Mathematics

Axoren

12:07 AM

@evinda That's why such a notation may be misleading and you should stick to standard big-o notation

Because the function itself may not be much-less than the function it's bigO of.

evinda

A ok.. But I could use << right? How can I explain it that with f<<g I mean $g=\Omega(f)$?

Stan Shunpike

@TedShifrin Once I become more familiar with differential geometry, will I then be able to understand the higher-dimensional forms of curl? I find curl in 3D very unintuitive.

Axoren

@evinda You would just state it.

$f(n) \ll g(n) \implies g(n) = \Omega(f(n))$

evinda

With f<<g I mean that $g=\Omega(f)$.
Like that @Axoren?

Axoren

You would simply say that, yes.

But the same ambiguity happens.

evinda

12:13 AM

@Axoren A ok..

Axoren

Consider now $0.0001f(n)^2 < f(n)$ on some interval.

Which it clearly is, on some interval

By your notation, we would write $f(n) \ll 0.0001f(n)^2$

evinda

$f(n) \ll g(n) \iff g(n) \in \Omega(f(n))$
So wouldn't this be: $0.0001f(n)^2 \ll f(n)$ ? @Axoren

Axoren

Oh, my mistake.

Yes, the lower-bound would be on the bottom.

But do you see how $0.0001f(n)^2$ isn't an assymptotic lower bound on $f(n)$?

Specifically, when $f(n) > 10000$

ugh, wait

evinda

If $0.0001f(n)^2$ is an assymptotic lower bound on $f(n)$ then $f(n)^2$ is also an assymptotic lower bound on $f(n)$. So $f(n)=\Omega(f(n)^2)$. That means that $\exists c>0, n_0 \in \mathbb{N}$ such that $f(n) \geq cf(n)^2 \Rightarrow 1 \geq f(n)$. @Axoren

Chris's sis

@robjohn Hope you won't miss this one (it's too nice)

6

Q: How to evaluate $I=\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$

Prima facie, this integral seems easy to calculate,but alas, this not's case $$I=\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$$ The numerical value is I=-1.122690024730644497584272... How to evaluate this integral? By against,I find: $$I=\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(2x)}\,d...

Axoren

12:21 AM

This notation is already confusing ME

@evinda I would ask your instructor if they're comfortable with you introducing your own notations in to your homework.

Because confusing the person grading your work is generally bad for all involved.

evinda

@Axoren Ok..

Axoren

As one of my math professors used to say "You make my job painfully easy."

evinda

@Axoren Did you also take a look at the table I sent you?

Axoren

I can't make heads or tails of it.

What are the values in the cells?

evinda

@Axoren The algorithm requires f(n) microseconds(1 microsecond=$10^{-6}$ second).
At each line we have a specific f(n) and for example if we are looking at 1 second, we use the fact that 1 second=$10^6$ microsecond and then we have $f(n)=10^6$ ms and so $\lg n=10^6 \Rightarrow n=2^{10^6}$

Axoren

12:27 AM

But if you look at the factorial

What does 9 mean in that?

evinda

the easiest way to find the $n$ for $f(n)=n!$, is by simply iterating through $n$ and check when it reaches the required time @Axoren

Axoren

@evinda What does 9 mean in this context?

Is it the answer?

This problem looks "completed".

Though I haven't checked anything.

@evinda What are they asking from you, since it seems like this problem isn't a problem?

dustin

Is anyone aware of any worked examples of equivalent metric spaces? I know this post exist but it is about general metrics $d$ and $d_1$.

evinda

It should be the answer because of this http://www.wolframalpha.com/input/?i=%28n%21%29%3D10%5E6
9 is the size of the problem so that it can be solved in time 10^6 microseconds.
I am given the array and I should fill it with the values of n.
I found that what I sent you in the web. I filled it by myself and wanted to verify it.. And I found for $f(n)=\lg n$ at 1 century that it is $2^{365 \cdot 24 \cdot 36 \cdot 10^{10}}=2^{31536 \cdot 10^{11}}$.
Am I wrong? @Axoren

Axoren

If it's the answer, I have no idea what you're expected to do with this problem

evinda

12:49 AM

@Axoren A ok...

@quid Do you maybe have an idea?

clrs.skanev.com/01/problems/01.html

The algorithm requires f(n) microseconds(1 microsecond=$10^{-6}$ second).
At each line we have a specific f(n) and for example if we are looking at 1 second, we use the fact that 1 second=$10^6$ microsecond and then we have $f(n)=10^6$ ms and so $\lg n=10^6 \Rightarrow n=2^{10^6}$

I found for $f(n)=\lg n$ at 1 century that it is $2^{365 \cdot 24 \cdot 36 \cdot 10^{10}}=2^{31536 \cdot 10^{11}}$.

Am I wrong?

quid

Hi @evinda Let me see.

Not sure. Let me try.

A century so 100 years times 365 days times 24 hours times 3600 second times 10^6 for micro is the same as 10^10 times 365 times 24 times 36. And this to a power of 2. Yes. Seems fine @evinda

evinda

So it should be $2^{31536 \cdot 10^{11}}$, right? @quid

quid

1:04 AM

@evinda yes.

