GRAFICA DE FUNCIONES1. Grafique con color fucsia la siguiente funcin.

Adems coloque como ttulo al grafico Ondas Sonoras en negrita y con tamao de letra 20. Agregue en el eje x la palabra Tiempo y en el eje y la palabra Amplitud, ambas en negrita.SOLUCION:clcclearx=[0:0.02:50];y1=sin(x/2)./[exp(x/100)];plot(x,y1,'m','linewidth',2);hold ongrid on title('\bf\fontsize{20} Ondas Sonoras')xlabel('\bf\fontsize{15} tiempo')ylabel('\bf\fontsize{15} amplitud')hold off

2. Graficar la siguiente funcin definida por tramos usando hold on. Ademas aumente un poco el grosor de la lnea y use un color diferente para cada tramo. (rojo, negro, fucsia)

SOLUCION:clcclearx=[-12:0.02:-7];y1=(x+9).^2-8;plot(x,y1,'r','linewidth',3);hold onx=[-7:0.02:0]; y2=-sqrt(25-(x+4).^2); plot(x,y2,'k','linewidth',3); hold on x=[0:0.002:6]; y3=(x-2).^2-7; plot(x,y3,'m','linewidth',3) grid on title('\bf\fontsize{20} Grafica De La Funcion f(x)') hold off

3. Grafique las siguientes funciones en un mismo cuadro.

La primera con color rojo y la segunda con color azul, adems, aumente un poco el grosor de la lnea, colquele como titulo en negrita Funciones de una Variable y encada k-simo punto de interseccin de ambas curvas escriba Pk.SOLUCION:clcclearx=-3:0.01:5;y1=-x.^3+4*x.^2-6*x-5;y2=x.^4./4-6*x.^2+13;plot(x,y1,'r','linewidth',2)hold on plot(x,y2,'b','linewidth',3)grid on title('\bf\fontsize{20}Funciones de una Varible')gtext('\bf\fontsize{20}p1')gtext('\bf\fontsize{20}p2')gtext('\bf\fontsize{20}p3')

4. Grafique las siguientes funciones en un mismo cuadro.

Ampliando el grosor de la lnea a 3, la primera con color rojo y la segunda con color azul. Adems colquele como ttulo Funciones de una Variable con tamao de letra 20 y en negrita. Modifique el eje Y usando axis de tal forma que verticalmente aparezca el intervalo [-4,2].Dentro de la regin de izquierda escribir Regin 1 en negrita y dentro de la regin derecha escribir Regin 2 en negrita.SOLUCION: clcclearx=-4:0.01:4;y1=2*abs(x)./(1+x.^4);y2=-4*abs(x)./(1+x.^4);plot(x,y1,'r','linewidth',3)hold onplot(x,y2,'b','linewidth',3)grid ontitle('\bf\fontsize{20}FUNCIONES DE UNA VARIABLE')axis([0 4 -4 2])axis equalgtext('\bf\fontsize{18}Region 1')gtext('\bf\fontsize{18}Region 2')hold off

METODO DE BISECCIN1. Dada la funcin:

a) Ubique grficamente los ceros de f entre dos enteros consecutivos.b) Hallar una solucin aproximada al mayor cero de f aplicando el mtodo de biseccin con una tolerancia de 10^(-8), es decir hasta que cumpla la condicin |Xn-Xn-1|> a=1, b=2, tol=10.^(-8)a = 1b = 2tol = 1.0000e-08>> [c]=bisec('f13',a,b,tol) 0 1.00000000000 2.00000000000 1.50000000000 2.37500000000 1 1.50000000000 2.00000000000 1.75000000000 -2.89062500000 2 1.50000000000 1.75000000000 1.62500000000 -0.27148437500 3 1.50000000000 1.62500000000 1.56250000000 1.04907226563 4 1.56250000000 1.62500000000 1.59375000000 0.38803100586 5 1.59375000000 1.62500000000 1.60937500000 0.05807113647 6 1.60937500000 1.62500000000 1.61718750000 -0.10675859451 7 1.60937500000 1.61718750000 1.61328125000 -0.02435654402 8 1.60937500000 1.61328125000 1.61132812500 0.01685411483 9 1.61132812500 1.61328125000 1.61230468750 -0.00375201274 10 1.61132812500 1.61230468750 1.61181640625 0.00655085186 11 1.61181640625 1.61230468750 1.61206054688 0.00139936972 12 1.61206054688 1.61230468750 1.61218261719 -0.00117633397 13 1.61206054688 1.61218261719 1.61212158203 0.00011151476 14 1.61212158203 1.61218261719 1.61215209961 -0.00053241039 15 1.61212158203 1.61215209961 1.61213684082 -0.00021044801 16 1.61212158203 1.61213684082 1.61212921143 -0.00004946667 17 1.61212158203 1.61212921143 1.61212539673 0.00003102403 18 1.61212539673 1.61212921143 1.61212730408 -0.00000922133 19 1.61212539673 1.61212730408 1.61212635040 0.00001090135 20 1.61212635040 1.61212730408 1.61212682724 0.00000084001 21 1.61212682724 1.61212730408 1.61212706566 -0.00000419066 22 1.61212682724 1.61212706566 1.61212694645 -0.00000167532 23 1.61212682724 1.61212694645 1.61212688684 -0.00000041765 24 1.61212682724 1.61212688684 1.61212685704 0.00000021118 25 1.61212685704 1.61212688684 1.61212687194 -0.00000010324c = 1.612126871943474

