       Re: How do I get Mathematica to Simplify this to 1?

• To: mathgroup at smc.vnet.net
• Subject: [mg104451] Re: [mg104437] How do I get Mathematica to Simplify this to 1?
• From: "David Park" <djmpark at comcast.net>
• Date: Sun, 1 Nov 2009 03:57:51 -0500 (EST)
• References: <16009656.1256974494025.JavaMail.root@n11>

```step1 = {(X - w Cos[k z])/
Sqrt[z^2 + (X - w Cos[k z])^2 + (Y - w Sin[k z])^2], (Y -
w Sin[k z])/
Sqrt[z^2 + (X - w Cos[k z])^2 + (Y - w Sin[k z])^2], -(z/
Sqrt[w^2 + X^2 + Y^2 + z^2 - 2 w X Cos[k z] - 2 w Y Sin[k z]])}

step2 = step1 /. Power[a_, -1/2] :> Power[Simplify[Expand[a]], -1/2]

step2.step2 // Simplify
1

David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/

From: dushan [mailto:dushanm at spinn.net]

After initially declaring that {w>0, k>0, {z,X,Y} el Reals}, a matrix-
vector multiplication produces the vector

{(X - w Cos[k z])/Sqrt[z^2 + (X - w Cos[k z])^2 + (Y - w Sin[k z])^2],
(Y - w Sin[k z])/Sqrt[z^2 + (X - w Cos[k z])^2 + (Y - w Sin[k z])^2],
-(z/Sqrt[  w^2 + X^2 + Y^2 + z^2 - 2 w X Cos[k z] - 2 w Y Sin[k z]])}

The denominators are in fact identical.  When I ask for Norm[%] I get

\[Sqrt](
Abs[z/Sqrt[w^2 + X^2 + Y^2 + z^2 - 2 w X Cos[k z] - 2 w Y Sin[k z]]]
^2  +
Abs[(X - w Cos[k z])/Sqrt[z^2 + (X - w Cos[k z])^2 + (Y - w Sin[k z])
^2]]^2 +
Abs[(Y - w Sin[k z])/Sqrt[z^2 + (X - w Cos[k z])^2 + (Y - w Sin[k z])
^2]]^2
)

and Simplify[%] reproduces this identical result instead of supplying

What am I doing wrong that prevents Mathematica from delivering the

A separate question: Is there available somewhere a short list of
symbols (such as '!!', '&&', "=.", '/@', etc.) and their meanings?  A
having it available as a 1-page crib-sheet would be very helpful to
newbies like me.

Thanks.

- Dushan

```

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