http://tinyurl.com/33pb6gq

I think this is a fairly stunning illustration of the way in which the 
difficulty of solving a polynomial equation skyrockets as the degree of 
the equation increases.

Post a reply to this message

Invisible wrote:
> http://tinyurl.com/33pb6gq

Actually, you know what?

http://tinyurl.com/3686x2y

Post a reply to this message

Le 10/08/2010 13:56, Invisible a écrit :
> Invisible wrote:
>> http://tinyurl.com/33pb6gq
> 
> Actually, you know what?
> 
> http://tinyurl.com/3686x2y

Solving Third & Fourth degres polynomials is known since a long time (I
have a book with that which was printed in the 60')

Now, the sad thing is it soon stop working for fifth and higher.


-- 
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Le_Forgeron wrote:

> Solving Third & Fourth degres polynomials is known since a long time (I
> have a book with that which was printed in the 60')

I'll bet to hell nobody actually *uses* the formulas above though. (!)

> Now, the sad thing is it soon stop working for fifth and higher.

I thought it only stops working in terms of "elementary" functions? As 
in, there are more specialised functions you can use to go a little bit 
higher still.

Also, I thought it was a case of being impossible to write a single 
closed-form formula for the solution to an arbitrary high-order 
polynomial. Like, if the polynomial has a special form, it might still 
be solvable (possibly easily).

Post a reply to this message

Invisible a écrit :
> Le_Forgeron wrote:
> 
>> Solving Third & Fourth degres polynomials is known since a long time (I
>> have a book with that which was printed in the 60')
> 
> I'll bet to hell nobody actually *uses* the formulas above though. (!)
> 

Probably not.  If you consider that most - if not all - computers 
calculate square roots by using an iterative method, it would probably 
be faster to use it on the original polynomial than to use it 8 times on 
smaller polynomials, as in the solution you linked to.

-- 
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Le 10/08/2010 14:17, Invisible a écrit :
> Le_Forgeron wrote:
> 
>> Solving Third & Fourth degres polynomials is known since a long time (I
>> have a book with that which was printed in the 60')
> 
> I'll bet to hell nobody actually *uses* the formulas above though. (!)
> 

Humans have a tendancy to make substitution of variables instead.

>> Now, the sad thing is it soon stop working for fifth and higher.
> 
> I thought it only stops working in terms of "elementary" functions? As
> in, there are more specialised functions you can use to go a little bit
> higher still.
> 
> Also, I thought it was a case of being impossible to write a single
> closed-form formula for the solution to an arbitrary high-order
> polynomial. Like, if the polynomial has a special form, it might still
> be solvable (possibly easily).

Yes, of course x^9 + a x^6 + b x^3 + c = 0 is just a rewrite of y^3 + a
y^2 + b y + c = 0, with y = x^3.

Solving fifth and higher requiers usually to get a first root, then
dividing the polynomial by (x-root) to reduce the degre (and start
over). Sometimes it is possible to extract a second degre polynomial
instead... that is where all the "(a+b)²=a²+2ab+b²" and the like come
handy when trying to move to the formula to a product instead of a sum)






-- 
A: Because it messes up the order in which people normally read text.<br/>
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A: Top-posting.<br/>
Q: What is the most annoying thing on usenet and in e-mail?

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Le_Forgeron <lef### [at] freefr> wrote:
> Now, the sad thing is it soon stop working for fifth and higher.

  More info on that subject:

http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

-- 
                                                          - Warp

Post a reply to this message

Warp wrote:

>   More info on that subject:
> 
> http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem

Hmm, interesting. I thought the theorum states that there is no single 
formula that covers all possible polynomials of a given degree. However, 
Wikipedia asserts something far stronger: It seems to claim that you can 
construct a single, fixed polynomial who's solutions (which are also 
fixed) cannot be expressed by radicals.

This indicates that there are some algebraic numbers that are not 
expressible by radicals. (And yet, the *definition* of an algebraic 
number is one expressible by radicals...)