2. Dada la funcin:

a)Ubique grficamente el cero de f entre dos enteros consecutivos.b)Hallar la solucin aproximada aplicando el mtodo de biseccin con una tolerancia de 10^(-8), es decir hasta que se cumpla la condicin: |Xn-Xn-1|> a=0,b=1,tol=10^(-8)a = 0b = 1tol = 1.000000000000000e-08>> [c]=bisec('f13',a,b,tol) 0 0.00000000000 1.00000000000 0.50000000000 -0.06013600695 1 0.50000000000 1.00000000000 0.75000000000 0.20737221164 2 0.50000000000 0.75000000000 0.62500000000 0.07190466341 3 0.50000000000 0.62500000000 0.56250000000 0.00564852643 4 0.50000000000 0.56250000000 0.53125000000 -0.02727721489 5 0.53125000000 0.56250000000 0.54687500000 -0.01082597264 6 0.54687500000 0.56250000000 0.55468750000 -0.00259203067 7 0.55468750000 0.56250000000 0.55859375000 0.00152737135 8 0.55468750000 0.55859375000 0.55664062500 -0.00053254261 9 0.55664062500 0.55859375000 0.55761718750 0.00049736036 10 0.55664062500 0.55761718750 0.55712890625 -0.00001760453 11 0.55712890625 0.55761718750 0.55737304688 0.00023987455 12 0.55712890625 0.55737304688 0.55725097656 0.00011113417 13 0.55712890625 0.55725097656 0.55718994141 0.00004676461 14 0.55712890625 0.55718994141 0.55715942383 0.00001457999 15 0.55712890625 0.55715942383 0.55714416504 -0.00000151228 16 0.55714416504 0.55715942383 0.55715179443 0.00000653385 17 0.55714416504 0.55715179443 0.55714797974 0.00000251078 18 0.55714416504 0.55714797974 0.55714607239 0.00000049925 19 0.55714416504 0.55714607239 0.55714511871 -0.00000050652 20 0.55714511871 0.55714607239 0.55714559555 -0.00000000364 21 0.55714559555 0.55714607239 0.55714583397 0.00000024781 22 0.55714559555 0.55714583397 0.55714571476 0.00000012209 23 0.55714559555 0.55714571476 0.55714565516 0.00000005923 24 0.55714559555 0.55714565516 0.55714562535 0.00000002779 25 0.55714559555 0.55714562535 0.55714561045 0.00000001208c = 0.557145610451698

3. Dada la funcin:

SOLUCION:a)clcclearx=1:0.02:2;y1=cos(x);y2=x-1;plot(x,y1,'g','linewidth',2);hold onplot(x,y2,'b','linewidth',2);grid on

b)>> a=1,b=2,tol=5*10^(-8)a = 1b = 2tol = 5.000000000000000e-08>> [c]=bisec('f13',a,b,tol) 0 1.00000000000 2.00000000000 1.50000000000 -0.42926279833 1 1.00000000000 1.50000000000 1.25000000000 0.06532236240 2 1.25000000000 1.50000000000 1.37500000000 -0.18045229201 3 1.25000000000 1.37500000000 1.31250000000 -0.05706623311 4 1.25000000000 1.31250000000 1.28125000000 0.00426746612 5 1.28125000000 1.31250000000 1.29687500000 -0.02636636309 6 1.28125000000 1.29687500000 1.28906250000 -0.01104096398 7 1.28125000000 1.28906250000 1.28515625000 -0.00338459919 8 1.28125000000 1.28515625000 1.28320312500 0.00044197448 9 1.28320312500 1.28515625000 1.28417968750 -0.00147117755 10 1.28320312500 1.28417968750 1.28369140625 -0.00051456778 11 1.28320312500 1.28369140625 1.28344726563 -0.00003628820 12 1.28320312500 1.28344726563 1.28332519531 0.00020284525 13 1.28332519531 1.28344726563 1.28338623047 0.00008327905 14 1.28338623047 1.28344726563 1.28341674805 0.00002349556 15 1.28341674805 1.28344726563 1.28343200684 -0.00000639629 16 1.28341674805 1.28343200684 1.28342437744 0.00000854964 17 1.28342437744 1.28343200684 1.28342819214 0.00000107668 18 1.28342819214 1.28343200684 1.28343009949 -0.00000265981 19 1.28342819214 1.28343009949 1.28342914581 -0.00000079157 20 1.28342819214 1.28342914581 1.28342866898 0.00000014256 21 1.28342866898 1.28342914581 1.28342890739 -0.00000032450 22 1.28342866898 1.28342890739 1.28342878819 -0.00000009097 23 1.28342866898 1.28342878819 1.28342872858 0.00000002579c = 1.283428728580475

c)aproximado: x=1.283428728580475

4. Usar el mtodo de biseccin para hallar la raz de f(x)=0 siendo:

SOLUCION:clcclearx=[-4:0.02:4];y1=x.^3;y2=-4*x.^2+10;plot(x,y1,'g','linewidth',2);hold onplot(x,y2,'b','linewidth',2);grid on

