Find a real root of \(f(x)=cos(x)-3x\) in the interval (0,1) by bisection method.

f=function(x) cos(x)-3*x
print(c(f(0),f(1),f(0)*f(1)))
## [1]  1.000000 -2.459698 -2.459698

Since \(f(a)\) and \(f(b)\) are of opposite signs, there is a real root in (a,b).

df=c()
a=0
b=1

for(i in 1:8)
{
  c=(a+b)/2
  df=rbind(df,c(i,a,b,c,f(a),f(c)))
  if(f(a)*f(c)<0) b=c else a=c
  }
df=data.frame(df)
names(df)=c('itr no.','a','b','c','f(a)','f(c)')
df

Regula Falsi Method

df=c()
a=0
b=1

for(i in 1:8)
{
  c=(a*f(b)-b*f(a))/(f(b)-f(a))
  df=rbind(df,c(i,a,b,f(a),f(b),c,f(c)))
  if(f(a)*f(c)<0) b=c else a=c
  }
df=data.frame(df)
names(df)=c('itr no.','a','b','f(a)','f(b)','c','f(c)')
df

Iteration Method

Find a real root of \(f(x)=cos(x)-3x\) near \(x=0\) by iteration method.

Sol: Rewrite the equation as \(x=e^{-x}\) so that \(\phi(x)=e^{-x}\).

phi=function(x) cos(x)/3
df=c()
x0=0

for(i in 1:8)
{
  x1=phi(x0)
  df=rbind(df,c(i,x0,x1))
  x0=x1
  }
df=data.frame(df)
names(df)=c('itr no.','x0','phi(x0)')
df

Newton Raphson Method

\[ x_{n+1}=x_n-\frac{f(x_n)}{f^\prime(x_n)} \]

Find a real root of \(f(x)=cos(x)-3x\) near \(x=0\).

Sol: Differentiating \(f^\prime(x)=-sin(x)-3\)

fprime=function(x) -sin(x)-3
df=c()
x0=0

for(i in 1:8)
{
  x1=x0-f(x0)/fprime(x0)
  df=rbind(df,c(i,x0,f(x0),fprime(x0),x1))
  x0=x1
  }
df=data.frame(df)
names(df)=c('itr no.','x0','f(x0)','fprime(x0)','x1')
df

Euler Method

Solve the following by Euler method \[ y^\prime=x^2−y,\: y(0)=1,\: h=0.1 \] Find the value of y(0.1) correct up to four decimal places.

f=function(x,y) x^2-y
x0=0
y0=1
h=0.1
y10=y0+h*f(x0,y0)
#y10

for (i in 1:3)
{
  y11=y0+h/2*(f(x0,y0)+f(x1,y10))
  print(round(y11,4))
  y10=y11
  x1=x0+h
}
## [1] 0.91
## [1] 0.905
## [1] 0.9053

RK 4th Order Method

Solve the following using RK forth order method \[ y\prime =y−x,\, y(0)=2,\, h=0.2 \] Find y(0.2).

f=function(x,y) y-x
x0=0
y0=2
h=0.2
k1=h*f(x0,y0)
k2=h*f(x0+h/2,y0+k1/2)
k3=h*f(x0+h/2,y0+k2/2)
k4=h*f(x0+h,y0+k3)
res=y0+1/6*(k1+2*k2+2*k3+k4)
data.frame(k1,k2,k3,k4,res)

\[ \begin{array}{|c|c|c|c|c|c|c|} \hline x&y &dy & d^2y &d^3y &d^4y &d^5y\\ \hline 0&7&&&&& \\ &&4&&&& \\ 5&11&&-1 &&&\\ & &3& &2&&\\ 10 &14 & &1& &-1&\\ & &4 & &1 & &0\\ 15&18 & &2 & &-1 &\\ & &6 & &0 & &\\ 20 &24 & &2 & & &\\ & &8 & & & &\\ 25&32 & & & & &\\ \hline \end{array} \]

