Ejercicio 1

Sea \(f(x)=\sqrt{x}-\cos x\). Usa el método de la bisección para encontrar \(x\in [0,1]\) tal que \(f(x)=0\).

Ejercicio 2

Usa el método de la bisección para encontrar una raíz con una precisión de \(10^{-2}\) para \(x^3-7x^2+14x-6=0\) en cada intervalo.

\[\begin{equation} ``` a) [0,1]\qquad\qquad ``` ``` ## $aprox ## [1] 0.5000000 0.7500000 0.6250000 0.5625000 0.5937500 0.5781250 0.5859375 ## ## $precision ## [1] 0.0078125 ## ## $iteraciones ## [1] 7 ``` El método de bisección después de 20 iteraciones y una precisión de 9.536743e^-07 la raíz es 0.5857859 ``` bisect(f_2, 0, 1) ``` ``` ## $root ## [1] 0.5857864 ## ## $f.root ## [1] -8.881784e-16 ## ## $iter ## [1] 54 ## ## $estim.prec ## [1] 1.110223e-16 ``` ``` b) [1, 3.2]\qquad\qquad ``` ``` ## $aprox ## [1] 2.100000 2.650000 2.925000 3.062500 2.993750 3.028125 3.010938 3.002344 ``` ``` ## ## $precision ## [1] 0.00859375 ## ## $iteraciones ## [1] 8 ``` El método de bisección después de 8 iteraciones y una precisión de 0.00859375 la raíz es 3.002344 ``` bisect(f_2, 1, 3.2) ``` ``` ## $root ## [1] 3 ## ## $f.root ## [1] 7.105427e-15 ## ## $iter ## [1] 54 ## ## $estim.prec ## [1] 4.440892e-16 ``` ``` c)[3.2, 4] \end{equation}\]

$aprox

[1] 3.60000 3.40000 3.50000 3.45000 3.42500 3.41250 3.41875

$precision

[1] 0.00625

$iteraciones

[1] 7

El método de bisección después de 7 iteraciones y una precisión de 0.000625 la raíz es 3.41875

bisect(f_2, 3.2, 4) ## $root ## [1] 3.414214 ## ## $f.root ## [1] -2.131628e-14 ## ## $iter ## [1] 52 ## ## $estim.prec ## [1] 4.440892e-16


# Ejercicio 3

Usa el metodo de la bisección para encontrar las soluciones con una precisión de $10^{-5}$ para los siguientes problemas.

a) $x-2^{-x}=0$ para $0\leq x\leq 1$


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metodo_biseccion(f_3a, 0, 1, 10^-5, 100)
## $aprox
##  [1] 0.5000000 0.7500000 0.6250000 0.6875000 0.6562500 0.6406250 0.6484375
##  [8] 0.6445312 0.6425781 0.6416016 0.6411133 0.6413574 0.6412354 0.6411743
## [15] 0.6412048 0.6411896 0.6411819
## 
## $precisión
## [1] 7.629395e-06
## 
## $iteraciones
## [1] 17

El método de bisección después de 17 iteraciones y una precisión de 7.629395e^-06, la raíz es 0.6411819

bisect(f_3a, 0, 1)
## $root
## [1] 0.6411857
## 
## $f.root
## [1] 0
## 
## $iter
## [1] 54
## 
## $estim.prec
## [1] 1.110223e-16
  1. \(e^x-x^2+3x-2=0\) para \(0\leq x\leq 1\)
metodo_biseccion(f_3b, 0, 1, 10^-5, 100)
## $aprox
##  [1] 0.5000000 0.2500000 0.3750000 0.3125000 0.2812500 0.2656250 0.2578125
##  [8] 0.2539062 0.2558594 0.2568359 0.2573242 0.2575684 0.2574463 0.2575073
## [15] 0.2575378 0.2575226 0.2575302
## 
## $precision
## [1] 7.629395e-06
## 
## $iteraciones
## [1] 17

