Mathematician Chat V.floor(π)

[mastoner20]

[QUOTE=Scoooooby;40622892]What's a good site to use for college liberal arts mathematics?
And yes you can rate me dumb, I failed it because I couldn't do mortgage stuff.
I could do algebra, but I'm scared and don't want to fail anymore so I'm retaking liberal arts.

And I want to make sure I pass, so any suggestions for sites?[/QUOTE]

PM me a list of all the general topics, and I could probably get together with you and tutor you.

As for individual websites; wolfram alpha is a pretty good online calculator for some pretty nifty things. Wikipedia can provide some decent 'dumbed-down' approaches to some of the topics. Just, as always with WP, be careful with the information. Quite a few of the math departments from various schools around the U.S. provide some of their lecture notes for certain classes online that are available for the public to view, this may be helpful. And I'm sure I'm missing a few other good sources, so if anyone else knows of any, please feel free to expand for me.

[agentalexandre]

Thanks guys, that's very helpful. I guess the rate of stupid mistakes grows exponentially depending on how late you stay up :v The problem with the inequality is not something I probably would have realised but that expansion of the sum was horrible. How shameful.

[Mr. Bleak]

Modeling with differential equations was a nightmare.

I have a profound respect for some of the guys here like Joey and Johnny, I probably couldn't comprehend half the stuff you guys know if I had a lifetime to learn it. Just goes to show that "class rank" has nothing to do with actual ability.

[Bradyns]

I saw this today.

[IMG]http://latex.codecogs.com/gif.latex?\lim_{n\rightarrow%20\infty}%20\frac{n}{\sqrt[n]{n!}}=%20e[/IMG]

Nifty!

[Mr. Bleak]

That's pretty cool, any background behind it? I've forgotten most of the stuff we did with limits.

[Krinkels]

-snip-

[Joey90]

[QUOTE=Mr. Bleak;40667471]That's pretty cool, any background behind it? I've forgotten most of the stuff we did with limits.[/QUOTE]
At the risk of stealing Bradyns' thunder...

It's a weak (rearranged) form of [url=http://en.wikipedia.org/wiki/Stirling's_approximation]Stirling's Approximation[/url], very useful for approximating things with factorials (or gamma functions) in.

To prove it you usually take the logs and approximate log(n!) by the integral of log(x)

As he's stated it though, it's quite a pretty identity, with a non-obvious answer!

Also, don't think that you can't comprehend stuff even if you don't get it at first sight! Not much of the stuff I know made sense the first few times I saw it... Just practice doing it a lot and hopefully things will start to fall into place.

[Bradyns]

[QUOTE=Joey90;40670993]At the risk of stealing Bradyns' thunder...

It's a weak (rearranged) form of [URL="http://en.wikipedia.org/wiki/Stirling's_approximation"]Stirling's Approximation[/URL], very useful for approximating things with factorials (or gamma functions) in.

To prove it you usually take the logs and approximate log(n!) by the integral of log(x)

As he's stated it though, it's quite a pretty identity, with a non-obvious answer!

Also, don't think that you can't comprehend stuff even if you don't get it at first sight! Not much of the stuff I know made sense the first few times I saw it... Just practice doing it a lot and hopefully things will start to fall into place.[/QUOTE]

There was no thunder to be stolen :P

Stirling was mentioned by our analysis lecturer the other day... We wanted to see how n! behaves.

As a class, we managed to get this:
[IMG]http://latex.codecogs.com/gif.latex?\frac{n^{n}}{e^n}e[/IMG]

Which, just by using some simple tools is relatively close to the actual thing:

[IMG]http://latex.codecogs.com/gif.latex?\frac{n^{n}}{e^n}\sqrt{2\pi n}[/IMG]

NB: For [I][U][B]really[/B][/U][/I] large values of 'n'.

[Mr. Bleak]

Found this post of mine from the last(?) thread:

[QUOTE= Me 10/1/12]Just started derivatives in calculus.

I have my 4 math credits, opting for statistics over calculus because I much prefer statistical analysis over these stupid abstract concepts, but I guess my counselor must hate me.

F-- here I come.[/QUOTE]

It's funny because I actually did very well once we got out of limit theory and started to work with actual differentiation and integration. I only need to score a 35 or so on my final to maintain an A in the class, but I'm certainly hoping to do much better.

I can't wait until I'm doing some college math and looking back at this, it feels nice to realize how far you can go in just 7 months or so.

