Answered: 38. V79

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Approximations with Taylor polynomials
a. Approximate the given quantities using Taylor polynomials with n = 3.
b. Compute the absolute error in the approximation, assuming the
exact value is given by a calculator.

Expert Solution

Step 1

The quantity is $\sqrt[4]{79}$ .

We have to find the approximate value of $\sqrt[4]{79}$ using the Taylor polynomial with $n = 3$ .

Step 2

Taylor polynomial of $f (x)$ centered at $x = a$ :

$f (x) = f (a) + \frac{f' (a)}{1!} (x - a) + \frac{f'' (a)}{2!} {(x - a)}^{2} + \frac{f''' (a)}{3!} {(x - a)}^{3} + \cdot \cdot \cdot$

For $n = 3$ ,

$f (x) = f (a) + \frac{f' (a)}{1!} (x - a) + \frac{f'' (a)}{2!} {(x - a)}^{2} + \frac{f''' (a)}{3!} {(x - a)}^{3}$ .

We have to find the approximate value of $\sqrt[4]{79}$ .

Let $f (x) = \sqrt[4]{x}$ and the nearest value $a = 81$ .

$\begin{array}{rcl} f (x) & = & x^{\frac{1}{4}}; f (81) = {(81)}^{\frac{1}{4}} = 3 \\ f' (x) & = & \frac{1}{4} x^{- \frac{3}{4}}; f' (81) = \frac{1}{4} {(81)}^{- \frac{3}{4}} = \frac{1}{108} \\ f'' (x) & = & - \frac{3}{16} x^{- \frac{7}{4}}; f'' (81) = - \frac{3}{16} {(81)}^{- \frac{7}{4}} = - \frac{1}{11664} \\ f''' (x) & = & \frac{21}{64} x^{- \frac{11}{4}}; f''' (81) = \frac{21}{64} {(81)}^{- \frac{11}{4}} = \frac{7}{3779163} \end{array}$

Now,

$P_{3} (x) = 3 + \frac{1}{108} (x - 81) - \frac{1}{2! \times 11664} {(x - 81)}^{2} + \frac{7}{3! \times 3779139} {(x - 81)}^{3}$

To approximate $\sqrt[4]{79}$ :

Put $x = 79$

$\begin{array}{rcl} P_{3} (79) & = & 3 + \frac{1}{108} (79 - 81) - \frac{1}{2! \times 11664} {(79 - 81)}^{2} + \frac{7}{3! \times 3779136} {(79 - 81)}^{3} \\ P_{3} (79) & = & 2.9813075440 \end{array}$

Therefore the approximate value is $P_{3} (79) \approx 2.9813075440$ .

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