Problems from 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: $$(0.6, 1.8221)$$.
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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:

$$0.0$$ $$1.0000$$    
$$0.2$$ $$1.2214$$   $$\dfrac{1.352-1.107}{0.4-0.0}=\dfrac{0.245}{0.4}=0.6125$$
$$0.4$$ $$1.4918$$    

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

2) 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:

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

3) 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:

$$0.0$$ $$1.0000$$      
$$0.2$$ $$1.2214$$   $$\dfrac{1.352-1.107}{0.4-0.0}=\dfrac{0.245}{0.4}=0.6125$$  
    $$\dfrac{1.4918-1.2214}{0.4-0.2}=\dfrac{0.2704}{0.2}=1.352$$   $$\dfrac{0.74875-0.6125}{0.6-0.0}=\dfrac{0.13625}{0.6}=0.22708$$
$$0.4$$ $$1.4918$$   $$\dfrac{1.6515-1.352}{0.6-0.2}=\dfrac{0.2995}{0.4}=0.74875$$  
$$0.6$$ $$1.8221$$      

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. $$1.3967$$
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