Finite Difference Methods

Forward Shift Operator E

If \(x_0,x_1,\ldots,x_n\) are n equally spaced points of width \(h\), then the shift operator E on the point \(x_i\) is \[ E(x_i)=x_{i+1} \] and \[ E(y_i)=y_{i+1}=y(x_i+h) \] Simlarly backward shift \(E^{-1}\) on \(y_i\) \[ E^{-1}(y_i)=y_{i-1}=y(x_i-h) \]

Forward difference operator \(\Delta\)

\[ \Delta y_i=y_{i+1}-y_i \] for \(i=0,\ldots,n-1\)

Forward difference operator \(\Delta\)

\[ \nabla y_i=y_i-y_{i-1} \] for \(i=1,\ldots,n\)

Forward difference Table

\[ \begin{array}{|c|c|c|c|c|c|c|} \hline x&y &dy & d^2y &d^3y &d^4y &d^5y\\ \hline x_0&y_0&&&&& \\ &&\Delta y_0&&&& \\ x_1&y_1&&\Delta^2 y_0 &&&\\ & &\Delta y_1& &\Delta^3 y_0 &&\\ x_2 &y_2 & & \Delta^2 y_1& &\Delta^4 y_0&\\ & &\Delta y_2 & &\Delta^3 y_1 & & \Delta^5 y_0\\ x_3&y_3 & &\Delta^2 y_2 & &\Delta^4 y_1 &\\ & &\Delta y_3 & &\Delta^3 y_2 & &\\x_4 &y_4 & &\Delta^2 y_3 & & &\\ & &\Delta y_4 & & & &\\ x_5&y_5 & & & & &\\ \hline \end{array} \]

\[ y(x_n-ph)=y_n+p\nabla y_n+\frac{p(p+1)}{2!}\nabla^2 y_n+ \frac{p(p+1)(p+2)}{3!}\nabla^3 y_n+\cdots \] where \[ p=\frac{x_n-x}{h} \] \[ h=x_1-x_0 \]

\[ \begin{array}{|c|c|c|c|c|c|c|} \hline x&y &dy & d^2y &d^3y &d^4y &d^5y\\ \hline x_0&y_0&&&&& \\ &&\nabla y_1&&&& \\ x_1&y_1&&\nabla^2 y_2 &&&\\ & &\nabla y_2& &\nabla^3 y_3 &&\\ x_2 &y_2 & & \nabla^2 y_3& &\nabla^4 y_4&\\ & &\nabla y_3 & &\nabla^3 y_4 & & \nabla^5 y_5\\ x_3&y_3 & &\nabla^2 y_4 & &\nabla^4 y_5 &\\ & &\nabla y_4 & &\nabla^3 y_5 & &\\x_4 &y_4 & &\nabla^2 y_5 & & &\\ & &\nabla y_5 & & & &\\ x_5&y_5 & & & & &\\ \hline \end{array} \]

\[ \begin{array}{|c|c|c|c|c|c|c|} \hline x&y &\delta y & \delta y &\delta ^3y &\delta ^4y &\delta ^5y\\ \hline x_0&y_0&&&&& \\ &&\delta y_{\frac{1}{2}}&&&& \\ x_1&y_1&&\delta^2 y_1 &&&\\ & &\delta y_{\frac{3}{2}}& &\delta^3 y_{\frac{3}{2}} &&\\ x_2 &y_2 & & \delta^2 y_2& &\delta^4 y_2&\\ & &\delta y_{\frac{5}{2}} & &\delta^3 y_{\frac{5}{2}} & & \delta^5 y_{\frac{5}{2}}\\ x_3&y_3 & &\delta^2 y_3 & &\delta^4 y_3 &\\ & &\delta y_{\frac{7}{2}} & &\delta^3 y_{\frac{7}{2}} & &\\x_4 &y_4 & &\delta^2 y_4 & & &\\ & &\delta y_{\frac{9}{2}} & & & &\\ x_5&y_5 & & & & &\\ \hline \end{array} \]

Lagrange Interpolation

\[ \begin{array}{rcl} L(x) &=& \frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\: y_0 +\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\: y_1\\ && +\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\: y_2 +\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\: y_3\\ \end{array} \]