El método de bisección después de 17 iteraciones y una precisión de 7.629395e^-06, la raíz es 0.2575302

bisect(f_3b, 0, 1)
## $root
## [1] 0.2575303
## 
## $f.root
## [1] -4.440892e-16
## 
## $iter
## [1] 55
## 
## $estim.prec
## [1] 5.551115e-17
  1. \(2x\cos (2x)-(x+1)^2=0\) para \(-3\leq x\leq -2\) y \(-1\leq x \leq 0\)
metodo_biseccion(f_3c, -3, -2, 10^-5, 100)
## $aprox
##  [1] -2.500000 -2.250000 -2.125000 -2.187500 -2.218750 -2.203125 -2.195312
##  [8] -2.191406 -2.189453 -2.190430 -2.190918 -2.191162 -2.191284 -2.191345
## [15] -2.191315 -2.191299 -2.191307
## 
## $precision
## [1] 7.629395e-06
## 
## $iteraciones
## [1] 17

El método de bisección después de 17 iteraciones y una precisión 7.629395e^-06, la raíz es -2.191307

bisect(f_3c, -3, -2)
## $root
## [1] -2.191308
## 
## $f.root
## [1] -3.108624e-15
## 
## $iter
## [1] 52
## 
## $estim.prec
## [1] 4.440892e-16
  1. \(x\cos x-2x^2+3x-1=0\) para \(0.2\leq x\leq 0.3\) y \(1.2\leq x \leq 1.3\)
metodo_biseccion(f_3d, 1.2, 1.3, 10^-5, 100)
## $aprox
##  [1] 1.250000 1.275000 1.262500 1.256250 1.259375 1.257812 1.257031 1.256641
##  [9] 1.256445 1.256543 1.256592 1.256616 1.256628 1.256622
## 
## $precision
## [1] 6.103516e-06
## 
## $iteraciones
## [1] 14

El método de bisección después de 14 iteraciones y una precisión 6.103516e^-06, la raíz es 1.256622

bisect(f_3d, .2, .3)
## $root
## [1] 0.2975302
## 
## $f.root
## [1] 0
## 
## $iter
## [1] 52
## 
## $estim.prec
## [1] 5.551115e-17

Ejercicio 4

Considera las funciones \(f(x)=x\) y \(g(x)=2 \sin x\). Usa el método de la bisección para encontrar una aproximación con una precisión de \(10^{-5}\) para el primer valor positivo \(x\) tal que \(f(x)=g(x)\).

g(x)=f(x) 2senx=x =====> 2senx-x=0 h(x)=2senx-x

metodo_biseccion(f_4, -0.2, -0.3, 10^-5, 100)
## $aprox
## [1] -0.25
## 
## $precision
## [1] -0.05
## 
## $iteraciones
## [1] 1

El método de bisección después de 1 iteracion y una precisión -0.05, la raíz es -0.25

bisect(f_4, -3, 2)
## $root
## [1] 0
## 
## $f.root
## [1] 0
## 
## $iter
## [1] 2
## 
## $estim.prec
## [1] 0

Ejercicio 5

Sea \(f(x)=(x+2)(x+1)x(x-1)^3(x-2)\). ¿A cuál raíz de \(f\) converge el método de la bisección cuando se aplica a los siguientes intervalos?