It seems there is a different number given on that page though.

Is this the source of your question @evinda

The reason might be that one can be more precise with a century.

In that one could count in the leap days.

evinda

No, it isn't the source of the question... @quid
I was given the table unfilled and I am asked to fill it with tha values of n.
I found this site and wanted to verify my results, but this was different.

@quid So how do they calculate it?

quid

Every fourth year has an extra day. So add 25 days, but then IIRC every 100 year is an exception so subtract 1. For adding 24 days. Then one has 36524 days times 24 times 36 times 10^8 this is then the number in the list.

But this seems a bit complicated to me.

Axoren

@quid It's easier to measure it as the average * 100

quid

@Axoren maybe. but what is the average?

Axoren

A little bit short of 365.25

Let me see if I can find the quantity

quid

1:14 AM

Well 365.24 my result divided by 100 :-)

Axoren

365.2425

Because every 4 centuries is a leap year

:P

quid

Right.

But this would not match the result for years. Where a standard year is assumed. So we might also assume a standard century.

Axoren

I wonder if there's a simple function of $d$ where $d$ is the number of days after January 1st, 0 AD/BC, that properly returns which day of the month it is.

evinda

If I assume that there 365 days, should I consider at the century the leap years? @quid @Axoren

Axoren

@evinda Express it in your answer that you're accounting or neglecting leap years.

Or provide both answers, either or.

Neither should be harder than the other.

quid

1:22 AM

@evinda I agree. It is not clear to me. Personally I would not bother with leap years, and the fact that every hundreds year is not a leap year is IMO not as widely known as to make it an implcit prerequisuty to answer such a question. The advice by @Axoren is really best. Do it with 365 and say that for simplicity you ignore leap years.

evinda

How do we calculate the average? :/ @Axoren @quid

Axoren

I just gave you the average

365.2425

Every year has that many days in it.

Any more accurate and you'd have to hook yourself up the NASA's observation servers

Let me see actually... do they have a setting for seeing how many days are ACTUALLY in a year?

ssd.jpl.nasa.gov/horizons.cgi

evinda

@quid A ok..So for $f(n)=\lg n$ it is $2^{31536 \cdot 10^{11}}$, right?

@Axoren :D

quid

@evinda yes I would give this answer, and perhaps add that I ignore leap years.

Axoren

Ugh, I can't figure out how to measure the observed solar-days, but I could find the number of lunar-days.

But it requires a lot more effort than I'm willing to put into it.

It's all sun-target-observer on the HORIZONs thing

When the sun is the target, it's confusing

evinda

1:37 AM

@quid @Axoren For n! at century, is the result equal to $31536 \cdot 10^{11}$ ? :/

Axoren

Well, factorial's harder.

You need the inverse factorial.

It's easier to use the inverse Gamma function and then find the nearest integer value.

But even that's tricky.

quid

@evinda no this is the result for n. For n! you need to find the n such that n! is still less than or equal to that number. It should be somewhere around 15. One would need to check with a caclulator.

Axoren

@quid's solution is fairly easier. Assuming the actual answer is near 15, simply trying 20+ factorials will net you the answer.

evinda

With what calculator could I check it? 'In wolfram I didn't get a result.

Axoren

http://www.wolframalpha.com/input/?i=n%21+%3C+100*365.2425*24+*+60+*+60+*1000

evinda

1:43 AM

@Axoren Why do we want n!<100?

Axoren

Use the new link

quid

@evinda it was just meant as illustration.

Axoren

Math.SE chat screwed up the link

I hate the * notation *.

Sometimes, you just want to say *

evinda

@Axoren Doesn't it has to hold that $n!=365 \cdot 24 \cdot 36 \cdot 10^8 \cdot 100$ ?

quid

Oh. It seems my guess was wrong.

evinda

1:44 AM

So don't we want the equality but an inequality @quid ?

:/

Axoren

@evinda It's unlikely that $n!$ will equal any integer. However, you do know that so long as $n! < T$ where $T$ is how much time you actually have, you can always solve a factorial problem in time $T$

quid

@evinda yes. because no factorial will fit exactly. It is like for powers of 2.

Axoren

@evinda So, you want to find the largest problem such that you can solve it in time $T$

That's going to be the largest $n$ for which $n! < T$

Funnily enough, $P(c = n! \text{ for some } n \in \mathbb N)$ is an interesting problem.

Just how likely is is for a natural number for be the factorial of a natural number?

evinda

Yes, it is interesting..

@Axoren @quid
Is the following right?
$\begin{matrix}
& 1 \text{ century }\\
- & -\\ \\
\lg n & 2^{31536 \cdot 10^{11}}\\ \\
\sqrt{n} & 994519296 \cdot 10^{22}\\ \\
n & 31536 \cdot 10^{11}\\ \\
n \lg n & 68610956750\\ \\
n^2 & 56156922\\ \\
n^3 & 146645\\ \\
2^n & 51\\ \\
n! & 16
\end{matrix}$

Axoren

The number of microseconds in a century are $3155695200000$

$2^{51} = 2251799813685248$

That's a much larger number, by 3 digits.

