Is $\frac{1}{(\textrm{can}-a-b\cdot i)^2}-\frac{1}{(\textrm{can}+\textrm{cod}-a-b \cdot i)^2}$ in any way simplifiable with Maple?

Question

So I wonder if Maple can reduce such lines like $$\frac{1}{(\textrm{can}-a-b\cdot i)^2}-\frac{1}{(\textrm{can}+\textrm{cod}-a-b \cdot i)^2}$$ (assuming all variables but $i$ are real)?

So, $can$ is the product of three variables? Or just a variable all its own? What would you like in a simplification? Real and Imaginary parts? — GEdgar, Sep 10 '11 at 21:45

score 4 · Accepted Answer · answered Sep 11 '11 at 01:07

Writing $x = can - a - i\cdot b$ and $y = cod$, your expression is equivalent to

$$\frac{1}{x^2} - \frac{1}{(x + y)^2}$$

You could try reducing it to a single fraction:

$$\frac{1}{x^2} - \frac{1}{(x + y)^2} = \frac{(x + y)^2 - x^2}{x^2 (x + y)^2} = \frac{2xy + y^2}{x^2 (x + y)^2} = \frac{y(2x + y)}{x^2 (x + y)^2}$$

I guess you'll have to decide which one looks cleanest or works best for your purposes, but it will not get much better than what you had originally.

score 1 · Answer 2 · answered Sep 10 '11 at 21:54

1

Let $\mathtt{A}$ be your expression. Both $\mathtt{simplify(A)}$ and $\mathtt{evalc(A)}$ are longer than $\mathtt{A}$, so probably most people would not call them "simplifications".

answered Sep 10 '11 at 21:54

GEdgar

111,679

Is $\frac{1}{(\textrm{can}-a-b\cdot i)^2}-\frac{1}{(\textrm{can}+\textrm{cod}-a-b \cdot i)^2}$ in any way simplifiable with Maple?

2 Answers2