\[\begin{equation} a) [-3,2.5]\qquad \qquad ```r metodo_biseccion(f_5, -3, 2.5, 10^-6, 100) ``` ``` ## $aprox ## [1] -0.250000 1.125000 1.812500 2.156250 1.984375 2.070312 2.027344 ## [8] 2.005859 1.995117 2.000488 1.997803 1.999146 1.999817 2.000153 ## [15] 1.999985 2.000069 2.000027 2.000006 1.999995 2.000000 1.999998 ## [22] 1.999999 2.000000 ## ## $precision ## [1] 6.556511e-07 ## ## $iteraciones ## [1] 23 ``` El método de bisección después de 23 iteraciones y una precisión 6.556511e^-07, la raíz es 2.000000 ```r bisect(f_4, -3, 2.5) ``` ``` ## $root ## [1] 0 ## ## $f.root ## [1] 0 ## ## $iter ## [1] 2 ## ## $estim.prec ## [1] 0 ``` b) [-2.5, 3]\qquad\qquad ```r metodo_biseccion(f_5, -2.5, 3, 10^-5, 100) ``` ``` ## $aprox ## [1] 0.250000 -1.125000 -1.812500 -2.156250 -1.984375 -2.070312 -2.027344 ## [8] -2.005859 -1.995117 -2.000488 -1.997803 -1.999146 -1.999817 -2.000153 ## [15] -1.999985 -2.000069 -2.000027 -2.000006 -1.999995 -2.000000 ## ## $precision ## [1] 5.245209e-06 ## ## $iteraciones ## [1] 20 ``` El método de bisección después de 20 iteraciones y una precisión 5.245209e^-067, la raíz es -2.000000 ```r bisect(f_4, -2.5, 3) ``` ``` ## $root ## [1] 0 ## ## $f.root ## [1] 0 ## ## $iter ## [1] 2 ## ## $estim.prec ## [1] 0 ``` c)[-1.75, 1.5]\qquad\qquad ```r metodo_biseccion(f_5, -1.75, 1.5, 10^-5, 100) ``` ``` ## $aprox ## [1] -0.1250000 -0.9375000 -1.3437500 -1.1406250 -1.0390625 -0.9882812 ## [7] -1.0136719 -1.0009766 -0.9946289 -0.9978027 -0.9993896 -1.0001831 ## [13] -0.9997864 -0.9999847 -1.0000839 -1.0000343 -1.0000095 -0.9999971 ## [19] -1.0000033 ## ## $precision ## [1] 6.198883e-06 ## ## $iteraciones ## [1] 19 ``` El método de bisección después de 19 iteraciones y una precisión 6.198883e^-06, la raíz es -1.0000033 ```r bisect(f_4, -1.75, 1.5) ``` ``` ## $root ## [1] 0 ## ## $f.root ## [1] 0 ## ## $iter ## [1] 2 ## ## $estim.prec ## [1] 0 ``` d) [-1.5, 1.75] \end{equation}\]

metodo_biseccion(f_5, -1.5, 1.75, 10^-7, 100)
## $aprox
##  [1] 0.1250000 0.9375000 1.3437500 1.1406250 1.0390625 0.9882812 1.0136719
##  [8] 1.0009766 0.9946289 0.9978027 0.9993896 1.0001831 0.9997864 0.9999847
## [15] 1.0000839 1.0000343 1.0000095 0.9999971 1.0000033 1.0000002 0.9999987
## [22] 0.9999995 0.9999999 1.0000000 0.9999999
## 
## $precision
## [1] 9.685755e-08
## 
## $iteraciones
## [1] 25

El método de bisección después de 25 iteraciones y una precisión 9.685755e-08, la raíz es 0.9999999

bisect(f_4, -1.5, 1.5)
## $root
## [1] 0
## 
## $f.root
## [1] 0
## 
## $iter
## [1] 2
## 
## $estim.prec
## [1] 0

Ejercicio 6

En cada una de las siguientes ecuaciones, determina un intervalo \([a,b]\) en que convergerá la iteración de punto fijo. Estima la cantidad de iteraciones necesarias para obtener aproximaciones con una exactitud de \(10^{-5}\) y realiza los cálculos.

it_pf <- function(g, q0, pr=1e-5, N=100){
  cond <- 1
  it <- 1
  q <- q0
  while(cond==1){
    if(it<=N){
      q[it+1] = g(q[it]) # iteración de la función 
      pr_it <- abs(q[it+1]-q[it]) # precisión en la iteración 
       if(pr_it<pr){
          resultados <- list(sucesion=q, precision=pr_it, iteraciones=it)
          return(resultados)
          cond <- 0
          }#final del segundo if
        else{it <- it+1}
    }#final del primer if
    else{
      print("Se alcanzo el maximo de iteraciones")
      cond <- 0
    }#fin del else
  }#final del while
}# final de la función
  1. \(\quad x=\frac{2-e^{x}+x^{2}}{3}\)
f_6 <- function(x){x}
g_6a<- function(x){(2-exp(x)+x^2)/3}
  
