# power series

• June 6th 2010, 02:37 AM
godai
power series
hi :)
I know the power series representation of sin x is Sum [ (-1)^n*(x)^(2n+1)/(2n+1)!] but what about (sinx)^2?

another quick question :
give an example of set A that has no supremum and why does it not violate the Completeness Axiom.
is {tanx : 0 < x < pi/2} correct?
• June 6th 2010, 05:58 AM
vincisonfire
To get the power series expansion of $sin^2(x)$, you can do 2 things :
1. Compute and evaluate the derivative up to some order and try to find a general expression for the coefficients;
2. $sin^2(x) = \sum_{i=0}^\infty c_nx^n = \left( \sum_{i=0}^\infty (-1)^{n} \frac{x^{2n+1}}{(2n+1)!} \right)^2$
Equate coefficient on both sides using this formula
$
\left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right)$

$= \sum_{i=0}^\infty \sum_{j=0}^\infty a_i b_j x^{i+j}$
$= \sum_{n=0}^\infty \left(\sum_{i=0}^n a_i b_{n-i}\right) x^n$
I suggest the second method.
• June 6th 2010, 09:14 AM
chisigma
A third way is the use of identity...

$\sin^{2} x = \frac{1}{2} - \frac{1}{2}\cdot \cos 2x$ (1)

... so that is...

$\sin^{2} x = \frac{1}{2} \cdot \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(2x)^{2n}}{(2n)!}$ (2)

Kind regards

$\chi$ $\sigma$
• June 6th 2010, 09:20 AM
vincisonfire
chisigma's trick is even better
• June 6th 2010, 09:51 AM
godai
lol why didn't I think of that!
thank you!!! (Rofl)