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Let $f(x) = e^{2x}\cos3x$. Estimate the value of $f(0.5)$ using Lagrange's interpolating polynomial of degree $3$ over the nodes $x=0, x=0.3, x=0.6$ and $x=1$. Also compute the error bound over the interval $[0,1]$ and the actual error $E(0.5)$.

I am not sure whether we should use $x$ in radians or degrees.

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  • $\begingroup$ The values of $x$ must be in radians, so ... $\endgroup$ – Nitin Uniyal Dec 24 '16 at 5:22
  • $\begingroup$ @Moo : Would it be possible for you to post the full solution? I am unable to figure out my mistake. $\endgroup$ – Pushpendra Pal Dec 24 '16 at 7:17
  • $\begingroup$ Default settings in my calculator was in degrees, hence the problem. Thank you @Moo $\endgroup$ – Pushpendra Pal Dec 24 '16 at 16:49
  • $\begingroup$ However, I am still unable to find the error bound. $\endgroup$ – Pushpendra Pal Dec 24 '16 at 16:51
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We are given $$f(x) = e^{2x}\cos3x, x_0 = 0.0, x_1 = 0.3, x_2 = 0.6, x_3 = 1.0$$

We are asked to construct the interpolation polynomial of degree three, to approximate $f(0.5)$, to compute the error bound over the interval $[0,1]$ and the actual error $E(0.5)$.

The cubic polynomial passing through these points can be expressed as:

$$P_3(x) = y_0 L_0(x) + y_1 L_1(x) + y_2 L_2(x) + y_3 L_3(x)$$

where:

  • $L_0(x) = \dfrac{(x − x_1)(x − x_2)(x-x_3)}{(x_0 − x_1)(x_0 − x_2)(x_0-x_3)} = -5.55556 (x-1) (x-0.6) (x-0.3)$
  • $L_1(x) = \dfrac{(x − x_0)(x − x_2)(x-x_3)}{(x_1 − x_0)(x_1 − x_2)(x_1-x_3)} = 15.873 (x-1.) (x-0.6) (x+0.) $
  • $L_2(x) = \dfrac{(x − x_0)(x − x_1)(x-x_3)}{(x_2 − x_0)(x_2 − x_1)(x_2-x_3)} = -13.8889 (x-1.) (x-0.3) (x+0.)$
  • $L_3(x) = \dfrac{(x − x_0)(x − x_1)(x-x_2)}{(x_3 − x_0)(x_3 − x_1)(x_3-x_2)} = 3.57143 (x-0.6) (x-0.3) (x+0.)$

The Lagrange Interpolating Polynomial is

$$P_3(x) = -3.22551 x^3-8.31722 x^2+3.22762 x+1.$$

The formula for the error bound is given by

$$E_n(x) = {f^{n+1}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$

Since we do not know where $\xi(x)$ is, we will find each error over the range and multiply those together, so we have

$$\max_{(x \in 0, 1)} |f^{(4)}(x)| = \max_{(x \in 0,1)} |120 e^{2 x} \sin 3 x-119 e^{2 x} \cos 3 x| = 998.291$$

Next, we need to find

$$\max_{(x \in 0, 1)} |(x-0)(x-0.3)(x-0.6)(x-1.0)| = 0.0175456$$

Thus we have an error bound of

$$E_3(x) = \dfrac{998.291}{24} \times 0.0175456 \le 0.729816$$

If we compute the actual error, we have

$$\mbox{Actual Error}~ = |f(0.5) - P_3(0.5)| = 0.0609676$$

We can also plot the actual function versus the Lagrange Interpolating Polynomial over the range

enter image description here

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  • $\begingroup$ Hi Moo, I am new to Langrange's interpolation. How did you find max(x∈0,1)|f(4)(x)|=max(x∈0,1)|120e2xsin3x−119e2xcos3x|=998.291 max(x∈0,1)|(x−0)(x−0.3)(x−0.6)(x−1.0)|=0.0175456 $\endgroup$ – Tortoise Jul 10 '17 at 10:35
  • $\begingroup$ @SurendraJ: For example, plot the function $(x-0)(x-0.3)(x-0.6)(x-1.)$ over $(x, 0, 1)$, find the critical points by taking the first derivative and setting it equal to zero - you will find three values of $x$ (see plot you just created), then recall we want the absolute value - so the max is clear. You will find the maximum as shown in the solution amongst those three values. Repeat this process for the other function $\endgroup$ – Moo Jul 10 '17 at 12:51

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