Mathematician Chat V.floor(π)

[Bradyns]

Found this awesome web-book on Proofs:
[url]http://www.people.vcu.edu/~rhammack/BookOfProof/index.html[/url] [IMG]http://www.giraffeboards.com/images/smilies/sa-added/emot-engleft.gif[/IMG]

[WubWubWompWomp]

Teacher today called negative acceleration deceleration today. [I]twitch[/I]

[JohnnyMo]

[QUOTE=WubWubWompWomp;39988751]Teacher today called negative acceleration deceleration today. [I]twitch[/I][/QUOTE]

So?

[Number-41]

As long as he doesn't use [URL="http://en.wikipedia.org/wiki/Jerk_(physics)"]jerk[/URL], [URL="http://en.wikipedia.org/wiki/Jounce"]jounce[/URL],[URL="http://en.wikipedia.org/wiki/Snap,_Crackle,_and_Pop#Physics"] snap crackle & pop[/URL], it's okay. :v:

[Guldfisk]

Guys I have a massive problem and I have no idea how to solve it, been sitting with it for around an hour now still no answer. Ok here goes;

A company will create 100.000 boxes with the volume 600cm^3. The boxes will have a cuboid shape with a square bottom.
The material price as is follows: (note kr is the currency)
the bottom: 1000kr/m^2
the sides: 600kr/m^2
the lid: 800kr/m^2
What measurements will yield the lowest possible cost?

I've not really gotten far but this is what I've got:
The height h = (600/x^2)

That's basically it, fast answers are really appreciated, even if you can't solve it but just so I know people has read this.
Also math test tomorrow, we've been working with derivatives and integrals

[Krinkels]

[QUOTE=IQ-Guldfisk;39992389]Guys I have a massive problem and I have no idea how to solve it, been sitting with it for around an hour now still no answer. Ok here goes;

A company will create 100.000 boxes with the volume 600cm^3. The boxes will have a cuboid shape with a square bottom.
The material price as is follows: (note kr is the currency)
the bottom: 1000kr/m^2
the sides: 600kr/m^2
the lid: 800kr/m^2
What measurements will yield the lowest possible cost?

I've not really gotten far but this is what I've got:
The height h = (600/x^2)

That's basically it, fast answers are really appreciated, even if you can't solve it but just so I know people has read this.
Also math test tomorrow, we've been working with derivatives and integrals[/QUOTE]

Keep in mind you have two measurements here, the lengths of the bottom (which is only one because it is square) and the height.
Let these be x and y respectively.
The volume measurement gives this first equation:
xy = 600
By dividing you can express y in terms of x and vice versa.
Use this to write the cost as a function for x or y but not both.
Then differentiate this function to find the minimum.

[timman]

Started La Place transforms in differential equations class!

They're pretty fun even if they do take a while to complete.

[account]

[QUOTE=Krinkels;39992919]Keep in mind you have two measurements here, the lengths of the bottom (which is only one because it is square) and the height.
Let these be x and y respectively.
The volume measurement gives this first equation:
xy = 600
By dividing you can express y in terms of x and vice versa.
Use this to write the cost as a function for x or y but not both.
Then differentiate this function to find the minimum.[/QUOTE]
x^2*y, not xy

[WubWubWompWomp]

[QUOTE=JohnnyMo1;39989927]So?[/QUOTE]
It's just a stupid pet peeve.

[Bradyns]

Handed my assignment in yesterday, was due in 3hours ago. :)

Question:
[QUOTE][IMG]http://i.imgur.com/wPi2pAg.png[/IMG][/QUOTE]

[B]Let[/B]
[IMG]http://latex.codecogs.com/gif.latex?f(x)=\frac{1}{x}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?g(x)=\frac{1}{x^{3}}[/IMG]

[B]Let
[/B]
[IMG]http://latex.codecogs.com/gif.latex?f(x)g(x)=\frac{1}{x}\cdot \frac{1}{x^{3}}=\frac{1}{x^{4}}[/IMG]

[B]=========================================[/B]