>> a=1,b=2,tol=10^(-8)a = 1b = 2tol = 1.000000000000000e-08>> [c]=bisec('f13',a,b,tol) 0 1.00000000000 2.00000000000 1.50000000000 2.37500000000 1 1.00000000000 1.50000000000 1.25000000000 -1.79687500000 2 1.25000000000 1.50000000000 1.37500000000 0.16210937500 3 1.25000000000 1.37500000000 1.31250000000 -0.84838867188 4 1.31250000000 1.37500000000 1.34375000000 -0.35098266602 5 1.34375000000 1.37500000000 1.35937500000 -0.09640884399 6 1.35937500000 1.37500000000 1.36718750000 0.03235578537 7 1.35937500000 1.36718750000 1.36328125000 -0.03214997053 8 1.36328125000 1.36718750000 1.36523437500 0.00007202476 9 1.36328125000 1.36523437500 1.36425781250 -0.01604669075 10 1.36425781250 1.36523437500 1.36474609375 -0.00798926281 11 1.36474609375 1.36523437500 1.36499023438 -0.00395910152 12 1.36499023438 1.36523437500 1.36511230469 -0.00194365901 13 1.36511230469 1.36523437500 1.36517333984 -0.00093584728 14 1.36517333984 1.36523437500 1.36520385742 -0.00043191880 15 1.36520385742 1.36523437500 1.36521911621 -0.00017994890 16 1.36521911621 1.36523437500 1.36522674561 -0.00005396254 17 1.36522674561 1.36523437500 1.36523056030 0.00000903099 18 1.36522674561 1.36523056030 1.36522865295 -0.00002246580 19 1.36522865295 1.36523056030 1.36522960663 -0.00000671741 20 1.36522960663 1.36523056030 1.36523008347 0.00000115679 21 1.36522960663 1.36523008347 1.36522984505 -0.00000278031 22 1.36522984505 1.36523008347 1.36522996426 -0.00000081176 23 1.36522996426 1.36523008347 1.36523002386 0.00000017251 24 1.36522996426 1.36523002386 1.36522999406 -0.00000031962 25 1.36522999406 1.36523002386 1.36523000896 -0.00000007356c = 1.365230008959770

5. Usar el mtodo de biseccin para hallar la raz f(x)=0, siendo:

SOLUCION:clcclearx=[0:0.02:1];y1=3*x-exp(x);y2=-sin(x);plot(x,y1,'g','linewidth',2);hold onplot(x,y2,'b','linewidth',2);grid on

>> a=0,b=1,tol=10^(-6)a = 0b = 1tol = 1.000000000000000e-06>> [c]=bisec('f13',a,b,tol) 0 0.00000000000 1.00000000000 0.50000000000 0.33070426790 1 0.00000000000 0.50000000000 0.25000000000 -0.28662145743 2 0.25000000000 0.50000000000 0.37500000000 0.03628111447 3 0.25000000000 0.37500000000 0.31250000000 -0.12189942659 4 0.31250000000 0.37500000000 0.34375000000 -0.04195596590 5 0.34375000000 0.37500000000 0.35937500000 -0.00261963457 6 0.35937500000 0.37500000000 0.36718750000 0.01688575295 7 0.35937500000 0.36718750000 0.36328125000 0.00714674163 8 0.35937500000 0.36328125000 0.36132812500 0.00226696530 9 0.35937500000 0.36132812500 0.36035156250 -0.00017548279 10 0.36035156250 0.36132812500 0.36083984375 0.00104595435 11 0.36035156250 0.36083984375 0.36059570313 0.00043528904 12 0.36035156250 0.36059570313 0.36047363281 0.00012991643 13 0.36035156250 0.36047363281 0.36041259766 -0.00002277985 14 0.36041259766 0.36047363281 0.36044311523 0.00005356912 15 0.36041259766 0.36044311523 0.36042785645 0.00001539484 16 0.36041259766 0.36042785645 0.36042022705 -0.00000369245 17 0.36042022705 0.36042785645 0.36042404175 0.00000585121 18 0.36042022705 0.36042404175 0.36042213440 0.00000107938c = 0.360422134399414

METODO DE NEWTON RAPHSON

SOLUCION:A)clcclearx=[-4:0.02:4];y1=x.^3;y2=x+6;plot(x,y1,'b','linewidth',2);hold onplot(x,y2,'r','linewidth',2);grid on

B.I)>> x0=1, tol=10.^(-6), N=50x0 = 1tol = 1.000000000000000e-06N = 50>> [x1]=newton('f15','df15',x0,tol,N) 1 4.000000000000 2 2.851063829787 3 2.238552899888 4 2.026265689754 5 2.000368955047 6 2.000000074231 7 2.000000000000 x1 = 2.000000000000003

B.II)>> x0=3, tol=10.^(-6), N=50x0 = 3tol = 1.000000000000000e-06N = 50>> [x1]=newton('f15','df15',x0,tol,N) 1 2.307692307692 2 2.041819894842 3 2.000924621582 4 2.000000465996 5 2.000000000000 x1 = 2.000000000000119