x_6a <- seq(from=-0.5, to=2.5, by=0.01)
y_f_6a <- f_6(x_6a)
y_g_6a <- g_6a(x_6a)

graf_6a <- ggplot()+
  geom_vline(xintercept = 0, linetype="dashed")+ #eje x
  geom_hline(yintercept = 0, linetype="dashed")+ #eje y
  geom_line(aes(x=x_6a, y=y_f_6a), color="orange", size=1)+
  geom_line(aes(x=x_6a, y=y_g_6a), color="forestgreen", size=1)+
  coord_fixed(ratio = 1)+ #misma escala en los ejes
  labs(x="x", y="f(x)", title="Punto fijo gráfica 6a ")+
  theme_bw()

#plot(graf_6a)

ggplotly(graf_6a)
it_pf(g_6a, 2.9, 1e-8, 50)
## $sucesion
##  [1]  2.9000000 -2.5880485  2.8742761 -2.4837113  2.6951296 -1.8479048
##  [7]  1.7523950 -0.2325044  0.4205040  0.2180316  0.2679705  0.2548332
## [13]  0.2582312  0.2573484  0.2575775  0.2575180  0.2575335  0.2575295
## [19]  0.2575305  0.2575302  0.2575303  0.2575303  0.2575303
## 
## $precision
## [1] 4.719076e-09
## 
## $iteraciones
## [1] 22

Después de 22 iteraciones y una precisión 4.719076e-09, el punto fijo de x=(2-ex+x2/3) es 0.2575303

  1. \(\quad x=\frac{5}{x^{2}}+2\)
f_6<- function(x){x}
g_6b<- function(x){(5/(x^2))+2}
  
x_6b <- seq(from=2, to=3, by=0.01)
y_f_6b <- f_6(x_6b)
y_g_6b <- g_6b(x_6b)

graf_6b <- ggplot()+
  #geom_vline(xintercept = 0, linetype="dashed")+ #eje x
  #geom_hline(yintercept = 0, linetype="dashed")+ #eje y
  geom_line(aes(x=x_6b, y=y_f_6b), color="orange", size=0.5)+
  geom_line(aes(x=x_6b, y=y_g_6b), color="forestgreen", size=0.5)+
  geom_polygon(aes(x=c(2.9,2.4,2.4,2.9), y=c (3, 3, 2.4, 2.4)),   size=0.5, color= "blue", fill=NA)+
  coord_fixed(ratio=1)+
  labs(x="x", y="f(x)", title="Punto fijo  ejercicio 6 b")+
  theme_bw()

#plot(graf_6b)

ggplotly(graf_6b)
it_pf(g_6b, 0.2, 1e-8, 50)
## $sucesion
##  [1]   0.200000 127.000000   2.000310   3.249613   2.473486   2.817243
##  [7]   2.629972   2.722882   2.674392   2.699069   2.686344   2.692862
## [13]   2.689512   2.691231   2.690348   2.690801   2.690569   2.690688
## [19]   2.690627   2.690658   2.690642   2.690650   2.690646   2.690648
## [25]   2.690647   2.690648   2.690647   2.690647   2.690647   2.690647
## [31]   2.690647   2.690647   2.690647
## 
## $precision
## [1] 5.412974e-09
## 
## $iteraciones
## [1] 32

Después de 32 iteraciones y una precisión 5.412974e^-09, el punto fijo de x=(5/x^2)+2 es 2.690647

  1. \(\quad x=\left(e^{x} / 3\right)^{1 / 2}\)
f_6<- function(x){x}
g_6c<- function(x){sqrt(exp(x)/3)}
  
x_6c <- seq(from=-1, to=2, by=0.01)
y_f_6c <- f_6(x_6c)
y_g_6c <- g_6c(x_6c)

graf_6c <- ggplot()+
  geom_vline(xintercept = 0, linetype="dashed")+ #eje x
  geom_hline(yintercept = 0, linetype="dashed")+ #eje y
  geom_line(aes(x=x_6c, y=y_f_6c), color="orange", size=0.5)+
  geom_line(aes(x=x_6c, y=y_g_6c), color="forestgreen", size=0.5)+
  geom_polygon(aes(x=c(0.5,1,1,0.5), y=c (0.5, 0.5, 1, 1)),   size=0.5, color= "blue", fill=NA)+
  coord_fixed(ratio=1)+
  labs(x="x", y="f(x)", title="Punto fijo  ejercicio 6 c")+
  theme_bw()