Post a reply to this message

Am 10.08.2010 13:56, schrieb Invisible:
> Invisible wrote:
>> http://tinyurl.com/33pb6gq
>
> Actually, you know what?
>
> http://tinyurl.com/3686x2y

Yeah, sure. I recite that daily from memory before I go to sleep...

Post a reply to this message

clipka wrote:

> Yeah, sure. I recite that daily from memory before I go to sleep...

Now there's a question: How do you unambiguously recite complex 
mathematical expressions out loud?

Post a reply to this message

On 10/08/2010 4:04 PM, Invisible wrote:
> clipka wrote:
>
>> Yeah, sure. I recite that daily from memory before I go to sleep...
>
> Now there's a question: How do you unambiguously recite complex
> mathematical expressions out loud?

With panache :-P

-- 

Best Regards,
	Stephen

Post a reply to this message

clipka wrote:
> Yeah, sure. I recite that daily from memory before I go to sleep...

This would be a better use of your time:

http://en.wikipedia.org/wiki/Feynman_point

-- 
Darren New, San Diego CA, USA (PST)
    C# - a language whose greatest drawback
    is that its best implementation comes
    from a company that doesn't hate Microsoft.

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Invisible wrote:
> (And yet, the *definition* of an algebraic 
> number is one expressible by radicals...)

You might try looking up the definition of an algebraic number.

Post a reply to this message

Kevin Wampler wrote:
> Invisible wrote:
>> (And yet, the *definition* of an algebraic number is one expressible 
>> by radicals...)
> 
> You might try looking up the definition of an algebraic number.

Heh, like I haven't already done *that*...

Post a reply to this message

Am 10.08.2010 17:31, schrieb Darren New:
> clipka wrote:
>> Yeah, sure. I recite that daily from memory before I go to sleep...
>
> This would be a better use of your time:
>
> http://en.wikipedia.org/wiki/Feynman_point

Ah - but that's not a serious challenge; that's as easy as memorizing 
E.A.Poe's "The Raven" (sort of), plus a few words...

<http://www.cadaeic.net/cadenza.htm>

Post a reply to this message

clipka wrote:

> Ah - but that's not a serious challenge; that's as easy as memorizing 
> E.A.Poe's "The Raven" (sort of), plus a few words...

Clearly you have never had The Rhyme of the Ancient Mariner inflicted 
upon you...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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On 10-8-2010 17:31, Darren New wrote:
> clipka wrote:
>> Yeah, sure. I recite that daily from memory before I go to sleep...
> 
> This would be a better use of your time:
> 
> http://en.wikipedia.org/wiki/Feynman_point
> 
Kate does 116 if I have counted correctly: 
http://www.youtube.com/watch?v=kZSHr5E7fZY

Post a reply to this message

Invisible <voi### [at] devnull> writes:

> Kevin Wampler wrote:
>> Invisible wrote:
>>> (And yet, the *definition* of an algebraic number is one expressible
>>> by radicals...)
>>
>> You might try looking up the definition of an algebraic number.
>
> Heh, like I haven't already done *that*...

To be explicit, that's *not* the definition of an algebraic number.

Post a reply to this message

>>>> (And yet, the *definition* of an algebraic number is one expressible
>>>> by radicals...)
>>> You might try looking up the definition of an algebraic number.
>> Heh, like I haven't already done *that*...
> 
> To be explicit, that's *not* the definition of an algebraic number.

Really? I was sure an algebraic number is any number computable using 
only addition, subtraction, multiplication, division and extraction of 
roots.

Post a reply to this message

> Hmm, interesting. I thought the theorum states that there is no single
 > formula that covers all possible polynomials of a given degree. However,
 > Wikipedia asserts something far stronger: It seems to claim that you can
 > construct a single, fixed polynomial who's solutions (which are also
 > fixed) cannot be expressed by radicals.