B.III)>> x0=5, tol=10.^(-6), N=50x0 = 5tol = 1.000000000000000e-06N = 50>> [x1]=newton('f15','df15',x0,tol,N) 1 3.459459459459 2 2.544284945850 3 2.114003432557 4 2.006524046564 5 2.000023102081 6 2.000000000291 7 2.000000000000 x1 = 2

B.IV)>> x0=20, tol=10.^(-6), N=50x0 = 20tol = 1.000000000000000e-06N = 50>> [x1]=newton('f15','df15',x0,tol,N) 1 13.349457881568 2 8.927560188750 3 6.001902115365 4 4.094678083500 5 2.906863139573 6 2.263906797478 7 2.031625413448 8 2.000532770404 9 2.000000154762 10 2.000000000000 x1 = 2.000000000000013

2.

SOLUCION:A)clcclearx=[-1:0.02:2];y1=x;y2=2.^(-x);plot(x,y1,'g','linewidth',2);hold onplot(x,y2,'b','linewidth',2);grid on

B.I) >> p0=0, tol=10^(-6), N=30

p0 = 0tol = 1.0000e-006N = 30

>> [p1]=newton('ff1','dff1',p0,tol,N) 1 1.442695040889 2 -2.772357397608 3 -0.744251166116 4 1.339435178364 5 -2.107842156924 6 0.040339712325 7 1.423186593712 8 -2.640397670985 9 -0.586753598839 10 1.419578926114 11 -2.616330162029 12 -0.558071097764 13 1.431473582448 14 -2.696077215670 15 -0.653167301994 16 1.388733560886 17 -2.414760341406 18 -0.318734281090 19 1.492645137605 20 -3.124512537366 21 -1.164943343704 22 1.027294526069 23 -0.550768748477 24 1.434359057416 25 -2.715594341518 26 -0.676462981713 27 1.376868792487 28 -2.339203036631 29 -0.229590396653 30 1.495602178309

p1 =

1.4956

B.II) >>p0=1, tol=10^(-6), N=40

p0 = 1tol = 1.0000e-006N = 40

>> [p1]=newton('ff1','dff1',p0,tol,N) 1 -0.442695040889 2 1.469910024362 3 -2.961643221456 4 -0.970464345855 5 1.186752103253 6 -1.268020599320 7 0.934280836455 8 -0.198743445293 9 1.493778734708 10 -3.132748142714 11 -1.174766488163 12 1.018662642987 13 -0.516150265890 14 1.447228474698 15 -2.803464977163 16 -0.781417784932 17 1.317162164214 18 -1.975133235470 19 0.192325760935 20 1.317985355601 21 -1.979973498764 22 0.186828136586 23 1.322721525111 24 -2.007917335586 25 0.155017347524 26 1.348700712116 27 -2.164127237012 28 -0.024821534355 29 1.453072572289 30 -2.843814511313 31 -0.829637761256 32 1.286525641323 33 -1.798460033930 34 0.390142403900 35 1.095201178833 36 -0.837749401446 37 1.281185811527 38 -1.768349508050 39 0.423232719859 40 1.047162291930

p1 = 1.0472

B.III)>> p0=-50, tol=10^(-6), N=100

p0 = -50tol = 1.0000e-006N = 100

>> [p1]=newton('ff1','dff1',p0,tol,N) 1 -48.557304959111 2 -47.114609918222 3 -45.671914877332 4 -44.229219836442 5 -42.786524795550 6 -41.343829754653 7 -39.901134713743 8 -38.458439672798 9 -37.015744631762 10 -35.573049590489 11 -34.130354548596 12 -32.687659505088 13 -31.244964457382 14 -29.802269398781 15 -28.359574311967 16 -26.916879152285 17 -25.474183804909 18 -24.031487975557 19 -22.588790912783 20 -21.146090705781 21 -19.703382518789 22 -18.260654180818 23 -16.817875254560 24 -15.374970193624 25 -13.931753163423 26 -12.487771933757 27 -11.041940246441 28 -9.591689679505 29 -8.131060382982 30 -6.646521779523 31 -5.108114703960 32 -3.451751688771 33 -1.553928926196 34 0.652296758979 35 0.615949311389 36 0.696764698904 37 0.510134622388 38 0.904675504485 39 -0.096070322516 40 1.476295980761 41 -3.006950330279 42 -1.024597377732 43 1.144694051531 44 -1.063941957765 45 1.112952494777 46 -0.917180276961 47 1.226212107849 48 -1.469819701683 49 0.738434548071 50 0.403761023530 51 1.075831034543 52 -0.753193095981 53 1.334186335530 54 -2.076239787608 55 0.076754892360 56 1.402665184469 57 -2.504874685593 58 -0.425504160913 59 1.474267318283 60 -2.992520285729 61 -1.007358307791 62 1.158295359932 63 -1.128723358537 64 1.058673585875 65 -0.680110816184 66 1.374963472022 67 -2.327170522936 68 -0.215431584455 69 1.494953701704 70 -3.141295895122 71 -1.184960548193 72 1.009651246338 73 -0.480445523803 74 1.459059342759 75 -2.885440699356 76 -0.879390220297 77 1.252962118544 78 -1.612501222595 79 0.590981295726 80 0.749421021712 81 0.374514794529 82 1.116749247496 83 -0.934413538685 84 1.213668025036 85 -1.404561567973 86 0.803554092023 87 0.222840769098 88 1.290346078787 89 -1.820126302718 90 0.366209952974 91 1.127908520231 92 -0.985568071081 93 1.175211568586 94 -1.210909691656 95 0.986471108189 96 -0.390620841358 97 1.481948379023 98 -3.047339117140 99 -1.072836103058 100 1.105645871894