[JohnnyMo]

Derivatives, abstract. :v:

[Bradyns]

So, I got this question in an assignment that I handed in a few hours back.. And I made next to no attempt at a solution, as I had no clue as to what was going on.

[QUOTE][IMG]http://i.imgur.com/N8VVIsn.png[/IMG][/QUOTE]

Situation is that we haven't touched Fourier series in our analysis part of the course (I think we start it next semester).

We've finished the course material and done as far as some applications of integrals (Rotations around an axis); proving all of the important/relevant theorems on the way.

In light of this, is there a solution that can be sought using only these tools?

[QUOTE]All I jotted down was:

[IMG]http://latex.codecogs.com/gif.latex?n \leq m \leq N[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?0[/IMG] when [IMG]http://latex.codecogs.com/gif.latex?m=n[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?1[/IMG] when [IMG]http://latex.codecogs.com/gif.latex?m\neq n[/IMG]

Now, this was not a legitimate answer, but I was at wits end.[/QUOTE]

[Number-41]

Just insert f(x) into the integral and you'll end up with a sum of integrals where each term has either m=n or m!=n.

You might have to transform those sines into their exponential form to actually solve those two cases, not sure.

[Bradyns]

-snip-

Essentially, I scribbled something down which was similar, but we haven't learnt how to evaluate summations like that.

I know how to bound a summation with integrals, but place a summation in an integrand and shit hits the mental fan.

[Joey90]

[QUOTE=Bradyns;40751796]-snip-

Essentially, I scribbled something down which was similar, but we haven't learnt how to evaluate summations like that.

I know how to bound a summation with integrals, but place a summation in an integrand and shit hits the mental fan.[/QUOTE]
What you posted before you snipped was essentially correct.

You (should) know that integral (f + g) = integral (f) + integral (g)

This extends to knowing that the integral of a [i]finite[/i] sum allows you to move the integral inside the sum:

sum integral (f[sub]n[/sub]) = integral sum (f[sub]n[/sub])

You are absolutely right that you need to be more careful if it's an infinite sum though! (Although 'usually' it's ok :v: )

What you end up with is a sum of integrals like a[sub]n[/sub]sin(nx)sin(mx) which equals what you wrote before.

One way of evaluating it is using what you wrote before, and noticing that if n is not equal to m, the integral is 0, but 1 otherwise. (As you guessed)

Thus the only nonzero term left in your sum is a[sub]m[/sub]

[ThisIsTheOne]

It's a finite summation so there's no worries about swapping the orders with limits that you'll get with infinite summations.

If you imagine sin(kx) as having its own dimension for each k, what this integral does is "decompose" f(x) into these dimensions. It's just like using the dot product with unit basis vectors in Rn to decompose a vector into its magnitude in each dimension. Instead of having a dot product over Rn, you have an inner product over a Hilbert space. The sin(kx) are like your basis "vectors".

[Number-41]

-snip-

[ThisIsTheOne]

-snip-

[Bradyns]

[QUOTE=Joey90;40751883]What you posted before you snipped was essentially correct.

You (should) know that integral (f + g) = integral (f) + integral (g)

This extends to knowing that the integral of a [i]finite[/i] sum allows you to move the integral inside the sum:

sum integral (f[sub]n[/sub]) = integral sum (f[sub]n[/sub])

You are absolutely right that you need to be more careful if it's an infinite sum though! (Although 'usually' it's ok :v: )

What you end up with is a sum of integrals like a[sub]n[/sub]sin(nx)sin(mx) which equals what you wrote before.

One way of evaluating it is using what you wrote before, and noticing that if n is not equal to m, the integral is 0, but 1 otherwise. (As you guessed)

Thus the only nonzero term left in your sum is a[sub]m[/sub][/QUOTE]

The way that I came to that was multiplying the summation by sin(mx), and using some trig identities.

Tried following my nose, but there was too much uncertainty in where I was going so I decided to just scribble down some of the things that occurred which seemed to make sense, I'll probably still get a mark.. :v:

[Mr. Bleak]

[QUOTE=JohnnyMo1;40745854]Derivatives, abstract. :v:[/QUOTE]

My point exactly. If it can be visualized on a simple graph, chances are it's not too abstract :v:

[agentalexandre]

Noob question on geometry here. I need to prove that given a line l and a point P then if m is a line passing through P and is parallel to l then it is the unique line passing through P and parallel to l. Is the following reasoning sufficient to prove this?

