Mathematics - 2014-12-20 (page 4 of 8)

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6:00 PM

@evinda So that you do not need to try all index pairs $(i,j)$

But this @ccorn is middle school scientific notation?

@ccorn I haven't understood it... And if y is a negative number?

@evinda Then $C$ is still sorted... because it's by absolute value...

It is not that big of a deal anyway @ccorn thanks for replying.

@skullpatrol Try it, perhaps they do something about it. Would be glad to hear that.

6:05 PM

@ccorn But then even if the quotient of two elements will be equal to the negative y we will not find it since we consider the elements by absolute value, or am I wrong?

@evinda Store the elements with the proper sign. Just sort by their absolute value.

If you find $C_j = B_i$ then since $C_j = y B_j$, you have $y B_j = B_i$, no sign mismatch

That's the basic idea. I assume that comparisons are taken as $O(1)$ here, otherwise you are not allowed to even sort, since the bit complexity is in the order of $m\log(m)^2$ for arbitrary-precision numbers.

Oops, replace $\log(m)^2$ with $\log(m)b$ where $b$ is some typical bitlength of a number

Well that's a first, I just answered my own question with a bounty on it :D

@evinda You need to increment both. "Under these conditions, increment that one, under those conditions, the other".

@DanielFischer Looks like you two have already discussed the workings of the comm algorithm

@ccorn What is the comm algorithm?

6:18 PM

comm man page

That page does not detail the algorithm. But the fact that the tool is there and works for stream inputs already suggests it's an $O(m)$ operation.

@DanielRust Did you place the bounty yourself?

@DanielFischer Nah, @BalarkaSen did

Can you even award a bounty to yourself?

@DanielRust Cool, question upvotes, answer upvotes plus bounty, if you're lucky.

@DanielRust No, you can't.

That's why it's good that somebody else put the bounty on your question.

@DanielFischer I haven't understood under which condition we have to increment i... :/

@DanielFischer should I complain to mathworld about the problem with their entry?

6:27 PM

@evinda You compare y*A[i] to A[j]. When do you increment j? And hence, what will likely be the point to increment i?

Since they do claim to be the world of math :P

@skullpatrol I have no opinion on that.

Thanks for the help @DanielFischer

What help, @skull?

If zero had a representation by that definition @DanielFischer

6:31 PM

Ah, that.

hi @DanielF and rat

From a learning point of view it is good to have a notation that can get arbitrarily close to zero and leave zero itself as a "special case". If you know what I mean

Hi @Ted. Who's rat?

That's what skull's avatar looks like to me.

@DanielFischer We compare at the beginning the first element of the second part with all the others. So, do we have to increment i if we haven't found after that two element of which the quotient is equal to y?

6:33 PM

hi @evinda

Hello @TedShifrin :)

@evinda No. You use two indices, $i$ and $j$, one for each sorted array. In each step, you decide which index to increment in order to (possibly) make the indexed elements come closer.

@evinda No, you don't compare the first element (of the second part) to all others. That's the point, since the array is sorted, you can know when you will not find a match if you go further.

$$\LARGE{\text{I found a new proof to the BASEL PROBLEM}}$$

odd that a simple "No" would be starred ...

@Chris'ssis: That is very obnoxious.

6:37 PM

@TedShifrin Why? :-)

Does anyone here know how to turn the power series $(1-2x+2x^2-2x^3+2x^4+....)$ into a power series representation?

@Mathy: Isn't it already?

It looks arrogant @Chris'ssis

@TedShifrin a representation like $\frac{1}{1-x}$

@skullpatrol What does it look arrogant?

6:38 PM

Well, suppose you had $1-x+x^2-x^3+x^4-\dots$. What would it be, @Mathy?

Guys, what are you talking there? :-)

@TedShifrin I agree.

@TedShifrin I know that $1+x+x^2+x^3+x^4+....$ is $\frac{1}{1-x}, so is it $\frac{x}{1-x}$?

No, look more carefully, @Mathy.

@TedShifrin $\frac{-x}{1-x}$?

6:39 PM

Instead of $x$ in your series, what do we have?

@Chris'ssis listen to the song first, I don't want you to get mad my friend :-)

Why are you multiplying by $x$?

@TedShifrin Are you still referring to $1-x+x^2-x^3+x^4-....$?