p1 =

1.105

B.IV) >> p0=-100, tol=10^(-6), N=100

p0 = -100tol = 1.0000e-006N = 100

>> [p1]=newton('ff1','dff1',p0,tol,N) 1 -98.557304959111 2 -97.114609918222 3 -95.671914877333 4 -94.229219836444 5 -92.786524795555 6 -91.343829754666 7 -89.901134713777 8 -88.458439672888 9 -87.015744631999 10 -85.573049591110 11 -84.130354550221 12 -82.687659509332 13 -81.244964468443 14 -79.802269427555 15 -78.359574386666 16 -76.916879345777 17 -75.474184304888 18 -74.031489263999 19 -72.588794223110 20 -71.146099182221 21 -69.703404141332 22 -68.260709100443 23 -66.818014059554 24 -65.375319018665 25 -63.932623977776 26 -62.489928936887 27 -61.047233895998 28 -59.604538855109 29 -58.161843814220 30 -56.719148773331 31 -55.276453732442 32 -53.833758691553 33 -52.391063650664 34 -50.948368609775 35 -49.505673568886 36 -48.062978527997 37 -46.620283487108 38 -45.177588446218 39 -43.734893405328 40 -42.292198364434 41 -40.849503323534 42 -39.406808282615 43 -37.964113241649 44 -36.521418200555 45 -35.078723159132 46 -33.636028116849 47 -32.193333072324 48 -30.750638021978 49 -29.307942956532 50 -27.865247852024 51 -26.422552646712 52 -24.979857182016 53 -23.537161052001 54 -22.094463221543 55 -20.651761062591 56 -19.209047936137 57 -17.766307167400 58 -16.323497157622 59 -14.880514955725 60 -13.437108191596 61 -11.992665286326 62 -10.545724653278 63 -9.092851399741 64 -7.626131899935 65 -6.127745938243 66 -4.558623861297 67 -2.836850707825 68 -0.821314970740 69 1.291949631576 70 -1.829251054499 71 0.356101220856 72 1.141221478410 73 -1.047586188895 74 1.126263823302 75 -0.977981341291 76 1.181027616743 77 -1.239585315901 78 0.960463160210 79 -0.293236540393 80 1.494697326833 81 -3.139429789595 82 -1.182735152378 83 1.011623084027 84 -0.488220433233 85 1.456610766214 86 -2.868379905910 87 -0.858998426271 88 1.266951293826 89 -1.689058299556 90 0.509358569160 91 0.906056791270 92 -0.100763887307 93 1.477495816934 94 -3.015501152671 95 -1.034811731910 96 1.136545963434 97 -1.025681800608 98 1.143832121035 99 -1.059875338210 100 1.116277377504

p1 =

1.1163

3.

SOLUCION:A)clcclearx=0:0.01:1;y1=exp(x);y2=4./(x+1);plot(x,y1,'b','linewidth',1);hold onplot(x,y2,'g','linewidth',1);grid on

B.I)>> x0=0, tol=10.^(-8), N=50x0 = 0tol = 1.000000000000000e-08N = 50>> [x1]=newton('f15','df15',x0,tol,N) 1 1.500000000000 2 1.040720183027 3 0.834216703888 4 0.799865471431 5 0.799041214555 6 0.799040753172 7 0.799040753172 x1 = 0.799040753171893

B.II)>> x0=10, tol=10.^(-8), N=50x0 = 10tol = 1.000000000000000e-08N = 50>> [x1]=newton('f15','df15',x0,tol,N) 1 9.083348466643 2 8.173614884427 3 7.272019237109 4 6.380170310305 5 5.500308601039 6 4.635815297612 7 3.792358652272 8 2.980566866471 9 2.222116920779 10 1.561642748089 11 1.078023740205 12 0.845096859906 13 0.800446764391 14 0.799042093768 15 0.799040753173 16 0.799040753172 x1 = 0.799040753171893

