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From: David Bernier on 15 Jan 2010 03:39
I've been trying to get a good picture of a challenge problem
posed by van Roomen around 1594 and soon after solved by Vieta.
What's still missing is the value of the constant 'c' in
Roomen's equation, among other things ...

According to Wikipedia
<
http://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te#The_Adriaan_van_Roomen_affair
>
and the MacTutor on the History of mathematics,
< http://www-history.mcs.st-and.ac.uk/Biographies/Roomen.html >

around 1594, the Dutch mathematician Adriaan van Roomen
proposed a problem which involved solving an equation of degree 45.

This was a challenge problem to European mathematicians, and Vieta
solved it.

It had something to do with an algebraic equation involving
sin(45x) and sin(x).

According to Eli Maor's book "Trigonometric Delights",
the equation was:

x^45 - 45 x^43 + 945 x^41 -12300 x^39 + ... - 3795x^3 + 45x = c,

for some constant c. Maor goes on, referring to F. Cajori,
if c is of the form 2 sin(phi) , x is of the form 2 sin(phi/45).

This ( c = 2 sin(phi) ) is consistent with c being a Chord
in the old trigonometry (Wikipedia uses a chord function crd):

crd(theta) = 2 sin(theta/2)
Cf.:
< http://en.wikipedia.org/wiki/Chord_(geometry)#Chords_in_trigonometry >

x_k = 2sin(phi + 2pi*k/45), for k = 0, 1 ... 44 [Maor's book]. (phi or theta?)


David Bernier
From: David Bernier on 15 Jan 2010 06:50
David Bernier wrote:
> I've been trying to get a good picture of a challenge problem
> posed by van Roomen around 1594 and soon after solved by Vieta.
> What's still missing is the value of the constant 'c' in
> Roomen's equation, among other things ...
>
> According to Wikipedia
> <
> http://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te#The_Adriaan_van_Roomen_affair
> >
> and the MacTutor on the History of mathematics,
> < http://www-history.mcs.st-and.ac.uk/Biographies/Roomen.html >
>
> around 1594, the Dutch mathematician Adriaan van Roomen
> proposed a problem which involved solving an equation of degree 45.
>
> This was a challenge problem to European mathematicians, and Vieta
> solved it.
>
> It had something to do with an algebraic equation involving
> sin(45x) and sin(x).
>
> According to Eli Maor's book "Trigonometric Delights",
> the equation was:
>
> x^45 - 45 x^43 + 945 x^41 -12300 x^39 + ... - 3795x^3 + 45x = c,
>
> for some constant c. Maor goes on, referring to F. Cajori,
> if c is of the form 2 sin(phi) , x is of the form 2 sin(phi/45).
>
> This ( c = 2 sin(phi) ) is consistent with c being a Chord
> in the old trigonometry (Wikipedia uses a chord function crd):
>
> crd(theta) = 2 sin(theta/2)
> Cf.:
> < http://en.wikipedia.org/wiki/Chord_(geometry)#Chords_in_trigonometry >
>
> x_k = 2sin(phi + 2pi*k/45), for k = 0, 1 ... 44 [Maor's book]. (phi or
> theta?)

From reading the Wikipedia (Dutch) article on Adriaan van Roomen,
it seems that

C = sqrt(2+sqrt(2+sqrt(2+sqrt(2))))
with C =? 2sin(15pi/32) ,

and one of the roots found by Vieta was:

x_0 = sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(3))))) ,
with x_0 =? 2sin(pi/96).
From: David Bernier on 15 Jan 2010 22:10
David Bernier wrote:
> David Bernier wrote:
>> I've been trying to get a good picture of a challenge problem
>> posed by van Roomen around 1594 and soon after solved by Vieta.
>> What's still missing is the value of the constant 'c' in
>> Roomen's equation, among other things ...
>>
>> According to Wikipedia
>> <
>> http://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te#The_Adriaan_van_Roomen_affair
>> >
>> and the MacTutor on the History of mathematics,
>> < http://www-history.mcs.st-and.ac.uk/Biographies/Roomen.html >
>>
>> around 1594, the Dutch mathematician Adriaan van Roomen
>> proposed a problem which involved solving an equation of degree 45.
[...]

>> According to Eli Maor's book "Trigonometric Delights",
>> the equation was:
>>
>> x^45 - 45 x^43 + 945 x^41 -12300 x^39 + ... - 3795x^3 + 45x = c,
>>
>> for some constant c. Maor goes on, referring to F. Cajori,
>> if c is of the form 2 sin(phi) , x is of the form 2 sin(phi/45).
>>
>> This ( c = 2 sin(phi) ) is consistent with c being a Chord
>> in the old trigonometry (Wikipedia uses a chord function crd):
>>
>> crd(theta) = 2 sin(theta/2)
>> Cf.:
>> < http://en.wikipedia.org/wiki/Chord_(geometry)#Chords_in_trigonometry >
>>
>> x_k = 2sin(phi + 2pi*k/45), for k = 0, 1 ... 44 [Maor's book]. (phi
>> or theta?)
>
> From reading the Wikipedia (Dutch) article on Adriaan van Roomen,
> it seems that
>
> C = sqrt(2+sqrt(2+sqrt(2+sqrt(2))))
> with C =? 2sin(15pi/32) ,
>
> and one of the roots found by Vieta was:
>
> x_0 = sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(3))))) ,
> with x_0 =? 2sin(pi/96).

