1}^\infty n \int\limits _n^{n+1} e^x dx \\ \sum_{n=1}^\infty n e^x_n^{n+1} \\ \sum_{n=1}^\infty n[e^{n+1}e^{n}]\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3actually I guess I don't need integration by parts

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thats looks very neat, but it doesn't converge right

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yeah wolfram says it doesn't diverge I thought this was suppose to converge those for some reason did i make a mistake above

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I mean it does diverge sorry

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you know what maybe if the exponent on the x was negative it would have converged

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2yeah exponential overtakes any polynomial, floor(x) is cheaper than a linear polynomial

freckles
 one year ago
Best ResponseYou've already chosen the best response.3oops didn't mean to delete that without copying it

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\sum_{n=1}^{\infty}n(e^{(n+1)}e^{n})\] whatever I know this converges I just need to find the sum now I was thinking this maybe a telescoping series

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{x} dx \\ =\sum_{n=1}^{\infty} \int\limits_{n}^{n+1} n e^{x} dx \\ =\sum_{n=1}^{\infty} n e ^{x}_n^{n+1} \\ \\ =\sum_{n=1}^{\infty} n[e^{(n+1)}e^{n}] \\ \text{ ratio test: } \\ \lim_{n \rightarrow \infty} \frac{(n+1)[e^{(n+2)}e^{(n+1)]}}{n [e^{(n+1)}e^{n}]}\] \[\lim_{n \rightarrow \infty}(1+\frac{1}{n}) \frac{ e^{(n+2)+(n+1)e^{(n+1)+(n+1)}}}{e^{(n+1)+(n+1)}e^{n+(n+1)} } \\ (1+0) \frac{e^{1}e^{0}}{e^{0}e^{1}} =1 \frac{e^{1}1}{1e} =1 \frac{1e}{ee^2} =\frac{e1}{ee^2} \\ =\frac{e1}{e(1e)}=\frac{1}{e}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[(e^{2}e^{1})+2(e^{3}e^{2})+3(e^{4}e^{3})+4(e^{5}e^{4}) +\cdots \] not exactly a telescoping series the way it is written at last

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i pulle that from ur deleted replies

freckles
 one year ago
Best ResponseYou've already chosen the best response.3thanks @ganeshie8 now people can know what problem I'm talking about I'm trying to evaluate: \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{x} dx\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2mimicing ur earlier work, \[\int\limits_{0}^{\infty} \lfloor x \rfloor e^{x} dx = (e1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1so the first floor (x) e^x diverge or not?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3i showed it diverge so I think it diverge (unless I made some mistake)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3well wolfram showed my sum diverged :p

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1\(\color{blue}{\text{Originally Posted by}}\) @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 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I could have use the ratio test to come up with that know

freckles
 one year ago
Best ResponseYou've already chosen the best response.3ues from that last line

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1ah ok i guess ration test can testify divergence

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2we can simply use the limit test

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1oh yes! that limit does not go to 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{nx}=\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{nx}= \frac{d}{dx}\frac{e^{x}}{1e^{x}}=\cdots\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and we know \[\frac{d}{dx} \frac{e^{x}}{1e^{x}} \\ \frac{d}{dx} \frac{1}{e^x1} \\ \frac{d}{dx}(e^x1)^{1} \\ 1(e^x1)^{2}(e^x) \\  \frac{e^x}{(e^x1)^2}\] but we should evaluate this at x=1 since we had that we really wanted to evaluate \[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{x} dx = (e1)\sum\limits_{n=1}^{\infty} \frac{n}{e^n} \] \[(e1) \frac{e}{(e1)^2}=\frac{e}{e1}\] but I think we are to get 1/(e1)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Oh right \[\int\limits\limits_{0}^{\infty} \lfloor x \rfloor e^{x} dx =\sum_{n=1}^\infty n[e^{(n+1)}e^{n}] = \frac{e1}{e}\sum\limits_{n=1}^{\infty} \frac{n}{e^n}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3thanks @ganeshie8 ok I was sorta failing at the algebra a little too but I see it now

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2i remember working this problem differently sometime back here in openstudy but google isn't helping at the moment..

freckles
 one year ago
Best ResponseYou've already chosen the best response.3And 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.

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1truly i could not see how you threw differentiation in there @gabylovesu how is that sum equal to that differentiation ?

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1wrong tagging lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Have 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\color{red}{\sum\limits_{n=1}^{\infty} \frac{n}{e^{nx}}=\sum\limits_{n=1}^{\infty} \frac{d}{dx}e^{nx}}=\frac{d}{dx}\sum\limits_{n=1}^{\infty} e^{nx}= \frac{d}{dx}\frac{e^{x}}{1e^{x}}=\cdots\] 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...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3hey @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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The idea would be to express \(\lfloor x\rfloor e^{x}\) in terms of the step function \[\theta(xc)=\begin{cases}1&\text{for }x\ge c\\0&\text{for }x<c\end{cases}\]

xapproachesinfinity
 one year ago
Best ResponseYou've already chosen the best response.1ooh i see it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lfloor x\rfloor e^{x}=\sum_{c=1}^\infty ce^{x}\Bigg(\theta(xc)\theta(xc1)\Bigg)\] So \[\begin{align*}\mathcal{L}\left\{\lfloor x\rfloor\right\}&=\int_0^\infty \lfloor x\rfloor e^{sx}\,dx\\ &=\int_0^\infty \sum_{c=1}^\infty ce^{sx}\Bigg(\theta(xc)\theta(xc1)\Bigg)\,dx\\ &=\sum_{c=1}^\infty c\int_0^\infty e^{sx}\Bigg(\theta(xc)\theta(xc1)\Bigg)\,dx\\ &=\sum_{c=1}^\infty c\mathcal{L}\{\theta(xc)\theta(xc1)\}\\ &=\sum_{c=1}^\infty c\frac{e^{cs}e^{(c+1)s}}{s} \end{align*}\] Setting \(s=1\) gives the series @ganeshie8 was working with. Nothing new I suppose :P

freckles
 one year ago
Best ResponseYou've already chosen the best response.3does seem similar but still pretty @SithsAndGiggles anyways this is making me hungry so peace and thanks for the fun

freckles
 one year ago
Best ResponseYou've already chosen the best response.3I will come back later though to analyze more deeply what you said

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As 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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