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I have been trying to solve the following equation, but I am still stuck after trying many different methods.

I have been given this equation to solve:

$$(1 + z^{-1})^4 (a + bz^{-1} + az^{-2}) - (1 - z^{-1})^4 (a - bz^{-1} + az^{-2}) = 2z^{-d}$$

I have simplified it down to $$a(4z^{-1} + 8z^{-3} + 4z^{-5}) + b(z^{-1} + 6z^{-3} + z^{-5}) = z^{-d}$$

The goal is to find the values for '$a$' and '$b$' for ANY value for '$d$'. I don't think I made any mistakes up to the simplifications.

EDIT: I thought of having $a = 1$ and $b = 4$, but this would give me

$$ -16z^{-3} = z^{-d} $$

This is not exactly equal since it should be $1$ in front, instead of $-16$

James the Great
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3 Answers3

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Presumably you are given a value for $z$ as well, but you don't have enough information. You can solve for either $a$ or $b$, but not both, with one equation.

Ross Millikan
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  • You might also simplify the equation by multiplying both sides by $z^6$. – John Wayland Bales Mar 18 '16 at 00:04
  • I am not solving for z. I am just trying to figure out some values for both a and b that would make the equation equal for ANY d – James the Great Mar 18 '16 at 06:10
  • Thanks for the tip on multiplying both sides by z^{6}, but since I am not solving for z, I can simply treat any z^{-n} as z^{n} – James the Great Mar 18 '16 at 06:11
  • You still can't solve for two variables in one equation unless something is special like $a^2+b^2=0$. There is nothing like that here. – Ross Millikan Mar 18 '16 at 20:31
  • Thanks for your reply. Hmm, I know the solutions would not be unique since there are two unknowns in one equation, but I was hoping for at least some non-unique solutions – James the Great Mar 19 '16 at 03:26
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    You can solve the equation for either $a$ or $b$. That is not hard, it is linear in both of them. That will give you a solution for one variable based on the other (and $d,z$). That is all you get. – Ross Millikan Mar 19 '16 at 03:29
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If you multiply both sides by $z^6$ you get \begin{equation} (z+1)(az^2+bz+a)-(z-1)(az^2-bz+a)=2z^{6-d} \end{equation} which simplifies to \begin{equation} (a+b)z^2+a=z^{6-d} \end{equation}

You may solve this for $a$, $b$, $d$ or $z$ in terms of the other three variables.

John Wayland Bales
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The answer would be:

a = -1/16

b = 1/4

That would give us $z^{-3}$, so d = 3

James the Great
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