# Power series

• Aug 24th 2012, 07:49 PM
Dragonkiller
Power series
Hi I'm having troubles with a question and was wondering if someone could guide me through it:
Given two mystery functions called anna and bob. Both have power series representations
that work for all x $\displaystyle \epsilon$ R
Anna(x) = $\displaystyle \sum_{n=0}^{\infty }a_{n}x^{n}$, bob(x)= $\displaystyle \sum_{n=0}^{\infty }b_{n}x^{n}$
Furthermore, anna(0) = 0; bob(0) = 1 and we know that deriving bob gives anna and deriving
anna gives bob.
a) Find anna and bob, i.e. nd the coefficients $\displaystyle a_{n}$ and $\displaystyle b_{n}$ of their power series.
b) Express the exponential function $\displaystyle e^{x}$ and $\displaystyle e^{-x}$ as a combination of anna and bob. Conversely,
express anna and bob as a combination of $\displaystyle e^{x}$ and $\displaystyle e^{-x}$
• Aug 24th 2012, 08:35 PM
Prove It
Re: Power series
Quote:

Originally Posted by Dragonkiller
Hi I'm having troubles with a question and was wondering if someone could guide me through it:
Given two mystery functions called anna and bob. Both have power series representations
that work for all x $\displaystyle \epsilon$ R
Anna(x) = $\displaystyle \sum_{n=0}^{\infty }a_{n}x^{n}$, bob(x)= $\displaystyle \sum_{n=0}^{\infty }b_{n}x^{n}$
Furthermore, anna(0) = 0; bob(0) = 1 and we know that deriving bob gives anna and deriving
anna gives bob.
a) Find anna and bob, i.e. nd the coefficients $\displaystyle a_{n}$ and $\displaystyle b_{n}$ of their power series.
b) Express the exponential function $\displaystyle e^{x}$ and $\displaystyle e^{-x}$ as a combination of anna and bob. Conversely,
express anna and bob as a combination of $\displaystyle e^{x}$ and $\displaystyle e^{-x}$

Here is the important information: You know that differentiating Anna will give Bob, and differentiating Bob will give Anna. So differentiating Anna TWICE will get back to Anna. This gives us the differential equation

\displaystyle \displaystyle \begin{align*} A'' &= A \\ A'' - A &= 0 \\ \textrm{Characteristic Equation: } m^2 - 1 &= 0 \\ m^2 &= 1 \\ m &= \pm 1 \\ \textrm{Therefore } A &= C_1 e^x + C_2 e^{-x} \end{align*}

We also know that \displaystyle \displaystyle \begin{align*} A(0) = 0 \end{align*} and \displaystyle \displaystyle \begin{align*} A'(0) = 1 \end{align*}, so we get

\displaystyle \displaystyle \begin{align*} 0 &= C_1 e^0 + C_2 e^{-0} \\ 0 &= C_1 + C_2 \\ \\ 1 &= C_1 e^0 - C_2 e^{-0} \\ 1 &= C_1 - C_2 \\ \\ C_1 = \frac{1}{2}, C_2 = -\frac{1}{2} \end{align*}

So therefore

\displaystyle \displaystyle \begin{align*} A = \frac{1}{2}e^x - \frac{1}{2}e^{-x} \end{align*} and \displaystyle \displaystyle \begin{align*} B = \frac{1}{2}e^x + \frac{1}{2}e^{-x} \end{align*}

Go from here.