Power series

**Dragonkiller** · August 24th 2012, 07:49 PM

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 $\epsilon$ R
Anna(x) = $\sum_{n=0}^{\infty }a_{n}x^{n}$ , bob(x)= $\sum_{n=0}^{\infty }b_{n}x^{n}$
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 $a_{n}$ and $b_{n}$ of their power series.
b) Express the exponential function $e^{x}$ and $e^{-x}$ as a combination of anna and bob. Conversely,
express anna and bob as a combination of $e^{x}$ and $e^{-x}$
Thanks in advance

**Prove It** · August 24th 2012, 08:35 PM

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 $\epsilon$ R
Anna(x) = $\sum_{n=0}^{\infty }a_{n}x^{n}$ , bob(x)= $\sum_{n=0}^{\infty }b_{n}x^{n}$
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 $a_{n}$ and $b_{n}$ of their power series.
b) Express the exponential function $e^{x}$ and $e^{-x}$ as a combination of anna and bob. Conversely,
express anna and bob as a combination of $e^{x}$ and $e^{-x}$
Thanks in advance

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

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

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

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

So therefore

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

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