Random Thoughts about Math

John Creighton · 2004-05-10 19:41:49

Say anything

Palomar · 2004-05-10 19:52:30

Say anything

*Okay.

My concept of Hell (eternal punishment) would be having to do math problems for eternity.

--Cindy

Byron · 2004-05-11 04:55:50

Say anything
*Okay.
My concept of Hell (eternal punishment) would be having to do math problems for eternity.
--Cindy

LOL...that's a good one, Cindy...

Funny to think that for those who excel in math, the opposite would be true...it'd be heaven to do math for eternity...lolol.

B

Palomar · 2004-05-11 05:45:34

Say anything
*Okay.
My concept of Hell (eternal punishment) would be having to do math problems for eternity.
--Cindy
LOL...that's a good one, Cindy...
Funny to think that for those who excel in math, the opposite would be true...it'd be heaven to do math for eternity...lolol.
B

*Hi Byron.

Would the mathematicians in Heaven be so kind as to pray me out of Hell? :laugh:

Yep, I remember my dismay when, around the 3rd (?) grade, math problems were no longer presented in the horizontal format (which was -so- comfy by that point). Now we've got to do math vertically. :-\ And then when *letters of the alphabet* got thrown into the mix! Geez. Talk about the ultimate mind-scramble. :hm:

--Cindy

Shaun Barrett · 2004-05-11 07:15:01

I think that, within certain limits, most people's grasp of mathematics depends very largely on the way in which it was presented to them in school.
A good mathematics teacher is very definitely born, not made!

bolbuyk · 2004-05-11 11:54:37

Does anybody know the way to solve a cubic equation, ti determining the unknown x from the formula ax^3+bx^2+cx+d=0? There most exist a straightforward way to do this, but I can't find.

Euler · 2004-05-11 12:32:15

Does anybody know the way to solve a cubic equation, ti determining the unknown x from the formula ax^3+bx^2+cx+d=0? There most exist a straightforward way to do this, but I can't find.

There is a cubic formula, but it is a lot more complicated than the quadratic fomula, and I would have to look it up. There is also a quartic formula, though there is no formula for quintics and higher.

Euler · 2004-05-11 13:04:05

Start out with x^3+px^2+qx+r=0 (if the first coefficient is not one then just divide everything by it). The first step is to turn it into a "depressed cubic" with the form x^3+ax+b=0. a= q - (p^2)/3 and b= 2(p^3)/27-pq/3+r. For the solution let A= cuberoot(-b/2+sqroot((b^2)/4+(a^3)/27)) and let B= cuberoot(-b/2-sqroot((b^2)/4+(a^3)/27)). The solutions to the cubic equation are x=A+B, x= -(A+B)/2+sqroot(-3)(A-B)/2, and x=-(A+B)/2-sqroot(-3)(A-B)/2

For a proof and some history read Journey Through Genius by William Dunham (one of the most interesting math books that I have read).

Mundaka · 2004-05-11 17:53:58

Palomar · 2004-05-11 19:02:15

*Hmmmmm. Interesting thoughts there, Mundaka.

Actually, I shouldn't feel -too- bad. I caught a doctor saying the other day, "one-third of a centimeter." So then I was like, "Um...how do I type that?" "0.33 cm"? But then that'd just total to 0.99 with 0.001 missing "in total." So I just typed it out "one-third of a centimeter" (his words, after all).

At least I hope I'm recalling correctly that metrics don't have fractions! :-\

Numbers...UGH. I hate them.

--Cindy

Cobra Commander · 2004-05-11 19:13:08

Actually, I shouldn't feel -too- bad. I caught a doctor saying the other day, "one-third of a centimeter." So then I was like, "Um...how do I type that?" "0.33 cm"? I just typed it out "one-third of a centimeter" (his words, after all).

Funny you should mention that. About a year at work I just casually estimated a measurement as being about a third of an inch. The rest of the engineering staff looked at me like I was reciting Shakespeare in Klingon. We argued the better part of the day about it, them claiming you can't have a third of an inch, .333333 ad infinitum.

I of course maintained that of course you can divide an inch into three equal parts.

Totally different ways of looking at the world.

Josh Cryer · 2004-05-11 20:11:07

Hmm, I'd think engineer's would realize that there is only so much accuracy that is useful. As a funny mathematician once said that 50 or 100 digits of pi would be able to essentially measure the ratio of the circumference of the universe to within one electron orbital or something ridiculous like that.

So, for all intents, .3333 is probably more useful than most engineering projects require.

My favorite mathematical curiosities are primes.

