Talk About Network

Google


Register and Login
Nick
Password
Register create new account Sign up is FREE and you can post replies, new topics, bookmark posts and more!
Recover lost password


Education > Math Recreational > ^
Latest [ Topics | Posts ] Archive Post A New Topic Post a Reply
<< Topic < Post Post 1 of 1 Topic 2777 of 2832
Post > Topic >>

^

by "Jon G." <jon8338@[EMAIL PROTECTED] > May 10, 2008 at 10:26 PM

Solve,

a_0*x^0 + a_1*x^1 + a_2*x^2 + ... +a_n*x^n = 0

x = ?

Let

Q=(a_0,a_1,a_2,...,a_n)
U=(1,1,1,...,1)
E=(1,1,1/2!,1/3!,1/4!,...,1/n!)
P=(1,x,x^2,x^3,....,x^n)

u=U/|U|

by vector analysis,

P*u=|Q|^2(E*P)/[(E*Q)(Q*u)]

express P as ratios of Q,U,E

qQ+rU+sE=P

dot both sides by Q,u,E and note that P*E=e^x.
Solve the matrix,

Q*Q  Q*U  Q*E  | 0
Q*u  U*u  E*u  | {|Q|^2/[(E*Q)(Q*u)]}e^x
Q*E  E*U  E*E  | e^x

P/|P| = (s_0,s_1,s_2...)  e^x cancels

P=P_u/s_0 because x^0=1

P=(p_0,p_1,p_2,p_3,...,p_n)=(1,x,x^2,x^3,..,x^n)

x=a+bi

(a+bi)^(1/n)
=
{a^2+b^2}^(.5/n)[cos((arctan(b/a)/n)] real
+{a^2+b^2}^(.5/n)[sin((arctan(b/a)/n)]i complex

by DeMoirve

x_n+1=x_n-
(a_0+a_1*x+a_2*x^2+a_3*x^3+...+a_n*x^n)
/
(a_1+2*a_2*x+3*a_3*x^2+...+n*a_n*x^(n-1)  )

by Newton

This strategy has some flaws.  You need n vectors to construct n
dimensions, 
but I only used 3.  Also, it is only useful for large n in order to
reflect 
the contribution from E*U=e^x.  It also generates numbers too large for 
computing.

Its advantage is that it does produce some numbers to play around with 
Newton's Method.  Their initial values may or may not be accurate, but
after 
7 iterations and 50 repeats, does it really matter?

=(a_1+MMULT(s_2*TRANSPOSE(qd_2)*((r_iter_7d*TRANSPOSE(s_2))^TRANSPOSE(n_1_2))*COS(ang_iter_7d*TRANSPOSE(n_1_2)),n_2))
 




 1 Posts in Topic:
^
"Jon G." <jo  2008-05-10 22:26:04 

Post A Reply:
  Go here to Signup

AddThis Feed Button


About - Advertising - Contact - Frequently Asked Questions - Privacy Policy - Terms of Use - Signup

Contact
tan12V112 Thu Jul 24 1:44:30 CDT 2008.