B.III)>> x0=100, tol=10.^(-8), N=200x0 = 100tol = 1.000000000000000e-08N = 200>> [x1]=newton('f15','df15',x0,tol,N) 1 99.009803921569 2 98.019703950686 3 97.029701980679 4 96.039799961184 5 95.049999900397 6 94.060303867435 7 93.070713994821 8 92.081232481092 9 91.091861593548 10 90.102603671135 11 89.113461127495 12 88.124436454166 13 87.135532223964 14 86.146751094542 15 85.158095812159 16 84.169569215641 17 83.181174240584 18 82.192913923781 19 81.204791407923 20 80.216809946557 21 79.228972909345 22 78.241283787646 23 77.253746200419 24 76.266363900502 25 75.279140781274 26 74.292080883738 27 73.305188404052 28 72.318467701550 29 71.331923307282 30 70.345559933119 31 69.359382481468 32 68.373396055650 33 67.387605970988 34 66.402017766668 35 65.416637218442 36 64.431470352239 37 63.446523458767 38 62.461803109192 39 61.477316171995 40 60.493069831102 41 59.509071605422 42 58.525329369909 43 57.541851378305 44 56.558646287715 45 55.575723185208 46 54.593091616635 47 53.610761617892 48 52.628743748884 49 51.647049130472 50 50.665689484718 51 49.684677178789 52 48.704025272936 53 47.723747572982 54 46.743858687868 55 45.764374092822 56 44.785310198843 57 43.806684429253 58 42.828515304210 59 41.850822534178 60 40.873627123515 61 39.896951485532 62 38.920819570549 63 37.945257008765 64 36.970291270024 65 35.995951842922 66 35.022270436127 67 34.049281205273 68 33.077021009412 69 32.105529701747 70 31.134850460250 71 30.165030164885 72 29.196119829495 73 28.228175098082 74 27.261256817265 75 26.295431699302 76 25.330773093300 77 24.367361886329 78 23.405287561428 79 22.444649446148 80 21.485558194008 81 20.528137552506 82 19.572526486235 83 18.618881743385 84 17.667380980407 85 16.718226595589 86 15.771650471744 87 14.827919897434 88 13.887345034912 89 12.950288448084 90 12.017177425217 91 11.088520187597 92 10.164927690730 93 9.247143879254 94 8.336089620230 95 7.432930772757 96 6.539193166745 97 5.656977434893 98 4.789402042964 99 3.941590661476 100 3.122967884621 101 2.352542862208 102 1.669715577632 103 1.147464444712 104 0.868604458304 105 0.802210511682 106 0.799047560526 107 0.799040753203 108 0.799040753172 x1 = 0.799040753171893

METODOS DE JACOBI Y GAUSS SEIDEL1 .SOLUCION: A) >> A=[1 2 -2;1 1 1;2 2 1]A = 1 2 -2 1 1 1 2 2 1

>> b=[7 2 5]'

b = 7 2 5

>> solu=inv(A)*bsolu =

1 2 -1

>> xo=[0 0 0]'xo =

0 0 0

>> tol=10^(-5)

tol = 1.0000e-005

>> max=50

max = 50

>> x=jacobi(A,b,xo,tol,max)

k = 1y = 7 2 5k = 2y = 13 -10 -13k = 3y = 1 2 -1k = 4y = 1 2 -1k = 4x = 1 2 -1

B) A=[1 2 -2;1 1 1;2 2 1]

A = 1 2 -2 1 1 1 2 2 1

>> b=[7 2 5]'

b = 7 2 5

>> solu=inv(A)*b

solu = 1 2 -1

>> xo=[0 0 0]'xo =

0 0 0

>> tol=10^(-5)

tol = 1.0000e-005

>> max=50

max = 50

>> x=gaussseidel(A,b,xo,tol,max)

k = 1y = 7 -5 1k = 2y = 19 -18 3k = 3y = 49 -50 7k = 4y = 121 -126 15k = 5y = 289 -302 31k = 6y = 673 -702 63k = 7y = 1537 -1598 127k = 8y = 3457 -3582 255k = 9y = 7681 -7934 511k = 10y = 16897 -17406 1023k = 11y = 36865 -37886 2047k = 12y = 79873 -81918 4095k = 13y = 172033 -176126 8191k = 14y = 368641 -376830 16383k = 15y = 786433 -802814 32767k = 16y = 1671169 -1703934 65535k = 17y = 3538945 -3604478 131071k = 18y = 7471105 -7602174 262143k = 19

y = 15728641 -15990782 524287k = 20y = 33030145 -33554430 1048575k = 21y = 69206017 -70254590 2097151k = 22y = 144703489 -146800638 4194303k = 23y = 301989889 -306184190 8388607k = 24y = 629145601 -637534206 16777215k = 25y = 1.0e+009 * 1.3086 -1.3254 0.0336k = 26y = 1.0e+009 * 2.7179 -2.7515 0.0671k = 27y = 1.0e+009 * 5.6371 -5.7043 0.1342k = 28y = 1.0e+010 * 1.1677 -1.1811 0.0268k = 29y = 1.0e+010 * 2.4159 -2.4428 0.0537k = 30y = 1.0e+010 * 4.9929 -5.0466 0.1074k = 31y = 1.0e+011 * 1.0308 -1.0415 0.0215k = 32y = 1.0e+011 * 2.1260 -2.1475 0.0429k = 33y = 1.0e+011 * 4.3809 -4.4238 0.0859k = 34y = 1.0e+011 * 9.0194 -9.1053 0.1718k = 35y = 1.0e+012 * 1.8554 -1.8726 0.0344k = 36y = 1.0e+012 * 3.8139 -3.8483 0.0687k = 37y = 1.0e+012 * 7.8340 -7.9027 0.1374k = 38y = 1.0e+013 * 1.6080 -1.6218 0.0275