That would take around 1000 centuries to perform, not 1.

quid

1:59 AM

@Axoren I think not.

evinda

For which f(n) do you mean? @Axoren

Axoren

The last one, evinda.

@quid What don't you think?

evinda

f(n)=n! ? @Axoren

Axoren

f(n) = 2^n

My mistake.

But also, for $n!$, $16!$ is larger than 3155695200000

evinda

So are the last two results wrong? @Axoren

quid

2:02 AM

@Axoren the number of seconds is (withou leap years) 31536 times 10^{11}

this is larger than what you claim.

The base 2 log of 31536 times 10^{11} is 51 plus a little so 51 is correct.

Axoren

100 years * 356.2425 days/year * 24 hours/day * 60 minutes/hour * 60 seconds/minute * 1000000 microseconds/second

Oh I see.

I'm short a magnitude of 1000

The factorial one is still wrong, unfortunately.

Consider 17!

evinda

So is f(n)=2^n right? @Axoren
http://www.wolframalpha.com/input/?i=n%21+%3C+100*365*24+*3600*10%5E6
So is it 18?

quid

@evinda it is 17 for the factorial

Axoren

@evinda The "number of solutions" includes $n = 0$

quid

Other than that the results seem correct. Exceot I do not see how you did the n lg n one

evinda

2:08 AM

@quid How do we find that it is 17?

Axoren

@evinda You check values of $n$ one at a time until they reach 17

quid

I calculate 31536 times 10^{11} - 17! this is positive and 31536 times 10^{11} - 18! which is negative.

Axoren

Then, you check 18 and see that it fails.

evinda

@quid So we find the differences till the difference we are looking at becomes negative, right?
As for $n \lg n$, I haven't calculated it since I didn't now how.. Which is the easiest way?

Axoren

You can take its inverse

Which is $\frac n {ProductLog(n)}$

mathworld.wolfram.com/LambertW-Function.html

evinda

2:13 AM

@Axoren Isn't there an other way?

Axoren

Exhaustive search, starting from $n = 1$ and counting up

Alternatively, you can pick a terribly large number for which you are sure it doesn't hold

And then binary search between it and 1 to find the value.

quid

@evinda for the first, yes. For the other I agree with @Axoren

Axoren

Personally, I like the inverse function idea because of it's $O(1)$ time complexity.

evinda

But how do we come to the number 68610956750?
Is it right? Because at the year, I found for f(n)=n*lgn the result 797633893349 which is bigger :/

@Axoren So can we compute the inverse without the use of the program?

And having found it, how do we use it?

Axoren

@evinda We can make Wolfram or Mathematic compute the inverse.

evinda

2:17 AM

wolframalpha.com/input/?i=inverse+n*lgn

and then?

Axoren

Example

The approximate form will give you the closest approximation.

Then, you pick the largest integer less than that

evinda

Ok, I will think about it tomorrow.. I will go to sleep now..
Thanks for your help @quid @Axoren

:)

Good night!!!

Axoren

Good night, @evinda

quid

You are welcome @evinda and good night!

David Wheeler

ppl keep upvoting my stupid answer. why me?

Axoren

2:24 AM

Don't like it, delete it.

Or edit it.

To remove the stupid

JMoravitz

2:38 AM

Out of curiousity, is there a mathjax Cyrillic alphabet? I'm wanting to do $\cyrb$ but its not displaying

David Wheeler

someday my brilliant idiocy will be recognized for what it is...n't

David Wheeler

4:13 AM

ping

JMoravitz

pong

David Wheeler

echo echo

JMoravitz

4:30 AM

$~_{~~\text{echo}_{~~\text{echo}_{~~\text{echo}}}}$

David Wheeler

i give up

i surender

i submit

Sayan Chattopadhyay

8:14 AM

Hi guys

Chris's sis

8:28 AM

Greetings

@SayanChattopadhyay I have a nice question for you

7

Q: How to evaluate $I=\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$

Prima facie, this integral seems easy to calculate,but alas, this not's case $$I=\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx$$ The numerical value is I=-1.122690024730644497584272... How to evaluate this integral? By against,I find: $$I=\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(2x)}\,d...

integration definite-integrals improper-integrals closed-form

You can start with the simpler one $$\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(2x)}\,dx=-\frac{\pi^3}{48}$$

@BalarkaSen ^^^

Balarka Sen

Don't do integrals.

Chris's sis

@BalarkaSen keep it there if one day you change your mind

Balarka Sen

I won't, I assure you.

I'd rather study complex analysis and learn how to compute integrals via residues. At least it's a new technique. What's the point in sticking to a bunch of substitution and clever tricks.

Chris's sis

@BalarkaSen Well, in the last period of times I did tons of integrals by using complex analysis, but not sure how it works here. Maybe you teach me your way (using CA).

Balarka Sen

I haven't even tried it, just speaking that I am against sticking to real analytic methods. Never saw you do stuff using complex analysis.

Chris's sis

8:38 AM

@BalarkaSen Aha ... (you don't know many things then)

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$Mathematics$ Mathematics