The following code:

In order to find $A_{0}^1$:
A_{0}^{1} = \lim_{z \rightarrow -1}\dfrac{d^0}{dz^0}\Bigg[\dfrac{(z+1)z}{(z+1)(z-2)}\Bigg]=\lim_{z \rightarrow -1}\dfrac{z}{(z-2)}=\dfrac{1}{3}
And in order to find $A_{0}^2$:
A_{0}^2 = \lim_{z \rightarrow 2} \dfrac{d^0}{dz^0}\Bigg[\dfrac{(z-2)z}{(z+1)(z-2)}\Bigg]=\lim_{z \rightarrow 2}\dfrac{z}{(z+1)}=\dfrac{2}{3}
From (38) we get:
where we have that $\Bigg[\dfrac{1}{z+1}\Bigg]$ can be rewritten as $\dfrac{1}{z}\dfrac{1}{1-\Big(-\dfrac{1}{z}\Big)}$

gives me the following:

enter image description here

The following is my preamble:

\usepackage[left=1.5in, right=1.5in, top=0.5in]{geometry}






\title{Homework Chapter 5}


How can I make the term shown after rewritten on the last line more proportionate?

  • (You don't need to provide 13 lines of code and 20 lines of preamble when your problem is in the 13th line.) I'm not quite sure what you mean by "more proportionate". But if you're wanting the final 1/z to be smaller, why not switch out of dfrac, which is "displaystyle frac"? – Teepeemm Apr 12 at 21:07
  • why are you using \dfrac in display mode it does nothing ( it is the same as \frac) and in the denominator it just forces it to use an over-sized fraction. – David Carlisle Apr 12 at 21:13
  • @Teepeemm: Next time I'll use 20 lines of preamble and 1 line of code. I ended up using 1/(1-(1/z)) – K.M Apr 12 at 21:16
  • @DavidCarlisle: Thanks for letting me know! – K.M Apr 12 at 21:17
  • Notice how the accepted answer only needed one package, and 2 lines of code for the "can be rewritten as part". So you could have presented your entire problem with 9 lines of code (3 of them blank). That's what I meant by providing too much code. – Teepeemm Apr 13 at 3:20

I'd use 1-(-z^{-1}).

I also made several fixes:

  • removed several redundant parentheses;
  • \Bigg is too big and wrong, it should be \biggl and \biggr;
  • also \Big is wrong in the same way and unnecessary;
  • the cross-reference should use \eqref and not the explicit number.


In order to find $A_{0}^1$
  = \lim_{z \rightarrow -1}\frac{d^0}{dz^0}\biggl[\frac{(z+1)z}{(z+1)(z-2)}\biggr]
  = \lim_{z \rightarrow -1}\frac{z}{z-2}=\frac{1}{3}
And in order to find $A_{0}^2$
  = \lim_{z \rightarrow 2} \frac{d^0}{dz^0}\biggl[\frac{(z-2)z}{(z+1)(z-2)}\biggr]
  = \lim_{z \rightarrow 2} \frac{z}{z+1}=\frac{2}{3}
From~\eqref{A01} we get
where we have that $\dfrac{1}{z+1}$ can be rewritten as 


enter image description here

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