discusison 14

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

\[f(0)= sin(0) = 0\]

\[f'(0)= cos(0) = 1\] \[f'''(0)= -sin(0) = 0\] \[f''''(0)= -cos(0)=-1\] + If we plug this into taylor expansion we know that even power coefficients are 0 and odd powers alternate in sign and are equal to 1. \[ x+0-\frac{x^3}{3!}+0++\frac{x^5}{5!}..\]

• To express that we want all odd coefficients from n =0 we can generalize to

\[ x^{2n+1}\]

\(n(0)=1, n(1)=3, n(2)=5, n(3)=7 ......\)

• To express the alternating signs, we can generalize to

\[(-1)^n\]

\(n(0)=pos,n(1)=neg, n(2)=pos, n(3)=neg ......\)

• Representing the nth numbers in the series our denominator would be \[(2n+1)!\] \(n(0)=1!, n(1)=3!, n(2)=5!, n(3)=7! ......\)

• All put together \[ \sum_{n=0}^\infty=(-1)^{n}* \frac{x^{2n+1}}{2n+1} \]