In this thread, I have asked about a way to calculate the coefficients of the power series for $\cos(\sin(x))$
What is wrong with my series expansion of $\cos(\sin(x))$
I received a reply from Willie Wong, in which he gives a rather terse formula for my calculation. I fail to grasp it since it is written in compact sigma notation so I need you to help me to understand a few points.
The relevant part is included in this image:
Firstly, I don't understand the red part $(n+1)(n+2)$ (in red underlined), is this the result from differentiation?
In the yellow underlined, I don't understand why he changes the subscript.
Could you please use this formula to calculate the first few terms for the series of $\cos(\sin(x))$. If possible, can you explain and rewrite the formula in more explicit forms the sigma notation, since I am still not used to compact notation.
Finally, is there a simpler method to derive the coefficients for this function's series?
Best Answer
We have \begin{eqnarray*} f''(x)=-f(x) \cos^2(x) -f'(x) \tan(x). \end{eqnarray*} The first few terms of $-\cos^2(x)$ are \begin{eqnarray*} -\cos^2(x) = -1+ x^2+ b_4 x^4 + \cdots \end{eqnarray*} and further terms can be calculated by squaring the power series for $ \cos(x)$.
The first few terms of $-\tan(x)$ are \begin{eqnarray*} -\tan(x) = -x+ \frac{1}{3} x^3+ c_5 x^5 + \cdots \end{eqnarray*} and further terms can found in equation $(32)$ here ... http://mathworld.wolfram.com/Tangent.html
We want to calculate $f(x)$ and we shall need its first & second derivatives \begin{eqnarray*} f(x)&=&1+a_2 x^2 +a_4 x^4+ \cdots \\ f'(x)&=&2a_2 x +4a_4 x^3+ \cdots \\ f''(x)&=&2a_2 +12a_4 x^2+ \cdots. \\ \end{eqnarray*} You see where the terms in the red boxes come from now ?
Plugging all these into the first equation gives \begin{eqnarray*} 2a_2 +12a_4 x^2+ \cdots = \left( a_0+a_2 x^2 +a_4 x^4+ \cdots \right) \left( -1+ x^2+ b_4 x^4 + \cdots \right)+ \left( 2a_2 x +4a_4 x^3+ \cdots \right) \left( -x+ \frac{1}{3} x^3+ c_5 x^5 + \cdots \right) \end{eqnarray*} Now just expand the brackets and equate coefficients of $x$.
We have \begin{eqnarray*} \cos(\sin(x))=1 -\frac{x^2}{2} +\frac{5x^4}{24}+\cdots. \end{eqnarray*}
Edit : Equating the $x^0$ terms (i.e constant terms) gives \begin{eqnarray*} 2a_2=-a_0 \\ a_2= -\frac{1}{2} \end{eqnarray*} Now collect all the $x^2$ terms and we have \begin{eqnarray*} 12a_4=a_0 -a_2 -2a_2 \\ a_4= \frac{5}{24}. \end{eqnarray*} You will need to include higher order terms if you wish to calculate the next order ... Good luck $ \ddot \smile$