For this week’s discussion, I will be solving problem 4.

\(f(x) = \sin x; f(0) = \sin(0) = 0\)

First derivative is \(f^{'}(x) = cos x; f^{'}(0) = cos(0) = 1\)

Second derivative is \(f^{''}(x) = -\sin x; f^{''}(0) = -\sin(0) = 0\)

Third derivative is \(f^{'''}(x) = -\cos x; f^{'''}(0) = -\cos(0) = -1\)

Fourth derivative is \(f^{(4)}(x) = \sin x; f^{(4)}(0) = \sin(0) = 0\)

Fifth derivative is \(f^{(5)}(x) = \cos x; f^{(5)}(0) = \cos(0) = 1\)

Sixth derivative is \(f^{(6)}(x) = -\sin x; f^{(6)}(0) = -\sin(0) = 0\)

Seventh derivative is \(f^{(7)}(x) = -\cos x; f^{(7)}(0) = -\cos(0) = -1\)

Continuing to \(n^{th}\) derivative, pattern identified is {0, 1, 0, -1}

Using the defination of Taylor and Maclaurin Series,

In our case Maclaurin Series will be

\(0\times \frac{x^0}{0!} + 1 \times \frac{x^1}{1!} + 0 \times \frac{x^2}{2!} - 1 \times \frac{x^3}{3!} + ...\)

= \(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - ....\)

= \(\sum_{n=0}^\infty (-x)^n\frac{x^{(2n+1)}}{(2n+1)!}\)

This proves that for Taylor Series of \(\sin x\), centered at \(x = 0\), Maclaurin Series is \(\sum_{n=0}^\infty (-x)^n\frac{x^{(2n+1)}}{(2n+1)!}\)

References