Here's the question you clicked on:
DoubleCook
What is the Taylor series for f(x)=sin^2(5x)? Hint: using the identity sin^2 x = 1/2 (1-cos(2x))
\[( \left( \sin \left( 5\,a \right) \right) ^{2}+10\,\sin \left( 5\,a \right) \cos \left( 5\,a \right) \left( x-a \right) + \left( -25\, \left( \sin \left( 5\,a \right) \right) ^{2}+25\, \left( \cos \left( 5\,a \right) \right) ^{2} \right) \left( x-a \right) ^{2}-{ \frac {500}{3}}\,\sin \left( 5\,a \right) \cos \left( 5\,a \right) \left( x-a \right) ^{3}+ \left( {\frac {625}{3}}\, \left( \sin \left( 5\,a \right) \right) ^{2}-{\frac {625}{3}}\, \left( \cos \left( 5\,a \right) \right) ^{2} \right) \left( x-a \right) ^{4}+{ \frac {2500}{3}}\,\sin \left( 5\,a \right) \cos \left( 5\,a \right) \left( x-a \right) ^{5}+O \left( \left( x-a \right) ^{6} \right) ) \]
Mmm what is it using the sigma notation? I was thinking something like \[\sum_{0}^{\infty} (1/2)- \frac{ 1/2 (-1)^k (10x)^{2k} }{ (2k)! }\] but my homework says it's wrong? :/
http://www.webassign.net/userimages/knownseries.jpg?db=v4net&id=154152 I used this
I am sorry i suck at sigma notation
Oh haha don't worry about it, thanks anyways! :]
I think the 1/2 and the minus are outside the sigma.
\[\frac{1}{2}-\frac{1}{2}\sum_{0}^{\infty}\frac{(-1)^{k}(10x)^{2k}}{(2k)!}\]
Oh outside? Great, thanks a lot! :]