k = 39y = 1.0e+013 * 3.2985 -3.3260 0.0550k = 40y = 1.0e+013 * 6.7620 -6.8170 0.1100k = 41y = 1.0e+014 * 1.3854 -1.3964 0.0220k = 42y = 1.0e+014 * 2.8367 -2.8587 0.0440k = 43y = 1.0e+014 * 5.8054 -5.8494 0.0880k = 44y = 1.0e+015 * 1.1875 -1.1963 0.0176k = 45y = 1.0e+015 * 2.4277 -2.4453 0.0352k = 46y = 1.0e+015 * 4.9610 -4.9962 0.0704k = 47y = 1.0e+016 * 1.0133 -1.0203 0.0141k = 48y = 1.0e+016 * 2.0688 -2.0829 0.0281k = 49y = 1.0e+016 * 4.2221 -4.2503 0.0563k = 50y = 1.0e+016 * 8.6131 -8.6694 0.1126k = 50x = 1.0e+016 * 8.6131 -8.6694 0.1126

2. SOLUCION:

a)>> A=[5 1 -1 2;1 6 2 2;2 1 -9 0; 2 3 1 8]

A = 5 1 -1 2 1 6 2 2 2 1 -9 0 2 3 1 8

>> b=[-6 19 -13 33]'

b = -6 19 -13 33

>> solu=inv(A)*b

solu = -3.0000 2.0000 1.0000 4.0000

>> xo=[2 -7 4 0]'

xo = 2 -7 4 0

>> tol=10^(-8)

tol = 1.0000e-008

>> max=100

max = 100

>> x=jacobi(A,b,xo,tol,max)k = 1y = 1.0000 1.5000 1.1111 5.7500k = 2y = -3.5778 0.7130 1.8333 3.1736k = 3y = -2.2454 2.0940 0.7286 4.5229k = 4y = -3.2822 1.7904 1.1781 3.8100k = 5y = -2.8465 2.0510 0.9140 4.1269k = 6y = -3.0782 1.9608 1.0398 3.9532k = 7y = -2.9655 2.0153 0.9783 4.0293k = 8y = -3.0191 1.9917 1.0094 3.9883k = 9y = -2.9918 2.0040 0.9948 4.0067k = 10y = -3.0045 1.9981 1.0023 3.9971k = 11y = -2.9980 2.0010 0.9988 4.0015k = 12y = -3.0011 1.9996 1.0005 3.9993k = 13y = -2.9995 2.0002 0.9997 4.0004k = 14y = -3.0002 1.9999 1.0001 3.9998k = 15y = -2.9999 2.0001 0.9999 4.0001k = 16y = -3.0001 2.0000 1.0000 4.0000k = 17y = -3.0000 2.0000 1.0000 4.0000k = 18y = -3.0000 2.0000 1.0000 4.0000k = 19y = -3.0000 2.0000 1.0000 4.0000k = 20y = -3.0000 2.0000 1.0000 4.0000k = 21y = -3.0000 2.0000 1.0000 4.0000k = 22y = -3.0000 2.0000 1.0000 4.0000k = 23y = -3.0000 2.0000 1.0000 4.0000k = 24y = -3.0000 2.0000 1.0000 4.0000k = 25y = -3.0000 2.0000 1.0000 4.0000k = 26y = -3.0000 2.0000 1.0000 4.0000k = 27y = -3.0000 2.0000 1.0000 4.0000k = 28y = -3.0000 2.0000 1.0000 4.0000k = 29y = -3.0000 2.0000 1.0000 4.0000k = 30y = -3.0000 2.0000 1.0000 4.0000k = 30x = -3.0000 2.0000 1.0000 4.0000

S HAY CONVERGENCIA

B)>> A=[5 1 -1 2;1 6 2 2;2 1 -9 0; 2 3 1 8]

A = 5 1 -1 2 1 6 2 2 2 1 -9 0 2 3 1 8

>> b=[-6 19 -13 33]'

b = -6 19 -13 33

>> solu=inv(A)*b

solu = -3.0000 2.0000 1.0000 4.0000

>> xo=[2 -7 4 0]'

xo = 2 -7 4 0

>> tol=10^(-8)

tol = 1.0000e-008

>> max=100

max = 100

>> x=gaussseidel(A,b,xo,tol,max)k = 1y = 1.0000 1.6667 1.8519 3.0185k = 2y = -2.3704 1.9383 1.1331 3.8491k = 3y = -2.9007 1.9894 1.0209 3.9765k = 4y = -2.9843 1.9982 1.0033 3.9963k = 5y = -2.9975 1.9997 1.0005 3.9994k = 6y = -2.9996 2.0000 1.0001 3.9999k = 7y = -2.9999 2.0000 1.0000 4.0000k = 8y = -3.0000 2.0000 1.0000 4.0000k = 9y = -3.0000 2.0000 1.0000 4.0000k = 10y = -3.0000 2.0000 1.0000 4.0000k = 11y = -3.0000 2.0000 1.0000 4.0000k = 12y = -3.0000 2.0000 1.0000 4.0000k = 13y = -3.0000 2.0000 1.0000 4.0000k = 13x = -3.0000 2.0000 1.0000 4.0000

S HAY CONVERGENCIA

3. SOLUCION:

A)A=[2 -1 1;2 2 2;-1 -1 2]

