Επίλυση εξίσωσης 3ου βαθμού...

[desolatorXT]

Prospathw na kanw to programma FS calculator na linei e3iswseis 3ou vathmou.... Prospathisa na vrw mia methologia kai kateli3a sta e3is:

 

theloume na lisoume to ax^3+bx^2+cx+d=0

 

ean d = 0, mia lisi einai opwsdipote i x =0

ean a = 0 tote linoume tin e3iswsi 2ou vathmou apla.

 

Se periptwsi twra pou kanenas oros den 0 (apo tous a,b,c,d)

 

exoume to e3is...

 

ax^3+bx^2+cx+d = 0 ->

x^2(ax+b )+cx+d = 0 ->

x^2(ax+b )+(cx+d)=0

 

Gia na einai mia prosthesi isi me to 0, prepei kai oi 2 oroi tis na einai eite 0, eite o enas antithetos apo ton allon. Opote gia na einai 0, prepei:

 

ax+b = 0 kai cx+d = 0, dld x= (-b )/a = (-d)/c opote ean (-b )/a = (-d)/c, tote exoume mia lisi, tin x = (-b )/a

 

Ean twra den isxiei ayto exoume:

 

x^2(ax+b )= - (cx+d)

 

(efoson to x den einai iso (-d)/c diairoume me (cx+d) kai exoume:

 

x^2 (ax+b )/(cx+d) = -1

 

Efoson to x^2 einai panta >0 gia na exei lisei mia e3iswsi tritou vathmou prepei na isxyei to

 

(ax+b )/(cx+d)<0

 

Gia na simvainei ayto, prepei:

 

1) To (ax+b ) na einai <0 KAI to (cx+d)>0 -> x<(-b )/a kai x>(-d)/c

 

2) To (ax+b ) na einai > 0 KAI to (cx+d)<0 -> x>(-b )/a kai x<(-d)/c

 

Se kathe ena apo ayta ta pedia, yparxei toulaxiston mia lisi ti e3iswseis!

 

Opote apo ayto gnwrizoume, oti mia lisi kimainete sto ena pedio, kai mia sto allo. Kai ta 2 pedia vevaia kaliptoun olo to sinolo R, alla, to ena pedio, einai poli mikro. Opote mporei eykola to pc, se liga deytera na vrei tin mia lisi pou anikei ekei.

 

παραδειγμα. -1x^3+2x^2-2x+1=0

 

a = -1

b = 2

c = -2

d = 1

 

to (-b )/a=-2/-1=2

 

to (-d)/c=-1/-2= 0,5

 

opote prepei x<2 kai x>0,5

 

i x>2 kai x<0,5

 

Sto pedio omws 0,5<x<2, mporoume eykola na vroume tin lisi... Vazontas loipon to pc na psa3ei olous tous arithmous sto poli mikro pedio ayto, de tha parei xrono, k to pc den tha argei na vrei tin lisi, pou einai to 1.

 

Opote me ayton ton tropo, mporei na vrethei, mia toulaxiston apo tis 3 liseis tis e3iswseis, se poli mikro xrono.

 

Edw kateli3a egw... An borei kapios na vrei mia kaliteri lisi, i na sinexisei ayton ton silogismo, isws mporesoume na ftasoume se ena simeio, opou apokleiontas times, tha ftasoume ston ypologismo lisewn apo to pc eykola k grigora. Tha itan eykolo na pw ston kwdika

 

for x=-999999 until x = 999999 me vima 0.0001

kane tis pra3eis kai

an i parastasi einai isi me to 0

tote to x ayto einai lisi

 

Alla ayto den einai katholou akrives, kai tha parei wres gia na ektelestei apo to pc.

 

Eyxaristw prokatavolika

[FarCry]

To kourazeis poli....

 

 

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

[desolatorXT]

To kourazeis poli....

 

 

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

 

ayti den einai mia geniki periptwsi...

 

o paragwntas tou x^3 einai isos me 1. Opote den mou kanei.

 

egw thelw na lisw tin e3iswsi ax^3+bx^2+cx+d=0

 

thnx pants

 

Tha pw ston kwdika oti an a=1, tote i liseis einai aytes pou dineis

[FarCry]

We first divide the standard equation by the leading coefficient to arrive at an equation of the form

 

.............................................................

Diabase ligo de blaptei

[desolatorXT]

.............................................................

Diabase ligo de blaptei

 

sry den prose3a to link, nomiza itan sto signature sou....

 

kai ektos aytou, kk mi fwnazeis de s eipa kai kampouri.

 

thnx pantws..

[FarCry]

De fonaksa. Aplos mi psaxneis na efeureis to troxo. Exete to sto mialo sou gia to mellon

[desolatorXT]

diavasa tis liseis...

 

Kamia apo aytes den einai katalili gia programmatismo...me e3eresi tin "Solution in terms of a, b, c, and d" kai tin "Lagrange resolvents"

 

pou kai aytes den kanoun gt thelw tis rizes mono sto sinolo twn pragmatikwn airhtmwn. Oi ypoloipes metholodogies, apaitoun dianoitikes ergasies apo ton xristi, pou o ypologistis den mporei na kanei... Paradeigma:

 

Suppose that we can find numbers u and v such that:

u^3-v^3=q uv=p/3

 

kati tetio den mporei na ylopoeithei apo ton ypologisti.

 

3erei mipws kaneis pws mporoume na xrisimopoiousoume tin diakrinousa wste na vroume tis liseis?

[desolatorXT]

Oswn afora ayta pou les stin eikona.... (in summary pou leei)

 

Exei kaneis kapion kwdika gia na ypologisw tin triti riza enos arithmou???

[FarCry]

http://www.java2s.com/Code/C/math.h/cbrtreturnsthecuberootofnum.htm

[alkisg]

> Exei kaneis kapion kwdika gia na ypologisw tin triti riza enos arithmou???

 

Αρκεί να υψώσεις στη δύναμη 1/3.

http://www.cplusplus.com/reference/clibrary/cmath/pow.html

[FarCry]

Epeidi i cbrt einai sto C99 protipo kai prepei na exeis enan simbato compiler opos einai o gcc, kaneis tin ilopoiisi opos anaferei kai o alkisg an de tin ipostirizei.

 

>#define  pow(x,y)  (exp(log(x)*(y)))
#define  cbrt(x)   (pow((x),1./3.))

 

The definitions given above are a "poor man's" solution to the problem but acceptable in many situations

[windward]

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

[desolatorXT]

thnx alla:

 

1) den grafw se C++ ()

2) vrika tin lisi : apla kaneis power(x,(1/3))

 

thnx gia oli tin voitheia