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kabbers
Find n=0(1n!)2+2n=0(m=n+11n!m!)\displaystyle\sum^{\infty}_{n=0} (\frac{1}{n!})^2 + 2\sum^{\infty}_{n=0}(\sum^{\infty}_{m=n+1} \frac{1}{n!m!})

edited :smile:

Spoiler

Reply 781
Dystopia

Spoiler



yep!
Apologies for this is off topic...

How to you embed maths characters into a TSR post? Do you do it on an external website and C+P?

I tried to search for a similar question, to no avail.
Overground
Apologies for this is off topic...

How to you embed maths characters into a TSR post? Do you do it on an external website and C+P?

I tried to search for a similar question, to no avail.


Nope, it's built in. It's called LaTeX and you can find about it > http://www.thestudentroom.co.uk/wiki/LaTex
Fun integral: sinx2sinx  dx\int \sin\frac{x}{2} \sqrt{\sin x} \; \mathrm{d}x
I think I have a method that will eventually lead to the answer, but I'm too tired to work it through to the end (and even more so, putting up a full worked answer)...

Spoiler

Okay, time for a full solution. This is the nastiest piece of work I've ever attempted to solve I think:s-smilie: Hope I'm right... I'll be very interested in seeing your method Dystopia, because I think there will be a variety of methods that will lead to the answer - and I don't think I've taken the shortest (and I've probably spent 3-4hrs on the integral!)

Half-finished Solution



sorry for the bad formatting, but am fed up with latex!
Well, cheating:

Spoiler

nota bene
Okay, time for a full solution. This is the nastiest piece of work I've ever attempted to solve I think:s-smilie: Hope I'm right... I'll be very interested in seeing your method Dystopia, because I think there will be a variety of methods that will lead to the answer - and I don't think I've taken the shortest (and I've probably spent 3-4hrs on the integral!)

[snip]

Erhm, now I should sub back for my t^2=tan(x/2) substitution, but I'll leave that for now I think!


sorry for the bad formatting, but am fed up with latex!
:eek:

You're so getting rep tomorrow!

What 'fun', eh? :wink:

Anyway, that was a pretty clever substitution (I certainly didn't spot it). However, you're right - definitely not the shortest way of doing it. My method isn't amazingly simple (one or two pages of work), but there are no nasty partial fractions at least.

Spoiler

Dystopia
What 'fun', eh? :wink:

Anyway, that was a pretty clever substitution (I certainly didn't spot it). However, you're right - definitely not the shortest way of doing it. My method isn't amazingly simple (one or two pages of work), but there are no nasty partial fractions at least.

Yes, great fun:biggrin: (At least I managed to avoid physics revision for a couple of hours!)

My substitution was efficient in getting it down to a rational function, but unfortunately it was a particularily nasty rational function (at some stage even mathematica couldn't do it, as I was cheking if I'd done it right).

Your method is obviously a lot more efficient, and I never spot the kpi+x substitutions sometimes required in STEP, and intergals like this - very nice approach!
Can we not update the wiki Latex page with some more basic funstions such as the probability ones because I'm fed up of teaching latex to S1'ers...
nota bene
Yes, great fun:biggrin: (At least I managed to avoid physics revision for a couple of hours!)

My substitution was efficient in getting it down to a rational function, but unfortunately it was a particularily nasty rational function (at some stage even mathematica couldn't do it, as I was cheking if I'd done it right).

Your method is obviously a lot more efficient, and I never spot the kpi+x substitutions sometimes required in STEP, and intergals like this - very nice approach!

Thanks. :smile:

Interestingly, if you substitute t=sinu2+cosu2t = \sin\frac{u}{2} + \cos\frac{u}{2} into 12(sinu2cosu2)cosu  du\frac{1}{\sqrt{2}} \int (\sin\frac{u}{2} - \cos\frac{u}{2})\sqrt{\cos u} \; \mathrm{d}u, you get 2(2t2t4)1/4  dt-\sqrt{2} \int (2t^{2} - t^{4})^{1/4} \; \mathrm{d}t, yet when you put this second integral into the Integrator, it gives the result in terms of a hypergeometric function.