It's an interesting fact. However, at one point, it occured to me that 
we take radicals for granted as computable. Consider sqrt(2). Somehow, 
we're OK with expressing a number as "the number which, when squared, 
equals two." Yet, we have no way of expressing it otherwise. We've all 
seen approximations to this number, we would recognize it if we saw the 
first few digits, yet we can't actually communicate it without 
describing it in terms of the solution to x*x = 2. (There may be other 
properties it has that we could use to describe it, but we still can't 
just write it down and point to it, like we can with, say, an integer. 
We can never find its precise value, we can only approximate it.)

So, it's kind of interesting that when we can't come up with a way to 
describe a number besides "one of the roots of this degree-5 
polynomial," we typically consider that less descriptive than the value 
of an expression with radicals.

  - Slime

Post a reply to this message

Slime wrote:
>  > Hmm, interesting. I thought the theorum states that there is no single
>  > formula that covers all possible polynomials of a given degree. However,
>  > Wikipedia asserts something far stronger: It seems to claim that you can
>  > construct a single, fixed polynomial who's solutions (which are also
>  > fixed) cannot be expressed by radicals.
> 
> It's an interesting fact. However, at one point, it occured to me that 
> we take radicals for granted as computable. Consider sqrt(2). Somehow, 
> we're OK with expressing a number as "the number which, when squared, 
> equals two." Yet, we have no way of expressing it otherwise. We've all 
> seen approximations to this number, we would recognize it if we saw the 
> first few digits, yet we can't actually communicate it without 
> describing it in terms of the solution to x*x = 2. (There may be other 
> properties it has that we could use to describe it, but we still can't 
> just write it down and point to it, like we can with, say, an integer. 
> We can never find its precise value, we can only approximate it.)
> 
> So, it's kind of interesting that when we can't come up with a way to 
> describe a number besides "one of the roots of this degree-5 
> polynomial," we typically consider that less descriptive than the value 
> of an expression with radicals.

I guess the point is, if you say sqrt(2), there are several simple 
algorithms that will allow you to compute exactly what number this is. 
Whereas if you say "the root of this complicated polynomial", there's no 
immediate way of computing it. (You'd have to say *which* root, for 
starters...)

The root notation is kind of a standardised way of mentioning the roots 
of particularly simple polynomials.

(Indeed, if you say Solve[x^5 - x + 1, x] to Mathematica, it replies 
with something like x = Root[1 - #1 + #1^5, 1], which simply means "the 
first root of x^5 - x + 1".)

Then again, A/B means "the solution to Bx - A = 0". And if you want to 
be pedantic about it, 768 means 7*10^2 + 6*10^1 + 8*10^0.

All of these are *representations of* numbers. The question is how 
useful they are, in a given context.

Standard roots are useful in that, for example, Sqrt(8) is clearly 
larger than Sqrt(2). But given Root[1 - #1 + #1^5, 1] and Root[1 - #1 + 
#1^5, 2], which one is larger? (Actually, at least one of those is 
complex, so it's probably not a meaningful question.) Decimal 
approximations are less accurate than fractions, yet vastly easier to 
compare and do arithmetic with. And so on.

Usually, the roots of an arbitrary polynomial aren't very easy to work 
with. But then, a 20-mile formula is no less easy to work on either.

Post a reply to this message

Slime wrote:
> It's an interesting fact. However, at one point, it occured to me that 
> we take radicals for granted as computable. 

http://en.wikipedia.org/wiki/Chaitin%27s_constant

How about a number that's trivial to describe in words, has an intuitive 
meaning, yet is provably impossible to compute?

-- 
Darren New, San Diego CA, USA (PST)
    Quoth the raven:
        Need S'Mores!

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Invisible wrote:
> 
> Really? I was sure an algebraic number is any number computable using 
> only addition, subtraction, multiplication, division and extraction of 
> roots.

http://lmgtfy.com/?q=algebraic+number&l=1


Also see the subsection:

http://en.wikipedia.org/wiki/Algebraic_number#Numbers_defined_by_radicals

Post a reply to this message

>> Really? I was sure an algebraic number is any number computable using 
>> only addition, subtraction, multiplication, division and extraction of 
>> roots.
> 
> http://en.wikipedia.org/wiki/Algebraic_number#Numbers_defined_by_radicals

Right. I think I see what happened here.