Maybe this is not of much general interest, but I thought I'd mention that
Google Books has a scan of a book from 1593 with van Roomen's equation.

As it turns out, van Roomen gave three examples and solved them, although
he made a mistake in his "C" value in one example. [ source is
a book on the history of Galois theory by Jean-Pierre Tignol].

These three are called something like:
Examplum primo datum, examplum duo datum and examplum tertio datum.


The fourth problem is "Examplum Quaestium" (pardon my Latin) with
a given C but no solution.

I figured bin. sqrt 2 + sqrt 3 means:
sqrt( sqrt(2) + sqrt(3) )
and
trinomiae sqrt 2 + sqrt 5 + sqrt 7
would mean:
sqrt( sqrt(2) + sqrt(5) + sqrt(7) ) ,
a square root of a trinomial.

The three examples and the challenge problem appear on a single page
at the end of the Praefatio (which is nearly 20 pages long).

Cf. "Ideae mathematicae pars prima, sive methodus polygonorum" (van Roomen)
< http://books.google.ca/books?id=YyA8AAAAcAAJ >
From: David Bernier on 19 Jan 2010 04:58
David Bernier wrote:
> David Bernier wrote:
>> David Bernier wrote:
[...]

>>> According to Eli Maor's book "Trigonometric Delights",
>>> the equation was:
>>>
>>> x^45 - 45 x^43 + 945 x^41 -12300 x^39 + ... - 3795x^3 + 45x = c,
>>>
>>> for some constant c. Maor goes on, referring to F. Cajori,
>>> if c is of the form 2 sin(phi) , x is of the form 2 sin(phi/45).
>>>
>>> This ( c = 2 sin(phi) ) is consistent with c being a Chord
>>> in the old trigonometry (Wikipedia uses a chord function crd):
>>>
>>> crd(theta) = 2 sin(theta/2)
>>> Cf.:
>>> < http://en.wikipedia.org/wiki/Chord_(geometry)#Chords_in_trigonometry >
>>>
>>> x_k = 2sin(phi + 2pi*k/45), for k = 0, 1 ... 44 [Maor's book]. (phi
>>> or theta?)
>>
>> From reading the Wikipedia (Dutch) article on Adriaan van Roomen,
>> it seems that
>>
>> C = sqrt(2+sqrt(2+sqrt(2+sqrt(2))))
>> with C =? 2sin(15pi/32) ,
>>
>> and one of the roots found by Vieta was:
>>
>> x_0 = sqrt(2-sqrt(2+sqrt(2+sqrt(2+sqrt(3))))) ,
>> with x_0 =? 2sin(pi/96).
>
> Maybe this is not of much general interest, but I thought I'd mention that
> Google Books has a scan of a book from 1593 with van Roomen's equation.
>
> As it turns out, van Roomen gave three examples and solved them, although
> he made a mistake in his "C" value in one example. [ source is
> a book on the history of Galois theory by Jean-Pierre Tignol].
>
> These three are called something like:
> Examplum primo datum, examplum duo datum and examplum tertio datum.
>
>
> The fourth problem is "Examplum Quaestium" (pardon my Latin) with
> a given C but no solution.
>
> I figured bin. sqrt 2 + sqrt 3 means:
> sqrt( sqrt(2) + sqrt(3) )
> and
> trinomiae sqrt 2 + sqrt 5 + sqrt 7
> would mean:
> sqrt( sqrt(2) + sqrt(5) + sqrt(7) ) ,
> a square root of a trinomial.
>
> The three examples and the challenge problem appear on a single page
> at the end of the Praefatio (which is nearly 20 pages long).
>
> Cf. "Ideae mathematicae pars prima, sive methodus polygonorum" (van Roomen)
> < http://books.google.ca/books?id=YyA8AAAAcAAJ >

Additionally, Google Books has a scan of Vieta's "Responsum" published
in 1595 in Latin.

The title in long form is:
<< Ad problema quod omnibus Mathematicis totius orbis
construendum proposuit Adrianus Romanus,
Francisci Vietae /Responsum/ . >>

Cf.:
< http://books.google.com/books?id=XxQ8AAAAcAAJ > .

I have trouble deciphering what is written, but others who
are knowledgeable in Latin and the history mathematics/algebra
will undoubtedly have less difficulty.

David Bernier
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