Yes. You start with $1+x+x^2+\dots$. How do you get $1-x+x^2-x^3+\dots$ from it?

@robjohn I feel guilty now that I found a terribly marvellous way of computing the Basel problem ... :-(

@skullpatrol :-)

Khallil Benyattou

6:41 PM

'Terribly marvellous'

:D

@TedShifrin $(1+x^2+x^4+x^6+.....) - (x+x^3+x^5+.....)$

@Chris'ssis When I found something that I was particularly fond of, it turned out to be at least 250 years old

Khallil Benyattou

A bit of modesty would go a long way, @Chris'ssis. :)

That doesn't help, @Mathy.

Although you can do it that way.

But to do it that way, you have to understand the basic principle, which is much easier in the first place.

6:42 PM

@ccorn I don't know what you found, but I found and published results that weren't known before. Besides that, I'm sure the proof I have to the Basel problem is new, but maybe @TedShifrin doesn't know that. :-)

Bonjourno, @Ted

If $f(x) = 1+x+x^2+x^3+\dots$, how do you express $1-x+x^2-x^3+x^4-\dots$ in terms of $f$?

@DanielFischer @ccorn So should we check the following?

`while (i<m and j<m and A[j]!=y*A[i]){
if (A[j]<y*A[i]) j++;
else i++;
}`

I don't think you spelled that right, @Mike :D

Bonjourn* @Mike

6:43 PM

@TedShifrin Somehow you negate every other term

So what do you need to do to $x$ to effect that, @Mathy?

No, @Khallil. Please don't.

@KhallilBenyattou Modesty for what? I mean anyone would be very happy for discovering such ways.

Khallil Benyattou

Sorry, @Ted!

@TedShifrin Oh, input $-x$?

Yeah!! @Mathy

6:44 PM

@TedShifrin is it $\frac{1}{1+x}$?

Now can you modify it slightly and figure out your original question, @Mathy?

@robjohn I found a very very neat way of solving the Basel problem. This is amazing to me (while I was working on a different problem).

Ok, @Ted, buon giorno

I had to ask my Aunt...

Yes, @Mathy. Now, how do you get $1-2x+2x^2-2x^3+\dots$?

Happiness is very different from obnoxiousness @Chris'ssis.

6:45 PM

Get her to teach you to speak Italian, @Mike :P

Sounds too hard...

@MikeMiller

1 - 2(x-x^2+x^3-x^4+...) = 1 - 2($\frac{1}{1+x}$) = $1 - \frac{2}{1+x}$ = $\frac{x-1}{1+x}$ @TedShifrin

Funny, Miller doesn't sound Italian :D

@evinda works for positive y and ascendingly sorted A

6:45 PM

Not quite, @Mathy.

You caught me, @Ted: I'm half English.

@ccorn: I think you mispinged :P

@MikeMiller

@TedShifrin yes, sorry, corrected

@TedShifrin Would (1-2x+2x^2-2x^3+2x^4+....) = $\frac{1}{1+x^2}$?

6:46 PM

@BalarkaSen and arrogant is very similar to obnoxious for some people :P

@BalarkaSen I often find your comments filled with a kind of inferiority complex, but I hope I'm wrong and I don't understand your profound thoughts on what I'm saying.

@skull: As I am both arrogant and obnoxious at times, I know whereof I speak :D

@Mathy: Stop guessing and write out the algebra carefully.

:D

@TedShifrin Honestly speaking, isn't it amazing to prove the Basel problem in a single line?

please unignore me, @Ted. C'mon.

6:48 PM

Are you referring to $\pi^2/6$? I've never heard it called the Basel problem. If so, stop writing everything in large print and tell us the one-line proof.

The Basel problem is a famous problem in mathematical analysis with relevance to number theory, first posed by Pietro Mengoli in 1644 and solved by Leonhard Euler in 1735. Since the problem had withstood the attacks of the leading mathematicians of the day, Euler's solution brought him immediate fame when he was twenty-eight. Euler generalised the problem considerably, and his ideas were taken up years later by Bernhard Riemann in his seminal 1859 paper On the Number of Primes Less Than a Given Magnitude, in which he defined his zeta function and proved its basic properties. The problem is named...

hi @Alizter

@TedShifrin ^^^

hi @TedShifrin

Yes, I already googled, @Chris'ssis.

