Wronski taken prisoner by the Russian Army??????

Józef Maria Hoene-Wroński was a Polish philosopher, mathematician, physicist, inventor, lawyer, occultist and economist. In 1794 he served in Poland's Kościuszko Uprising as a second lieutenant of artillery, was taken prisoner, and remained until 1797 in the Russian Army. After resigning in the rank of lieutenant colonel in 1798, he studied in the Holy Roman Empire until 1800, when he enlisted in the Polish Legion at Marseille. In 1803, Wroński joined the Marseille Observatory but was forced to leave the observatory after his theories were dismissed as grandiose rubbish.

Although most of his inflated claims were groundless, his mathematical work contains flashes of deep insight and many important intermediary results, the most significant of which was his work on series. The coefficients in Wroński's new series proved important after his death, forming a determinant now known as the Wronskian (the name which Thomas Muir had given them in 1882).

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tf

You looked up the name Wronski so that you could make a bait with his name??? Please get a hobby friend :<

soraol [#2]

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As suggested by the equations in (13), the determination of the constants c1and c2 in (12) depends on a certain 2 2 determinant of values of y1, y2, and their derivatives. Given two functions f and g, the Wronskian of f and g is the determinant

W = [ [ f , g ] , [ f ', g ' ] ] = fg - f ' g '

We write either W(f,g) or W(x) depending on whether we wish to emphasize the two functions or the point x at which the Wronskian is to be evaluated. For example, W(cosx, sinx) = 1.

These are examples of linearly independent pairs of solutions of differential equations (see Examples 1 and 2). Note that in both cases the Wronskian is everywhere nonzero. On the other hand, if the functions f and g are linearly dependent, with f = kg:

W(f,g) = 0

Neonfreak [#4]

You looked up the name Wronski so that you could make a bait with his name??? Please get a hobby friend :<

THEOREM 3 Wronskians of Solutions
Suppose that y1 and y2 are two solutions of the homogeneous second-order linear equation (Eq. (9))
y'' + p(x)y' + q(x)y = 0
on an open interval I on which p and q are continuous.
(a) If y1 and y2 are linearly dependent, then W(y1,y2) = 0
(b) If y1 and y2 are linearly independent, then W(y1, y2) != 0 over I

Thus, given two solutions of Eq. (9), there are just two possibilities: The Wronskian W is identically zero if the solutions are linearly dependent; the Wronskian is never zero if the solutions are linearly independent. The latter fact is what we need to show that y = c1y1 + c2y2 is the general solution of Eq. (9) if y1 and y2 are linearly independent solutions.

sentinelmain377 [#3]

tf

my linear algebra teacher kinda sucks ass so im reading a textbook. I didn't expect to find wronksi in the textbook, so I dove into his history! Then I made a post on vlr.gg because I don't want to row reduce 15 matrices by hand!

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