0

I'm given that

$$w=z+\sum_{i=2}^\infty a_iz^i$$ $$z=w+\sum_{i=2}^\infty b_iw^i$$ $$y=z-\sum_{i=2}^\infty (-1)^ia_iz^i$$ And that those series are all convergent (in particular I'm not given that they are absolutely convergent). I'm asked to express $z$ as a power series in $y$. However, since the series are not absolutely convergent I can't do something like collect powers of $z$ in an expression of the form $\sum_{i=0}^\infty c_iy^i$ and than equate coefficients, because that requires me re-arranging the order of the summation right?

So considering that we cannot re-arrange the terms, how can we solve this?

Thanks

user2520938
  • 7,235

1 Answers1

1

It's very simple: At stake are the three functions $$f:\quad z\mapsto w:=z+\sum_{k=2}^\infty a_k z^k\ ,\qquad g:\quad w\mapsto z:=w+\sum_{k=2}^\infty b_k w^k\ ,$$ and $$h:\quad z\mapsto y:=z-\sum_{k=2}(-1)^k a_k z^k$$ relating the variables $z$, $w$, and $y$ among each other. From the naming of the variables one surmises that $f$ and $g$ are inverses of each other, whereby $f(0)=g(0)=0$, $\>f'(0)=g'(0)=1$. Furthermore we have $y=h(z)=-f(-z)$ for all $z$ in a neighborhood $U$ of $0$, and therefore $$g(-y)=g\bigl(f(-z)\bigr)=-z\qquad\forall\>z\in U\ .$$ This implies $$z=-g(-y)=y-\sum_{k=0}^\infty(-1)^k b_k y^k\ .$$

Christian Blatter
  • 226,825
  • That's very nice! It didn't occur to me to regard these as actual functions, I was just trying to manipulate terms and equate coefficients and such. Thanks – user2520938 May 04 '15 at 14:08