[B]For f(x):
[/B]
[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{-}}\frac{1}{x}=-\infty[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{+}}\frac{1}{x}=\infty[/IMG]

[B]=========================================[/B]

[B]For g(x):[/B]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{-}}\frac{1}{x^{3}}=-\infty[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{+}}\frac{1}{x^{3}}=\infty[/IMG]

[B]=========================================[/B]

[B]f(x) · g(x):
[/B]
[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{-}}\frac{1}{x^{4}}=\infty[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{+}}\frac{1}{x^{4}}=\infty[/IMG]

[B]=========================================
[/B]
Given that the product of f(x) and g(x) converges at infinity, the limit at 0 exists at infinity, even though a limit does not exists for f(x) or g(x) at 0.

Yes/no/maybe?

[Cooty]

[QUOTE=Bradyns;39998129]Handed my assignment in yesterday, was due in 3hours ago. :)

Question:

[B]Let[/B]
[IMG]http://latex.codecogs.com/gif.latex?f(x)=\frac{1}{x}[/IMG]
[IMG]http://latex.codecogs.com/gif.latex?g(x)=\frac{1}{x^{3}}[/IMG]

[B]Let
[/B]
[IMG]http://latex.codecogs.com/gif.latex?f(x)g(x)=\frac{1}{x}\cdot \frac{1}{x^{3}}=\frac{1}{x^{4}}[/IMG]

[B]=========================================[/B]

[B]For f(x):
[/B]
[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{-}}\frac{1}{x}=-\infty[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{+}}\frac{1}{x}=\infty[/IMG]

[B]=========================================[/B]

[B]For g(x):[/B]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{-}}\frac{1}{x^{3}}=-\infty[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{+}}\frac{1}{x^{3}}=\infty[/IMG]

[B]=========================================[/B]

[B]f(x) · g(x):
[/B]
[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{-}}\frac{1}{x^{4}}=\infty[/IMG]

[IMG]http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow 0^{+}}\frac{1}{x^{4}}=\infty[/IMG]

[B]=========================================
[/B]
Given that the product of f(x) and g(x) converges at infinity, the limit at 0 exists at infinity, even though a limit does not exists for f(x) or g(x) at 0.

Yes/no/maybe?[/QUOTE]

Looks convincing enough to me - you probably could find an easier example, but it seems to work. If both sides converge to the same value, then the limit at least exists, even if it is infinite.

[Joey90]

[QUOTE=Bradyns;39998129]
Given that the product of f(x) and g(x) converges at infinity, the limit at 0 exists at infinity, even though a limit does not exists for f(x) or g(x) at 0.

Yes/no/maybe?[/QUOTE]
It depends whether you're treating a limit of 'infinity' as a genuine limit or not... Generally in real analysis I wouldn't*, (whereas in complex analysis I'd be more inclined to, but then +infinity = -infinity so your example wouldn't work anyway!)

A better example, assuming you can pick any functions you like, would have a discontinuity at a, but not go to infinity... Something like a step function going up and a step function going down.

*e.g. You say the harmonic series diverges, not that it converges to infinity. I mean, it's fine to write it etc. but if the question specifically asks for a function to [i]converge[/i], that usually means to a finite value.

[Bradyns]

[QUOTE=Joey90;39998313]It depends whether you're treating a limit of 'infinity' as a genuine limit or not... Generally in real analysis I wouldn't*, (whereas in complex analysis I'd be more inclined to, but then +infinity = -infinity so your example wouldn't work anyway!)

A better example, assuming you can pick any functions you like, would have a discontinuity at a, but not go to infinity... Something like a step function going up and a step function going down.

*e.g. You say the harmonic series diverges, not that it converges to infinity. I mean, it's fine to write it etc. but if the question specifically asks for a function to [i]converge[/i], that usually means to a finite value.[/QUOTE]

I didn't know if we could pick piecewise functions or not. :/

So, I stuck with infinity.

[Joey90]

[QUOTE=Bradyns;39998355]I didn't know if we could pick piecewise functions or not. :/

So, I stuck with infinity.[/QUOTE]
I mean, if you want it to not converge, you're going to need some kind of discontinuous function...

