Newton's method

Considering the table of the function $f(x)=e^x$:

$x$ $0.0$ $0.2$ $0.4$
$f(x)$ $1.0000$ $1.2214$ $1.4918$
  1. Find the value of $\sqrt[3]{e}$ by interpolation.
  2. Give a bound for the error.
  3. Find the value of $\sqrt[3]{e}$ by interpolation if we add a new point: $(0.6, 1.8221)$.
  1. We are going to interpolate the function from the given information and later we will find an approximation for $\sqrt[3]{e}$. We write the table:
$0.0$ $1.0000$    
    $\dfrac{1.2214-1.0000}{0.2-0.0}=\dfrac{0.2214}{0.2}=1.107$  
$0.2$ $1.2214$   $\dfrac{1.352-1.107}{0.4-0.0}=\dfrac{0.245}{0.4}=0.6125$
    $\dfrac{1.4918-1.2214}{0.4-0.2}=\dfrac{0.2704}{0.2}=1.352$  
$0.4$ $1.4918$    

Thus the interpolating polynomial is

$$ \begin{array}{rl} P_2(x)=& 1.000+1.107\cdot(x-0)+0.6125\cdot(x-0)\cdot(x-0.2) \\ =& 1.000+1.107x-0.1225x+0.6125x^2\\ =& 1.000+0.9845x+0.6125x^2 \end{array}$$

Then, $\sqrt[3]{e}=e^{\frac{1}{3}}\approx P_2\Big( \dfrac{1}{3}\Big) = 1+0.9845\cdot\dfrac{1}{3}+0.6125\cdot\dfrac{1}{9}=1.3962$

  1. Since we already know the calculation that we have done is an approximation, we are going to try to give a bound for the error. To do so, we will have to know the third derivative of the function and a bound for its module knowing that $x\in(0,0.4)$:

$$f^{(3)}(x)=e^x \Rightarrow |f^{(3)}(x)|\leqslant e^{0.4} $$

since the function is increasing. Therefore:

$$\begin{array}{rl} |\text{error}|=& \Big| f\Big( \dfrac{1}{3} \Big) - P_2 \Big( \dfrac{1}{3} \Big) \Big| = \Big| \dfrac{f^{(3)}(\xi(x))}{3!}\cdot\Big( \dfrac{1}{3}-0\Big)\cdot \Big( \dfrac{1}{3}-0.2\Big)\cdot\Big( \dfrac{1}{3}-0.4\Big) \Big| \\ \leqslant & \Big| \dfrac{e^{0.4}}{3!}\cdot \dfrac{1}{3} \cdot \cdot \Big( \dfrac{1}{3}-0.2\Big)\cdot\Big( \dfrac{1}{3}-0.4\Big) \Big| = 0.36\cdot 10^{-4} \end{array} $$

  1. Bearing in mind that we have calculated the interpolating polynomial using Newton's method, when adding another point we will use the calculations we already did. We add a new line with the new point in the previous table:
$0.0$ $1.0000$      
    $\dfrac{1.2214-1.0000}{0.2-0.0}=\dfrac{0.2214}{0.2}=1.107$    
$0.2$ $1.2214$   $\dfrac{1.352-1.107}{0.4-0.0}=\dfrac{0.245}{0.4}=0.6125$  
    $\dfrac{1.4918-1.2214}{0.4-0.2}=\dfrac{0.2704}{0.2}=1.352$   $\dfrac{0.74875-0.6125}{0.6-0.0}=\dfrac{0.13625}{0.6}=0.22708$
$0.4$ $1.4918$   $\dfrac{1.6515-1.352}{0.6-0.2}=\dfrac{0.2995}{0.4}=0.74875$  
    $\dfrac{1.8221-1.4918}{0.6-0.4}=\dfrac{0.3303}{0.2}=1.6515$    
$0.6$ $1.8221$      

With the new calculations. Now, the interpolating polynomial (now of degree $3$) is:

$$ \begin{array}{rl} P_3(x)=& 1+1.107\cdot(x-0)+0.6125\cdot(x-0)\cdot(x-0.2)\\ & +0.22708\cdot(x-0)\cdot(x-0.2)\cdot(x-0.4)\\ =& 1+1.107x-0.1225x+0.6125x^2+0.22708x^3\\ &-0.0908x^2-0.0454x^2+0.0182x\\ =& 1+1.0027x+0.4853x^2 +0.2271x^3 \end{array}$$

So, the value of $\sqrt[3]{e}$ will be: $$\sqrt[3]{e}=e^{\frac{1}{3}}\approx P_3\Big( \dfrac{1}{3}\Big) = 1+1.0027\cdot\dfrac{1}{3}+0.4853\cdot\dfrac{1}{9}+0.2271\dfrac{1}{27}=1.3967$$

  1. $1.3962$
  2. $7.36\cdot 10^{-4}$
  3. $1.3967$
Back to topic