Suppose there exists another line n that is parallel to l and contains P. Then l is parallel to m and l is parallel to n. However we can deduce from this that m is parallel to n because transitivity is a property of parallel lines. However, m and n have a common point, namely P. Parallel lines must either have no points in common or must be equal hence m=n. This is a contradiction as we said n was distinct from m hence m is the unique parallel line to l passing through P.

This doesn't seem like its enough and my lecture notes does some complicated stuff with altitudes, can anyone tell me why my reasoning is wrong?

Edit: Just realised that "Parallel lines must either have no points in common or must be equal hence m=n" is the exact thing I'm trying to prove. woops.

[JohnnyMo]

Yep. You can drop the contradiction entirely and do it directly. Other than that, your reasoning seems solid.

[Xakoro]

All faces of an elongated pentagonal bifrustum are of equal area.
A) Given that there exists a sphere perfectly circumscribing the elongated pentagonal bifrustum what is the ratio of their volumes?
B) Find the value of the ratio for an elongated conic bifrustum where all faces are of equal area.

[Krinkels]

Say you compose sin(x)/2 + x with sin(x)+x and keep composing left until sin(x)/k + x. What function is that?
And say you did the same thing going right? Do you get the same answer?

[JohnnyMo]

Wow. According to his homepage,([url]http://www.dmmm.uniroma1.it/~agostino.prastaro/HOMEPAGEPRAS.htm[/url]) Dr. Agostino Prastaro has solved Goldbach's conjecture, the twin primes conjecture, the Navier-Stokes existence and smoothness problem, the Yang-Mills existence and mass gap problem, and the Riemann hypothesis. Why has this man not received a Fields medal yet?

Oh right. Because he's nuts.

[Joey90]

[QUOTE=Krinkels;40835364]Say you compose sin(x)/2 + x with sin(x)+x and keep composing left until sin(x)/k + x. What function is that?
And say you did the same thing going right? Do you get the same answer?[/QUOTE]
Do you mean compose or multiply?

Either way it sounds pretty horrible...

[ThisIsTheOne]

[QUOTE=JohnnyMo1;40836734]Wow. According to his homepage,([url]http://www.dmmm.uniroma1.it/~agostino.prastaro/HOMEPAGEPRAS.htm[/url]) Dr. Agostino Prastaro has solved Goldbach's conjecture, the twin primes conjecture, the Navier-Stokes existence and smoothness problem, the Yang-Mills existence and mass gap problem, and the Riemann hypothesis. Why has this man not received a Fields medal yet?

Oh right. Because he's nuts.[/QUOTE]

His Riemann hypothesis 'proof': [url]http://arxiv.org/pdf/1305.6845v1.pdf[/url]

I don't know what to make of it. I'd like to know what's wrong with it. It all seems like pretty standard properties of the zeta function, then some results I haven't seen before, then 'quantum mappings'.

[Joey90]

[QUOTE=ThisIsTheOne;40839350]His Riemann hypothesis 'proof': [url]http://arxiv.org/pdf/1305.6845v1.pdf[/url]

I don't know what to make of it. I'd like to know what's wrong with it. It all seems like pretty standard properties of the zeta function, then some results I haven't seen before, then 'quantum mappings'.[/QUOTE]
It looks like he's trying to extend the zeta function to a (meromorphic) function from the Riemann Sphere to itself. Unfortunately the zeta function has an essential singularity at infinity, which means that the resulting function is not meromorphic (or continuous) and so the key thing (the result about the sum of zeroes and poles with multiplicity is zero) is completely false.

I mean he basically deduces that there are only 2 (i.e. one mirrored pair of) zeroes on the critical line, which is definitely false.

[zenby]

I'm finished with my Precalculus class and I'm moving on to AP Calc BC. High school math classes are painfully slow for me and I've heard BC is difficult but I really doubt it will be any better than every other class I've taken.

[Number-41]

Aww yeah, aced my metric spaces/topology crash course exam. Took me only half a day to study it (so that's why it's a crash course and not full-blown course).

I like how the proofs suddenly become very neat when you go from metric spaces to topologies. Sets are awesome.

[Bradyns]

[B]LaTeX extension for Google Chrome.[/B]

[QUOTE][IMG]http://i.imgur.com/Itcw9li.png[/IMG][/QUOTE]

[URL]http://chrome.google.com/webstore/detail/tex-the-world-for-chromiu/mbfninnbhfepghkkcgdnmfmhhbjmhggn[/URL]

NB: mind the spaces used inside [; ;].

[; x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ;]