#plot(graf_6a)

ggplotly(graf_6c)
it_pf (g_6c, 1, 1e-8, 50)
## $sucesion
##  [1] 1.0000000 0.9518897 0.9292650 0.9188121 0.9140225 0.9118362 0.9108400
##  [8] 0.9103864 0.9101800 0.9100860 0.9100433 0.9100238 0.9100150 0.9100109
## [15] 0.9100091 0.9100083 0.9100079 0.9100077 0.9100076 0.9100076 0.9100076
## [22] 0.9100076
## 
## $precision
## [1] 7.397953e-09
## 
## $iteraciones
## [1] 21

Después de 21 iteraciones con una preicisión de 7.397953e^-09 el punto fijo de x=((ex/3))1/2 es 0.9100076

  1. \(\quad x=5^{-x}\)
f_6<- function(x){x}
g_6d<- function(x){5^(-x)}
  
x_6d <- seq(from=-.3, to=4, by=0.01)
y_f_6d <- f_6(x_6d)
y_g_6d <- g_6d(x_6d)

graf_6d <- ggplot()+
  geom_vline(xintercept = 0, linetype="dashed")+ #eje x
  geom_hline(yintercept = 0, linetype="dashed")+ #eje y
  geom_line(aes(x=x_6d, y=y_f_6d), color="orange", size=0.5)+
  geom_line(aes(x=x_6d, y=y_g_6d), color="forestgreen", size=0.5)+
  geom_polygon(aes(x=c(.3, .6, .6, .3), y=c (.3, .3, .6, .6)),   size=0.5, color= "blue", fill=NA)+
  coord_fixed(ratio=1)+
  labs(x="x", y="f(x)", title="Punto fijo  ejercicio 6 d")+
  theme_bw()

#plot(graf_6a)

ggplotly(graf_6d)
it_pf(g_6d, 0.3, 1e-8, 100)
## $sucesion
##  [1] 0.3000000 0.6170339 0.3704349 0.5509056 0.4120345 0.5152290 0.4363856
##  [8] 0.4954269 0.4505173 0.4842860 0.4586682 0.4779745 0.4633511 0.4743856
## [15] 0.4660352 0.4723407 0.4675715 0.4711743 0.4684501 0.4705085 0.4689523
## [22] 0.4701283 0.4692393 0.4699112 0.4694034 0.4697872 0.4694971 0.4697163
## [29] 0.4695506 0.4696758 0.4695812 0.4696527 0.4695986 0.4696395 0.4696086
## [36] 0.4696320 0.4696143 0.4696277 0.4696176 0.4696252 0.4696194 0.4696238
## [43] 0.4696205 0.4696230 0.4696211 0.4696225 0.4696215 0.4696223 0.4696217
## [50] 0.4696221 0.4696218 0.4696220 0.4696218 0.4696220 0.4696219 0.4696220
## [57] 0.4696219 0.4696219 0.4696219 0.4696219 0.4696219 0.4696219 0.4696219
## [64] 0.4696219
## 
## $precision
## [1] 9.208277e-09
## 
## $iteraciones
## [1] 63

Después de 63 iteraciones con una preicisión de 9.208277e^-09 el punto fijo de x=5^-x es 0.4696219