I've been learning calculus using the infintesimal approach, it feels more relaxing than that of limits (actually, calculus simply didn't make sense at all before I found this particular textbook), but I keep getting stuck and having to reference my algebra books, so I only read it when I really feel like wasting time, on the weekends.

Mundaka · 2004-05-11 21:04:07

bolbuyk · 2004-05-12 05:50:19

Start out with x^3+px^2+qx+r=0 (if the first coefficient is not one then just divide everything by it). The first step is to turn it into a "depressed cubic" with the form x^3+ax+b=0. a= q - (p^2)/3 and b= 2(p^3)/27-pq/3+r. For the solution let A= cuberoot(-b/2+sqroot((b^2)/4+(a^3)/27)) and let B= cuberoot(-b/2-sqroot((b^2)/4+(a^3)/27)). The solutions to the cubic equation are x=A+B, x= -(A+B)/2+sqroot(-3)(A-B)/2, and x=-(A+B)/2-sqroot(-3)(A-B)/2
For a proof and some history read Journey Through Genius by William Dunham (one of the most interesting math books that I have read).

Thank you very much!

If I´m right, some solutions can be complex. Is that right?

Cobra Commander · 2004-05-12 05:59:13

Hmm, I'd think engineer's would realize that there is only so much accuracy that is useful.

Yeah, but at least in my experience throwing non-nominal numbers into the mix screws with 'em.

One of these days I'll draw up something in thirds and sevenths.

Euler · 2004-05-12 14:26:54

If I´m right, some solutions can be complex. Is thhat right?

If (b^2)/4+(a^3)/27>0, there is a real root and 2 complex conjugate roots.
If (b^2)/4+(a^3)/27=0, there are 3 real roots, at least 2 of which are equal.
If (b^2)/4+(a^3)/27<0, there are 3 real and unequal roots.

Euler · 2004-05-12 16:01:57

Am I the only math major here?

Shaun Barrett · 2004-05-12 18:33:00

Have mercy, Euler!
Some of us haven't dealt with cubics and quartics for over 30 years.
(Now, if I could just find my highschool class notes! )

Mundaka · 2004-05-13 02:51:06

bolbuyk · 2004-05-13 07:55:54

I've an arithmetic problem for many many years. It's the following:

Consider a natural number (not 0), call it n(i). If it's odd, then : n(i+1)=3*n(i)+1
Of it's even, then n(i+1)=n(i)/2

Continuing this, I'm nearly sure every natural number will end to 1. Try it with eg 27 (without calculator!!!).

But can I prove this? I'm trying for years. I have proven that every number will lead to a number expressed as x+4 where x is a 6-time number, except binary numbers (1,2,4,8,16,eg). When proving that every x+4 number leads to 1, we're there.

But intuition says, it shall turn out to be a proof as follow:
First proove that by doing this no number will return to itself (except 1,2 and 4)
Then proove no number will reach another number that is a certain amount bigger than the number you start with.

In that case it must follow every number will reach 1.

Another possibility is show it for 2 and proove that it works for some n+1 if it works for n.

Any ideas? :band:

Euler · 2004-05-13 11:57:47

I think I have seen this problem before. If I recall correctly, it is a fairly well known unsolved problem. So I will work on it, but there is no guarantee that I will be able to solve it.

Euler · 2004-05-13 12:56:23

http://en.wikipedia.org/wiki/Collatz_conjecture]Collatz conjecture(wikipedia)

Paul Erdos said "Mathematics is not yet ready for such problems." What does he know? I haven't given up.

Mark Friedenbach · 2004-05-13 15:30:09

Simplified a little bit, what you need to prove is that for any prime number n, n = 3n+1 can be applied finite number of times to get a power of two.

What do you mean by "6-time number"?

I think Paul Erdos is being a bit hasty... if it hasn't been proven yet that just means that the solution is non-trivial. But I'm a engineering major not a math major, so what do I know. :;):

Euler · 2004-05-13 16:00:54

Simplified a little bit, what you need to prove is that for any prime number n, n = 3n+1 can be applied finite number of times to get a power of two.

No that is not an equivalent problem, the +1 part makes it so that the order in which the steps happen matters.

Mark Friedenbach · 2004-05-13 22:57:48

Simplified a little bit, what you need to prove is that for any prime number n, n = 3n+1 can be applied finite number of times to get a power of two.
No that is not an equivalent problem, the +1 part makes it so that the order in which the steps happen matters.

I'm pretty sure if you replace "any prime number n" with "any number n larger than 2" it is equivalent.

Here's a related site: http://personal.computrain.nl/eric/wondrous/]On 3x + 1 Problem

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