       Re: Related quastions. Question 2

• To: mathgroup at smc.vnet.net
• Subject: [mg57531] Re: [mg57499] Related quastions. Question 2
• From: "David Park" <djmp at earthlink.net>
• Date: Tue, 31 May 2005 04:59:58 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Vlad,

Some users don't like to use % and %% statement references because they do
not work if the cells are not evaluated in the correct order. I agree with
that, but in the example I gave below all the statements were within a
SINGLE cell and refer to other statements within the SAME cell, so there is
no possibility that they won't be evaluated in the correct order. This is a
good way to do a calculation because you can keep adding and modifying until
you get what you want. There is no problem in evaluating the cell over and
over. If you don't want the Print statements you can just leave them out. (I
often find them useful to make it clear what is being done at each step.)
Sometimes when using several steps to get a result I put a ";" after some of
the steps to suppress that output. If you wish, you can also give
intermediate output names (like step1 =..., step2 =..., et ceteria) and then
refer to them by name.

Provided the individual steps evaluate quickly, this is an extremely
convenient method for carrying out a calculation. You can see exactly what
is happening and yet everything is in a single cell. If some step takes a
long time, then you may wish to give a name to the output and end the cell
there. Then you can continue on with a new cell starting with the
intermediate name.

If you want the same thing done to both sides of an equation, why not just
Map the function to both sides as I did?

equation = a x + b == c;

step1 = # - b & /@ equation
a x == -b + c

solution = Distribute[#/a] & /@ step1
x == -(b/a) + c/a

if that's the form you want.

But I find it easier to have as much as I can in one cell because if I make
a mistake at an early stage I can just correct it and do one reevaluation of
the cell.

David Park

From: Kazimir [mailto:kazimir04 at yahoo.co.uk]
To: mathgroup at smc.vnet.net

Dear Bob,

you gave the example

Print["Equation to solve for x"]
a x + b == c
Print["Subtract b from each side"]
# - b & /@ %%
Print["Divide each side by a"]
#/a & /@ %%

It works but i do not like very much expressions with  %%, as I should
not insert more lines before it or I have to modify this symbol. I

expr= (a x + b == c)

and then make a transform like expr1 =  expr - b. It clearly does not
work. Could you advise me a mathematical operartion over expression
which would return a x == c - b .

The best thing I found was
expr = (a x + b == c)
Distribute[#1 - b &@expr, Equal]
however, it does not work in all cases that I need.