I think that's the first time I've come across an integral that it expresses in terms of non-elementary functions when it isn't necessary to do so. (Assuming I haven't made a mistake.)
When I used the sub t=tan(x/2) and got the integral t3/2(1+t2)2dt\int \frac{t^{3/2}}{(1+t^2)^2}dt where mathematica gave me a non-elementary function as well.

This stuff gets interesting when you start realising what that these non-elementary functions are actually pretty useful, and occasionally you can find explicit values for them:smile:

And no, I don't think you have made a mistake, I get the same.

edit: Incidentally, if you want a simple problem to play with, see if you can find a recurrence with L{fn(t)}\displaystyle\mathcal{L} \left\{f^n(t)\right\} where f^n is the nth derivative.(And then you can solve it explicitly, and prove by induction if you feel like it) I stumbled upon that one yesterday, but to my disappointment I found it on wikipedia:p:

edit2

Spoiler

nota bene
edit: Incidentally, if you want a simple problem to play with, see if you can find a recurrence with L{fn(t)}\displaystyle\mathcal{L} \left\{f^n(t)\right\} where f^n is the nth derivative.(And then you can solve it explicitly, and prove by induction if you feel like it) I stumbled upon that one yesterday, but to my disappointment I found it on wikipedia:p:

I don't know anything about Laplace transforms unfortunately, so I'm afraid I don't really understand what you're asking me to do. :s-smilie:

(I didn't even know what L{fn(t)}\displaystyle\mathcal{L} \left\{f^n(t)\right\} actually meant, but I remembered hearing about something called a Laplace transform and looked it up...)


However it appears I have contemplated something vaguely similar in the past. Basically, I noticed that (subject to a couple of conditions of convergence) 0exf(x)  dx=f(0)+f(0)+f(0)+\displaystyle \int^{\infty}_{0} e^{-x} f(x) \; \mathrm{d}x = f(0) + f'(0) + f''(0) + \cdots which, along with Maclaurin's theorem, I used to find functions which satisfy, for example, 0exf(x)  dx=ln2\displaystyle \int^{\infty}_{0} e^{-x} f(x) \; \mathrm{d}x = \ln 2
I don't know much about Laplace transformations either; you only need the definition of a laplace transform i.e. L{f(t)}=0estf(t)dt\displaystyle\mathcal{L}\left\{f(t)\right\}= \displaystyle\int_0^{\infty} e^{-st}f(t)dt and then you can (quite easily) show what is in my spoiler. Then, play around with how that spoiler generalises (start by looking at L{f''(t)}=? to get an idea of what is happening).

I only found this as an exercise in my book (the part in the spoiler and the L{f''(t)} case) - it's not that interesting.

Oh, and I think that nice result you have there is very closely related to the one I am after, both seem to rely on repeated integration by parts (except on the other integrand)...
nota bene
I don't know much about Laplace transformations either; you only need the definition of a laplace transform i.e. L{f(t)}=0estf(t)dt\displaystyle\mathcal{L}\left\{f(t)\right\}= \displaystyle\int_0^{\infty} e^{-st}f(t)dt and then you can (quite easily) show what is in my spoiler. Then, play around with how that spoiler generalises (start by looking at L{f''(t)}=? to get an idea of what is happening).

I only found this as an exercise in my book (the part in the spoiler and the L{f''(t)} case) - it's not that interesting.

Oh, and I think that nice result you have there is very closely related to the one I am after, both seem to rely on repeated integration by parts (except on the other integrand)...

Is this what you were looking for:

L{fn(t)}=snL{f(t)}r=1nsnrfr1(0)\displaystyle \mathcal{L}\left\{f^{n} (t)\right\} = s^{n} \mathcal{L}\left\{f(t)\right\} - \sum_{r=1}^{n} s^{n-r} f^{r-1}(0)

If so, you're right - setting s = 1, and letting n tend to infinity gives the same result (again, assuming various things tend to zero).

