| Author |
 Topic  |
|
|
uberclay
Advanced Member
Canada
159 Posts |
 Posted - 08/21/2007 : 23:54:01
|
Heya guys! Long time, no...
Anyway, I am reading "Unknown Quantity" by John Derbyshire. His book states that it is for the non mathematician but, as part of this prestigious group, I am baffled by the first cubic he presents.
He first gives a general cubic, x + Px + Qx + R = 0
And then presents it as a depressed cubic by:
(x + P/3) + (Q - P/3)(x + P/3) + (R - QP/3 + 2P/27)
I can see that he has substituted (x + p/3) for x in the original but I can't resolve the steps he used to eliminate Px.
My experience with cubics is thus far limited to decomposition - can anyone find the time to explain to me how he got from the general from to the depressed form he presented?
Thank you for your time.
Clay
P.S. If anyone has the book - it's page 58. |
 |
|
|
tkhunny
Advanced Member
USA
1001 Posts |
Posted - 08/22/2007 : 00:15:44
|
It's a piece of cake, Uber. You'll kick yourself.
From x + Px + Qx + R = 0
Just make a simple substitution and MAKE IT get rid of the quadratic term. Substitute x = y + A
(y+A) + P(y+A) + Q(y+A) + R = 0
Expand all that. Collect the y terms. You should get y(3A + P). The point of the exercise was to eliminate the quadratic term, so pick 3A + P = 0, or A = -P/3.
Therefore, if you make the substitution x = y - P/3, the quadratic term will vanish when the expression is in powers of y. |
|
|
|
uberclay
Advanced Member
Canada
159 Posts |
Posted - 08/22/2007 : 15:44:32
|
| Thanks tk. I'll give it a go. |
|
|
|
uberclay
Advanced Member
Canada
159 Posts |
Posted - 08/22/2007 : 23:35:06
|
| Thanks David. The section on Cardano's method was extremely helpful. |
|
|
| |
Topic |
|
Interactive Math Goodies Software
Topic Locked
