# Using an implicit assumption to simplify

How can I simplify below expression knowing that r12>>z? Is there any way to expand below expression?

-(1/(128 eta \[Pi] r12^3 z^3))(-6 r12^4 z Cos[the] +
4 H r12 (-2 r12^3 + z^3) Cos[the - tp] +
12 H r12 z^3 Cos[3 the + tp] + 4 r12^5 Sin[2 the] +
6 H^2 z^3 Sin[2 the] - 8 r12^2 z^3 Sin[2 the] +
6 H r12^3 z Sin[tp] + 4 H^2 r12^3 Sin[2 tp] + H^2 z^3 Sin[2 tp] +
15 H^2 z^3 Sin[4 the + 2 tp])

• How about Series[%, {z, 0, 1}], which assumes that z is small. Alternative expansions might be more appropriate, depending on the relative magnitude of the other variables. Commented Jun 30, 2018 at 9:18
• There are things to do after your question is answered. You received two answers a week ago. While it's a good idea to stay vigilant for some time and wait 24 hours before accepting** the best one, one week is enough wait. Participation, that includes voting and accepting, is essential for the site, please do your part. Commented Jul 7, 2018 at 9:52

Your expression is a rational function of z so that Series won't actually do anything to it. To check this, define the given expression as

expr = -(1/(128 eta Pi r12^3 z^3)) (-6 r12^4 z Cos[the] +
4 H r12 (-2 r12^3 + z^3) Cos[the - tp] +
12 H r12 z^3 Cos[3 the + tp] + 4 r12^5 Sin[2 the] +
6 H^2 z^3 Sin[2 the] - 8 r12^2 z^3 Sin[2 the] +
6 H r12^3 z Sin[tp] + 4 H^2 r12^3 Sin[2 tp] + H^2 z^3 Sin[2 tp] +
15 H^2 z^3 Sin[4 the + 2 tp])


And then check that it's already in the form that Series would produce by itself:

Simplify[expr == Normal@Series[expr, {z, 0, 1}]]


True

You could make a list of the powers of z appearing in the expression and decide which of the powers can be neglected by inspecting the coefficients:

Expand@SeriesCoefficient[expr, {z, 0, n}]


$$\begin{cases} \frac{3 \text{r12} \cos (\text{the})}{64 \pi \eta }-\frac{3 H \sin (\text{tp})}{64 \pi \eta } & n=-2 \\ -\frac{H^2 \sin (2 \text{tp})}{32 \pi \eta }+\frac{H \text{r12} \cos (\text{the}-\text{tp})}{16 \pi \eta }-\frac{\text{r12}^2 \sin (2 \text{the})}{32 \pi \eta } & n=-3 \\ -\frac{15 H^2 \sin (4 \text{the}+2 \text{tp})}{128 \pi \eta \text{r12}^3}-\frac{3 H^2 \sin (2 \text{the})}{64 \pi \eta \text{r12}^3}-\frac{H^2 \sin (2 \text{tp})}{128 \pi \eta \text{r12}^3}-\frac{H \cos (\text{the}-\text{tp})}{32 \pi \eta \text{r12}^2}-\frac{3 H \cos (3 \text{the}+\text{tp})}{32 \pi \eta \text{r12}^2}+\\+\frac{\sin (2 \text{the})}{16 \pi \eta \text{r12}} & n=0 \end{cases}$$

If you know that z<<tp just substitute z->eps tp in your expression

expr=-(1/(128 eta \[Pi] r12^3 z^3))(-6 r12^4 z Cos[the] +
4 H r12 (-2 r12^3 + z^3) Cos[the - tp] +
12 H r12 z^3 Cos[3 the + tp] + 4 r12^5 Sin[2 the] +
6 H^2 z^3 Sin[2 the] - 8 r12^2 z^3 Sin[2 the] +
6 H r12^3 z Sin[tp] + 4 H^2 r12^3 Sin[2 tp] + H^2 z^3 Sin[2 tp] +15 H^2 z^3   Sin[4 the + 2 tp]) /. z->eps tp ;


series expansion relating to eps gives

seps = Normal[Series[expr, {eps, 0,-2(*examplary*)}]] /.eps -> z/tp
(*(3 (r12 Cos[the] - H Sin[tp]))/(64 eta \[Pi] z^2) + (2 H r12 Cos[the - tp]- r12^2 Sin[2 the] - H^2 Sin[2 tp])/(32 eta \[Pi] z^3)*)


the approximation you are looking for.