  1. \(\quad x=6^{-x}\)
f_6<- function(x){x}
g_6e<- function(x){6^(-x)}
  
x_6e <- seq(from=-0, to=4, by=0.01)
y_f_6e <- f_6(x_6e)
y_g_6e <- g_6e(x_6e)

graf_6e <- ggplot()+
  geom_vline(xintercept = 0, linetype="dashed")+ #eje x
  geom_hline(yintercept = 0, linetype="dashed")+ #eje y
  geom_line(aes(x=x_6e, y=y_f_6e), color="orange", size=0.5)+
  geom_line(aes(x=x_6e, y=y_g_6e), color="forestgreen", size=0.5)+
  geom_polygon(aes(x=c(.3, .6, .3, .6), y=c (.3, .3, .6, .6)),   size=0.5, color= "blue", fill=NA)+
  coord_fixed(ratio=1)+
  labs(x="x", y="f(x)", title="Punto fijo  ejercicio 6 e")+
  theme_bw()

#plot(graf_6a)

ggplotly(graf_6e)
it_pf(g_6e, .3, 1e-8, 100)
## $sucesion
##  [1] 0.3000000 0.5841907 0.3510842 0.5330935 0.3847447 0.5018922 0.4068665
##  [8] 0.4823878 0.4213367 0.4700416 0.4307612 0.4621710 0.4368789 0.4571325
## [15] 0.4408408 0.4538990 0.4434023 0.4518205 0.4450567 0.4504832 0.4461244
## [22] 0.4496222 0.4468131 0.4490677 0.4472573 0.4487105 0.4475436 0.4484803
## [29] 0.4477283 0.4483320 0.4478473 0.4482364 0.4479240 0.4481748 0.4479734
## [36] 0.4481351 0.4480053 0.4481095 0.4480258 0.4480930 0.4480391 0.4480823
## [43] 0.4480476 0.4480755 0.4480531 0.4480711 0.4480566 0.4480682 0.4480589
## [50] 0.4480664 0.4480604 0.4480652 0.4480614 0.4480645 0.4480620 0.4480640
## [57] 0.4480624 0.4480637 0.4480626 0.4480634 0.4480628 0.4480633 0.4480629
## [64] 0.4480632 0.4480630 0.4480632 0.4480630 0.4480631 0.4480630 0.4480631
## [71] 0.4480630 0.4480631 0.4480631 0.4480631 0.4480631 0.4480631 0.4480631
## [78] 0.4480631 0.4480631 0.4480631 0.4480631
## 
## $precision
## [1] 8.248957e-09
## 
## $iteraciones
## [1] 80

Después de 80 iteraciones con una precisión de 8.248957e^-09 el punto fijo de x=6^-x es 0.4480631

  1. \(\quad x=0.5(\sin x+\cos x)\)
f_6<- function(x){x}
g_6f<- function(x){0.5*(sin(x)+cos(x))}
  
x_6f <- seq(from=-5, to=5, by=0.01)
y_f_6f <- f_6(x_6f)
y_g_6f <- g_6f(x_6f)

graf_6f <- ggplot()+
  geom_vline(xintercept = 0, linetype="dashed")+ #eje x
  geom_hline(yintercept = 0, linetype="dashed")+ #eje y
  geom_line(aes(x=x_6f, y=y_f_6f), color="orange", size=0.5)+
  geom_line(aes(x=x_6f, y=y_g_6f), color="forestgreen", size=0.5)+
  geom_polygon(aes(x=c(0,1,1,0), y=c (0, 0, 1, 1)),   size=0.5, color= "blue", fill=NA)+
  coord_fixed(ratio=1)+
  labs(x="x", y="f(x)", title="Punto fijo  ejercicio 6 f")+
  theme_bw()

#plot(graf_6a)

ggplotly(graf_6f)
it_pf(g_6a, 0.2, 1e-8, 50)
## $sucesion
##  [1] 0.2000000 0.2728657 0.2535773 0.2585582 0.2572636 0.2575995 0.2575123
##  [8] 0.2575349 0.2575291 0.2575306 0.2575302 0.2575303 0.2575303 0.2575303
## 
## $precision
## [1] 6.919711e-09
## 
## $iteraciones
## [1] 13

Después de 13 iteraciones con una precisión de 6.919711e^-09 el punto fijo de x=0.5(sin(x)+cos(x)) es 0.2575303