A = 2 -1 1 2 2 2 -1 -1 2

>> b=[-1 4 -5]'

b = -1 4 -5

>> solu=inv(A)*b

solu = 1.0000 2.0000 -1.0000

>> xo=[0 0 0]'

xo = 0 0 0

>> tol=10^(-6)

tol = 1.0000e-006

>> max=50

max = 50

>> x=jacobi(A,b,xo,tol,max)k = 1y = -0.5000 2.0000 -2.5000k = 2y = 1.7500 5.0000 -1.7500k = 3y = 2.8750 2.0000 0.8750k = 4y = 0.0625 -1.7500 -0.0625k = 5y = -1.3438 2.0000 -3.3438k = 6y = 2.1719 6.6875 -2.1719k = 7y = 3.9297 2.0000 1.9297k = 8y = -0.4648 -3.8594 0.4648k = 9y = -2.6621 2.0000 -4.6621k = 10y = 2.8311 9.3242 -2.8311k = 11y = 5.5776 2.0000 3.5776k = 12y = -1.2888 -7.1553 1.2888k = 13y = -4.7220 2.0000 -6.7220k = 14y = 3.8610 13.4441 -3.8610k = 15y = 8.1526 2.0000 6.1526k = 16y = -2.5763 -12.3051 2.5763k = 17y = -7.9407 2.0000 -9.9407k = 18y = 5.4703 19.8814 -5.4703k = 19y = 12.1759 2.0000 10.1759k = 20y = -4.5879 -20.3517 4.5879k = 21y = -12.9698 2.0000 -14.9698k = 22y = 7.9849 29.9397 -7.9849k = 23y = 18.4623 2.0000 16.4623k = 24y = -7.7311 -32.9246 7.7311k = 25y = -20.8279 2.0000 -22.8279k = 26y = 11.9139 45.6557 -11.9139k = 27y = 28.2848 2.0000 26.2848k = 28y = -12.6424 -52.5697 12.6424k = 29y = -33.1061 2.0000 -35.1061k = 30y = 18.0530 70.2121 -18.0530k = 31y = 43.6326 2.0000 41.6326

k = 32y = -20.3163 -83.2651 20.3163k = 33y = -52.2907 2.0000 -54.2907k = 34y = 27.6454 108.5814 -27.6454k = 35y = 67.6134 2.0000 65.6134k = 36y = -32.3067 -131.2268 32.3067k = 37y = -82.2667 2.0000 -84.2667k = 38y = 42.6334 168.5335 -42.6334k = 39y = 105.0834 2.0000 103.0834k = 40y = -51.0417 -206.1668 51.0417k = 41y = -129.1043 2.0000 -131.1043k = 42y = 66.0521 262.2085 -66.0521k = 43y = 163.6303 2.0000 161.6303k = 44y = -80.3152 -323.2607 80.3152k = 45y = -202.2879 2.0000 -204.2879k = 46y = 102.6440 408.5758 -102.6440k = 47y = 255.1099 2.0000 253.1099k = 48y = -126.0549 -506.2198 126.0549k = 49y = -316.6374 2.0000 -318.6374k = 50y = 159.8187 637.2747 -159.8187k = 50x = 159.8187 637.2747 -159.8187

NO HAY CONVERGENCIA

B)A=[2 -1 1;2 2 2;-1 -1 2]

A = 2 -1 1 2 2 2 -1 -1 2

>> b=[-1 4 -5]'

b = -1 4 -5

>> solu=inv(A)*b

solu = 1.0000 2.0000 -1.0000

>> xo=[0 0 0]'

xo = 0 0 0

>> tol=10^(-6)

tol = 1.0000e-006

>> max=50

max = 50

x=gaussseidel(A,b,xo,tol,max)k = 1y = -0.5000 2.5000 -1.5000k = 2y = 1.5000 2.0000 -0.7500k = 3y = 0.8750 1.8750 -1.1250k = 4y = 1.0000 2.1250 -0.9375k = 5y = 1.0313 1.9063 -1.0313k = 6y = 0.9688 2.0625 -0.9844k = 7y = 1.0234 1.9609 -1.0078k = 8y = 0.9844 2.0234 -0.9961k = 9y = 1.0098 1.9863 -1.0020k = 10y = 0.9941 2.0078 -0.9990k = 11y = 1.0034 1.9956 -1.0005k = 12y = 0.9980 2.0024 -0.9998k = 13y = 1.0011 1.9987 -1.0001k = 14y = 0.9994 2.0007 -0.9999k = 15y = 1.0003 1.9996 -1.0000k = 16y = 0.9998 2.0002 -1.0000k = 17y = 1.0001 1.9999 -1.0000k = 18y = 0.9999 2.0001 -1.0000k = 19y = 1.0000 2.0000 -1.0000k = 20y = 1.0000 2.0000 -1.0000k = 21y = 1.0000 2.0000 -1.0000k = 22y = 1.0000 2.0000 -1.0000k = 23y = 1.0000 2.0000 -1.0000k = 24y = 1.0000 2.0000 -1.0000k = 25y = 1.0000 2.0000 -1.0000k = 26y = 1.0000 2.0000 -1.0000k = 27y = 1.0000 2.0000 -1.0000k = 27x = 1.0000 2.0000 -1.0000

S HAY CONVERGENCIA