By the way, do you know anything about double integrals? When I tried to find a function such that 0exf(x)  dx=113+15=π4\displaystyle \int^{\infty}_{0} e^{-x} f(x) \; \mathrm{d}x = 1 - \frac{1}{3} + \frac{1}{5} - \cdots = \frac{\pi}{4}, I think I ended up solving a differential equation to get f(x)=12x0xett  dt\displaystyle f(x) = \frac{1}{2\sqrt{x}} \int^{x}_{0} \frac{e^{-t}}{\sqrt{t}} \; \mathrm{d}t. I don't know anything about double integrals at all, but I assume that this is related to the result 0ex2  dx=π2\displaystyle \int^{\infty}_{0} e^{-x^{2}} \; \mathrm{d}x = \frac{\sqrt{\pi}}{2}?

Oh, and when I was avoiding doing physics, I noticed that considering the real and imaginary parts of ln(1+eix)\ln(1 + e^{ix}) gives you a couple of nice identities. One of them provides yet another incredibly dodgy way of finding the sum of the reciprocals of the squares, whereas the other provides an interesting way of finding the classic integral 0π2ln(cosx)  dx\displaystyle \int^{\frac{\pi}{2}}_{0} \ln(\cos x) \; \mathrm{d}x :smile:
Dystopia
Is this what you were looking for:

L{fn(t)}=snL{f(t)}r=1nsnrfr1(0)\displaystyle \mathcal{L}\left\{f^{n} (t)\right\} = s^{n} \mathcal{L}\left\{f(t)\right\} - \sum_{r=1}^{n} s^{n-r} f^{r-1}(0)

If so, you're right - setting s = 1, and letting n tend to infinity gives the same result (again, assuming various things tend to zero).

Yes that was what I was looking for:smile:
By the way, do you know anything about double integrals? When I tried to find a function such that 0exf(x)  dx=113+15=π4\displaystyle \int^{\infty}_{0} e^{-x} f(x) \; \mathrm{d}x = 1 - \frac{1}{3} + \frac{1}{5} - \cdots = \frac{\pi}{4}, I think I ended up solving a differential equation to get f(x)=12x0xett  dt\displaystyle f(x) = \frac{1}{2\sqrt{x}} \int^{x}_{0} \frac{e^{-t}}{\sqrt{t}} \; \mathrm{d}t. I don't know anything about double integrals at all, but I assume that this is related to the result 0ex2  dx=π2\displaystyle \int^{\infty}_{0} e^{-x^{2}} \; \mathrm{d}x = \frac{\sqrt{\pi}}{2}?

My knowledge of double integrals is limited, and I don't see how to do that one... It felt more familiar with 12x0x2eu22udu\frac{1}{2\sqrt{x}}\int_0^{x^2} \frac{e^{-u^2}}{2u}du, but that one is not possible to evaluate I think. (you can't play the trick with polar coordinates because u*v will be nasty in the denominator) Mathematica tells me that this is what is called an exponential integral.
What you can do is write the series expansion for e^(stuff) and divide by x and integrate, apply limits etc. But that will only leave you with two (or one, depending on if the lower one is zero/constant) infinite series that are fairly useless I suspect?
edit: Are you going to let your x go to infinity? If so there's more hope in trying to find similarities to the 0.5π0.5\sqrt{\pi} integral.
Oh, and when I was avoiding doing physics, I noticed that considering the real and imaginary parts of ln(1+eix)\ln(1 + e^{ix}) gives you a couple of nice identities. One of them provides yet another incredibly dodgy way of finding the sum of the reciprocals of the squares, whereas the other provides an interesting way of finding the classic integral 0π2ln(cosx)  dx\displaystyle \int^{\frac{\pi}{2}}_{0} \ln(\cos x) \; \mathrm{d}x :smile:

Looks fun - will have a look after my physics exams next week (I don't know anything yet!)
Reply 797
looks interesting
I had no idea there was a maths society on tsr. Will be joining me up :smile:
Hey, I'm just about to go into my 2nd year of uni and we all have to do a 45 minute maths talk for the first years (and second years and tutors) - it can be on anything mathematical - a mathematician, a theorem, a number, a whole branch of mathematics, etc.

I was wondering if anyone has any good/interesting ideas, because I am drawing a blank. Thanks

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