If you want an example that works in pretty much any situation (real or complex) and doesn't have [i]any[/i] limit from either direction, try sin(1/x) and 1/sin(1/x)

You should check your definition of convergence to see if infinite values are allowed, if they are then of course it's fine (and even if they're not, you're at least getting the general idea of the question!)

[Jo The Shmo]

[media]http://www.youtube.com/watch?v=bFNjA9LOPsg[/media]

only in indiana...

[titopei]

So, my friend says 1*10^-781 is not greater than 0 because a number that small cannot be applied to real life. What are your guys' thoughts

[harryh11]

Assume 10^-781 = 0
Invent a new unit of measurement: the Harbit (Hb).

1 Hb = 10^781 metres.

Now, if I'm one meter away from you, I'm 10^-781 Harbits away from you.
But 10^-781 = 0 by your friend, so I'm 0 Hb = 0 meters away from you.
CONTRADICTION!!!!!!

Alternate proof:
It is obvious that 10^-781 > 0.

[Bradyns]

[QUOTE=titopei;40036402]So, my friend says 1*10^-781 is not greater than 0 because a number that small cannot be applied to real life. What are your guys' thoughts[/QUOTE]

Google agrees with him..

[IMG]http://i.imgur.com/URxhazF.png[/IMG]

I wonder what Wolfram has to say?

[IMG]http://i.imgur.com/ZtFwg4K.png[/IMG]

Conclusion:
Your friend is an idiot.

[JohnnyMo]

[QUOTE=titopei;40036402]So, my friend says 1*10^-781 is not greater than 0 because a number that small cannot be applied to real life. What are your guys' thoughts[/QUOTE]

Your friend does not seem terribly sharp.

[editline]25th March 2013[/editline]

[QUOTE=harryh11;40036673]Alternate proof:
It is obvious that 10^-781 > 0.[/QUOTE]

Alternative alternative proof:
The proof is left as an exercise to the reader.

[editline]25th March 2013[/editline]

Alternative alternative alternative proof:
Ask if your friend agrees that 0 < 1.

If 10^-781 > 0, is it less than zero? I doubt he'd concede that, so likely he'd say it's equal to zero. In that case, isn't 2*10^-781 = 10^-781, since the difference between them is the same as the difference between 10^-781 and 0? Then continue that process to find that 10^781 * 10^-781 = 1 = 0.

Then apply the principle of explosion using the contradiction "0 < 1 and 0 = 1" to prove the statement, "your friend is fucking retarded."

[Krinkels]

Wasn't ultrafinitism a little like this?

[mastoner20]

This reminds me of a debate I had with a pretty good friend of mine who's a Gen-ed student at my school about why .999... = 1. He told me I was full of crap, even after showing him the formal proof and two 'Engineering Proofs' as my Structures teacher calls them.

[JohnnyMo]

[QUOTE=mastoner20;40038326]This reminds me of a debate I had with a pretty good friend of mine who's a Gen-ed student at my school about why .999... = 1. He told me I was full of crap, even after showing him the formal proof and two 'Engineering Proofs' as my Structures teacher calls them.[/QUOTE]

Maybe all universities should require every major to take a class called "Introduction to Math Not Giving a Shit About Your Intuition."

[mastoner20]

All I know was it was highly annoying. I finally got him with 'what is 1/3 plus 1/3 in decimal form.' Of course, he answered .666.... So I asked him to add another 1/3. He finally realized that .999... was in fact 1. oed.

[JohnnyMo]

Oxford English Dictionary?

[mastoner20]

Haha. Oper edei deixai. Greek form of QED.

[Mr Skeltal]

If I were a vampire, calculus would be my stake

[mastoner20]

It's Greek... it's everywhere in math... how is that pretentious, or out of place?

[mastoner20]

Ah. I've always done that for the end of my formal proofs (using the Greek letters, which might seem a touch more pretentious), and an upside-down question mark as an inside joke with my proff for cases/induction steps.

[AxeI]

For some reason, to me it just makes one seems like a dick when one concludes a demonstration by QED.

[JohnnyMo]

I use a square