Question

What is the Taylor series for f(x)=sin^2(5x)?
Hint: using the identity sin^2 x = 1/2 (1-cos(2x))

Answer 1

\[( \left( \sin \left( 5\,a \right) \right) ^{2}+10\,\sin \left( 5\,a \right) \cos \left( 5\,a \right) \left( x-a \right) + \left( -25\, \left( \sin \left( 5\,a \right) \right) ^{2}+25\, \left( \cos \left( 5\,a \right) \right) ^{2} \right) \left( x-a \right) ^{2}-{ \frac {500}{3}}\,\sin \left( 5\,a \right) \cos \left( 5\,a \right) \left( x-a \right) ^{3}+ \left( {\frac {625}{3}}\, \left( \sin \left( 5\,a \right) \right) ^{2}-{\frac {625}{3}}\, \left( \cos \left( 5\,a \right) \right) ^{2} \right) \left( x-a \right) ^{4}+{ \frac {2500}{3}}\,\sin \left( 5\,a \right) \cos \left( 5\,a \right) \left( x-a \right) ^{5}+O \left( \left( x-a \right) ^{6} \right) ) \]

Answer 2

Something like that?

Answer 3

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? :/

Answer 4

http://www.webassign.net/userimages/knownseries.jpg?db=v4net&id=154152
I used this

Answer 5

I am sorry i suck at sigma notation

Answer 6

Oh haha don't worry about it, thanks anyways! :]

Answer 7

I think the 1/2 and the minus are outside the sigma.

Answer 8

\[\frac{1}{2}-\frac{1}{2}\sum_{0}^{\infty}\frac{(-1)^{k}(10x)^{2k}}{(2k)!}\]

Answer 9

Oh outside? Great, thanks a lot! :]