An algebraic number is any number that is the solution to a polynomial 
equation. And since I thought that all such numbers are constructable by 
radicals, I assumed "solution of a polynomial" and "constructable by 
radicals" were equivilent definitions for an algebraic number. 
Apparently they aren't equivilent. (Which still seems deeply weird to me...)

As I say, I thought the problem was that no single formula could 
encompass the solutions to every possible 5th-degree polynomial. I 
didn't realise that there are single, fixed polynomials who's solutions 
are simply inexpressible.

Post a reply to this message

Slime wrote:
> 
> So, it's kind of interesting that when we can't come up with a way to 
> describe a number besides "one of the roots of this degree-5 
> polynomial," we typically consider that less descriptive than the value 
> of an expression with radicals.
> 

I suspect that the reason here might be largely historical.  IIRC long 
ago, the concept of what number could be defined "exactly" was 
essentially derived from the Greek notion of which number could be 
constructed using only a straightedge and a compass.  It turns out that 
everything you can write with integers, addition, subtraction, 
multiplication, division and square roots can be written in this form. 
Thus if you had a number written in this form, you knew that it could be 
created with a compass-straightedge construction, but other things like 
pi, cube roots, etc. were iffier -- particularly so with roots of 
general polynomials.  This probably set the stage for looking at 
"allowed numbers" as being build by the combination of a set of basic 
functions, even when the set of allowed elementary functions was 
extended to allow things like nth-roots, exp and log which don't 
correspond to circle-straightedge constructions.  Nowadays computers 
allow us to compute roots to arbitrary polynomials relatively 
efficiently, so this way of thinking about things isn't as useful as it 
used to be.

Caveat: The above is pure conjecture.

Post a reply to this message

Invisible wrote:
>>> Really? I was sure an algebraic number is any number computable using 
>>> only addition, subtraction, multiplication, division and extraction 
>>> of roots.
>>
>> http://en.wikipedia.org/wiki/Algebraic_number#Numbers_defined_by_radicals
> 
> Right. I think I see what happened here.
> 
> An algebraic number is any number that is the solution to a polynomial 
> equation. And since I thought that all such numbers are constructable by 
> radicals, I assumed "solution of a polynomial" and "constructable by 
> radicals" were equivilent definitions for an algebraic number. 
> Apparently they aren't equivilent. (Which still seems deeply weird to 
> me...)

You'll get used to it soon enough.  I'm sure the result was surprising 
to many people when it came out.  Beautiful bit of math too.

Post a reply to this message

> I guess the point is, if you say sqrt(2), there are several simple
> algorithms that will allow you to compute exactly what number this is.

Really? Exactly? I would like to see that algorithm! =)


> Whereas if you say "the root of this complicated polynomial", there's no
> immediate way of computing it. (You'd have to say *which* root, for
> starters...)

To my knowledge (which may have big holes in this area), calculating a 
square root and finding the root of a polynomial are both typically done 
with iterative algorithms that converge on a result (and can't give you 
the "true" precise answer). In fact, I think the most common way of 
approximating either one is with Newton's method. You illustrate my 
point: even though calculating a square root and the root of a 
polynomial are similar tasks with similar limitations, you believe that 
sqrt(2) can be "computed exactly" whereas you agree that a root of a 
polynomial cannot.

The problem of *which* root does complicate things some in the 
polynomial case.


> All of these are *representations of* numbers. The question is how
> useful they are, in a given context.

Yeah. We just seem to culturally draw the line between what we consider 
useful and not useful somewhere between radicals and roots of 
polynomials, which aren't very different. I suppose there are reasons to 
have the line there, but I don't think many people realize that it's an 
arbitrary line.

Post a reply to this message

Slime <pov### [at] slimelandcom> writes:

> To my knowledge (which may have big holes in this area), calculating a
> square root and finding the root of a polynomial are both typically done
> with iterative algorithms that converge on a result (and can't give you
> the "true" precise answer). In fact, I think the most common way of

Yes. Once you're in the vicinity of a real root (to keep it simple), you
can calculate it to arbitrary precision.