6:49 PM

@TedShifrin Would 1- 2 (x-x^2+x^3+..) help at all?

I have been covering myself in mechanics and electromagnetism

@Chris'ssis I have no reason to feel inferiority complex towards you, I am just saying that writing out stuff in huge fonts is superbly annoying.

@Pedro isn't the only one who knows how to google :D

I love that stuff, @alizter :)

I have one of my favorite physics questions to give you.

@BalarkaSen I never feel that when other users do it like that. :-)

Are you going to have a Basel chapter in your book @Chris'ssis?

6:50 PM

Easier, @mathy: $2(1-x+x^2-x^3+x^4 -\dots) = $?

@TedShifrin OK give. please.

@skullpatrol I don't know, at the moment I'm only shocked ...

I need to see on this site the proofs provided to this problem ....

@TedShifrin that equals $\frac{2}{1+x}$

OK, now what do you need to do to that to get your original, @Mathy?

@Chris'ssis have fun :-)

6:52 PM

@skullpatrol I'm shocked!!!

@Alizter Can you please tell Ted to unignore me? I'm even prepared to do differential geometry if he does.

- 1 to get from 2 to 1

@Alizter: Have you worked the standard question about a point mass starting at some height $h<R$ above the equator on a sphere of radius $R$? It starts to slide along the sphere and flies off at what height?

Bingo @Mathy :)

44 secs ago, by Balarka Sen

@Alizter Can you please tell Ted to unignore me? I'm even prepared to do differential geometry if he does.

Enjoy the moment of discovery @Chris'ssis

6:53 PM

@Alizter: That's the warm-up for my real question :)

@TedShifrin Final answer is: $\frac{1-x}{1+x}$?

Haha @Alizter

@evinda Yes. Assuming $y > 1$.

Looks right to me, @Mathy.

@TedShifrin I have never seen something like that

6:54 PM

@BalarkaSen Did you see I answered my question that you placed a bounty on?

P.S. @Balarka: No one tells me to do anything.

Oh, @Alizter, that's a standard force diagram/conservation of energy question. Work it. Then I'll give you the interesting question :)

Learn^

@DanielRust No, will look. glad you answered it.

Actually I have never done any 3d mechanics.

@DanielFischer Why do we have to assume that $y>1$ and not just $y>0$ ?

6:54 PM

@TedShifrin Alright! Thank you for all your help!

It's just a 2-D problem, @Alizter. Do it on a circle.

You're welcome, @Mathy. Remember the techniques I tried to get you to follow :)

@BalarkaSen I left out the rather tedious end part, but all of the crucial justification is there.

@Ted I'm serious about doing differential geometry.

I need to understand the problem properly first.

ah

I get it

The point mass slides with no friction along the circle, starting at a point below the top. There is gravity, @Alizter.

6:56 PM

OK let me think

Gravity makes it slide ...

Darn, now I'll need to rework the problem :P

hi Bronze @Jasper

Don't do the problem for him, @Khallil :P

You just can't resist ... natural enthusiasm overwhelms you :D

@TedShifrin Hello! I am still feeling down these few days. Need to get my mood up and prepare to study soon.

@Ted Funny story, by the way: I was reading Milnor/Stasheff before the final on Thursday, and when the professor saw me, cringed and said she thinks it's more Stasheff's style than Milnor's...

Khallil Benyattou

Ah, I got rid of it! I haven't done any mechanics in a while so I just felt drawn to the question, @Ted!

@BalarkaSen ~~tell~~ ASK Ted to unignore

6:57 PM

@Jasper: I'm afraid you're afraid of the pressure you've put on yourself. That's why I advised to ease it a bit.

I understand, @Khallil. It's a cool question, but the better one follows.

@skullpatrol He cannot do it if he is already being ignored.

doffs his hat to Jasper

but the lamp is awkward to doff

@evinda If $y = 1$, then the question is whether the two array elements with quotient $y$ need to be different. If not, all you need is a nonzero element. If they need to be different, you'd just check in the sorted array whether there are two successive equal nonzero elements, and the loop as it stands would give you a false positive with $i = j$. You could then use the same loop if you add a check for i < j.

@TedShifrin I turned hats off again, so I can't see any hats.

@DanielF: Are you getting nervous about the impending doom? :)

Ah, @Jasper: I didn't know we could turn viewing them off and on.

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