DATA_605_Discussion_14_DThonn_8.8_#20

** DATA_605_Discussion_14_Taylor_Series_Chapter_8.8_Problem_20 **

** Taylor_Series **

Chapter 8.8 , # 20

20). Use the Taylor series give in Key Idea 32 to verify the given identity.

Reference: see Example 271 in Chapter 8.8 of the Apex calculus text, Hartman (page 483)

Verify: d/dx(cos(x)) = - sin(x)

start with cos(x)

\[cos(x) = \Sigma_{i=1}^{\infty} (-1)^n * \frac {x^{2n}}{(2n)!}\]

take the derivative of cos(x):

\[\frac {d(cos(x))}{dx} = \frac {d}{dx} \Sigma_{i=1}^{\infty} (-1)^n * \frac {x^{2n}}{(2n)!}\] evaluate the derivative of the right side:

\[ = \Sigma_{i=1}^{\infty} (-1)^n * 2n * \frac {x^{2n-1}}{(2n)!}\]

expand the series:

\[ = -x + \frac {x^3}{3!} - \frac {x^5}{5!} + \frac {x^7}{7!} - ... \] take the negative of the series:

\[ = - [ x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac {x^7}{7!} + ...] \] substitute the series in the brackets for sin(x), and verification complete:

\[ = -sin(x)\]

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