Post a reply to this message

>> I guess the point is, if you say sqrt(2), there are several simple
>> algorithms that will allow you to compute exactly what number this is.
> 
> Really? Exactly? I would like to see that algorithm! =)

You can compute it to as much precision as you can compute any other 
result. In practise, that's enough.

>> Whereas if you say "the root of this complicated polynomial", there's no
>> immediate way of computing it. (You'd have to say *which* root, for
>> starters...)
> 
> To my knowledge (which may have big holes in this area), calculating a 
> square root and finding the root of a polynomial are both typically done 
> with iterative algorithms that converge on a result (and can't give you 
> the "true" precise answer). In fact, I think the most common way of 
> approximating either one is with Newton's method. You illustrate my 
> point: even though calculating a square root and the root of a 
> polynomial are similar tasks with similar limitations, you believe that 
> sqrt(2) can be "computed exactly" whereas you agree that a root of a 
> polynomial cannot.

To some extent, it's like the difference between a fraction and a 
decimal. The latter is readily comparable and easy to operate on. The 
former... less so. (E.g., is 17/21 > 1247/1918?)

If I say Sqrt(5), it is immediately obvious that one such number exists, 
is strictly positive, and is somewhere between 1 and 5. If I say "x^5 - 
3x^3 + 4x - 6 = 0", it is not immediately obvious how many unique 
solutions exist, whether any of them are purely real (and if so, what 
their sign is), what the magnitudes of these solutions are, or anything 
else. Deducing even these basic facts is quite a bit of work.

In this respect, the Sqrt() function is a kind of "standard form" for 
equation solutions. And by putting things into standard forms of one 
kind or another, they become more useful.

> Yeah. We just seem to culturally draw the line between what we consider 
> useful and not useful somewhere between radicals and roots of 
> polynomials, which aren't very different. I suppose there are reasons to 
> have the line there, but I don't think many people realize that it's an 
> arbitrary line.

Arbitrary, perhaps. (It depends on your definition of "useful", 
obviously.) In this case, it doesn't look especially arbitrary to me.

Post a reply to this message

Kevin Wampler wrote:

> Caveat: The above is pure conjecture.

Heh, all the best ones are. ;-)

I'm still trying to figure out whether there exists a geometric way to 
construct logarithms...

Post a reply to this message

> I'm still trying to figure out whether there exists a geometric way to 
> construct logarithms...

Move your pen along at a speed inversely proportional to the distance from 
where you started for X seconds.  Or something like that :-)

Post a reply to this message

On 8/10/2010 6:32 PM, andrel wrote:
> On 10-8-2010 17:31, Darren New wrote:
>> clipka wrote:
>>> Yeah, sure. I recite that daily from memory before I go to sleep...
>>
>> This would be a better use of your time:
>>
>> http://en.wikipedia.org/wiki/Feynman_point
>>
> Kate does 116 if I have counted correctly:
> http://www.youtube.com/watch?v=kZSHr5E7fZY

How many digits of pi do you need before your calculation of the volume 
of the universe is off by no more than the average volume of a water 
molecule?

Regards,
John

Post a reply to this message

On 22-8-2010 20:38, John VanSickle wrote:
> On 8/10/2010 6:32 PM, andrel wrote:
>> On 10-8-2010 17:31, Darren New wrote:
>>> clipka wrote:
>>>> Yeah, sure. I recite that daily from memory before I go to sleep...
>>>
>>> This would be a better use of your time:
>>>
>>> http://en.wikipedia.org/wiki/Feynman_point
>>>
>> Kate does 116 if I have counted correctly:
>> http://www.youtube.com/watch?v=kZSHr5E7fZY
> 
> How many digits of pi do you need before your calculation of the volume 
> of the universe is off by no more than the average volume of a water 
> molecule?
> 
> Regards,
> John

one mol of water is 18 gram is 18 ml hence volume of one molecule of 
water is (18×10^-3 /(6.02214179×10^23))=
universe 3×10^80
therefore ratio of these volumes:
3×10^80/(18×10^-3 /(6.02214179×10^23))= 10^106

:(

Post a reply to this message

>> Kate does 116 if I have counted correctly:
>> http://www.youtube.com/watch?v=kZSHr5E7fZY
> 
> How many digits of pi do you need before your calculation of the volume 
> of the universe is off by no more than the average volume of a water 
> molecule?

