*                           ONOMA:ELVIS PAPA                                  *
*                           AEM: 952									      *
*                           email:papa@uth.gr                                 *
*                           Tmhma: A				                          *
*********************************QUIZ 2****************************************

1) ->D 
    Dikaiologhsh: 
	Θεώρημα Ενδιάμεσης Τιμής(1-D):
    Αν f συνεχής στο [α, b] και sign(f(α))? sign(f(b)) τότε υπάρχει ένα τουλάχιστον x* στο [α, b] τ.ω. f(x*)=0.
	
2) -> B
    Dikaiologhsh: 
	f(x)=xe-x - 0.3
				    a  f(a)    b    f(b)     c    f(c)
	1h epanaphpsh:	1  0,0678  5    -0,2664  3    -0,1507
	2h epanaphpsh:	1  0,0678  3    -0,1507  2    -.0,0293
	3h epanaphpsh:  1  0,0678  2    -0,0293  1,5  0,0347
	
3) -> D

4) -> C
	Dikaiologhsh: 
	f(x)=x^2
	Den boroume na efarmosoume thn methodo tou Newton afou h synarthsh mas einai pada mh arnhtikh me monadikh
	riza x t.w. f(x)=0 na einai to x=0.Synepws h methodos tou Newton den borei na efarmostei.
	
5) -> C 
	Dikaiologhsh: 
	F(X)=X^2-R,apo methodo tou newton exoume:Xi+1=Xi-F(Xi)/F'(Xi) =) Xi+1=Xi-Xi^2-R/2Xi =)
	Xi+1=(2Xi^2-Xi^2+R)/2Xi=1/2(Xi+R/Xi)
	
6) -> C
    Dikaiologhsh: 
	F(X)=X^2-4 , X1=3 , F(X1)=5 , F'(X1)=6
	
	X2=X1-F(X1)/F'(X2) =) X2=3-6/5=2,167
	
7) -> B
	Dikaiologhsh: 
	Apo methodo tou newton exoume: X3+1=X3-F(3)/F'(3)=X3-F(3)/tan57=3-5/1.5328=-0.247
	
8) -> Pinakas
	Dikaiologhsh: 
	Gia arxikh timh 0 h methodos tou Newton adynatei na lusei to F(x)=0 giati F'(0)=0 kai to F'(0)=0
	vriskete ston paranomasth.
	Gia arxikh timh 5 exoume:
	
		           x		F(x)		F'(x)		h
				   5		121			75			1.613333
				   3.38867  34.843422	34.408601   1.012636
				   2.374034 9.380144    16.908112   0.554771
				   1.819263 2.021247    9.929153    0.203566
				   1.615697 0.217739    7.831430    0.027803
				   1.587894 0.003727    7.564222	0.0004927
				   
	Gia arxikh timh -2 exoume:
	
					x		  F(x)		  F'(x)		    h
					-2		  4			  12			0.333333
					-2.333333 -16.703698  -16.333328	1.022675
					-3.356008 -41.798012  33.788369 	-1.237053
					-2.118955 -13.514045  13.469910 	-1.003276
					-1.115685 -5.388752	  3.734259		-1.443057
					0.327372  -3.964914	  0.321517		-12.33189
					
	Parathroume pws o algorithmos Newton gia thn sygkekrimenh periptwsh apotygxanei logw mh kalou arxikou shmeiou
	afou to h apoklinei sinexws apo to 0.
	
				   
  9) -> D
	Dikaiologhsh: 
	Parathrwntas tis times twn parapanw pinakwn vlepoume oti me prosegish 4 epitygxabetai akriveia se 2 
	shmadika pshfia.
	
  10) -> A
	Dikaiologhsh:
	F(x)x^2-R , F'(x)=2x
	Apo methodo temnousas exoume : xi+1=xi-F(xi)/[(F(xi)-F(xi-1)))/(xi-xi-1)] =) xi+1=xi-xi^3-xi^2*(xi-1)-R*xi+R*xi-1 ... prakseis.....
	=) xi+1=(R+xi*xi-1)/xi+xi-1
	
  11) -> A
	Dikaiologhsh:
	F(x)=x^2-4 , xn=3 , xn-1=4
	Apo mthodo temnousas exoume : xn+1=xn-F(xn)/[F(xn)-F(xn-1)/xn-(xn-1)] = xn - ((xn^2)-4)*(xn-(xn-1))/xn^2-(xn-1)^2
	= 3 - (3^2)-4*(3-4)/9-16= 0.5714
	
  12) -> B
    Dikaiologhsh:
	Xo=3 , F'(3)=tan57=1.5398
	Apo methodo temnousas kai lamvanwdas ypopsh tis parapanw synthhkes exoume:
	xn+1=xn-F(xn)/1.5398 = 3 - 5/1.5398 = -0.24704
	
  13) -> B
     Dikaiologhsh:
	 Parathrwdas thn grafikh parastash ths synarthshs F(x)=cosx parathroume pws sta shmeia π/4 kai 3π/4 isyei
	 F(x)!=0 kai epipleon F'(π/4)=0 kai F'(3π/4)=0.Synepws oi arxikes prosegiseis π/4 kai 3π/4 den einai katallhles
	 me thn methodo ths temnousas.