Re: new user help
- To: mathgroup at smc.vnet.net
- Subject: [mg13406] Re: new user help
- From: phbrf at t-online.de (Peter Breitfeld)
- Date: Thu, 23 Jul 1998 03:32:50 -0400
- Organization: das ist ein weites Feld ...
- References: <6okkvj$1md@smc.vnet.net> <6ous3s$jh3@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Allan Hayes <hay at haystack.demon.cc.uk> schrieb/wrote:
:
: John M. Dlugosz wrote in message <6okkvj$1md at smc.vnet.net>...
: >I'm lost. I just don't know how to get started...
: >
: >The excersize I've chosen for myself is to start with
: >
: > x^2+y^2+4x-6y+4==0
: >
: >and manipulate it into the form (x-h)^2+(y-k)^2==r^2
: >
: >So... how do I "manipulate" the equasion? The functions like Expand,
: >Factor, etc. don't help much. I my calculator (an HP48) I can point to
: >specific subexpressions and apply operations to them, like factor,
: >distribute, changing forms, etc.
: >
: >How do I collect the x's together in parens, the y's in parens, and
: >complete the squares? Doing it on paper defeats the point! I want to
: >learn how to "do math" using this tool. That's more than just asking
: >"OK, what's X?". It means manipulating things and arranging them,
: >getting to know how the symbols all fit together.
: >
: >--John
:
Try to use a replacement-rule:
eq = x^2+y^2+4x-6y+4==0
quadRule = { x_^2 + x_ -> (x+1/2)^2-1/4, [1]
x_^2 + b_ x_ -> (x+b/2)^2-b^2/4,
a_ x^2 + x_ -> a(x-1/(2a))^2-1/(4a),
a_ x_^2 + b_ x_ ->a(x+b/(2a))^2-b^2/(4a) }
Then do
In: qf=eq //. quadRule [2] Out: -9 +
(2+x)^2 +(-3+y)^2==0
In: (#-qf[[1,1]])& /@ qf [3] Out: (2+x)^2
+ (-3+y)^2 == 9
[1] To make this work in the general case, you have to give all the
four rules, because the FullForm of x^2+4x doesn't match
a_ x_^2 + b_ x_ because there is no "a" etc.
[2] If you don't use ReplaceRepeated here, you only get the x-term in
quadratic form
[3] Mathematica orders term always with numbers in the first place, so
qf[[1,1] represents the "-9" in this example. (#-qf[[1]])& is a pure
function, which subtracts qf[[1]] from the given expression. Here I
map this function on the equation (via /@) so the "-9" will be sub-
tracted from both sides of the equation qf.
I don't think there is a simple way to write the 9 as 3^2, because
Mathematica will simplify 3^2 to 9 immediately.
es gruesst
Peter
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