50 digits allows you to compute the radius of the observable universe to 
within the diammeter of a hydrogen atom.

As for whole water molecules, and volumes rather than lengths, I can't 
say off the top of my head.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Le 22/08/2010 22:01, andrel a écrit :
> On 22-8-2010 20:38, John VanSickle wrote:

>> How many digits of pi do you need before your calculation of the
>> volume of the universe is off by no more than the average volume of a
>> water molecule?
>>
>> Regards,
>> John
> 
> one mol of water is 18 gram is 18 ml hence volume of one molecule of
> water is (18×10^-3 /(6.02214179×10^23))=
> universe 3×10^80

How did you measure that ?
What's the speed of the expansion everyone is claiming to happen ?


> therefore ratio of these volumes:
> 3×10^80/(18×10^-3 /(6.02214179×10^23))= 10^106

So, we would be fine with integer of 353 bits, at least for volume
expressed in water molecule... what about an hydrogen atom instead ?

Could computer technology get past the wall of 384 bits for integer one
day ? (and what would they compute ?)

-- 
Real software engineers work from 9 to 5, because that is<br/>
the way the job is described in the formal spec.  Working<br/>
late would feel like using an undocumented external procedure.

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Am 23.08.2010 09:18, schrieb Le_Forgeron:

> Could computer technology get past the wall of 384 bits for integer one
> day ?

Computer technology already does that. Typically not in hardware, but in 
software it's state of the art.

> (and what would they compute ?)

Cryptographic stuff for instance. Current encryption and digital 
signature technology typically require integer computations with 1024 
bits or more.

It's not really difficult to implement, actually. All you need to do is 
think of a processor's "native" integer types as digits. For instance, 
on an 8-bit processor you'd perform your computations in base 256.

Post a reply to this message

On 23-8-2010 9:49, clipka wrote:
> Am 23.08.2010 09:18, schrieb Le_Forgeron:
> 
>> Could computer technology get past the wall of 384 bits for integer one
>> day ?
> 
> Computer technology already does that. Typically not in hardware, but in 
> software it's state of the art.

correction: it was state of the art 40 odd years ago.

Post a reply to this message

On 23-8-2010 9:18, Le_Forgeron wrote:
> Le 22/08/2010 22:01, andrel a écrit :
>> On 22-8-2010 20:38, John VanSickle wrote:
> 
>>> How many digits of pi do you need before your calculation of the
>>> volume of the universe is off by no more than the average volume of a
>>> water molecule?
>>>
>>> Regards,
>>> John
>> one mol of water is 18 gram is 18 ml hence volume of one molecule of
>> water is (18×10^-3 /(6.02214179×10^23))=
>> universe 3×10^80
> 
> How did you measure that ?

I didn't, but it is the value both wolfram and wikipedia give. I assume 
it is the best I can do for now.

> What's the speed of the expansion everyone is claiming to happen ?
> 
> 
>> therefore ratio of these volumes:
>> 3×10^80/(18×10^-3 /(6.02214179×10^23))= 10^106
> 
> So, we would be fine with integer of 353 bits, at least for volume
> expressed in water molecule... what about an hydrogen atom instead ?

That wasn't the question ;)

Post a reply to this message

andrel wrote:
> correction: it was state of the art 40 odd years ago.

"This machine is state of the art."

"But it doesn't work."

"Well, yes. That's the state of the art, I'm afraid."


-- 
Darren New, San Diego CA, USA (PST)
    Quoth the raven:
        Need S'Mores!

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