Taylor Polynomials! I would appreciate any assistance! The more explanation the better. Use a second degree Taylor polynomial to approximate the two solutions to cos(x) = 3/4 inthe interval (-"/2, "/2).

Given that, The function cosx=34 We want to find two solutions of cosx=34 using second-degree Taylor…

Answered: Use a second degree Taylor polynomial…

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Taylor Polynomials! I would appreciate any assistance! The more explanation the better.

Use a second degree Taylor polynomial to approximate the two solutions to cos(x) = 3/4 in
the interval (-"/2, "/2).

Expert Solution

Step 1

Given that,

The function $\cos x = \frac{3}{4}$

We want to find two solutions of $\cos x = \frac{3}{4}$ using second-degree Taylor polynomial

Take $f (x) = \cos x - \frac{3}{4}$

We need to find Taylor's polynomial at x =0.

The Taylor series approximation is given by formula,

$f (x) = f (0) + \frac{f' (0)}{1!} x + \frac{f'' (0)}{2!} x^{2} + . . . + \frac{f^{n} (0)}{n!}$

Step 2

$f (x) = \cos x - \frac{3}{4}$

$f (0) = \cos 0 - \frac{3}{4} = 1 - \frac{3}{4} = \frac{1}{4}$

$f' (x) = - \sin x$

f'(0) =-sin(0) =0

$f'' (x) = - \cos x f'' (0) = - \cos (0) = - 1$

Thus the second-degree Taylors approximation is,

$f (x) = \frac{1}{4} + \frac{0}{1!} x + \frac{- 1}{2!} x^{2} = \frac{1}{4} - \frac{1}{2} x^{2}$

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Use a second degree Taylor polynomial to approximate the two solutions to cos(x) = 3/4 in the interval (-"/2, "/2).